XPost: comp.theory, sci.logic, sci.math

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax
[000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx
[000012f2](05)  e880feffff      call 00001177 // call H
[000012f7](03)  83c408          add esp,+08
[000012fa](02)  85c0            test eax,eax
[000012fc](02)  7402            jz 00001300
[000012fe](02)  ebfe            jmp 000012fe
[00001300](01)  5d              pop ebp
[00001301](01)  c3              ret
Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 // push P
[0000130f](05)  e8d3ffffff      call 000012e7 // call P
[00001314](03)  83c404          add esp,+04
[00001317](02)  33c0            xor eax,eax
[00001319](01)  5d              pop ebp
[0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a]

  machine   stack     stack     machine    assembly
  address   address   data      code       language
  ========  ========  ========  =========  =============
[00001307][00102190][00000000] 55         push ebp
[00001308][00102190][00000000] 8bec       mov ebp,esp
[0000130a][0010218c][000012e7] 68e7120000 push 000012e7 // push P
[0000130f][00102188][00001314] e8d3ffffff call 000012e7 // call P
[000012e7][00102184][00102190] 55         push ebp      // enter
executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp
[000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push eax      // push P
[000012ee][00102180][000012e7] 8b4d08     mov ecx,[ebp+08]
[000012f1][0010217c][000012e7] 51         push ecx      // push P
[000012f2][00102178][000012f7] e880feffff call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace Stored at:212244
[000012e7][00212230][00212234] 55          push ebp      // enter
emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp
[000012ea][00212230][00212234] 8b4508      mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push eax      // push P
[000012ee][0021222c][000012e7] 8b4d08      mov ecx,[ebp+08]
[000012f1][00212228][000012e7] 51          push ecx      // push P
[000012f2][00212224][000012f7] e880feffff  call 00001177 // call H

So, by what instruction reference manual is a call 00001177 followedby
the execution of the instruction at 000012e7.

Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push ebp      // enter
emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp
[000012ea][0025cc58][0025cc5c] 8b4508      mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push eax      // push P
[000012ee][0025cc54][000012e7] 8b4d08      mov ecx,[ebp+08]
[000012f1][0025cc50][000012e7] 51          push ecx      // push P
[000012f2][0025cc4c][000012f7] e880feffff  call 00001177 // call H
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly emulates its input
that it must emulate the first seven instructions of P. Because the
seventh instruction of P repeats this process we know with complete
certainty that the correct and complete emulation of P by H would
never reach its final “ret” instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the procedure", because
that H always has the option to abort its simulation, just like this
onne did, and return to its P and see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL REBUTTAL
AT ALL:

The partial correct x86 emulation of the input to H(P,P) conclusively
proves that the complete and correct x86 emulation would never stop
running.


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
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XPost: comp.theory, sci.logic, sci.math

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
  if (H(x, x))
    HERE: goto HERE;
  return;
}

int main()
{
  P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax
[000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx
[000012f2](05)  e880feffff      call 00001177 // call H [000012f7](03)  83c408          add esp,+08
[000012fa](02)  85c0            test eax,eax
[000012fc](02)  7402            jz 00001300
[000012fe](02)  ebfe            jmp 000012fe
[00001300](01)  5d              pop ebp
[00001301](01)  c3              ret
Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 // push P [0000130f](05)  e8d3ffffff      call 000012e7 // call P [00001314](03)  83c404          add esp,+04
[00001317](02)  33c0            xor eax,eax
[00001319](01)  5d              pop ebp
[0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a]

 machine   stack     stack     machine    assembly
 address   address   data      code       language
 ========  ========  ========  =========  ============= [00001307][00102190][00000000] 55         push ebp [00001308][00102190][00000000] 8bec       mov ebp,esp [0000130a][0010218c][000012e7] 68e7120000 push 000012e7 // push P [0000130f][00102188][00001314] e8d3ffffff call 000012e7 // call P [000012e7][00102184][00102190] 55         push ebp      // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp [000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08] [000012ed][00102180][000012e7] 50         push eax      // push P
[000012ee][00102180][000012e7] 8b4d08     mov ecx,[ebp+08] [000012f1][0010217c][000012e7] 51         push ecx      // push P
[000012f2][00102178][000012f7] e880feffff call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace Stored at:212244 [000012e7][00212230][00212234] 55          push ebp      // enter
emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp [000012ea][00212230][00212234] 8b4508      mov eax,[ebp+08] [000012ed][0021222c][000012e7] 50          push eax      // push P
[000012ee][0021222c][000012e7] 8b4d08      mov ecx,[ebp+08] [000012f1][00212228][000012e7] 51          push ecx      // push P
[000012f2][00212224][000012f7] e880feffff  call 00001177 // call H

So, by what instruction reference manual is a call 00001177 followedby
the execution of the instruction at 000012e7.

Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push ebp      // enter
emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp [000012ea][0025cc58][0025cc5c] 8b4508      mov eax,[ebp+08] [000012ed][0025cc54][000012e7] 50          push eax      // push P
[000012ee][0025cc54][000012e7] 8b4d08      mov ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51          push ecx      // push P
[000012f2][0025cc4c][000012f7] e880feffff  call 00001177 // call H
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly emulates its input
that it must emulate the first seven instructions of P. Because the
seventh instruction of P repeats this process we know with complete
certainty that the correct and complete emulation of P by H would never
reach its final “ret” instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the procedure", because
that H always has the option to abort its simulation, just like this
onne did, and return to its P and see it halt.

All you have shown is that either:
H isn't a computation, not behaving the same for all copies given the
same input, and thus, by definition, nat a halt decider, or

H doesn't CORRECT analysis its simulation of its input.

[000012f7][00102184][00102190] 83c408      add esp,+08 [000012fa][00102184][00102190] 85c0        test eax,eax [000012fc][00102184][00102190] 7402        jz 00001300 [00001300][00102188][00001314] 5d          pop ebp [00001301][0010218c][000012e7] c3          ret           // executed P
halts
[00001314][00102190][00000000] 83c404      add esp,+04 [00001317][00102190][00000000] 33c0        xor eax,eax [00001319][00102194][00100000] 5d          pop ebp [0000131a][00102198][00000000] c3          ret          // main() halts
Number of Instructions Executed(15900)

And thus at the end, your trace shows that P(P) Halted, and thus the
CORRECT answer for H(P,P) needs to be 1.

Your logic, as pointed out above, is UNSOUND.

YOU FAIL.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax
[000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx
[000012f2](05)  e880feffff      call 00001177 // call H
[000012f7](03)  83c408          add esp,+08
[000012fa](02)  85c0            test eax,eax
[000012fc](02)  7402            jz 00001300
[000012fe](02)  ebfe            jmp 000012fe
[00001300](01)  5d              pop ebp
[00001301](01)  c3              ret
Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 // push P
[0000130f](05)  e8d3ffffff      call 000012e7 // call P
[00001314](03)  83c404          add esp,+04
[00001317](02)  33c0            xor eax,eax
[00001319](01)  5d              pop ebp
[0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a]

  machine   stack     stack     machine    assembly
  address   address   data      code       language
  ========  ========  ========  =========  =============
[00001307][00102190][00000000] 55         push ebp
[00001308][00102190][00000000] 8bec       mov ebp,esp
[0000130a][0010218c][000012e7] 68e7120000 push 000012e7 // push P
[0000130f][00102188][00001314] e8d3ffffff call 000012e7 // call P
[000012e7][00102184][00102190] 55         push ebp      // enter
executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp
[000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push eax      // push P
[000012ee][00102180][000012e7] 8b4d08     mov ecx,[ebp+08]
[000012f1][0010217c][000012e7] 51         push ecx      // push P
[000012f2][00102178][000012f7] e880feffff call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace Stored at:212244 >>> [000012e7][00212230][00212234] 55          push ebp      // enter
emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp
[000012ea][00212230][00212234] 8b4508      mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push eax      // push P
[000012ee][0021222c][000012e7] 8b4d08      mov ecx,[ebp+08]
[000012f1][00212228][000012e7] 51          push ecx      // push P
[000012f2][00212224][000012f7] e880feffff  call 00001177 // call H

So, by what instruction reference manual is a call 00001177 followedby
the execution of the instruction at 000012e7.

Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push ebp      // enter
emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp
[000012ea][0025cc58][0025cc5c] 8b4508      mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push eax      // push P
[000012ee][0025cc54][000012e7] 8b4d08      mov ecx,[ebp+08]
[000012f1][0025cc50][000012e7] 51          push ecx      // push P
[000012f2][0025cc4c][000012f7] e880feffff  call 00001177 // call H
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly emulates its
input that it must emulate the first seven instructions of P. Because
the seventh instruction of P repeats this process we know with
complete certainty that the correct and complete emulation of P by H
would never reach its final “ret” instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the procedure",
because that H always has the option to abort its simulation, just
like this onne did, and return to its P and see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL REBUTTAL
AT ALL:

The partial correct x86 emulation of the input to H(P,P) conclusively
proves that the complete and correct x86 emulation would never stop
running.

You SAY that, but you don't answer the actual questions about HOW.

This shows you are just an idiot, parroting ideas that you don't
actually understand.

It really looks like you have gaslighted yourself into believing your
own lies.

HOW is the correct simulation of a CALL 00001177 instruction not
followed by the instruction at 00001177 ?

If you can't explain that, you just prove that you don't understand what
you are saying.

the actual FACTS show you are wrong.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax
[000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx
[000012f2](05)  e880feffff      call 00001177 // call H
[000012f7](03)  83c408          add esp,+08
[000012fa](02)  85c0            test eax,eax
[000012fc](02)  7402            jz 00001300
[000012fe](02)  ebfe            jmp 000012fe
[00001300](01)  5d              pop ebp
[00001301](01)  c3              ret
Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 // push P
[0000130f](05)  e8d3ffffff      call 000012e7 // call P
[00001314](03)  83c404          add esp,+04
[00001317](02)  33c0            xor eax,eax
[00001319](01)  5d              pop ebp
[0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a]

  machine   stack     stack     machine    assembly
  address   address   data      code       language >>>>   ========  ========  ========  =========  =============
[00001307][00102190][00000000] 55         push ebp
[00001308][00102190][00000000] 8bec       mov ebp,esp
[0000130a][0010218c][000012e7] 68e7120000 push 000012e7 // push P
[0000130f][00102188][00001314] e8d3ffffff call 000012e7 // call P
[000012e7][00102184][00102190] 55         push ebp      // enter
executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp
[000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push eax      // push P
[000012ee][00102180][000012e7] 8b4d08     mov ecx,[ebp+08]
[000012f1][0010217c][000012e7] 51         push ecx      // push P
[000012f2][00102178][000012f7] e880feffff call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace Stored
at:212244 [000012e7][00212230][00212234] 55          push ebp >>>>   // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp
[000012ea][00212230][00212234] 8b4508      mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push eax      // push
P [000012ee][0021222c][000012e7] 8b4d08      mov ecx,[ebp+08]
[000012f1][00212228][000012e7] 51          push ecx      // push
P [000012f2][00212224][000012f7] e880feffff  call 00001177 //
call H

So, by what instruction reference manual is a call 00001177
followedby the execution of the instruction at 000012e7.

Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp
[000012ea][0025cc58][0025cc5c] 8b4508      mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push eax      // push
P [000012ee][0025cc54][000012e7] 8b4d08      mov ecx,[ebp+08]
[000012f1][0025cc50][000012e7] 51          push ecx      // push
P [000012f2][0025cc4c][000012f7] e880feffff  call 00001177 //
call H Local Halt Decider: Infinite Recursion Detected
Simulation Stopped

It is completely obvious that when H(P,P) correctly emulates its
input that it must emulate the first seven instructions of P.
Because the seventh instruction of P repeats this process we
know with complete certainty that the correct and complete
emulation of P by H would never reach its final “ret”
instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the procedure",
because that H always has the option to abort its simulation,
just like this onne did, and return to its P and see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P)
conclusively proves that the complete and correct x86 emulation
would never stop running.

You SAY that, but you don't answer the actual questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO EVIDENCE WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE CORRECT PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY PROVES THAT THE CORRECT
AND COMPLETE X86 EMULATION OF THE INPUT TO H(P,P) WOULD NEVER STOP
RUNNING.

It is completely obvious that when H(P,P) correctly emulates its
input that it must emulate the first seven instructions of P. Because
the seventh instruction of P repeats this process we know with
complete certainty that the correct and complete emulation of P by H
would never reach its final “ret” instruction, thus never halts.

If P should have halted (i.e. no infinite loop) then your simulation
detector, S (not H), gets the answer wrong. You S is NOT a halting
decider.

/Flibble

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax
[000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx
[000012f2](05)  e880feffff      call 00001177 // call H
[000012f7](03)  83c408          add esp,+08
[000012fa](02)  85c0            test eax,eax
[000012fc](02)  7402            jz 00001300
[000012fe](02)  ebfe            jmp 000012fe
[00001300](01)  5d              pop ebp
[00001301](01)  c3              ret
Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 // push P
[0000130f](05)  e8d3ffffff      call 000012e7 // call P
[00001314](03)  83c404          add esp,+04
[00001317](02)  33c0            xor eax,eax
[00001319](01)  5d              pop ebp
[0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a]

  machine   stack     stack     machine    assembly
  address   address   data      code       language
  ========  ========  ========  =========  =============
[00001307][00102190][00000000] 55         push ebp
[00001308][00102190][00000000] 8bec       mov ebp,esp
[0000130a][0010218c][000012e7] 68e7120000 push 000012e7 // push P
[0000130f][00102188][00001314] e8d3ffffff call 000012e7 // call P
[000012e7][00102184][00102190] 55         push ebp      // enter
executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp
[000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push eax      // push P
[000012ee][00102180][000012e7] 8b4d08     mov ecx,[ebp+08]
[000012f1][0010217c][000012e7] 51         push ecx      // push P
[000012f2][00102178][000012f7] e880feffff call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace Stored at:212244 >>>> [000012e7][00212230][00212234] 55          push ebp      // enter
emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp
[000012ea][00212230][00212234] 8b4508      mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push eax      // push P
[000012ee][0021222c][000012e7] 8b4d08      mov ecx,[ebp+08]
[000012f1][00212228][000012e7] 51          push ecx      // push P
[000012f2][00212224][000012f7] e880feffff  call 00001177 // call H

So, by what instruction reference manual is a call 00001177
followedby the execution of the instruction at 000012e7.

Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push ebp      // enter
emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp
[000012ea][0025cc58][0025cc5c] 8b4508      mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push eax      // push P
[000012ee][0025cc54][000012e7] 8b4d08      mov ecx,[ebp+08]
[000012f1][0025cc50][000012e7] 51          push ecx      // push P
[000012f2][0025cc4c][000012f7] e880feffff  call 00001177 // call H
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly emulates its
input that it must emulate the first seven instructions of P.
Because the seventh instruction of P repeats this process we know
with complete certainty that the correct and complete emulation of P
by H would never reach its final “ret” instruction, thus never halts. >>>

Problem, the 7th intruction DOESN't "Just repeat the procedure",
because that H always has the option to abort its simulation, just
like this onne did, and return to its P and see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P) conclusively
proves that the complete and correct x86 emulation would never stop
running.

You SAY that, but you don't answer the actual questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE CORRECT PARTIAL
EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY PROVES THAT THE CORRECT
AND COMPLETE X86 EMULATION OF THE INPUT TO H(P,P) WOULD NEVER STOP
RUNNING.

It is completely obvious that when H(P,P) correctly emulates its input
that it must emulate the first seven instructions of P. Because the
seventh instruction of P repeats this process we know with complete
certainty that the correct and complete emulation of P by H would never
reach its final “ret” instruction, thus never halts.




--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax
[000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx
[000012f2](05)  e880feffff      call 00001177 // call H
[000012f7](03)  83c408          add esp,+08
[000012fa](02)  85c0            test eax,eax
[000012fc](02)  7402            jz 00001300
[000012fe](02)  ebfe            jmp 000012fe
[00001300](01)  5d              pop ebp
[00001301](01)  c3              ret
Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 // push P
[0000130f](05)  e8d3ffffff      call 000012e7 // call P
[00001314](03)  83c404          add esp,+04
[00001317](02)  33c0            xor eax,eax
[00001319](01)  5d              pop ebp
[0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a]

  machine   stack     stack     machine    assembly >>>>>>   address   address   data      code       language >>>>>>   ========  ========  ========  =========  =============
[00001307][00102190][00000000] 55         push ebp
[00001308][00102190][00000000] 8bec       mov ebp,esp
[0000130a][0010218c][000012e7] 68e7120000 push 000012e7 // push P
[0000130f][00102188][00001314] e8d3ffffff call 000012e7 // call P
[000012e7][00102184][00102190] 55         push ebp      // enter
executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp
[000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push eax      // push P
[000012ee][00102180][000012e7] 8b4d08     mov ecx,[ebp+08]
[000012f1][0010217c][000012e7] 51         push ecx      // push P
[000012f2][00102178][000012f7] e880feffff call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace Stored
at:212244 [000012e7][00212230][00212234] 55          push ebp >>>>>>   // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp
[000012ea][00212230][00212234] 8b4508      mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push eax      // push
P [000012ee][0021222c][000012e7] 8b4d08      mov ecx,[ebp+08] >>>>>> [000012f1][00212228][000012e7] 51          push ecx      // push
P [000012f2][00212224][000012f7] e880feffff  call 00001177 //
call H

So, by what instruction reference manual is a call 00001177
followedby the execution of the instruction at 000012e7.

Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp
[000012ea][0025cc58][0025cc5c] 8b4508      mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push eax      // push
P [000012ee][0025cc54][000012e7] 8b4d08      mov ecx,[ebp+08] >>>>>> [000012f1][0025cc50][000012e7] 51          push ecx      // push
P [000012f2][0025cc4c][000012f7] e880feffff  call 00001177 //
call H Local Halt Decider: Infinite Recursion Detected
Simulation Stopped

It is completely obvious that when H(P,P) correctly emulates its
input that it must emulate the first seven instructions of P.
Because the seventh instruction of P repeats this process we
know with complete certainty that the correct and complete
emulation of P by H would never reach its final “ret”
instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the procedure",
because that H always has the option to abort its simulation,
just like this onne did, and return to its P and see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P)
conclusively proves that the complete and correct x86 emulation
would never stop running.

You SAY that, but you don't answer the actual questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE CORRECT PARTIAL
EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY PROVES THAT THE CORRECT
AND COMPLETE X86 EMULATION OF THE INPUT TO H(P,P) WOULD NEVER STOP
RUNNING.

It is completely obvious that when H(P,P) correctly emulates its
input that it must emulate the first seven instructions of P. Because
the seventh instruction of P repeats this process we know with
complete certainty that the correct and complete emulation of P by H
would never reach its final “ret” instruction, thus never halts.

If P should have halted (i.e. no infinite loop) then your simulation detector, S (not H), gets the answer wrong. You S is NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL REBUTTAL
AT ALL.

_P()
[00001352](01) 55 push ebp
[00001353](02) 8bec mov ebp,esp
[00001355](03) 8b4508 mov eax,[ebp+08]
[00001358](01) 50 push eax // push P
[00001359](03) 8b4d08 mov ecx,[ebp+08]
[0000135c](01) 51 push ecx // push P
[0000135d](05) e840feffff call 000011a2 // call H
[00001362](03) 83c408 add esp,+08
[00001365](02) 85c0 test eax,eax
[00001367](02) 7402 jz 0000136b
[00001369](02) ebfe jmp 00001369
[0000136b](01) 5d pop ebp
[0000136c](01) c3 ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly emulates its input
that it must emulate the first seven instructions of P. Because the
seventh instruction of P repeats this process we know with complete
certainty that the correct and complete emulation of P by H would never
reach its final “ret” instruction, thus never halts.


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax
[000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx
[000012f2](05)  e880feffff      call 00001177 // call H
[000012f7](03)  83c408          add esp,+08
[000012fa](02)  85c0            test eax,eax
[000012fc](02)  7402            jz 00001300
[000012fe](02)  ebfe            jmp 000012fe
[00001300](01)  5d              pop ebp
[00001301](01)  c3              ret
Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 // push P
[0000130f](05)  e8d3ffffff      call 000012e7 // call P
[00001314](03)  83c404          add esp,+04
[00001317](02)  33c0            xor eax,eax
[00001319](01)  5d              pop ebp
[0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a]

  machine   stack     stack     machine    assembly >>>>>>   address   address   data      code       language >>>>>>   ========  ========  ========  =========  =============
[00001307][00102190][00000000] 55         push ebp
[00001308][00102190][00000000] 8bec       mov ebp,esp
[0000130a][0010218c][000012e7] 68e7120000 push 000012e7 //
push P [0000130f][00102188][00001314] e8d3ffffff call 000012e7
// call P [000012e7][00102184][00102190] 55         push ebp >>>>>>    // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp
[000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov
ecx,[ebp+08] [000012f1][0010217c][000012e7] 51         push >>>>>> ecx      // push P [000012f2][00102178][000012f7] e880feffff >>>>>> call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace Stored
at:212244 [000012e7][00212230][00212234] 55          push ebp >>>>>>   // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp
[000012ea][00212230][00212234] 8b4508      mov eax,[ebp+08] >>>>>> [000012ed][0021222c][000012e7] 50          push eax      //
push P [000012ee][0021222c][000012e7] 8b4d08      mov
ecx,[ebp+08] [000012f1][00212228][000012e7] 51          push >>>>>> ecx      // push P [000012f2][00212224][000012f7] e880feffff >>>>>> call 00001177 // call H

So, by what instruction reference manual is a call 00001177
followedby the execution of the instruction at 000012e7.

Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp
[000012ea][0025cc58][0025cc5c] 8b4508      mov eax,[ebp+08] >>>>>> [000012ed][0025cc54][000012e7] 50          push eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08      mov
ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51          push >>>>>> ecx      // push P [000012f2][0025cc4c][000012f7] e880feffff >>>>>> call 00001177 // call H Local Halt Decider: Infinite Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly emulates
its input that it must emulate the first seven instructions of
P. Because the seventh instruction of P repeats this process we
know with complete certainty that the correct and complete
emulation of P by H would never reach its final “ret”
instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the procedure",
because that H always has the option to abort its simulation,
just like this onne did, and return to its P and see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P)
conclusively proves that the complete and correct x86 emulation
would never stop running.

You SAY that, but you don't answer the actual questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly emulates its
input that it must emulate the first seven instructions of P.
Because the seventh instruction of P repeats this process we know
with complete certainty that the correct and complete emulation of
P by H would never reach its final “ret” instruction, thus never
halts.

If P should have halted (i.e. no infinite loop) then your simulation detector, S (not H), gets the answer wrong. You S is NOT a halting decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01) 55 push ebp
[00001353](02) 8bec mov ebp,esp
[00001355](03) 8b4508 mov eax,[ebp+08]
[00001358](01) 50 push eax // push P
[00001359](03) 8b4d08 mov ecx,[ebp+08]
[0000135c](01) 51 push ecx // push P
[0000135d](05) e840feffff call 000011a2 // call H
[00001362](03) 83c408 add esp,+08
[00001365](02) 85c0 test eax,eax
[00001367](02) 7402 jz 0000136b
[00001369](02) ebfe jmp 00001369
[0000136b](01) 5d pop ebp
[0000136c](01) c3 ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly emulates its
input that it must emulate the first seven instructions of P. Because
the seventh instruction of P repeats this process we know with
complete certainty that the correct and complete emulation of P by H
would never reach its final “ret” instruction, thus never halts.

We are going around and around and around in circles. I will try again:

If you replace the opcodes "EB FE" at 00001369 with the opcodes "90 90"
then your H gets the answer wrong: P should have halted.

/Flibble

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax
[000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx
[000012f2](05)  e880feffff      call 00001177 // call H
[000012f7](03)  83c408          add esp,+08
[000012fa](02)  85c0            test eax,eax
[000012fc](02)  7402            jz 00001300
[000012fe](02)  ebfe            jmp 000012fe
[00001300](01)  5d              pop ebp
[00001301](01)  c3              ret
Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 // push P
[0000130f](05)  e8d3ffffff      call 000012e7 // call P
[00001314](03)  83c404          add esp,+04
[00001317](02)  33c0            xor eax,eax
[00001319](01)  5d              pop ebp
[0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a]

  machine   stack     stack     machine    assembly >>>>>>>>   address   address   data      code       language >>>>>>>>   ========  ========  ========  =========  ============= >>>>>>>> [00001307][00102190][00000000] 55         push ebp
[00001308][00102190][00000000] 8bec       mov ebp,esp
[0000130a][0010218c][000012e7] 68e7120000 push 000012e7 //
push P [0000130f][00102188][00001314] e8d3ffffff call 000012e7 >>>>>>>> // call P [000012e7][00102184][00102190] 55         push ebp >>>>>>>>    // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp
[000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08] >>>>>>>> [000012ed][00102180][000012e7] 50         push eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov
ecx,[ebp+08] [000012f1][0010217c][000012e7] 51         push >>>>>>>> ecx      // push P [000012f2][00102178][000012f7] e880feffff >>>>>>>> call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace Stored >>>>>>>> at:212244 [000012e7][00212230][00212234] 55          push ebp >>>>>>>>   // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp
[000012ea][00212230][00212234] 8b4508      mov eax,[ebp+08] >>>>>>>> [000012ed][0021222c][000012e7] 50          push eax      //
push P [000012ee][0021222c][000012e7] 8b4d08      mov
ecx,[ebp+08] [000012f1][00212228][000012e7] 51          push >>>>>>>> ecx      // push P [000012f2][00212224][000012f7] e880feffff >>>>>>>> call 00001177 // call H

So, by what instruction reference manual is a call 00001177
followedby the execution of the instruction at 000012e7.

Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp
[000012ea][0025cc58][0025cc5c] 8b4508      mov eax,[ebp+08] >>>>>>>> [000012ed][0025cc54][000012e7] 50          push eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08      mov
ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51          push >>>>>>>> ecx      // push P [000012f2][0025cc4c][000012f7] e880feffff >>>>>>>> call 00001177 // call H Local Halt Decider: Infinite Recursion >>>>>>>> Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly emulates
its input that it must emulate the first seven instructions of >>>>>>>> P. Because the seventh instruction of P repeats this process we >>>>>>>> know with complete certainty that the correct and complete
emulation of P by H would never reach its final “ret”
instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the procedure", >>>>>>> because that H always has the option to abort its simulation,
just like this onne did, and return to its P and see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P)
conclusively proves that the complete and correct x86 emulation
would never stop running.

You SAY that, but you don't answer the actual questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly emulates its
input that it must emulate the first seven instructions of P.
Because the seventh instruction of P repeats this process we know
with complete certainty that the correct and complete emulation of
P by H would never reach its final “ret” instruction, thus never
halts.

If P should have halted (i.e. no infinite loop) then your simulation
detector, S (not H), gets the answer wrong. You S is NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01) 55 push ebp
[00001353](02) 8bec mov ebp,esp
[00001355](03) 8b4508 mov eax,[ebp+08]
[00001358](01) 50 push eax // push P
[00001359](03) 8b4d08 mov ecx,[ebp+08]
[0000135c](01) 51 push ecx // push P
[0000135d](05) e840feffff call 000011a2 // call H
[00001362](03) 83c408 add esp,+08
[00001365](02) 85c0 test eax,eax
[00001367](02) 7402 jz 0000136b
[00001369](02) ebfe jmp 00001369
[0000136b](01) 5d pop ebp
[0000136c](01) c3 ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly emulates its
input that it must emulate the first seven instructions of P. Because
the seventh instruction of P repeats this process we know with
complete certainty that the correct and complete emulation of P by H
would never reach its final “ret” instruction, thus never halts.

We are going around and around and around in circles. I will try again:

If you replace the opcodes "EB FE" at 00001369 with the opcodes "90 90"
then your H gets the answer wrong: P should have halted.

/Flibble

As I already said before this is merely your cluelessness that when
H(P,P) is invoked the correct x86 emulation of the input to H(P,P) makes
and code after [0000135d] unreachable.

https://en.wikipedia.org/wiki/Unreachable_code

_P()
[00001352](01) 55 push ebp
[00001353](02) 8bec mov ebp,esp
[00001355](03) 8b4508 mov eax,[ebp+08]
[00001358](01) 50 push eax // push P
[00001359](03) 8b4d08 mov ecx,[ebp+08]
[0000135c](01) 51 push ecx // push P
[0000135d](05) e840feffff call 000011a2 // call H
[00001362](03) 83c408 add esp,+08
[00001365](02) 85c0 test eax,eax
[00001367](02) 7402 jz 0000136b
[00001369](02) ebfe jmp 00001369
[0000136b](01) 5d pop ebp
[0000136c](01) c3 ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly emulates its input
that it must emulate the first seven instructions of P. Because the
seventh instruction of P repeats this process we know with complete
certainty that the correct and complete emulation of P by H would never
reach its final “ret” instruction, thus never halts.

People that are not dumber than a box of rocks will understand this to
mean that the correct x86 emulation of the input to H(P,P) by H cannot
possibly reach any instruction after [0000135d].


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax
[000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx
[000012f2](05)  e880feffff      call 00001177 // call H >>>>>>>>> [000012f7](03)  83c408          add esp,+08
[000012fa](02)  85c0            test eax,eax
[000012fc](02)  7402            jz 00001300
[000012fe](02)  ebfe            jmp 000012fe
[00001300](01)  5d              pop ebp
[00001301](01)  c3              ret
Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 // push P >>>>>>>>> [0000130f](05)  e8d3ffffff      call 000012e7 // call P >>>>>>>>> [00001314](03)  83c404          add esp,+04
[00001317](02)  33c0            xor eax,eax
[00001319](01)  5d              pop ebp
[0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine    assembly >>>>>>>>>     address   address   data      code       language
    ========  ========  ========  =========  ============= >>>>>>>>> [00001307][00102190][00000000] 55         push ebp
[00001308][00102190][00000000] 8bec       mov ebp,esp
[0000130a][0010218c][000012e7] 68e7120000 push 000012e7 //
push P [0000130f][00102188][00001314] e8d3ffffff call 000012e7 >>>>>>>>> // call P [000012e7][00102184][00102190] 55         push ebp >>>>>>>>>     // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp
[000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08] >>>>>>>>> [000012ed][00102180][000012e7] 50         push eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov
ecx,[ebp+08] [000012f1][0010217c][000012e7] 51         push >>>>>>>>> ecx      // push P [000012f2][00102178][000012f7] e880feffff >>>>>>>>> call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace Stored >>>>>>>>> at:212244 [000012e7][00212230][00212234] 55          push ebp
    // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp >>>>>>>>> [000012ea][00212230][00212234] 8b4508      mov eax,[ebp+08] >>>>>>>>> [000012ed][0021222c][000012e7] 50          push eax      //
push P [000012ee][0021222c][000012e7] 8b4d08      mov
ecx,[ebp+08] [000012f1][00212228][000012e7] 51          push >>>>>>>>> ecx      // push P [000012f2][00212224][000012f7] e880feffff >>>>>>>>> call 00001177 // call H

So, by what instruction reference manual is a call 00001177
followedby the execution of the instruction at 000012e7.

Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp >>>>>>>>> [000012ea][0025cc58][0025cc5c] 8b4508      mov eax,[ebp+08] >>>>>>>>> [000012ed][0025cc54][000012e7] 50          push eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08      mov
ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51          push >>>>>>>>> ecx      // push P [000012f2][0025cc4c][000012f7] e880feffff >>>>>>>>> call 00001177 // call H Local Halt Decider: Infinite Recursion >>>>>>>>> Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly emulates >>>>>>>>> its input that it must emulate the first seven instructions of >>>>>>>>> P. Because the seventh instruction of P repeats this process we >>>>>>>>> know with complete certainty that the correct and complete
emulation of P by H would never reach its final “ret”
instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the procedure", >>>>>>>> because that H always has the option to abort its simulation,
just like this onne did, and return to its P and see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P)
conclusively proves that the complete and correct x86 emulation
would never stop running.

You SAY that, but you don't answer the actual questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly emulates its
input that it must emulate the first seven instructions of P.
Because the seventh instruction of P repeats this process we know
with complete certainty that the correct and complete emulation of
P by H would never reach its final “ret” instruction, thus never >>>>> halts.

If P should have halted (i.e. no infinite loop) then your simulation
detector, S (not H), gets the answer wrong.  You S is NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P >>> [00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P >>> [0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly emulates its
input that it must emulate the first seven instructions of P. Because
the seventh instruction of P repeats this process we know with
complete certainty that the correct and complete emulation of P by H
would never reach its final “ret” instruction, thus never halts.

We are going around and around and around in circles. I will try again:

If you replace the opcodes "EB FE" at 00001369 with the opcodes "90 90"
then your H gets the answer wrong: P should have halted.

/Flibble

As I already said before this is merely your cluelessness that when
H(P,P) is invoked the correct x86 emulation of the input to H(P,P) makes
and code after [0000135d] unreachable.

Wrong, because when that H return the value 0, it will get there.

0000135d is only unreachable if H never returns.

If H never returns, then if fails to be a decider, and thus can't be a
halt decider.

You need to get you definition of H straight, and until you do, you
can't make your arguement.

Either H does abort its simulation of it input of P,P, and return 0 and
thus P(P) is halting because that code is reached, or

H doesn't abort but just emulates as you othertimes claim, and this does
make P(P) non-halting, but it also make H(P,P) non-answering and thus H
is not a decider.

Your argument fails because you try to make be both options be true at
the same time but they are mutually exclusive of each other. You have
created a Liar's paradox, and are claiming that it is simultaniously
True and False and thus resolved.

You H is just not an actually definite function.

https://en.wikipedia.org/wiki/Unreachable_code

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P [00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P [0000135d](05)  e840feffff      call 000011a2 // call H [00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly emulates its input
that it must emulate the first seven instructions of P. Because the
seventh instruction of P repeats this process we know with complete
certainty that the correct and complete emulation of P by H would never
reach its final “ret” instruction, thus never halts.

The 7th instruction doesn't just repeat the process, unless H is defined
to be an unconditional execution of its input.

Since that is NOT your definition of H, that is a false statement, so
your logic is UNSOUND.

People that are not dumber than a box of rocks will understand this to
mean that the correct x86 emulation of the input to H(P,P) by H cannot possibly reach any instruction after [0000135d].

The correct x86 emulation of the input to H(P,P) will reach the final
ret instruction if H(P,P) returns 0, because it aborted its simulation
of its input (and thus doesn't meet the defintion of a 'correct' emulation).

If H DOES do a correct and complete emulation of its input, then it
never answers for H(P,P) so fails to be the decider it is claimed to be.

You are just caught in assuming something that isn't true, because you
claim to sides of a contradiction are both true.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>> [000012ed](01)  50              push eax
[000012ee](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>> [000012f1](01)  51              push ecx
[000012f2](05)  e880feffff      call 00001177 // call H >>>>>>>>>>> [000012f7](03)  83c408          add esp,+08
[000012fa](02)  85c0            test eax,eax >>>>>>>>>>> [000012fc](02)  7402            jz 00001300
[000012fe](02)  ebfe            jmp 000012fe >>>>>>>>>>> [00001300](01)  5d              pop ebp
[00001301](01)  c3              ret
Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 // push P >>>>>>>>>>> [0000130f](05)  e8d3ffffff      call 000012e7 // call P >>>>>>>>>>> [00001314](03)  83c404          add esp,+04
[00001317](02)  33c0            xor eax,eax
[00001319](01)  5d              pop ebp
[0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine    assembly
    address   address   data      code       language
    ========  ========  ========  =========  ============= >>>>>>>>>>> [00001307][00102190][00000000] 55         push ebp >>>>>>>>>>> [00001308][00102190][00000000] 8bec       mov ebp,esp >>>>>>>>>>> [0000130a][0010218c][000012e7] 68e7120000 push 000012e7 // >>>>>>>>>>> push P [0000130f][00102188][00001314] e8d3ffffff call 000012e7 >>>>>>>>>>> // call P [000012e7][00102184][00102190] 55         push ebp
    // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp >>>>>>>>>>> [000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08] >>>>>>>>>>> [000012ed][00102180][000012e7] 50         push eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov >>>>>>>>>>> ecx,[ebp+08] [000012f1][0010217c][000012e7] 51         push >>>>>>>>>>> ecx      // push P [000012f2][00102178][000012f7] e880feffff >>>>>>>>>>> call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace Stored >>>>>>>>>>> at:212244 [000012e7][00212230][00212234] 55          push ebp
    // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp >>>>>>>>>>> [000012ea][00212230][00212234] 8b4508      mov eax,[ebp+08] >>>>>>>>>>> [000012ed][0021222c][000012e7] 50          push eax      //
push P [000012ee][0021222c][000012e7] 8b4d08      mov >>>>>>>>>>> ecx,[ebp+08] [000012f1][00212228][000012e7] 51          push
ecx      // push P [000012f2][00212224][000012f7] e880feffff >>>>>>>>>>> call 00001177 // call H

So, by what instruction reference manual is a call 00001177 >>>>>>>>>> followedby the execution of the instruction at 000012e7.

Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp >>>>>>>>>>> [000012ea][0025cc58][0025cc5c] 8b4508      mov eax,[ebp+08] >>>>>>>>>>> [000012ed][0025cc54][000012e7] 50          push eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08      mov >>>>>>>>>>> ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51          push
ecx      // push P [000012f2][0025cc4c][000012f7] e880feffff >>>>>>>>>>> call 00001177 // call H Local Halt Decider: Infinite Recursion >>>>>>>>>>> Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly emulates >>>>>>>>>>> its input that it must emulate the first seven instructions of >>>>>>>>>>> P. Because the seventh instruction of P repeats this process we >>>>>>>>>>> know with complete certainty that the correct and complete >>>>>>>>>>> emulation of P by H would never reach its final “ret” >>>>>>>>>>> instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the procedure", >>>>>>>>>> because that H always has the option to abort its simulation, >>>>>>>>>> just like this onne did, and return to its P and see it halt. >>>>>>>>> THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL >>>>>>>>> REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P)
conclusively proves that the complete and correct x86 emulation >>>>>>>>> would never stop running.

You SAY that, but you don't answer the actual questions about HOW. >>>>>>>

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO EVIDENCE >>>>>>> WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY PROVES THAT >>>>>>> THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly emulates its >>>>>>> input that it must emulate the first seven instructions of P.
Because the seventh instruction of P repeats this process we know >>>>>>> with complete certainty that the correct and complete emulation of >>>>>>> P by H would never reach its final “ret” instruction, thus never >>>>>>> halts.

If P should have halted (i.e. no infinite loop) then your simulation >>>>>> detector, S (not H), gets the answer wrong.  You S is NOT a halting >>>>>> decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly emulates its
input that it must emulate the first seven instructions of P. Because >>>>> the seventh instruction of P repeats this process we know with
complete certainty that the correct and complete emulation of P by H >>>>> would never reach its final “ret” instruction, thus never halts.

We are going around and around and around in circles. I will try again: >>>>
If you replace the opcodes "EB FE" at 00001369 with the opcodes "90 90" >>>> then your H gets the answer wrong: P should have halted.

/Flibble

As I already said before this is merely your cluelessness that when
H(P,P) is invoked the correct x86 emulation of the input to H(P,P)
makes and code after [0000135d] unreachable.

Wrong, because when that H return the value 0, it will get there.

Like I said people that are dumber than a box of rocks won't be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to H(P,P) cannot
possibly reach any instruction beyond [0000135d].

So, you are defining that you H(P,P) never returns because it is caught
in the infinite rcursion.

Thats fine, just says it can't be the correctly answering decider you
claim it to be.

People that are very stupid might believe that once the emulation of the input to H(P,P) has been aborted that this dead process will magically
leap to [0000136c] even though it is a dead process.

What "Dead Process". We aren't asking about the partial simulation that
H is doing, we are asking about the ACTUAL machine that the input
represents.

H can't abort that, because nothing can. P. once designed instantly
creates a mapping from all input to results, and if H(P,P) returns 0,
that mapping is P(P) will Halt.

Maybe we just need to wait for the ultimate master to abort YOUR
emulation, leaving just a "dead process" and ending this debacle.

It is clear you don't understand what a Turing Machine is.

You can abort your emulation of the Turing Machine, you can't abort the
actual Turing Machine. Thats like defining a sequence of the numbers 1
to 10, but trying to stop in at 5.

Once you define the sequnce, it is there. Once you derine the Machine,
its behavior for all inputs is defined. Doesn't matter if you abort your simulation of it.

It seems your reasoning aborted some point a couple of decades ago.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>> [000012ed](01)  50              push eax
[000012ee](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>> [000012f1](01)  51              push ecx
[000012f2](05)  e880feffff      call 00001177 // call H >>>>>>>>>> [000012f7](03)  83c408          add esp,+08
[000012fa](02)  85c0            test eax,eax
[000012fc](02)  7402            jz 00001300
[000012fe](02)  ebfe            jmp 000012fe
[00001300](01)  5d              pop ebp
[00001301](01)  c3              ret
Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 // push P >>>>>>>>>> [0000130f](05)  e8d3ffffff      call 000012e7 // call P >>>>>>>>>> [00001314](03)  83c404          add esp,+04
[00001317](02)  33c0            xor eax,eax
[00001319](01)  5d              pop ebp
[0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine    assembly
    address   address   data      code       language
    ========  ========  ========  =========  ============= >>>>>>>>>> [00001307][00102190][00000000] 55         push ebp >>>>>>>>>> [00001308][00102190][00000000] 8bec       mov ebp,esp >>>>>>>>>> [0000130a][0010218c][000012e7] 68e7120000 push 000012e7 // >>>>>>>>>> push P [0000130f][00102188][00001314] e8d3ffffff call 000012e7 >>>>>>>>>> // call P [000012e7][00102184][00102190] 55         push ebp >>>>>>>>>>     // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp >>>>>>>>>> [000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08] >>>>>>>>>> [000012ed][00102180][000012e7] 50         push eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov
ecx,[ebp+08] [000012f1][0010217c][000012e7] 51         push >>>>>>>>>> ecx      // push P [000012f2][00102178][000012f7] e880feffff >>>>>>>>>> call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace Stored >>>>>>>>>> at:212244 [000012e7][00212230][00212234] 55          push ebp
    // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp >>>>>>>>>> [000012ea][00212230][00212234] 8b4508      mov eax,[ebp+08] >>>>>>>>>> [000012ed][0021222c][000012e7] 50          push eax      //
push P [000012ee][0021222c][000012e7] 8b4d08      mov >>>>>>>>>> ecx,[ebp+08] [000012f1][00212228][000012e7] 51          push
ecx      // push P [000012f2][00212224][000012f7] e880feffff >>>>>>>>>> call 00001177 // call H

So, by what instruction reference manual is a call 00001177
followedby the execution of the instruction at 000012e7.

Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp >>>>>>>>>> [000012ea][0025cc58][0025cc5c] 8b4508      mov eax,[ebp+08] >>>>>>>>>> [000012ed][0025cc54][000012e7] 50          push eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08      mov >>>>>>>>>> ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51          push
ecx      // push P [000012f2][0025cc4c][000012f7] e880feffff >>>>>>>>>> call 00001177 // call H Local Halt Decider: Infinite Recursion >>>>>>>>>> Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly emulates >>>>>>>>>> its input that it must emulate the first seven instructions of >>>>>>>>>> P. Because the seventh instruction of P repeats this process we >>>>>>>>>> know with complete certainty that the correct and complete >>>>>>>>>> emulation of P by H would never reach its final “ret”
instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the procedure", >>>>>>>>> because that H always has the option to abort its simulation, >>>>>>>>> just like this onne did, and return to its P and see it halt. >>>>>>>> THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL >>>>>>>> REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P)
conclusively proves that the complete and correct x86 emulation >>>>>>>> would never stop running.

You SAY that, but you don't answer the actual questions about HOW. >>>>>>

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly emulates its
input that it must emulate the first seven instructions of P.
Because the seventh instruction of P repeats this process we know
with complete certainty that the correct and complete emulation of >>>>>> P by H would never reach its final “ret” instruction, thus never >>>>>> halts.

If P should have halted (i.e. no infinite loop) then your simulation >>>>> detector, S (not H), gets the answer wrong.  You S is NOT a halting >>>>> decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P >>>> [00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P >>>> [0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly emulates its
input that it must emulate the first seven instructions of P. Because
the seventh instruction of P repeats this process we know with
complete certainty that the correct and complete emulation of P by H
would never reach its final “ret” instruction, thus never halts.

We are going around and around and around in circles. I will try again:

If you replace the opcodes "EB FE" at 00001369 with the opcodes "90 90"
then your H gets the answer wrong: P should have halted.

/Flibble

As I already said before this is merely your cluelessness that when
H(P,P) is invoked the correct x86 emulation of the input to H(P,P)
makes and code after [0000135d] unreachable.

Wrong, because when that H return the value 0, it will get there.

Like I said people that are dumber than a box of rocks won't be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to H(P,P) cannot
possibly reach any instruction beyond [0000135d].

People that are very stupid might believe that once the emulation of the
input to H(P,P) has been aborted that this dead process will magically
leap to [0000136c] even though it is a dead process.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp >>>>>>>>>>>> [000012ea](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>> [000012ed](01)  50              push eax >>>>>>>>>>>> [000012ee](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>> [000012f1](01)  51              push ecx >>>>>>>>>>>> [000012f2](05)  e880feffff      call 00001177 // call H >>>>>>>>>>>> [000012f7](03)  83c408          add esp,+08
[000012fa](02)  85c0            test eax,eax >>>>>>>>>>>> [000012fc](02)  7402            jz 00001300 >>>>>>>>>>>> [000012fe](02)  ebfe            jmp 000012fe >>>>>>>>>>>> [00001300](01)  5d              pop ebp
[00001301](01)  c3              ret
Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp >>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp >>>>>>>>>>>> [0000130a](05)  68e7120000      push 000012e7 // push P >>>>>>>>>>>> [0000130f](05)  e8d3ffffff      call 000012e7 // call P >>>>>>>>>>>> [00001314](03)  83c404          add esp,+04
[00001317](02)  33c0            xor eax,eax >>>>>>>>>>>> [00001319](01)  5d              pop ebp
[0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine    assembly
    address   address   data      code       language
    ========  ========  ========  =========  ============= >>>>>>>>>>>> [00001307][00102190][00000000] 55         push ebp >>>>>>>>>>>> [00001308][00102190][00000000] 8bec       mov ebp,esp >>>>>>>>>>>> [0000130a][0010218c][000012e7] 68e7120000 push 000012e7 // >>>>>>>>>>>> push P [0000130f][00102188][00001314] e8d3ffffff call 000012e7 >>>>>>>>>>>> // call P [000012e7][00102184][00102190] 55         push ebp
    // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp >>>>>>>>>>>> [000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08] >>>>>>>>>>>> [000012ed][00102180][000012e7] 50         push eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov >>>>>>>>>>>> ecx,[ebp+08] [000012f1][0010217c][000012e7] 51         push
ecx      // push P [000012f2][00102178][000012f7] e880feffff >>>>>>>>>>>> call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace Stored >>>>>>>>>>>> at:212244 [000012e7][00212230][00212234] 55          push ebp
    // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp >>>>>>>>>>>> [000012ea][00212230][00212234] 8b4508      mov eax,[ebp+08] >>>>>>>>>>>> [000012ed][0021222c][000012e7] 50          push eax      //
push P [000012ee][0021222c][000012e7] 8b4d08      mov >>>>>>>>>>>> ecx,[ebp+08] [000012f1][00212228][000012e7] 51          push
ecx      // push P [000012f2][00212224][000012f7] e880feffff >>>>>>>>>>>> call 00001177 // call H

So, by what instruction reference manual is a call 00001177 >>>>>>>>>>> followedby the execution of the instruction at 000012e7. >>>>>>>>>>>
Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp >>>>>>>>>>>> [000012ea][0025cc58][0025cc5c] 8b4508      mov eax,[ebp+08] >>>>>>>>>>>> [000012ed][0025cc54][000012e7] 50          push eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08      mov >>>>>>>>>>>> ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51          push
ecx      // push P [000012f2][0025cc4c][000012f7] e880feffff >>>>>>>>>>>> call 00001177 // call H Local Halt Decider: Infinite Recursion >>>>>>>>>>>> Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly emulates >>>>>>>>>>>> its input that it must emulate the first seven instructions of >>>>>>>>>>>> P. Because the seventh instruction of P repeats this process we >>>>>>>>>>>> know with complete certainty that the correct and complete >>>>>>>>>>>> emulation of P by H would never reach its final “ret” >>>>>>>>>>>> instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the procedure", >>>>>>>>>>> because that H always has the option to abort its simulation, >>>>>>>>>>> just like this onne did, and return to its P and see it halt. >>>>>>>>>> THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL >>>>>>>>>> REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P)
conclusively proves that the complete and correct x86 emulation >>>>>>>>>> would never stop running.

You SAY that, but you don't answer the actual questions about HOW. >>>>>>>>

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO EVIDENCE >>>>>>>> WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY PROVES THAT >>>>>>>> THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly emulates its >>>>>>>> input that it must emulate the first seven instructions of P.
Because the seventh instruction of P repeats this process we know >>>>>>>> with complete certainty that the correct and complete emulation of >>>>>>>> P by H would never reach its final “ret” instruction, thus never >>>>>>>> halts.

If P should have halted (i.e. no infinite loop) then your simulation >>>>>>> detector, S (not H), gets the answer wrong.  You S is NOT a halting >>>>>>> decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly emulates its
input that it must emulate the first seven instructions of P. Because >>>>>> the seventh instruction of P repeats this process we know with
complete certainty that the correct and complete emulation of P by H >>>>>> would never reach its final “ret” instruction, thus never halts. >>>>>

We are going around and around and around in circles. I will try
again:

If you replace the opcodes "EB FE" at 00001369 with the opcodes "90
90"
then your H gets the answer wrong: P should have halted.

/Flibble

As I already said before this is merely your cluelessness that when
H(P,P) is invoked the correct x86 emulation of the input to H(P,P)
makes and code after [0000135d] unreachable.

Wrong, because when that H return the value 0, it will get there.

Like I said people that are dumber than a box of rocks won't be able
to correctly understand this.

When H(P,P) is invoked the correctly emulated input to H(P,P) cannot
possibly reach any instruction beyond [0000135d].

So, you are defining that you H(P,P) never returns because it is caught
in the infinite rcursion.

Thats fine, just says it can't be the correctly answering decider you
claim it to be.

I have corrected you on this too many times.


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp >>>>>>>>>>>>> [000012ea](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>> [000012ed](01)  50              push eax >>>>>>>>>>>>> [000012ee](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>> [000012f1](01)  51              push ecx >>>>>>>>>>>>> [000012f2](05)  e880feffff      call 00001177 // call H >>>>>>>>>>>>> [000012f7](03)  83c408          add esp,+08 >>>>>>>>>>>>> [000012fa](02)  85c0            test eax,eax >>>>>>>>>>>>> [000012fc](02)  7402            jz 00001300 >>>>>>>>>>>>> [000012fe](02)  ebfe            jmp 000012fe >>>>>>>>>>>>> [00001300](01)  5d              pop ebp >>>>>>>>>>>>> [00001301](01)  c3              ret
Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp >>>>>>>>>>>>> [0000130a](05)  68e7120000      push 000012e7 // push P >>>>>>>>>>>>> [0000130f](05)  e8d3ffffff      call 000012e7 // call P >>>>>>>>>>>>> [00001314](03)  83c404          add esp,+04 >>>>>>>>>>>>> [00001317](02)  33c0            xor eax,eax >>>>>>>>>>>>> [00001319](01)  5d              pop ebp >>>>>>>>>>>>> [0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine    assembly
    address   address   data      code       language
    ========  ========  ========  =========  ============= >>>>>>>>>>>>> [00001307][00102190][00000000] 55         push ebp >>>>>>>>>>>>> [00001308][00102190][00000000] 8bec       mov ebp,esp >>>>>>>>>>>>> [0000130a][0010218c][000012e7] 68e7120000 push 000012e7 // >>>>>>>>>>>>> push P [0000130f][00102188][00001314] e8d3ffffff call 000012e7 >>>>>>>>>>>>> // call P [000012e7][00102184][00102190] 55         push ebp
    // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp >>>>>>>>>>>>> [000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08] >>>>>>>>>>>>> [000012ed][00102180][000012e7] 50         push eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov >>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0010217c][000012e7] 51         push
ecx      // push P [000012f2][00102178][000012f7] e880feffff >>>>>>>>>>>>> call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace Stored >>>>>>>>>>>>> at:212244 [000012e7][00212230][00212234] 55          push ebp
    // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp >>>>>>>>>>>>> [000012ea][00212230][00212234] 8b4508      mov eax,[ebp+08] >>>>>>>>>>>>> [000012ed][0021222c][000012e7] 50          push eax      //
push P [000012ee][0021222c][000012e7] 8b4d08      mov >>>>>>>>>>>>> ecx,[ebp+08] [000012f1][00212228][000012e7] 51          push
ecx      // push P [000012f2][00212224][000012f7] e880feffff >>>>>>>>>>>>> call 00001177 // call H

So, by what instruction reference manual is a call 00001177 >>>>>>>>>>>> followedby the execution of the instruction at 000012e7. >>>>>>>>>>>>
Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp >>>>>>>>>>>>> [000012ea][0025cc58][0025cc5c] 8b4508      mov eax,[ebp+08] >>>>>>>>>>>>> [000012ed][0025cc54][000012e7] 50          push eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08      mov >>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51          push
ecx      // push P [000012f2][0025cc4c][000012f7] e880feffff >>>>>>>>>>>>> call 00001177 // call H Local Halt Decider: Infinite Recursion >>>>>>>>>>>>> Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly emulates >>>>>>>>>>>>> its input that it must emulate the first seven instructions of >>>>>>>>>>>>> P. Because the seventh instruction of P repeats this >>>>>>>>>>>>> process we
know with complete certainty that the correct and complete >>>>>>>>>>>>> emulation of P by H would never reach its final “ret” >>>>>>>>>>>>> instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the
procedure",
because that H always has the option to abort its simulation, >>>>>>>>>>>> just like this onne did, and return to its P and see it halt. >>>>>>>>>>> THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL >>>>>>>>>>> REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P) >>>>>>>>>>> conclusively proves that the complete and correct x86 emulation >>>>>>>>>>> would never stop running.

You SAY that, but you don't answer the actual questions about >>>>>>>>>> HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO EVIDENCE >>>>>>>>> WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE CORRECT >>>>>>>>> PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY PROVES THAT >>>>>>>>> THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT TO H(P,P) >>>>>>>>> WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly emulates its >>>>>>>>> input that it must emulate the first seven instructions of P. >>>>>>>>> Because the seventh instruction of P repeats this process we know >>>>>>>>> with complete certainty that the correct and complete emulation of >>>>>>>>> P by H would never reach its final “ret” instruction, thus never >>>>>>>>> halts.

If P should have halted (i.e. no infinite loop) then your
simulation
detector, S (not H), gets the answer wrong.  You S is NOT a halting >>>>>>>> decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly emulates its >>>>>>> input that it must emulate the first seven instructions of P.
Because
the seventh instruction of P repeats this process we know with
complete certainty that the correct and complete emulation of P by H >>>>>>> would never reach its final “ret” instruction, thus never halts. >>>>>>

We are going around and around and around in circles. I will try
again:

If you replace the opcodes "EB FE" at 00001369 with the opcodes
"90 90"
then your H gets the answer wrong: P should have halted.

/Flibble

As I already said before this is merely your cluelessness that when
H(P,P) is invoked the correct x86 emulation of the input to H(P,P)
makes and code after [0000135d] unreachable.

Wrong, because when that H return the value 0, it will get there.

Like I said people that are dumber than a box of rocks won't be able
to correctly understand this.

When H(P,P) is invoked the correctly emulated input to H(P,P) cannot
possibly reach any instruction beyond [0000135d].

So, you are defining that you H(P,P) never returns because it is
caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering decider you
claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

If H(P,P) actually returns 0, then P(P) will Halt, even you have agreed
to that.

If H(P,P) does a correct simulation, and not abort, then while P(P)
never halts, H(P,P) never returns 0, so it fails.

You just keep claiming it does both at the same time.

Let see you actually program a computation to do that. I will ask you
what it the first instruction where the execution of the two cases
differ, and why did it differ.

All it shows is that your argument is caught in the Liar's paradox
because you are a Liar.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp >>>>>>>>>>>>>> [000012ea](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>> [000012ed](01)  50              push eax >>>>>>>>>>>>>> [000012ee](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>> [000012f1](01)  51              push ecx >>>>>>>>>>>>>> [000012f2](05)  e880feffff      call 00001177 // call H >>>>>>>>>>>>>> [000012f7](03)  83c408          add esp,+08 >>>>>>>>>>>>>> [000012fa](02)  85c0            test eax,eax >>>>>>>>>>>>>> [000012fc](02)  7402            jz 00001300 >>>>>>>>>>>>>> [000012fe](02)  ebfe            jmp 000012fe >>>>>>>>>>>>>> [00001300](01)  5d              pop ebp >>>>>>>>>>>>>> [00001301](01)  c3              ret
Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp >>>>>>>>>>>>>> [0000130a](05)  68e7120000      push 000012e7 // push P >>>>>>>>>>>>>> [0000130f](05)  e8d3ffffff      call 000012e7 // call P >>>>>>>>>>>>>> [00001314](03)  83c404          add esp,+04 >>>>>>>>>>>>>> [00001317](02)  33c0            xor eax,eax >>>>>>>>>>>>>> [00001319](01)  5d              pop ebp >>>>>>>>>>>>>> [0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine    assembly
    address   address   data      code       language
    ========  ========  ========  =========  =============
[00001307][00102190][00000000] 55         push ebp >>>>>>>>>>>>>> [00001308][00102190][00000000] 8bec       mov ebp,esp >>>>>>>>>>>>>> [0000130a][0010218c][000012e7] 68e7120000 push 000012e7 // >>>>>>>>>>>>>> push P [0000130f][00102188][00001314] e8d3ffffff call >>>>>>>>>>>>>> 000012e7
// call P [000012e7][00102184][00102190] 55         push ebp
    // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp >>>>>>>>>>>>>> [000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08] >>>>>>>>>>>>>> [000012ed][00102180][000012e7] 50         push eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov >>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0010217c][000012e7] 51         push
ecx      // push P [000012f2][00102178][000012f7] e880feffff
call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace Stored >>>>>>>>>>>>>> at:212244 [000012e7][00212230][00212234] 55          push ebp
    // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp >>>>>>>>>>>>>> [000012ea][00212230][00212234] 8b4508      mov eax,[ebp+08] >>>>>>>>>>>>>> [000012ed][0021222c][000012e7] 50          push eax      //
push P [000012ee][0021222c][000012e7] 8b4d08      mov >>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][00212228][000012e7] 51          push
ecx      // push P [000012f2][00212224][000012f7] e880feffff
call 00001177 // call H

So, by what instruction reference manual is a call 00001177 >>>>>>>>>>>>> followedby the execution of the instruction at 000012e7. >>>>>>>>>>>>>
Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp >>>>>>>>>>>>>> [000012ea][0025cc58][0025cc5c] 8b4508      mov eax,[ebp+08] >>>>>>>>>>>>>> [000012ed][0025cc54][000012e7] 50          push eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08      mov >>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51          push
ecx      // push P [000012f2][0025cc4c][000012f7] e880feffff
call 00001177 // call H Local Halt Decider: Infinite >>>>>>>>>>>>>> Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly emulates >>>>>>>>>>>>>> its input that it must emulate the first seven
instructions of
P. Because the seventh instruction of P repeats this >>>>>>>>>>>>>> process we
know with complete certainty that the correct and complete >>>>>>>>>>>>>> emulation of P by H would never reach its final “ret” >>>>>>>>>>>>>> instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the >>>>>>>>>>>>> procedure",
because that H always has the option to abort its simulation, >>>>>>>>>>>>> just like this onne did, and return to its P and see it halt. >>>>>>>>>>>> THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL >>>>>>>>>>>> REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P) >>>>>>>>>>>> conclusively proves that the complete and correct x86 emulation >>>>>>>>>>>> would never stop running.

You SAY that, but you don't answer the actual questions about >>>>>>>>>>> HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO EVIDENCE >>>>>>>>>> WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE CORRECT >>>>>>>>>> PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY PROVES THAT >>>>>>>>>> THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT TO H(P,P) >>>>>>>>>> WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly emulates its >>>>>>>>>> input that it must emulate the first seven instructions of P. >>>>>>>>>> Because the seventh instruction of P repeats this process we know >>>>>>>>>> with complete certainty that the correct and complete
emulation of
P by H would never reach its final “ret” instruction, thus never >>>>>>>>>> halts.

If P should have halted (i.e. no infinite loop) then your
simulation
detector, S (not H), gets the answer wrong.  You S is NOT a >>>>>>>>> halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL >>>>>>>> REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly emulates its >>>>>>>> input that it must emulate the first seven instructions of P.
Because
the seventh instruction of P repeats this process we know with >>>>>>>> complete certainty that the correct and complete emulation of P >>>>>>>> by H
would never reach its final “ret” instruction, thus never halts. >>>>>>>

We are going around and around and around in circles. I will try >>>>>>> again:

If you replace the opcodes "EB FE" at 00001369 with the opcodes
"90 90"
then your H gets the answer wrong: P should have halted.

/Flibble

As I already said before this is merely your cluelessness that
when H(P,P) is invoked the correct x86 emulation of the input to
H(P,P) makes and code after [0000135d] unreachable.

Wrong, because when that H return the value 0, it will get there.

Like I said people that are dumber than a box of rocks won't be able
to correctly understand this.

When H(P,P) is invoked the correctly emulated input to H(P,P) cannot
possibly reach any instruction beyond [0000135d].

So, you are defining that you H(P,P) never returns because it is
caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering decider you
claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know that H
performs a correct x86 emulation of its input and then examine the
execution trace.

There is no sense explaining HOW H determines that H(P,P)==0 is correct
until after there is a universal consensus that H(P,P)==0 is correct.




--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>> [000012ea](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>> [000012ed](01)  50              push eax >>>>>>>>>>>>>>> [000012ee](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>>> [000012f1](01)  51              push ecx >>>>>>>>>>>>>>> [000012f2](05)  e880feffff      call 00001177 // call H >>>>>>>>>>>>>>> [000012f7](03)  83c408          add esp,+08 >>>>>>>>>>>>>>> [000012fa](02)  85c0            test eax,eax >>>>>>>>>>>>>>> [000012fc](02)  7402            jz 00001300 >>>>>>>>>>>>>>> [000012fe](02)  ebfe            jmp 000012fe >>>>>>>>>>>>>>> [00001300](01)  5d              pop ebp >>>>>>>>>>>>>>> [00001301](01)  c3              ret >>>>>>>>>>>>>>> Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>> [0000130a](05)  68e7120000      push 000012e7 // push P >>>>>>>>>>>>>>> [0000130f](05)  e8d3ffffff      call 000012e7 // call P >>>>>>>>>>>>>>> [00001314](03)  83c404          add esp,+04 >>>>>>>>>>>>>>> [00001317](02)  33c0            xor eax,eax >>>>>>>>>>>>>>> [00001319](01)  5d              pop ebp >>>>>>>>>>>>>>> [0000131a](01)  c3              ret >>>>>>>>>>>>>>> Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine    assembly
    address   address   data      code       language
    ========  ========  ========  =========  =============
[00001307][00102190][00000000] 55         push ebp >>>>>>>>>>>>>>> [00001308][00102190][00000000] 8bec       mov ebp,esp >>>>>>>>>>>>>>> [0000130a][0010218c][000012e7] 68e7120000 push 000012e7 // >>>>>>>>>>>>>>> push P [0000130f][00102188][00001314] e8d3ffffff call >>>>>>>>>>>>>>> 000012e7
// call P [000012e7][00102184][00102190] 55         push ebp
    // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp >>>>>>>>>>>>>>> [000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08] >>>>>>>>>>>>>>> [000012ed][00102180][000012e7] 50         push eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov >>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0010217c][000012e7] 51         push
ecx      // push P [000012f2][00102178][000012f7] e880feffff
call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace Stored >>>>>>>>>>>>>>> at:212244 [000012e7][00212230][00212234] 55          push
ebp
    // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp >>>>>>>>>>>>>>> [000012ea][00212230][00212234] 8b4508      mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push eax      //
push P [000012ee][0021222c][000012e7] 8b4d08      mov >>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][00212228][000012e7] 51          push
ecx      // push P [000012f2][00212224][000012f7] e880feffff
call 00001177 // call H

So, by what instruction reference manual is a call 00001177 >>>>>>>>>>>>>> followedby the execution of the instruction at 000012e7. >>>>>>>>>>>>>>
Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp >>>>>>>>>>>>>>> [000012ea][0025cc58][0025cc5c] 8b4508      mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08      mov >>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51          push
ecx      // push P [000012f2][0025cc4c][000012f7] e880feffff
call 00001177 // call H Local Halt Decider: Infinite >>>>>>>>>>>>>>> Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly emulates >>>>>>>>>>>>>>> its input that it must emulate the first seven
instructions of
P. Because the seventh instruction of P repeats this >>>>>>>>>>>>>>> process we
know with complete certainty that the correct and complete >>>>>>>>>>>>>>> emulation of P by H would never reach its final “ret” >>>>>>>>>>>>>>> instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the >>>>>>>>>>>>>> procedure",
because that H always has the option to abort its simulation, >>>>>>>>>>>>>> just like this onne did, and return to its P and see it halt. >>>>>>>>>>>>> THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL >>>>>>>>>>>>> REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P) >>>>>>>>>>>>> conclusively proves that the complete and correct x86 >>>>>>>>>>>>> emulation
would never stop running.

You SAY that, but you don't answer the actual questions >>>>>>>>>>>> about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO EVIDENCE >>>>>>>>>>> WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE CORRECT >>>>>>>>>>> PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY PROVES >>>>>>>>>>> THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT TO H(P,P) >>>>>>>>>>> WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly emulates its >>>>>>>>>>> input that it must emulate the first seven instructions of P. >>>>>>>>>>> Because the seventh instruction of P repeats this process we >>>>>>>>>>> know
with complete certainty that the correct and complete
emulation of
P by H would never reach its final “ret” instruction, thus never
halts.

If P should have halted (i.e. no infinite loop) then your
simulation
detector, S (not H), gets the answer wrong.  You S is NOT a >>>>>>>>>> halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL >>>>>>>>> REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H >>>>>>>>> [00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly emulates its >>>>>>>>> input that it must emulate the first seven instructions of P. >>>>>>>>> Because
the seventh instruction of P repeats this process we know with >>>>>>>>> complete certainty that the correct and complete emulation of P >>>>>>>>> by H
would never reach its final “ret” instruction, thus never halts. >>>>>>>>

We are going around and around and around in circles. I will try >>>>>>>> again:

If you replace the opcodes "EB FE" at 00001369 with the opcodes >>>>>>>> "90 90"
then your H gets the answer wrong: P should have halted.

/Flibble

As I already said before this is merely your cluelessness that
when H(P,P) is invoked the correct x86 emulation of the input to >>>>>>> H(P,P) makes and code after [0000135d] unreachable.

Wrong, because when that H return the value 0, it will get there.

Like I said people that are dumber than a box of rocks won't be
able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to H(P,P)
cannot possibly reach any instruction beyond [0000135d].

So, you are defining that you H(P,P) never returns because it is
caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering decider
you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know that H
performs a correct x86 emulation of its input and then examine the
execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) returns 0, which
it can only do if it does not actually do a correct emulation

There is no sense explaining HOW H determines that H(P,P)==0 is correct
until after there is a universal consensus that H(P,P)==0 is correct.

It is correct ONLY if H(P,P) doesn't return 0.

Note, by definition, a CORRECT emuluation never ends for a non-halting
input.

Thus, for H to say its input is non-halting, H can NOT have actually
done a correct emulation.

It might have been able to prove the input was non-halting from its
partial simulation, as you show with infinite loop, but its emulation is
not BY DEFINITION correct.

For P, once you stipulate that H(P,P) will return 0, we can actually
prove that the correct emulation of that input will Halt. so the return
value of 0 can not be correct.

To claim that H can do BOTH a correct emulation and return the value 0
means, in effect, that you are declairing that infinity is a finite
number. You are stuck in your own liar's paradox.

The fact that you refuse to see this, just shows the limitations of your
mental capability.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>> [000012ea](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>>> [000012ed](01)  50              push eax >>>>>>>>>>>>>>>> [000012ee](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>>>> [000012f1](01)  51              push ecx >>>>>>>>>>>>>>>> [000012f2](05)  e880feffff      call 00001177 // call H >>>>>>>>>>>>>>>> [000012f7](03)  83c408          add esp,+08 >>>>>>>>>>>>>>>> [000012fa](02)  85c0            test eax,eax >>>>>>>>>>>>>>>> [000012fc](02)  7402            jz 00001300 >>>>>>>>>>>>>>>> [000012fe](02)  ebfe            jmp 000012fe >>>>>>>>>>>>>>>> [00001300](01)  5d              pop ebp >>>>>>>>>>>>>>>> [00001301](01)  c3              ret >>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>> [0000130a](05)  68e7120000      push 000012e7 // push P >>>>>>>>>>>>>>>> [0000130f](05)  e8d3ffffff      call 000012e7 // call P >>>>>>>>>>>>>>>> [00001314](03)  83c404          add esp,+04 >>>>>>>>>>>>>>>> [00001317](02)  33c0            xor eax,eax >>>>>>>>>>>>>>>> [00001319](01)  5d              pop ebp >>>>>>>>>>>>>>>> [0000131a](01)  c3              ret >>>>>>>>>>>>>>>> Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine    assembly
    address   address   data      code       language
    ========  ========  ========  =========  =============
[00001307][00102190][00000000] 55         push ebp >>>>>>>>>>>>>>>> [00001308][00102190][00000000] 8bec       mov ebp,esp >>>>>>>>>>>>>>>> [0000130a][0010218c][000012e7] 68e7120000 push 000012e7 // >>>>>>>>>>>>>>>> push P [0000130f][00102188][00001314] e8d3ffffff call >>>>>>>>>>>>>>>> 000012e7
// call P [000012e7][00102184][00102190] 55         push
ebp
    // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp >>>>>>>>>>>>>>>> [000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08] >>>>>>>>>>>>>>>> [000012ed][00102180][000012e7] 50         push eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov >>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0010217c][000012e7] 51         push
ecx      // push P [000012f2][00102178][000012f7] >>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace >>>>>>>>>>>>>>>> Stored
at:212244 [000012e7][00212230][00212234] 55
push ebp
    // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp >>>>>>>>>>>>>>>> [000012ea][00212230][00212234] 8b4508      mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push eax      //
push P [000012ee][0021222c][000012e7] 8b4d08      mov >>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][00212228][000012e7] 51 >>>>>>>>>>>>>>>> push
ecx      // push P [000012f2][00212224][000012f7] >>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H

So, by what instruction reference manual is a call 00001177 >>>>>>>>>>>>>>> followedby the execution of the instruction at 000012e7. >>>>>>>>>>>>>>>
Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp >>>>>>>>>>>>>>>> [000012ea][0025cc58][0025cc5c] 8b4508      mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08      mov >>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51 >>>>>>>>>>>>>>>> push
ecx      // push P [000012f2][0025cc4c][000012f7] >>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H Local Halt Decider: Infinite >>>>>>>>>>>>>>>> Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>> emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>> process we
know with complete certainty that the correct and complete >>>>>>>>>>>>>>>> emulation of P by H would never reach its final “ret” >>>>>>>>>>>>>>>> instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the >>>>>>>>>>>>>>> procedure",
because that H always has the option to abort its >>>>>>>>>>>>>>> simulation,
just like this onne did, and return to its P and see it >>>>>>>>>>>>>>> halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO >>>>>>>>>>>>>> ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P) >>>>>>>>>>>>>> conclusively proves that the complete and correct x86 >>>>>>>>>>>>>> emulation
would never stop running.

You SAY that, but you don't answer the actual questions >>>>>>>>>>>>> about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO >>>>>>>>>>>> EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE CORRECT >>>>>>>>>>>> PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY PROVES >>>>>>>>>>>> THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT TO H(P,P) >>>>>>>>>>>> WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly emulates >>>>>>>>>>>> its
input that it must emulate the first seven instructions of P. >>>>>>>>>>>> Because the seventh instruction of P repeats this process we >>>>>>>>>>>> know
with complete certainty that the correct and complete
emulation of
P by H would never reach its final “ret” instruction, thus >>>>>>>>>>>> never
halts.

If P should have halted (i.e. no infinite loop) then your >>>>>>>>>>> simulation
detector, S (not H), gets the answer wrong.  You S is NOT a >>>>>>>>>>> halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL >>>>>>>>>> REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>> [00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>> [0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H >>>>>>>>>> [00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly emulates its >>>>>>>>>> input that it must emulate the first seven instructions of P. >>>>>>>>>> Because
the seventh instruction of P repeats this process we know with >>>>>>>>>> complete certainty that the correct and complete emulation of >>>>>>>>>> P by H
would never reach its final “ret” instruction, thus never halts. >>>>>>>>>

We are going around and around and around in circles. I will >>>>>>>>> try again:

If you replace the opcodes "EB FE" at 00001369 with the opcodes >>>>>>>>> "90 90"
then your H gets the answer wrong: P should have halted.

/Flibble

As I already said before this is merely your cluelessness that >>>>>>>> when H(P,P) is invoked the correct x86 emulation of the input to >>>>>>>> H(P,P) makes and code after [0000135d] unreachable.

Wrong, because when that H return the value 0, it will get there. >>>>>> Like I said people that are dumber than a box of rocks won't be

able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to H(P,P)
cannot possibly reach any instruction beyond [0000135d].

So, you are defining that you H(P,P) never returns because it is
caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering decider
you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know that H
performs a correct x86 emulation of its input and then examine the
execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) returns 0, which
it can only do if it does not actually do a correct emulation

The correctly emulated input to H(P,P) never gets past its machine
address [0000135d].


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>> [000012ea](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>>>> [000012ed](01)  50              push eax >>>>>>>>>>>>>>>>> [000012ee](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>>>>> [000012f1](01)  51              push ecx >>>>>>>>>>>>>>>>> [000012f2](05)  e880feffff      call 00001177 // call H >>>>>>>>>>>>>>>>> [000012f7](03)  83c408          add esp,+08 >>>>>>>>>>>>>>>>> [000012fa](02)  85c0            test eax,eax >>>>>>>>>>>>>>>>> [000012fc](02)  7402            jz 00001300 >>>>>>>>>>>>>>>>> [000012fe](02)  ebfe            jmp 000012fe >>>>>>>>>>>>>>>>> [00001300](01)  5d              pop ebp >>>>>>>>>>>>>>>>> [00001301](01)  c3              ret >>>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>> [0000130a](05)  68e7120000      push 000012e7 // push P >>>>>>>>>>>>>>>>> [0000130f](05)  e8d3ffffff      call 000012e7 // call P >>>>>>>>>>>>>>>>> [00001314](03)  83c404          add esp,+04 >>>>>>>>>>>>>>>>> [00001317](02)  33c0            xor eax,eax >>>>>>>>>>>>>>>>> [00001319](01)  5d              pop ebp >>>>>>>>>>>>>>>>> [0000131a](01)  c3              ret >>>>>>>>>>>>>>>>> Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine    assembly
    address   address   data      code       language
    ========  ========  ========  =========  =============
[00001307][00102190][00000000] 55         push ebp >>>>>>>>>>>>>>>>> [00001308][00102190][00000000] 8bec       mov ebp,esp >>>>>>>>>>>>>>>>> [0000130a][0010218c][000012e7] 68e7120000 push 000012e7 // >>>>>>>>>>>>>>>>> push P [0000130f][00102188][00001314] e8d3ffffff call >>>>>>>>>>>>>>>>> 000012e7
// call P [000012e7][00102184][00102190] 55
push ebp
    // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp >>>>>>>>>>>>>>>>> [000012ea][00102184][00102190] 8b4508     mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov >>>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0010217c][000012e7] 51 >>>>>>>>>>>>>>>>> push
ecx      // push P [000012f2][00102178][000012f7] >>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace >>>>>>>>>>>>>>>>> Stored
at:212244 [000012e7][00212230][00212234] 55 push ebp >>>>>>>>>>>>>>>>>     // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp >>>>>>>>>>>>>>>>> [000012ea][00212230][00212234] 8b4508      mov >>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push >>>>>>>>>>>>>>>>> eax      //
push P [000012ee][0021222c][000012e7] 8b4d08      mov >>>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][00212228][000012e7] 51 push >>>>>>>>>>>>>>>>> ecx      // push P [000012f2][00212224][000012f7] >>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H

So, by what instruction reference manual is a call 00001177 >>>>>>>>>>>>>>>> followedby the execution of the instruction at 000012e7. >>>>>>>>>>>>>>>>
Your "CPU" is broken, or emulation incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55          push >>>>>>>>>>>>>>>>> ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp >>>>>>>>>>>>>>>>> [000012ea][0025cc58][0025cc5c] 8b4508      mov >>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push >>>>>>>>>>>>>>>>> eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08      mov >>>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51 push >>>>>>>>>>>>>>>>> ecx      // push P [000012f2][0025cc4c][000012f7] >>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H Local Halt Decider: Infinite >>>>>>>>>>>>>>>>> Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>> emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>> process we
know with complete certainty that the correct and complete >>>>>>>>>>>>>>>>> emulation of P by H would never reach its final “ret” >>>>>>>>>>>>>>>>> instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the >>>>>>>>>>>>>>>> procedure",
because that H always has the option to abort its >>>>>>>>>>>>>>>> simulation,
just like this onne did, and return to its P and see it >>>>>>>>>>>>>>>> halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO >>>>>>>>>>>>>>> ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P) >>>>>>>>>>>>>>> conclusively proves that the complete and correct x86 >>>>>>>>>>>>>>> emulation
would never stop running.

You SAY that, but you don't answer the actual questions >>>>>>>>>>>>>> about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO >>>>>>>>>>>>> EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE CORRECT >>>>>>>>>>>>> PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY >>>>>>>>>>>>> PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT TO H(P,P) >>>>>>>>>>>>> WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly
emulates its
input that it must emulate the first seven instructions of P. >>>>>>>>>>>>> Because the seventh instruction of P repeats this process >>>>>>>>>>>>> we know
with complete certainty that the correct and complete >>>>>>>>>>>>> emulation of
P by H would never reach its final “ret” instruction, thus >>>>>>>>>>>>> never
halts.

If P should have halted (i.e. no infinite loop) then your >>>>>>>>>>>> simulation
detector, S (not H), gets the answer wrong.  You S is NOT a >>>>>>>>>>>> halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL >>>>>>>>>>> REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>> [00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>> [0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H >>>>>>>>>>> [00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax >>>>>>>>>>> [00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369 >>>>>>>>>>> [0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly emulates its >>>>>>>>>>> input that it must emulate the first seven instructions of P. >>>>>>>>>>> Because
the seventh instruction of P repeats this process we know with >>>>>>>>>>> complete certainty that the correct and complete emulation of >>>>>>>>>>> P by H
would never reach its final “ret” instruction, thus never halts.

We are going around and around and around in circles. I will >>>>>>>>>> try again:

If you replace the opcodes "EB FE" at 00001369 with the
opcodes "90 90"
then your H gets the answer wrong: P should have halted.

/Flibble

As I already said before this is merely your cluelessness that >>>>>>>>> when H(P,P) is invoked the correct x86 emulation of the input >>>>>>>>> to H(P,P) makes and code after [0000135d] unreachable.

Wrong, because when that H return the value 0, it will get there. >>>>>>> Like I said people that are dumber than a box of rocks won't be

able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to H(P,P)
cannot possibly reach any instruction beyond [0000135d].

So, you are defining that you H(P,P) never returns because it is
caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering decider
you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know that H
performs a correct x86 emulation of its input and then examine the
execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) returns 0,
which it can only do if it does not actually do a correct emulation

The correctly emulated input to H(P,P) never gets past its machine
address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly return 0 if
it correctly emulates its input, but you can't drop that requirement.

You can state a requirement that make this true several ways, but
ultimately, you need something that constrains the behavior of H.

It can be that H doesn't return 0
It can be that H actaully does a correct simulation, and thus doesn't
abort its simulation.

You have a history of shading the truth, so you don't get a pass with
sloppy wording.

All statements must include the conditions that apply to them.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>> [000012ea](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>>>>> [000012ed](01)  50              push eax >>>>>>>>>>>>>>>>>> [000012ee](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>>>>>> [000012f1](01)  51              push ecx >>>>>>>>>>>>>>>>>> [000012f2](05)  e880feffff      call 00001177 // call H
[000012f7](03)  83c408          add esp,+08 >>>>>>>>>>>>>>>>>> [000012fa](02)  85c0            test eax,eax >>>>>>>>>>>>>>>>>> [000012fc](02)  7402            jz 00001300 >>>>>>>>>>>>>>>>>> [000012fe](02)  ebfe            jmp 000012fe >>>>>>>>>>>>>>>>>> [00001300](01)  5d              pop ebp >>>>>>>>>>>>>>>>>> [00001301](01)  c3              ret >>>>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>> [0000130a](05)  68e7120000      push 000012e7 // push P
[0000130f](05)  e8d3ffffff      call 000012e7 // call P
[00001314](03)  83c404          add esp,+04 >>>>>>>>>>>>>>>>>> [00001317](02)  33c0            xor eax,eax >>>>>>>>>>>>>>>>>> [00001319](01)  5d              pop ebp >>>>>>>>>>>>>>>>>> [0000131a](01)  c3              ret >>>>>>>>>>>>>>>>>> Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine    assembly
    address   address   data      code       language
    ========  ========  ========  ========= >>>>>>>>>>>>>>>>>> =============
[00001307][00102190][00000000] 55         push ebp >>>>>>>>>>>>>>>>>> [00001308][00102190][00000000] 8bec       mov ebp,esp >>>>>>>>>>>>>>>>>> [0000130a][0010218c][000012e7] 68e7120000 push >>>>>>>>>>>>>>>>>> 000012e7 //
push P [0000130f][00102188][00001314] e8d3ffffff call >>>>>>>>>>>>>>>>>> 000012e7
// call P [000012e7][00102184][00102190] 55 push ebp >>>>>>>>>>>>>>>>>>     // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp >>>>>>>>>>>>>>>>>> [000012ea][00102184][00102190] 8b4508     mov >>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push >>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov >>>>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0010217c][000012e7] 51 push >>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][00102178][000012f7] >>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace >>>>>>>>>>>>>>>>>> Stored
at:212244 [000012e7][00212230][00212234] 55 push ebp >>>>>>>>>>>>>>>>>>     // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp
[000012ea][00212230][00212234] 8b4508      mov >>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push >>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][0021222c][000012e7] 8b4d08      mov >>>>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][00212228][000012e7] 51 push >>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][00212224][000012f7] >>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H

So, by what instruction reference manual is a call >>>>>>>>>>>>>>>>> 00001177
followedby the execution of the instruction at 000012e7. >>>>>>>>>>>>>>>>>
Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55          push >>>>>>>>>>>>>>>>>> ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp
[000012ea][0025cc58][0025cc5c] 8b4508      mov >>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push >>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08      mov >>>>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51 push >>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][0025cc4c][000012f7] >>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H Local Halt Decider: Infinite >>>>>>>>>>>>>>>>>> Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>> emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>>> process we
know with complete certainty that the correct and >>>>>>>>>>>>>>>>>> complete
emulation of P by H would never reach its final “ret” >>>>>>>>>>>>>>>>>> instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the >>>>>>>>>>>>>>>>> procedure",
because that H always has the option to abort its >>>>>>>>>>>>>>>>> simulation,
just like this onne did, and return to its P and see it >>>>>>>>>>>>>>>>> halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO >>>>>>>>>>>>>>>> ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P) >>>>>>>>>>>>>>>> conclusively proves that the complete and correct x86 >>>>>>>>>>>>>>>> emulation
would never stop running.

You SAY that, but you don't answer the actual questions >>>>>>>>>>>>>>> about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO >>>>>>>>>>>>>> EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE CORRECT >>>>>>>>>>>>>> PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY >>>>>>>>>>>>>> PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT TO H(P,P) >>>>>>>>>>>>>> WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>> emulates its
input that it must emulate the first seven instructions of P. >>>>>>>>>>>>>> Because the seventh instruction of P repeats this process >>>>>>>>>>>>>> we know
with complete certainty that the correct and complete >>>>>>>>>>>>>> emulation of
P by H would never reach its final “ret” instruction, thus >>>>>>>>>>>>>> never
halts.

If P should have halted (i.e. no infinite loop) then your >>>>>>>>>>>>> simulation
detector, S (not H), gets the answer wrong.  You S is NOT a >>>>>>>>>>>>> halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL >>>>>>>>>>>> REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>> [00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>> [0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H >>>>>>>>>>>> [00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax >>>>>>>>>>>> [00001367](02)  7402            jz 0000136b >>>>>>>>>>>> [00001369](02)  ebfe            jmp 00001369 >>>>>>>>>>>> [0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly emulates >>>>>>>>>>>> its
input that it must emulate the first seven instructions of >>>>>>>>>>>> P. Because
the seventh instruction of P repeats this process we know with >>>>>>>>>>>> complete certainty that the correct and complete emulation >>>>>>>>>>>> of P by H
would never reach its final “ret” instruction, thus never >>>>>>>>>>>> halts.

We are going around and around and around in circles. I will >>>>>>>>>>> try again:

If you replace the opcodes "EB FE" at 00001369 with the
opcodes "90 90"
then your H gets the answer wrong: P should have halted. >>>>>>>>>>>
/Flibble

As I already said before this is merely your cluelessness that >>>>>>>>>> when H(P,P) is invoked the correct x86 emulation of the input >>>>>>>>>> to H(P,P) makes and code after [0000135d] unreachable.

Wrong, because when that H return the value 0, it will get there. >>>>>>>> Like I said people that are dumber than a box of rocks won't be >>>>>>>> able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to H(P,P)
cannot possibly reach any instruction beyond [0000135d].

So, you are defining that you H(P,P) never returns because it is >>>>>>> caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering decider >>>>>>> you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know that H
performs a correct x86 emulation of its input and then examine the
execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) returns 0,
which it can only do if it does not actually do a correct emulation

The correctly emulated input to H(P,P) never gets past its machine
address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly return 0 if
it correctly emulates its input, but you can't drop that requirement.

void P(u32 x)
{
if (H(x, x))
HERE: goto HERE;
return;
}

int main()
{
Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to main().
--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote:

On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>> [000012ea](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>>>>>> [000012ed](01)  50              push eax >>>>>>>>>>>>>>>>>>> [000012ee](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>>>>>>> [000012f1](01)  51              push ecx >>>>>>>>>>>>>>>>>>> [000012f2](05)  e880feffff      call 00001177 // call H
[000012f7](03)  83c408          add esp,+08 >>>>>>>>>>>>>>>>>>> [000012fa](02)  85c0            test eax,eax >>>>>>>>>>>>>>>>>>> [000012fc](02)  7402            jz 00001300 >>>>>>>>>>>>>>>>>>> [000012fe](02)  ebfe            jmp 000012fe >>>>>>>>>>>>>>>>>>> [00001300](01)  5d              pop ebp >>>>>>>>>>>>>>>>>>> [00001301](01)  c3              ret >>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>> [0000130a](05)  68e7120000      push 000012e7 // push P
[0000130f](05)  e8d3ffffff      call 000012e7 // call P
[00001314](03)  83c404          add esp,+04 >>>>>>>>>>>>>>>>>>> [00001317](02)  33c0            xor eax,eax >>>>>>>>>>>>>>>>>>> [00001319](01)  5d              pop ebp >>>>>>>>>>>>>>>>>>> [0000131a](01)  c3              ret >>>>>>>>>>>>>>>>>>> Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine    assembly
    address   address   data      code       language
    ========  ========  ========  ========= >>>>>>>>>>>>>>>>>>> =============
[00001307][00102190][00000000] 55         push ebp >>>>>>>>>>>>>>>>>>> [00001308][00102190][00000000] 8bec       mov ebp,esp >>>>>>>>>>>>>>>>>>> [0000130a][0010218c][000012e7] 68e7120000 push >>>>>>>>>>>>>>>>>>> 000012e7 //
push P [0000130f][00102188][00001314] e8d3ffffff call >>>>>>>>>>>>>>>>>>> 000012e7
// call P [000012e7][00102184][00102190] 55 push ebp >>>>>>>>>>>>>>>>>>>     // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp >>>>>>>>>>>>>>>>>>> [000012ea][00102184][00102190] 8b4508     mov >>>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push >>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov >>>>>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0010217c][000012e7] 51 push >>>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][00102178][000012f7] >>>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H

Begin Local Halt Decider Simulation   Execution Trace >>>>>>>>>>>>>>>>>>> Stored
at:212244 [000012e7][00212230][00212234] 55 push ebp >>>>>>>>>>>>>>>>>>>     // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp
[000012ea][00212230][00212234] 8b4508      mov >>>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push >>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][0021222c][000012e7] 8b4d08      mov >>>>>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][00212228][000012e7] 51 push >>>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][00212224][000012f7] >>>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H

So, by what instruction reference manual is a call >>>>>>>>>>>>>>>>>> 00001177
followedby the execution of the instruction at 000012e7. >>>>>>>>>>>>>>>>>>
Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55          push >>>>>>>>>>>>>>>>>>> ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp
[000012ea][0025cc58][0025cc5c] 8b4508      mov >>>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push >>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08      mov >>>>>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51 push >>>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][0025cc4c][000012f7] >>>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H Local Halt Decider: Infinite >>>>>>>>>>>>>>>>>>> Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>>> emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>>>> process we
know with complete certainty that the correct and >>>>>>>>>>>>>>>>>>> complete
emulation of P by H would never reach its final “ret” >>>>>>>>>>>>>>>>>>> instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the >>>>>>>>>>>>>>>>>> procedure",
because that H always has the option to abort its >>>>>>>>>>>>>>>>>> simulation,
just like this onne did, and return to its P and see >>>>>>>>>>>>>>>>>> it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO >>>>>>>>>>>>>>>>> ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P) >>>>>>>>>>>>>>>>> conclusively proves that the complete and correct x86 >>>>>>>>>>>>>>>>> emulation
would never stop running.

You SAY that, but you don't answer the actual questions >>>>>>>>>>>>>>>> about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO >>>>>>>>>>>>>>> EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE >>>>>>>>>>>>>>> CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY >>>>>>>>>>>>>>> PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT TO >>>>>>>>>>>>>>> H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven instructions >>>>>>>>>>>>>>> of P.
Because the seventh instruction of P repeats this process >>>>>>>>>>>>>>> we know
with complete certainty that the correct and complete >>>>>>>>>>>>>>> emulation of
P by H would never reach its final “ret” instruction, >>>>>>>>>>>>>>> thus never
halts.

If P should have halted (i.e. no infinite loop) then your >>>>>>>>>>>>>> simulation
detector, S (not H), gets the answer wrong.  You S is NOT >>>>>>>>>>>>>> a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO ACTUAL >>>>>>>>>>>>> REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>> [00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>> [0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H >>>>>>>>>>>>> [00001362](03)  83c408          add esp,+08 >>>>>>>>>>>>> [00001365](02)  85c0            test eax,eax >>>>>>>>>>>>> [00001367](02)  7402            jz 0000136b >>>>>>>>>>>>> [00001369](02)  ebfe            jmp 00001369 >>>>>>>>>>>>> [0000136b](01)  5d              pop ebp >>>>>>>>>>>>> [0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly
emulates its
input that it must emulate the first seven instructions of >>>>>>>>>>>>> P. Because
the seventh instruction of P repeats this process we know with >>>>>>>>>>>>> complete certainty that the correct and complete emulation >>>>>>>>>>>>> of P by H
would never reach its final “ret” instruction, thus never >>>>>>>>>>>>> halts.

We are going around and around and around in circles. I will >>>>>>>>>>>> try again:

If you replace the opcodes "EB FE" at 00001369 with the >>>>>>>>>>>> opcodes "90 90"
then your H gets the answer wrong: P should have halted. >>>>>>>>>>>>
/Flibble

As I already said before this is merely your cluelessness >>>>>>>>>>> that when H(P,P) is invoked the correct x86 emulation of the >>>>>>>>>>> input to H(P,P) makes and code after [0000135d] unreachable. >>>>>>>>>>

Wrong, because when that H return the value 0, it will get there. >>>>>>>>> Like I said people that are dumber than a box of rocks won't be >>>>>>>>> able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to H(P,P) >>>>>>>>> cannot possibly reach any instruction beyond [0000135d].

So, you are defining that you H(P,P) never returns because it is >>>>>>>> caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering
decider you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know that H
performs a correct x86 emulation of its input and then examine the
execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) returns 0,
which it can only do if it does not actually do a correct emulation

The correctly emulated input to H(P,P) never gets past its machine
address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly return 0
if it correctly emulates its input, but you can't drop that requirement.

void P(u32 x)
{
  if (H(x, x))
    HERE: goto HERE;
  return;
}

int main()
{
  Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to main().

But it also return 0 to the computation P(P), maybe not the copy that it
is simulating, since it aborts that before it get to it, but it will in
the CORRECT SIMULATION of its input, which is the code of P with an
input of the code of P, which IS P(P). (Remember, correct simluation
don't abort, since programs don't just disappear).

Until you can prove that H can have two different results for identical
code running with identical inputs, which violates fundamental proofs in compuatation theory, you can't make that claim.

You like to claim that this isn't true, and if you COULD prove it, that
would make you famous, but you can't.

To prove your assertion, what is the first x86 instruction in the
operation of the H(P,P) called from main behaves differently then that
exact same x86 instruction in the operation o the H(P,P) call from P(P),
which generates a different resultant state.

If there is no first difference, there can't be ANY difference if H(P,P)
is a finite compuatiom, and thus H(P,P) does return 0 to P(P).

Go on, show your proof and make yourself famous, or show you have just
been lying (or stupid) and just ignoring this fundamental property of Computations.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/2022 4:54 PM, Richard Damon wrote:

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>> [000012ea](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>>>>>>> [000012ed](01)  50              push eax >>>>>>>>>>>>>>>>>>>> [000012ee](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>>>>>>>> [000012f1](01)  51              push ecx >>>>>>>>>>>>>>>>>>>> [000012f2](05)  e880feffff      call 00001177 // call H
[000012f7](03)  83c408          add esp,+08 >>>>>>>>>>>>>>>>>>>> [000012fa](02)  85c0            test eax,eax >>>>>>>>>>>>>>>>>>>> [000012fc](02)  7402            jz 00001300 >>>>>>>>>>>>>>>>>>>> [000012fe](02)  ebfe            jmp 000012fe >>>>>>>>>>>>>>>>>>>> [00001300](01)  5d              pop ebp >>>>>>>>>>>>>>>>>>>> [00001301](01)  c3              ret >>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>> [0000130a](05)  68e7120000      push 000012e7 // push P
[0000130f](05)  e8d3ffffff      call 000012e7 // call P
[00001314](03)  83c404          add esp,+04 >>>>>>>>>>>>>>>>>>>> [00001317](02)  33c0            xor eax,eax >>>>>>>>>>>>>>>>>>>> [00001319](01)  5d              pop ebp >>>>>>>>>>>>>>>>>>>> [0000131a](01)  c3              ret >>>>>>>>>>>>>>>>>>>> Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine    assembly
    address   address   data      code       language
    ========  ========  ========  ========= >>>>>>>>>>>>>>>>>>>> =============
[00001307][00102190][00000000] 55         push ebp >>>>>>>>>>>>>>>>>>>> [00001308][00102190][00000000] 8bec       mov ebp,esp
[0000130a][0010218c][000012e7] 68e7120000 push >>>>>>>>>>>>>>>>>>>> 000012e7 //
push P [0000130f][00102188][00001314] e8d3ffffff >>>>>>>>>>>>>>>>>>>> call 000012e7
// call P [000012e7][00102184][00102190] 55 push ebp >>>>>>>>>>>>>>>>>>>>     // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp
[000012ea][00102184][00102190] 8b4508     mov >>>>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push >>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov >>>>>>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0010217c][000012e7] 51 push >>>>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][00102178][000012f7] >>>>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H

Begin Local Halt Decider Simulation   Execution >>>>>>>>>>>>>>>>>>>> Trace Stored
at:212244 [000012e7][00212230][00212234] 55 push ebp >>>>>>>>>>>>>>>>>>>>     // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp
[000012ea][00212230][00212234] 8b4508      mov >>>>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push >>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][0021222c][000012e7] 8b4d08      mov >>>>>>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][00212228][000012e7] 51 push >>>>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][00212224][000012f7] >>>>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H

So, by what instruction reference manual is a call >>>>>>>>>>>>>>>>>>> 00001177
followedby the execution of the instruction at 000012e7. >>>>>>>>>>>>>>>>>>>
Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55          push >>>>>>>>>>>>>>>>>>>> ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp
[000012ea][0025cc58][0025cc5c] 8b4508      mov >>>>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push >>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08      mov >>>>>>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51 push >>>>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][0025cc4c][000012f7] >>>>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H Local Halt Decider: Infinite >>>>>>>>>>>>>>>>>>>> Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>>>> emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>>>>> process we
know with complete certainty that the correct and >>>>>>>>>>>>>>>>>>>> complete
emulation of P by H would never reach its final “ret” >>>>>>>>>>>>>>>>>>>> instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the >>>>>>>>>>>>>>>>>>> procedure",
because that H always has the option to abort its >>>>>>>>>>>>>>>>>>> simulation,
just like this onne did, and return to its P and see >>>>>>>>>>>>>>>>>>> it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>>> NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P) >>>>>>>>>>>>>>>>>> conclusively proves that the complete and correct x86 >>>>>>>>>>>>>>>>>> emulation
would never stop running.

You SAY that, but you don't answer the actual questions >>>>>>>>>>>>>>>>> about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO >>>>>>>>>>>>>>>> EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE >>>>>>>>>>>>>>>> CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY >>>>>>>>>>>>>>>> PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT TO >>>>>>>>>>>>>>>> H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven instructions >>>>>>>>>>>>>>>> of P.
Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>> process we know
with complete certainty that the correct and complete >>>>>>>>>>>>>>>> emulation of
P by H would never reach its final “ret” instruction, >>>>>>>>>>>>>>>> thus never
halts.

If P should have halted (i.e. no infinite loop) then your >>>>>>>>>>>>>>> simulation
detector, S (not H), gets the answer wrong.  You S is NOT >>>>>>>>>>>>>>> a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO >>>>>>>>>>>>>> ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>> [00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>> [0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H >>>>>>>>>>>>>> [00001362](03)  83c408          add esp,+08 >>>>>>>>>>>>>> [00001365](02)  85c0            test eax,eax >>>>>>>>>>>>>> [00001367](02)  7402            jz 0000136b >>>>>>>>>>>>>> [00001369](02)  ebfe            jmp 00001369 >>>>>>>>>>>>>> [0000136b](01)  5d              pop ebp >>>>>>>>>>>>>> [0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>> emulates its
input that it must emulate the first seven instructions of >>>>>>>>>>>>>> P. Because
the seventh instruction of P repeats this process we know >>>>>>>>>>>>>> with
complete certainty that the correct and complete emulation >>>>>>>>>>>>>> of P by H
would never reach its final “ret” instruction, thus never >>>>>>>>>>>>>> halts.

We are going around and around and around in circles. I >>>>>>>>>>>>> will try again:

If you replace the opcodes "EB FE" at 00001369 with the >>>>>>>>>>>>> opcodes "90 90"
then your H gets the answer wrong: P should have halted. >>>>>>>>>>>>>
/Flibble

As I already said before this is merely your cluelessness >>>>>>>>>>>> that when H(P,P) is invoked the correct x86 emulation of the >>>>>>>>>>>> input to H(P,P) makes and code after [0000135d] unreachable. >>>>>>>>>>>

Wrong, because when that H return the value 0, it will get >>>>>>>>>>> there.

Like I said people that are dumber than a box of rocks won't >>>>>>>>>> be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to H(P,P) >>>>>>>>>> cannot possibly reach any instruction beyond [0000135d].

So, you are defining that you H(P,P) never returns because it >>>>>>>>> is caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering
decider you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know that
H performs a correct x86 emulation of its input and then examine
the execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) returns 0,
which it can only do if it does not actually do a correct emulation

The correctly emulated input to H(P,P) never gets past its machine
address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly return 0
if it correctly emulates its input, but you can't drop that requirement.

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to main().

But it also return 0 to the computation P(P), maybe not the copy that it
is simulating, since it aborts that before it get to it,

Finally you are not stupid or deceptive.



--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On Thu, 9 Jun 2022 17:09:08 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 4:54 PM, Richard Damon wrote:

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>> On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>> [000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax >>>>>>>>>>>>>>>>>>>> [000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx >>>>>>>>>>>>>>>>>>>> [000012f2](05)  e880feffff      call 00001177 // >>>>>>>>>>>>>>>>>>>> call H [000012f7](03)  83c408          add >>>>>>>>>>>>>>>>>>>> esp,+08 [000012fa](02)  85c0            test >>>>>>>>>>>>>>>>>>>> eax,eax [000012fc](02)  7402            jz >>>>>>>>>>>>>>>>>>>> 00001300 [000012fe](02)  ebfe            jmp >>>>>>>>>>>>>>>>>>>> 000012fe [00001300](01)  5d              pop ebp
[00001301](01)  c3              ret >>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>> [0000130a](05)  68e7120000      push 000012e7 // >>>>>>>>>>>>>>>>>>>> push P [0000130f](05)  e8d3ffffff      call >>>>>>>>>>>>>>>>>>>> 000012e7 // call P [00001314](03)  83c404 >>>>>>>>>>>>>>>>>>>>   add esp,+04 [00001317](02)  33c0 >>>>>>>>>>>>>>>>>>>> xor eax,eax [00001319](01)  5d              pop
ebp [0000131a](01)  c3              ret >>>>>>>>>>>>>>>>>>>> Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine >>>>>>>>>>>>>>>>>>>> assembly address   address   data      code >>>>>>>>>>>>>>>>>>>> language ========  ========  ======== >>>>>>>>>>>>>>>>>>>> ========= =============
[00001307][00102190][00000000] 55         push >>>>>>>>>>>>>>>>>>>> ebp [00001308][00102190][00000000] 8bec >>>>>>>>>>>>>>>>>>>> mov ebp,esp [0000130a][0010218c][000012e7] >>>>>>>>>>>>>>>>>>>> 68e7120000 push 000012e7 //
push P [0000130f][00102188][00001314] e8d3ffffff >>>>>>>>>>>>>>>>>>>> call 000012e7
// call P [000012e7][00102184][00102190] 55 push >>>>>>>>>>>>>>>>>>>> ebp // enter executed P
[000012e8][00102184][00102190] 8bec       mov >>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00102184][00102190] 8b4508 >>>>>>>>>>>>>>>>>>>> mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push >>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][00102180][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08] [000012f1][0010217c][000012e7] >>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P
[000012f2][00102178][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>> call 00001177 // call H

Begin Local Halt Decider Simulation   Execution >>>>>>>>>>>>>>>>>>>> Trace Stored
at:212244 [000012e7][00212230][00212234] 55 push >>>>>>>>>>>>>>>>>>>> ebp // enter emulated P
[000012e8][00212230][00212234] 8bec        mov >>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00212230][00212234] 8b4508 >>>>>>>>>>>>>>>>>>>>   mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push
eax      //
push P [000012ee][0021222c][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08] [000012f1][00212228][000012e7] >>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P
[000012f2][00212224][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>> call 00001177 // call H

So, by what instruction reference manual is a >>>>>>>>>>>>>>>>>>> call 00001177
followedby the execution of the instruction at >>>>>>>>>>>>>>>>>>> 000012e7.

Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55          push
ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov >>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][0025cc58][0025cc5c] 8b4508 >>>>>>>>>>>>>>>>>>>>   mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push
eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08] [000012f1][0025cc50][000012e7] >>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P
[000012f2][0025cc4c][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>> call 00001177 // call H Local Halt Decider: >>>>>>>>>>>>>>>>>>>> Infinite Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>> correctly emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats >>>>>>>>>>>>>>>>>>>> this process we
know with complete certainty that the correct >>>>>>>>>>>>>>>>>>>> and complete
emulation of P by H would never reach its final >>>>>>>>>>>>>>>>>>>> “ret” instruction, thus never halts. >>>>>>>>>>>>>>>>>>>

Problem, the 7th intruction DOESN't "Just repeat >>>>>>>>>>>>>>>>>>> the procedure",
because that H always has the option to abort its >>>>>>>>>>>>>>>>>>> simulation,
just like this onne did, and return to its P and >>>>>>>>>>>>>>>>>>> see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS >>>>>>>>>>>>>>>>>> IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to >>>>>>>>>>>>>>>>>> H(P,P) conclusively proves that the complete and >>>>>>>>>>>>>>>>>> correct x86 emulation
would never stop running.

You SAY that, but you don't answer the actual >>>>>>>>>>>>>>>>> questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>> NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT >>>>>>>>>>>>>>>> THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P)
CONCLUSIVELY PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT >>>>>>>>>>>>>>>> TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven
instructions of P.
Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>> process we know
with complete certainty that the correct and >>>>>>>>>>>>>>>> complete emulation of
P by H would never reach its final “ret” >>>>>>>>>>>>>>>> instruction, thus never
halts.

If P should have halted (i.e. no infinite loop) then >>>>>>>>>>>>>>> your simulation
detector, S (not H), gets the answer wrong.  You S is >>>>>>>>>>>>>>> NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>> NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>> [00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>> [0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H >>>>>>>>>>>>>> [00001362](03)  83c408          add esp,+08 >>>>>>>>>>>>>> [00001365](02)  85c0            test eax,eax >>>>>>>>>>>>>> [00001367](02)  7402            jz 0000136b >>>>>>>>>>>>>> [00001369](02)  ebfe            jmp 00001369 >>>>>>>>>>>>>> [0000136b](01)  5d              pop ebp >>>>>>>>>>>>>> [0000136c](01)  c3              ret >>>>>>>>>>>>>> Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>> emulates its
input that it must emulate the first seven
instructions of P. Because
the seventh instruction of P repeats this process we >>>>>>>>>>>>>> know with
complete certainty that the correct and complete >>>>>>>>>>>>>> emulation of P by H
would never reach its final “ret” instruction, thus >>>>>>>>>>>>>> never halts.

We are going around and around and around in circles. I >>>>>>>>>>>>> will try again:

If you replace the opcodes "EB FE" at 00001369 with the >>>>>>>>>>>>> opcodes "90 90"
then your H gets the answer wrong: P should have halted. >>>>>>>>>>>>>
/Flibble

As I already said before this is merely your
cluelessness that when H(P,P) is invoked the correct x86 >>>>>>>>>>>> emulation of the input to H(P,P) makes and code after >>>>>>>>>>>> [0000135d] unreachable.

Wrong, because when that H return the value 0, it will >>>>>>>>>>> get there.

Like I said people that are dumber than a box of rocks
won't be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to
H(P,P) cannot possibly reach any instruction beyond
[0000135d].

So, you are defining that you H(P,P) never returns because >>>>>>>>> it is caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering >>>>>>>>> decider you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know
that H performs a correct x86 emulation of its input and then
examine the execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) returns
0, which it can only do if it does not actually do a correct
emulation

The correctly emulated input to H(P,P) never gets past its
machine address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly
return 0 if it correctly emulates its input, but you can't drop
that requirement.

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to
main().

But it also return 0 to the computation P(P), maybe not the copy
that it is simulating, since it aborts that before it get to it,

Finally you are not stupid or deceptive.

If H never returns a value to P then H is not a halting decider; if H
returns a value of 0 to main() when P halts then H is not a halting
decider.

H is not a halting decider; H is a simulation detector, S.

/Flibble

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On Thu, 9 Jun 2022 17:18:38 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 5:15 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:09:08 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 4:54 PM, Richard Damon wrote:

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 6/9/22 11:47 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>> void P(u32 x)

{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov >>>>>>>>>>>>>>>>>>>>>> eax,[ebp+08] [000012ed](01)  50 >>>>>>>>>>>>>>>>>>>>>> push eax [000012ee](03)  8b4d08          mov >>>>>>>>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1](01)  51 >>>>>>>>>>>>>>>>>>>>>> push ecx [000012f2](05)  e880feffff      call >>>>>>>>>>>>>>>>>>>>>> 00001177 // call H [000012f7](03)  83c408 >>>>>>>>>>>>>>>>>>>>>>     add esp,+08 [000012fa](02)  85c0 >>>>>>>>>>>>>>>>>>>>>> test eax,eax [000012fc](02)  7402 >>>>>>>>>>>>>>>>>>>>>> jz 00001300 [000012fe](02)  ebfe            jmp
000012fe [00001300](01)  5d              pop
ebp [00001301](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 >>>>>>>>>>>>>>>>>>>>>> // push P [0000130f](05)  e8d3ffffff      call >>>>>>>>>>>>>>>>>>>>>> 000012e7 // call P [00001314](03)  83c404 >>>>>>>>>>>>>>>>>>>>>>   add esp,+04 [00001317](02)  33c0 >>>>>>>>>>>>>>>>>>>>>> xor eax,eax [00001319](01)  5d              pop
ebp [0000131a](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine
assembly address   address   data      code >>>>>>>>>>>>>>>>>>>>>> language ========  ========  ======== >>>>>>>>>>>>>>>>>>>>>> ========= =============
[00001307][00102190][00000000] 55         push >>>>>>>>>>>>>>>>>>>>>> ebp [00001308][00102190][00000000] 8bec >>>>>>>>>>>>>>>>>>>>>> mov ebp,esp [0000130a][0010218c][000012e7] >>>>>>>>>>>>>>>>>>>>>> 68e7120000 push 000012e7 //
push P [0000130f][00102188][00001314] >>>>>>>>>>>>>>>>>>>>>> e8d3ffffff call 000012e7
// call P [000012e7][00102184][00102190] 55 >>>>>>>>>>>>>>>>>>>>>> push ebp // enter executed P
[000012e8][00102184][00102190] 8bec       mov >>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00102184][00102190] 8b4508 >>>>>>>>>>>>>>>>>>>>>> mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push >>>>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][00102180][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08] [000012f1][0010217c][000012e7] >>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P
[000012f2][00102178][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H

Begin Local Halt Decider Simulation   Execution >>>>>>>>>>>>>>>>>>>>>> Trace Stored
at:212244 [000012e7][00212230][00212234] 55 >>>>>>>>>>>>>>>>>>>>>> push ebp // enter emulated P
[000012e8][00212230][00212234] 8bec        mov >>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00212230][00212234] 8b4508 >>>>>>>>>>>>>>>>>>>>>>   mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push
eax      //
push P [000012ee][0021222c][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08]
[000012f1][00212228][000012e7] 51 push ecx >>>>>>>>>>>>>>>>>>>>>> // push P [000012f2][00212224][000012f7] >>>>>>>>>>>>>>>>>>>>>> e880feffff call 00001177 // call H >>>>>>>>>>>>>>>>>>>>>

So, by what instruction reference manual is a >>>>>>>>>>>>>>>>>>>>> call 00001177
followedby the execution of the instruction at >>>>>>>>>>>>>>>>>>>>> 000012e7.

Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55          push
ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov >>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][0025cc58][0025cc5c] 8b4508 >>>>>>>>>>>>>>>>>>>>>>   mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push
eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08]
[000012f1][0025cc50][000012e7] 51 push ecx >>>>>>>>>>>>>>>>>>>>>> // push P [000012f2][0025cc4c][000012f7] >>>>>>>>>>>>>>>>>>>>>> e880feffff call 00001177 // call H Local Halt >>>>>>>>>>>>>>>>>>>>>> Decider: Infinite Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>>>> correctly emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats >>>>>>>>>>>>>>>>>>>>>> this process we
know with complete certainty that the correct >>>>>>>>>>>>>>>>>>>>>> and complete
emulation of P by H would never reach its final >>>>>>>>>>>>>>>>>>>>>> “ret” instruction, thus never halts. >>>>>>>>>>>>>>>>>>>>>

Problem, the 7th intruction DOESN't "Just repeat >>>>>>>>>>>>>>>>>>>>> the procedure",
because that H always has the option to abort >>>>>>>>>>>>>>>>>>>>> its simulation,
just like this onne did, and return to its P and >>>>>>>>>>>>>>>>>>>>> see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS >>>>>>>>>>>>>>>>>>>> IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to >>>>>>>>>>>>>>>>>>>> H(P,P) conclusively proves that the complete and >>>>>>>>>>>>>>>>>>>> correct x86 emulation
would never stop running.

You SAY that, but you don't answer the actual >>>>>>>>>>>>>>>>>>> questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS >>>>>>>>>>>>>>>>>> IS NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT >>>>>>>>>>>>>>>>>> THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) >>>>>>>>>>>>>>>>>> CONCLUSIVELY PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT >>>>>>>>>>>>>>>>>> TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>> instructions of P.
Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>>> process we know
with complete certainty that the correct and >>>>>>>>>>>>>>>>>> complete emulation of
P by H would never reach its final “ret” >>>>>>>>>>>>>>>>>> instruction, thus never
halts.

If P should have halted (i.e. no infinite loop) then >>>>>>>>>>>>>>>>> your simulation
detector, S (not H), gets the answer wrong.  You S >>>>>>>>>>>>>>>>> is NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>> NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>>> [00001358](01)  50              push eax      //
push P [00001359](03)  8b4d08          mov >>>>>>>>>>>>>>>> ecx,[ebp+08] [0000135c](01)  51              push
ecx      // push P [0000135d](05)  e840feffff >>>>>>>>>>>>>>>> call 000011a2 // call H [00001362](03)  83c408 >>>>>>>>>>>>>>>>    add esp,+08 [00001365](02)  85c0            test
eax,eax [00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369 >>>>>>>>>>>>>>>> [0000136b](01)  5d              pop ebp >>>>>>>>>>>>>>>> [0000136c](01)  c3              ret >>>>>>>>>>>>>>>> Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven
instructions of P. Because
the seventh instruction of P repeats this process we >>>>>>>>>>>>>>>> know with
complete certainty that the correct and complete >>>>>>>>>>>>>>>> emulation of P by H
would never reach its final “ret” instruction, thus >>>>>>>>>>>>>>>> never halts.

We are going around and around and around in circles. >>>>>>>>>>>>>>> I will try again:

If you replace the opcodes "EB FE" at 00001369 with >>>>>>>>>>>>>>> the opcodes "90 90"
then your H gets the answer wrong: P should have >>>>>>>>>>>>>>> halted.

/Flibble

As I already said before this is merely your
cluelessness that when H(P,P) is invoked the correct >>>>>>>>>>>>>> x86 emulation of the input to H(P,P) makes and code >>>>>>>>>>>>>> after [0000135d] unreachable.

Wrong, because when that H return the value 0, it will >>>>>>>>>>>>> get there.

Like I said people that are dumber than a box of rocks >>>>>>>>>>>> won't be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to >>>>>>>>>>>> H(P,P) cannot possibly reach any instruction beyond
[0000135d].

So, you are defining that you H(P,P) never returns because >>>>>>>>>>> it is caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering >>>>>>>>>>> decider you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know >>>>>>>> that H performs a correct x86 emulation of its input and then >>>>>>>> examine the execution trace.

And a CORRECT emulation of the code will Halt if H(P,P)
returns 0, which it can only do if it does not actually do a
correct emulation

The correctly emulated input to H(P,P) never gets past its
machine address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly
return 0 if it correctly emulates its input, but you can't drop
that requirement.

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to
main().

But it also return 0 to the computation P(P), maybe not the copy
that it is simulating, since it aborts that before it get to it,

Finally you are not stupid or deceptive.

If H never returns a value to P then H is not a halting decider; if
H returns a value of 0 to main() when P halts then H is not a
halting decider.

H is not a halting decider; H is a simulation detector, S.

/Flibble

In other words you are saying the infintely nested emulation only
excutes for a few steps and then it stops on its own.

I am saying that your H is not a halting decider; I am saying that your
H is just a simulation detector, S.

/Flibble

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/22 6:09 PM, olcott wrote:

On 6/9/2022 4:54 PM, Richard Damon wrote:

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote:

On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>>> [000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax >>>>>>>>>>>>>>>>>>>>> [000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx >>>>>>>>>>>>>>>>>>>>> [000012f2](05)  e880feffff      call 00001177 // >>>>>>>>>>>>>>>>>>>>> call H
[000012f7](03)  83c408          add esp,+08 >>>>>>>>>>>>>>>>>>>>> [000012fa](02)  85c0            test eax,eax >>>>>>>>>>>>>>>>>>>>> [000012fc](02)  7402            jz 00001300 >>>>>>>>>>>>>>>>>>>>> [000012fe](02)  ebfe            jmp 000012fe >>>>>>>>>>>>>>>>>>>>> [00001300](01)  5d              pop ebp >>>>>>>>>>>>>>>>>>>>> [00001301](01)  c3              ret >>>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>>> [0000130a](05)  68e7120000      push 000012e7 // >>>>>>>>>>>>>>>>>>>>> push P
[0000130f](05)  e8d3ffffff      call 000012e7 // >>>>>>>>>>>>>>>>>>>>> call P
[00001314](03)  83c404          add esp,+04 >>>>>>>>>>>>>>>>>>>>> [00001317](02)  33c0            xor eax,eax >>>>>>>>>>>>>>>>>>>>> [00001319](01)  5d              pop ebp >>>>>>>>>>>>>>>>>>>>> [0000131a](01)  c3              ret >>>>>>>>>>>>>>>>>>>>> Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine    assembly
    address   address   data      code       language
    ========  ========  ========  ========= >>>>>>>>>>>>>>>>>>>>> =============
[00001307][00102190][00000000] 55         push ebp
[00001308][00102190][00000000] 8bec       mov ebp,esp
[0000130a][0010218c][000012e7] 68e7120000 push >>>>>>>>>>>>>>>>>>>>> 000012e7 //
push P [0000130f][00102188][00001314] e8d3ffffff >>>>>>>>>>>>>>>>>>>>> call 000012e7
// call P [000012e7][00102184][00102190] 55 push ebp >>>>>>>>>>>>>>>>>>>>>     // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp
[000012ea][00102184][00102190] 8b4508     mov >>>>>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push >>>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov >>>>>>>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0010217c][000012e7] 51 push >>>>>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][00102178][000012f7] >>>>>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H

Begin Local Halt Decider Simulation   Execution >>>>>>>>>>>>>>>>>>>>> Trace Stored
at:212244 [000012e7][00212230][00212234] 55 push ebp >>>>>>>>>>>>>>>>>>>>>     // enter emulated P
[000012e8][00212230][00212234] 8bec        mov ebp,esp
[000012ea][00212230][00212234] 8b4508      mov >>>>>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push >>>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][0021222c][000012e7] 8b4d08      mov
ecx,[ebp+08] [000012f1][00212228][000012e7] 51 push >>>>>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][00212224][000012f7] >>>>>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H

So, by what instruction reference manual is a call >>>>>>>>>>>>>>>>>>>> 00001177
followedby the execution of the instruction at >>>>>>>>>>>>>>>>>>>> 000012e7.

Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55          push >>>>>>>>>>>>>>>>>>>>> ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov ebp,esp
[000012ea][0025cc58][0025cc5c] 8b4508      mov >>>>>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push >>>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08      mov
ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51 push >>>>>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][0025cc4c][000012f7] >>>>>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H Local Halt Decider: >>>>>>>>>>>>>>>>>>>>> Infinite Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>>>>> emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats >>>>>>>>>>>>>>>>>>>>> this process we
know with complete certainty that the correct and >>>>>>>>>>>>>>>>>>>>> complete
emulation of P by H would never reach its final “ret” >>>>>>>>>>>>>>>>>>>>> instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat the >>>>>>>>>>>>>>>>>>>> procedure",
because that H always has the option to abort its >>>>>>>>>>>>>>>>>>>> simulation,
just like this onne did, and return to its P and see >>>>>>>>>>>>>>>>>>>> it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>>>> NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to H(P,P) >>>>>>>>>>>>>>>>>>> conclusively proves that the complete and correct x86 >>>>>>>>>>>>>>>>>>> emulation
would never stop running.

You SAY that, but you don't answer the actual >>>>>>>>>>>>>>>>>> questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO >>>>>>>>>>>>>>>>> EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE >>>>>>>>>>>>>>>>> CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY >>>>>>>>>>>>>>>>> PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT TO >>>>>>>>>>>>>>>>> H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven instructions >>>>>>>>>>>>>>>>> of P.
Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>> process we know
with complete certainty that the correct and complete >>>>>>>>>>>>>>>>> emulation of
P by H would never reach its final “ret” instruction, >>>>>>>>>>>>>>>>> thus never
halts.

If P should have halted (i.e. no infinite loop) then >>>>>>>>>>>>>>>> your simulation
detector, S (not H), gets the answer wrong.  You S is >>>>>>>>>>>>>>>> NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO >>>>>>>>>>>>>>> ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>> [00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>>> [0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H >>>>>>>>>>>>>>> [00001362](03)  83c408          add esp,+08 >>>>>>>>>>>>>>> [00001365](02)  85c0            test eax,eax >>>>>>>>>>>>>>> [00001367](02)  7402            jz 0000136b >>>>>>>>>>>>>>> [00001369](02)  ebfe            jmp 00001369 >>>>>>>>>>>>>>> [0000136b](01)  5d              pop ebp >>>>>>>>>>>>>>> [0000136c](01)  c3              ret >>>>>>>>>>>>>>> Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven instructions >>>>>>>>>>>>>>> of P. Because
the seventh instruction of P repeats this process we know >>>>>>>>>>>>>>> with
complete certainty that the correct and complete >>>>>>>>>>>>>>> emulation of P by H
would never reach its final “ret” instruction, thus never >>>>>>>>>>>>>>> halts.

We are going around and around and around in circles. I >>>>>>>>>>>>>> will try again:

If you replace the opcodes "EB FE" at 00001369 with the >>>>>>>>>>>>>> opcodes "90 90"
then your H gets the answer wrong: P should have halted. >>>>>>>>>>>>>>
/Flibble

As I already said before this is merely your cluelessness >>>>>>>>>>>>> that when H(P,P) is invoked the correct x86 emulation of >>>>>>>>>>>>> the input to H(P,P) makes and code after [0000135d]
unreachable.

Wrong, because when that H return the value 0, it will get >>>>>>>>>>>> there.

Like I said people that are dumber than a box of rocks won't >>>>>>>>>>> be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to H(P,P) >>>>>>>>>>> cannot possibly reach any instruction beyond [0000135d].

So, you are defining that you H(P,P) never returns because it >>>>>>>>>> is caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering
decider you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know that >>>>>>> H performs a correct x86 emulation of its input and then examine >>>>>>> the execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) returns 0, >>>>>> which it can only do if it does not actually do a correct emulation >>>>>>

The correctly emulated input to H(P,P) never gets past its machine
address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly return 0
if it correctly emulates its input, but you can't drop that
requirement.

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to main().

But it also return 0 to the computation P(P), maybe not the copy that
it is simulating, since it aborts that before it get to it,

Finally you are not stupid or deceptive.

So, you agree that your H(P,P) that returns a 0, claiming it to be
correct, also returns that 0 to P(P) and thus P(P) Halts, and thus
H(P,P) is INCORRECT in saying that its input represents a non-halting computation?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/2022 5:15 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:09:08 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 4:54 PM, Richard Damon wrote:

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>>>> [000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax >>>>>>>>>>>>>>>>>>>>>> [000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx >>>>>>>>>>>>>>>>>>>>>> [000012f2](05)  e880feffff      call 00001177 // >>>>>>>>>>>>>>>>>>>>>> call H [000012f7](03)  83c408          add >>>>>>>>>>>>>>>>>>>>>> esp,+08 [000012fa](02)  85c0            test >>>>>>>>>>>>>>>>>>>>>> eax,eax [000012fc](02)  7402            jz >>>>>>>>>>>>>>>>>>>>>> 00001300 [000012fe](02)  ebfe            jmp >>>>>>>>>>>>>>>>>>>>>> 000012fe [00001300](01)  5d              pop ebp
[00001301](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>>>> [0000130a](05)  68e7120000      push 000012e7 // >>>>>>>>>>>>>>>>>>>>>> push P [0000130f](05)  e8d3ffffff      call >>>>>>>>>>>>>>>>>>>>>> 000012e7 // call P [00001314](03)  83c404 >>>>>>>>>>>>>>>>>>>>>>   add esp,+04 [00001317](02)  33c0 >>>>>>>>>>>>>>>>>>>>>> xor eax,eax [00001319](01)  5d              pop
ebp [0000131a](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine >>>>>>>>>>>>>>>>>>>>>> assembly address   address   data      code >>>>>>>>>>>>>>>>>>>>>> language ========  ========  ======== >>>>>>>>>>>>>>>>>>>>>> ========= =============
[00001307][00102190][00000000] 55         push >>>>>>>>>>>>>>>>>>>>>> ebp [00001308][00102190][00000000] 8bec >>>>>>>>>>>>>>>>>>>>>> mov ebp,esp [0000130a][0010218c][000012e7] >>>>>>>>>>>>>>>>>>>>>> 68e7120000 push 000012e7 //
push P [0000130f][00102188][00001314] e8d3ffffff >>>>>>>>>>>>>>>>>>>>>> call 000012e7
// call P [000012e7][00102184][00102190] 55 push >>>>>>>>>>>>>>>>>>>>>> ebp // enter executed P
[000012e8][00102184][00102190] 8bec       mov >>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00102184][00102190] 8b4508 >>>>>>>>>>>>>>>>>>>>>> mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push >>>>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][00102180][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08] [000012f1][0010217c][000012e7] >>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P
[000012f2][00102178][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H

Begin Local Halt Decider Simulation   Execution >>>>>>>>>>>>>>>>>>>>>> Trace Stored
at:212244 [000012e7][00212230][00212234] 55 push >>>>>>>>>>>>>>>>>>>>>> ebp // enter emulated P
[000012e8][00212230][00212234] 8bec        mov >>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00212230][00212234] 8b4508 >>>>>>>>>>>>>>>>>>>>>>   mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push >>>>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][0021222c][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08] [000012f1][00212228][000012e7] >>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P
[000012f2][00212224][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H

So, by what instruction reference manual is a >>>>>>>>>>>>>>>>>>>>> call 00001177
followedby the execution of the instruction at >>>>>>>>>>>>>>>>>>>>> 000012e7.

Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55          push >>>>>>>>>>>>>>>>>>>>>> ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov >>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][0025cc58][0025cc5c] 8b4508 >>>>>>>>>>>>>>>>>>>>>>   mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push >>>>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08] [000012f1][0025cc50][000012e7] >>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P
[000012f2][0025cc4c][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H Local Halt Decider: >>>>>>>>>>>>>>>>>>>>>> Infinite Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>>>> correctly emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats >>>>>>>>>>>>>>>>>>>>>> this process we
know with complete certainty that the correct >>>>>>>>>>>>>>>>>>>>>> and complete
emulation of P by H would never reach its final >>>>>>>>>>>>>>>>>>>>>> “ret” instruction, thus never halts. >>>>>>>>>>>>>>>>>>>>>

Problem, the 7th intruction DOESN't "Just repeat >>>>>>>>>>>>>>>>>>>>> the procedure",
because that H always has the option to abort its >>>>>>>>>>>>>>>>>>>>> simulation,
just like this onne did, and return to its P and >>>>>>>>>>>>>>>>>>>>> see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS >>>>>>>>>>>>>>>>>>>> IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to >>>>>>>>>>>>>>>>>>>> H(P,P) conclusively proves that the complete and >>>>>>>>>>>>>>>>>>>> correct x86 emulation
would never stop running.

You SAY that, but you don't answer the actual >>>>>>>>>>>>>>>>>>> questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>>> NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT >>>>>>>>>>>>>>>>>> THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P)
CONCLUSIVELY PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT >>>>>>>>>>>>>>>>>> TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>> instructions of P.
Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>>> process we know
with complete certainty that the correct and >>>>>>>>>>>>>>>>>> complete emulation of
P by H would never reach its final “ret” >>>>>>>>>>>>>>>>>> instruction, thus never
halts.

If P should have halted (i.e. no infinite loop) then >>>>>>>>>>>>>>>>> your simulation
detector, S (not H), gets the answer wrong.  You S is >>>>>>>>>>>>>>>>> NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>> NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>>> [00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>>>> [0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H >>>>>>>>>>>>>>>> [00001362](03)  83c408          add esp,+08 >>>>>>>>>>>>>>>> [00001365](02)  85c0            test eax,eax >>>>>>>>>>>>>>>> [00001367](02)  7402            jz 0000136b >>>>>>>>>>>>>>>> [00001369](02)  ebfe            jmp 00001369 >>>>>>>>>>>>>>>> [0000136b](01)  5d              pop ebp >>>>>>>>>>>>>>>> [0000136c](01)  c3              ret >>>>>>>>>>>>>>>> Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven
instructions of P. Because
the seventh instruction of P repeats this process we >>>>>>>>>>>>>>>> know with
complete certainty that the correct and complete >>>>>>>>>>>>>>>> emulation of P by H
would never reach its final “ret” instruction, thus >>>>>>>>>>>>>>>> never halts.

We are going around and around and around in circles. I >>>>>>>>>>>>>>> will try again:

If you replace the opcodes "EB FE" at 00001369 with the >>>>>>>>>>>>>>> opcodes "90 90"
then your H gets the answer wrong: P should have halted. >>>>>>>>>>>>>>>
/Flibble

As I already said before this is merely your
cluelessness that when H(P,P) is invoked the correct x86 >>>>>>>>>>>>>> emulation of the input to H(P,P) makes and code after >>>>>>>>>>>>>> [0000135d] unreachable.

Wrong, because when that H return the value 0, it will >>>>>>>>>>>>> get there.

Like I said people that are dumber than a box of rocks >>>>>>>>>>>> won't be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to >>>>>>>>>>>> H(P,P) cannot possibly reach any instruction beyond
[0000135d].

So, you are defining that you H(P,P) never returns because >>>>>>>>>>> it is caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering >>>>>>>>>>> decider you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know
that H performs a correct x86 emulation of its input and then
examine the execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) returns
0, which it can only do if it does not actually do a correct
emulation

The correctly emulated input to H(P,P) never gets past its
machine address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly
return 0 if it correctly emulates its input, but you can't drop
that requirement.

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to
main().

But it also return 0 to the computation P(P), maybe not the copy
that it is simulating, since it aborts that before it get to it,

Finally you are not stupid or deceptive.

If H never returns a value to P then H is not a halting decider; if H
returns a value of 0 to main() when P halts then H is not a halting
decider.

H is not a halting decider; H is a simulation detector, S.

/Flibble

In other words you are saying the infintely nested emulation only
excutes for a few steps and then it stops on its own.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/22 6:18 PM, olcott wrote:

On 6/9/2022 5:15 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:09:08 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 4:54 PM, Richard Damon wrote:

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 6/9/22 11:47 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> void P(u32 x)

{
      if (H(x, x))
        HERE: goto HERE;
      return;
}

int main()
{
      P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>>>>> [000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax >>>>>>>>>>>>>>>>>>>>>>> [000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx >>>>>>>>>>>>>>>>>>>>>>> [000012f2](05)  e880feffff      call 00001177 // >>>>>>>>>>>>>>>>>>>>>>> call H [000012f7](03)  83c408          add >>>>>>>>>>>>>>>>>>>>>>> esp,+08 [000012fa](02)  85c0            test
eax,eax [000012fc](02)  7402            jz >>>>>>>>>>>>>>>>>>>>>>> 00001300 [000012fe](02)  ebfe            jmp
000012fe [00001300](01)  5d              pop ebp
[00001301](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>>>>> [0000130a](05)  68e7120000      push 000012e7 // >>>>>>>>>>>>>>>>>>>>>>> push P [0000130f](05)  e8d3ffffff      call >>>>>>>>>>>>>>>>>>>>>>> 000012e7 // call P [00001314](03)  83c404 >>>>>>>>>>>>>>>>>>>>>>>    add esp,+04 [00001317](02)  33c0 >>>>>>>>>>>>>>>>>>>>>>> xor eax,eax [00001319](01)  5d              pop
ebp [0000131a](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0020) [0000131a]

     machine   stack     stack     machine
assembly address   address   data      code >>>>>>>>>>>>>>>>>>>>>>>   language ========  ========  ======== >>>>>>>>>>>>>>>>>>>>>>> ========= =============
[00001307][00102190][00000000] 55         push >>>>>>>>>>>>>>>>>>>>>>> ebp [00001308][00102190][00000000] 8bec >>>>>>>>>>>>>>>>>>>>>>> mov ebp,esp [0000130a][0010218c][000012e7] >>>>>>>>>>>>>>>>>>>>>>> 68e7120000 push 000012e7 //
push P [0000130f][00102188][00001314] e8d3ffffff >>>>>>>>>>>>>>>>>>>>>>> call 000012e7
// call P [000012e7][00102184][00102190] 55 push >>>>>>>>>>>>>>>>>>>>>>> ebp // enter executed P
[000012e8][00102184][00102190] 8bec       mov >>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00102184][00102190] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>   mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push >>>>>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][00102180][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08] [000012f1][0010217c][000012e7] >>>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P
[000012f2][00102178][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H

Begin Local Halt Decider Simulation   Execution >>>>>>>>>>>>>>>>>>>>>>> Trace Stored
at:212244 [000012e7][00212230][00212234] 55 push >>>>>>>>>>>>>>>>>>>>>>> ebp // enter emulated P
[000012e8][00212230][00212234] 8bec        mov >>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00212230][00212234] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>    mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push
eax      //
push P [000012ee][0021222c][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>   mov ecx,[ebp+08] [000012f1][00212228][000012e7] >>>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P
[000012f2][00212224][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H

So, by what instruction reference manual is a >>>>>>>>>>>>>>>>>>>>>> call 00001177
followedby the execution of the instruction at >>>>>>>>>>>>>>>>>>>>>> 000012e7.

Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55          push
ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov >>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][0025cc58][0025cc5c] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>    mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push
eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>   mov ecx,[ebp+08] [000012f1][0025cc50][000012e7] >>>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P
[000012f2][0025cc4c][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H Local Halt Decider: >>>>>>>>>>>>>>>>>>>>>>> Infinite Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>>>>> correctly emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats >>>>>>>>>>>>>>>>>>>>>>> this process we
know with complete certainty that the correct >>>>>>>>>>>>>>>>>>>>>>> and complete
emulation of P by H would never reach its final >>>>>>>>>>>>>>>>>>>>>>> “ret” instruction, thus never halts. >>>>>>>>>>>>>>>>>>>>>>

Problem, the 7th intruction DOESN't "Just repeat >>>>>>>>>>>>>>>>>>>>>> the procedure",
because that H always has the option to abort its >>>>>>>>>>>>>>>>>>>>>> simulation,
just like this onne did, and return to its P and >>>>>>>>>>>>>>>>>>>>>> see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS >>>>>>>>>>>>>>>>>>>>> IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to >>>>>>>>>>>>>>>>>>>>> H(P,P) conclusively proves that the complete and >>>>>>>>>>>>>>>>>>>>> correct x86 emulation
would never stop running.

You SAY that, but you don't answer the actual >>>>>>>>>>>>>>>>>>>> questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>>>> NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT >>>>>>>>>>>>>>>>>>> THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) >>>>>>>>>>>>>>>>>>> CONCLUSIVELY PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT >>>>>>>>>>>>>>>>>>> TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>>> instructions of P.
Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>>>> process we know
with complete certainty that the correct and >>>>>>>>>>>>>>>>>>> complete emulation of
P by H would never reach its final “ret” >>>>>>>>>>>>>>>>>>> instruction, thus never
halts.

If P should have halted (i.e. no infinite loop) then >>>>>>>>>>>>>>>>>> your simulation
detector, S (not H), gets the answer wrong.  You S is >>>>>>>>>>>>>>>>>> NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>> NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>>>> [00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>>>>> [0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H >>>>>>>>>>>>>>>>> [00001362](03)  83c408          add esp,+08 >>>>>>>>>>>>>>>>> [00001365](02)  85c0            test eax,eax >>>>>>>>>>>>>>>>> [00001367](02)  7402            jz 0000136b >>>>>>>>>>>>>>>>> [00001369](02)  ebfe            jmp 00001369 >>>>>>>>>>>>>>>>> [0000136b](01)  5d              pop ebp >>>>>>>>>>>>>>>>> [0000136c](01)  c3              ret >>>>>>>>>>>>>>>>> Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven
instructions of P. Because
the seventh instruction of P repeats this process we >>>>>>>>>>>>>>>>> know with
complete certainty that the correct and complete >>>>>>>>>>>>>>>>> emulation of P by H
would never reach its final “ret” instruction, thus >>>>>>>>>>>>>>>>> never halts.

We are going around and around and around in circles. I >>>>>>>>>>>>>>>> will try again:

If you replace the opcodes "EB FE" at 00001369 with the >>>>>>>>>>>>>>>> opcodes "90 90"
then your H gets the answer wrong: P should have halted. >>>>>>>>>>>>>>>>
/Flibble

As I already said before this is merely your
cluelessness that when H(P,P) is invoked the correct x86 >>>>>>>>>>>>>>> emulation of the input to H(P,P) makes and code after >>>>>>>>>>>>>>> [0000135d] unreachable.

Wrong, because when that H return the value 0, it will >>>>>>>>>>>>>> get there.

Like I said people that are dumber than a box of rocks >>>>>>>>>>>>> won't be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to >>>>>>>>>>>>> H(P,P) cannot possibly reach any instruction beyond
[0000135d].

So, you are defining that you H(P,P) never returns because >>>>>>>>>>>> it is caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering >>>>>>>>>>>> decider you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know >>>>>>>>> that H performs a correct x86 emulation of its input and then >>>>>>>>> examine the execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) returns >>>>>>>> 0, which it can only do if it does not actually do a correct
emulation

The correctly emulated input to H(P,P) never gets past its
machine address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly
return 0 if it correctly emulates its input, but you can't drop
that requirement.

void P(u32 x)
{
    if (H(x, x))
      HERE: goto HERE;
    return;
}

int main()
{
    Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to
main().

But it also return 0 to the computation P(P), maybe not the copy
that it is simulating, since it aborts that before it get to it,

Finally you are not stupid or deceptive.

If H never returns a value to P then H is not a halting decider; if H
returns a value of 0 to main() when P halts then H is not a halting
decider.

H is not a halting decider; H is a simulation detector, S.

/Flibble

In other words you are saying the infintely nested emulation only
excutes for a few steps and then it stops on its own.

No, the "infinitely nested emulation" halts when the first H(P,P) that
P(P) called aborts it simulation.

--- SoupGate-Win32 v1.05
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XPost: comp.theory, sci.logic, sci.math

On 6/9/2022 5:17 PM, Richard Damon wrote:

On 6/9/22 6:09 PM, olcott wrote:

On 6/9/2022 4:54 PM, Richard Damon wrote:

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 6/9/22 11:47 AM, olcott wrote:

void P(u32 x)
{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>>>> [000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax >>>>>>>>>>>>>>>>>>>>>> [000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx >>>>>>>>>>>>>>>>>>>>>> [000012f2](05)  e880feffff      call 00001177 // >>>>>>>>>>>>>>>>>>>>>> call H
[000012f7](03)  83c408          add esp,+08 >>>>>>>>>>>>>>>>>>>>>> [000012fa](02)  85c0            test eax,eax >>>>>>>>>>>>>>>>>>>>>> [000012fc](02)  7402            jz 00001300 >>>>>>>>>>>>>>>>>>>>>> [000012fe](02)  ebfe            jmp 000012fe >>>>>>>>>>>>>>>>>>>>>> [00001300](01)  5d              pop ebp >>>>>>>>>>>>>>>>>>>>>> [00001301](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>>>> [0000130a](05)  68e7120000      push 000012e7 // >>>>>>>>>>>>>>>>>>>>>> push P
[0000130f](05)  e8d3ffffff      call 000012e7 // >>>>>>>>>>>>>>>>>>>>>> call P
[00001314](03)  83c404          add esp,+04 >>>>>>>>>>>>>>>>>>>>>> [00001317](02)  33c0            xor eax,eax >>>>>>>>>>>>>>>>>>>>>> [00001319](01)  5d              pop ebp >>>>>>>>>>>>>>>>>>>>>> [0000131a](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine    assembly
    address   address   data      code       language
    ========  ========  ========  ========= >>>>>>>>>>>>>>>>>>>>>> =============
[00001307][00102190][00000000] 55         push ebp
[00001308][00102190][00000000] 8bec       mov ebp,esp
[0000130a][0010218c][000012e7] 68e7120000 push >>>>>>>>>>>>>>>>>>>>>> 000012e7 //
push P [0000130f][00102188][00001314] e8d3ffffff >>>>>>>>>>>>>>>>>>>>>> call 000012e7
// call P [000012e7][00102184][00102190] 55 push ebp >>>>>>>>>>>>>>>>>>>>>>     // enter executed P
[000012e8][00102184][00102190] 8bec       mov ebp,esp
[000012ea][00102184][00102190] 8b4508     mov >>>>>>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push >>>>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov >>>>>>>>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1][0010217c][000012e7] 51 push >>>>>>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][00102178][000012f7] >>>>>>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H

Begin Local Halt Decider Simulation   Execution >>>>>>>>>>>>>>>>>>>>>> Trace Stored
at:212244 [000012e7][00212230][00212234] 55 push ebp >>>>>>>>>>>>>>>>>>>>>>     // enter emulated P
[000012e8][00212230][00212234] 8bec        mov >>>>>>>>>>>>>>>>>>>>>> ebp,esp
[000012ea][00212230][00212234] 8b4508      mov >>>>>>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push >>>>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][0021222c][000012e7] 8b4d08      mov
ecx,[ebp+08] [000012f1][00212228][000012e7] 51 push >>>>>>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][00212224][000012f7] >>>>>>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H

So, by what instruction reference manual is a call >>>>>>>>>>>>>>>>>>>>> 00001177
followedby the execution of the instruction at >>>>>>>>>>>>>>>>>>>>> 000012e7.

Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55          push >>>>>>>>>>>>>>>>>>>>>> ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov >>>>>>>>>>>>>>>>>>>>>> ebp,esp
[000012ea][0025cc58][0025cc5c] 8b4508      mov >>>>>>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push >>>>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08      mov
ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51 push >>>>>>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][0025cc4c][000012f7] >>>>>>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H Local Halt Decider: >>>>>>>>>>>>>>>>>>>>>> Infinite Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>>>> correctly emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats >>>>>>>>>>>>>>>>>>>>>> this process we
know with complete certainty that the correct and >>>>>>>>>>>>>>>>>>>>>> complete
emulation of P by H would never reach its final “ret”
instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat >>>>>>>>>>>>>>>>>>>>> the procedure",
because that H always has the option to abort its >>>>>>>>>>>>>>>>>>>>> simulation,
just like this onne did, and return to its P and >>>>>>>>>>>>>>>>>>>>> see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>>>>> NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to >>>>>>>>>>>>>>>>>>>> H(P,P)
conclusively proves that the complete and correct >>>>>>>>>>>>>>>>>>>> x86 emulation
would never stop running.

You SAY that, but you don't answer the actual >>>>>>>>>>>>>>>>>>> questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>>> NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE >>>>>>>>>>>>>>>>>> CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY >>>>>>>>>>>>>>>>>> PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT TO >>>>>>>>>>>>>>>>>> H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>> instructions of P.
Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>>> process we know
with complete certainty that the correct and complete >>>>>>>>>>>>>>>>>> emulation of
P by H would never reach its final “ret” instruction, >>>>>>>>>>>>>>>>>> thus never
halts.

If P should have halted (i.e. no infinite loop) then >>>>>>>>>>>>>>>>> your simulation
detector, S (not H), gets the answer wrong.  You S is >>>>>>>>>>>>>>>>> NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO >>>>>>>>>>>>>>>> ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>>> [00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>>>> [0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H >>>>>>>>>>>>>>>> [00001362](03)  83c408          add esp,+08 >>>>>>>>>>>>>>>> [00001365](02)  85c0            test eax,eax >>>>>>>>>>>>>>>> [00001367](02)  7402            jz 0000136b >>>>>>>>>>>>>>>> [00001369](02)  ebfe            jmp 00001369 >>>>>>>>>>>>>>>> [0000136b](01)  5d              pop ebp >>>>>>>>>>>>>>>> [0000136c](01)  c3              ret >>>>>>>>>>>>>>>> Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven instructions >>>>>>>>>>>>>>>> of P. Because
the seventh instruction of P repeats this process we >>>>>>>>>>>>>>>> know with
complete certainty that the correct and complete >>>>>>>>>>>>>>>> emulation of P by H
would never reach its final “ret” instruction, thus >>>>>>>>>>>>>>>> never halts.

We are going around and around and around in circles. I >>>>>>>>>>>>>>> will try again:

If you replace the opcodes "EB FE" at 00001369 with the >>>>>>>>>>>>>>> opcodes "90 90"
then your H gets the answer wrong: P should have halted. >>>>>>>>>>>>>>>
/Flibble

As I already said before this is merely your cluelessness >>>>>>>>>>>>>> that when H(P,P) is invoked the correct x86 emulation of >>>>>>>>>>>>>> the input to H(P,P) makes and code after [0000135d] >>>>>>>>>>>>>> unreachable.

Wrong, because when that H return the value 0, it will get >>>>>>>>>>>>> there.

Like I said people that are dumber than a box of rocks won't >>>>>>>>>>>> be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to >>>>>>>>>>>> H(P,P) cannot possibly reach any instruction beyond [0000135d]. >>>>>>>>>>>

So, you are defining that you H(P,P) never returns because it >>>>>>>>>>> is caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering >>>>>>>>>>> decider you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know
that H performs a correct x86 emulation of its input and then
examine the execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) returns
0, which it can only do if it does not actually do a correct
emulation

The correctly emulated input to H(P,P) never gets past its machine >>>>>> address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly return
0 if it correctly emulates its input, but you can't drop that
requirement.

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to main().

But it also return 0 to the computation P(P), maybe not the copy that
it is simulating, since it aborts that before it get to it,

Finally you are not stupid or deceptive.

So, you agree that your H(P,P) that returns a 0, claiming it to be
correct, also returns that 0 to P(P) and thus P(P) Halts, and thus
H(P,P) is INCORRECT in saying that its input represents a non-halting computation?

That you are waaaay too stupid to understand that I correctly proved
that P(P) and the correct simulatuon of the input to H(P,P) actually
have different halting behavior is no actual rebuttal at all.


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/2022 5:27 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:18:38 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 5:15 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:09:08 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 4:54 PM, Richard Damon wrote:

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 6/9/22 11:47 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>> void P(u32 x)

{
     if (H(x, x))
       HERE: goto HERE; >>>>>>>>>>>>>>>>>>>>>>>>      return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov >>>>>>>>>>>>>>>>>>>>>>>> eax,[ebp+08] [000012ed](01)  50 >>>>>>>>>>>>>>>>>>>>>>>> push eax [000012ee](03)  8b4d08          mov >>>>>>>>>>>>>>>>>>>>>>>> ecx,[ebp+08] [000012f1](01)  51 >>>>>>>>>>>>>>>>>>>>>>>> push ecx [000012f2](05)  e880feffff      call >>>>>>>>>>>>>>>>>>>>>>>> 00001177 // call H [000012f7](03)  83c408 >>>>>>>>>>>>>>>>>>>>>>>>     add esp,+08 [000012fa](02)  85c0 >>>>>>>>>>>>>>>>>>>>>>>> test eax,eax [000012fc](02)  7402 >>>>>>>>>>>>>>>>>>>>>>>> jz 00001300 [000012fe](02)  ebfe            jmp
000012fe [00001300](01)  5d              pop
ebp [00001301](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301] >>>>>>>>>>>>>>>>>>>>>>>>
_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 >>>>>>>>>>>>>>>>>>>>>>>> // push P [0000130f](05)  e8d3ffffff      call >>>>>>>>>>>>>>>>>>>>>>>> 000012e7 // call P [00001314](03)  83c404 >>>>>>>>>>>>>>>>>>>>>>>>   add esp,+04 [00001317](02)  33c0 >>>>>>>>>>>>>>>>>>>>>>>> xor eax,eax [00001319](01)  5d              pop
ebp [0000131a](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0020) [0000131a] >>>>>>>>>>>>>>>>>>>>>>>>
    machine   stack     stack     machine
assembly address   address   data      code >>>>>>>>>>>>>>>>>>>>>>>> language ========  ========  ======== >>>>>>>>>>>>>>>>>>>>>>>> ========= =============
[00001307][00102190][00000000] 55         push >>>>>>>>>>>>>>>>>>>>>>>> ebp [00001308][00102190][00000000] 8bec >>>>>>>>>>>>>>>>>>>>>>>> mov ebp,esp [0000130a][0010218c][000012e7] >>>>>>>>>>>>>>>>>>>>>>>> 68e7120000 push 000012e7 //
push P [0000130f][00102188][00001314] >>>>>>>>>>>>>>>>>>>>>>>> e8d3ffffff call 000012e7
// call P [000012e7][00102184][00102190] 55 >>>>>>>>>>>>>>>>>>>>>>>> push ebp // enter executed P
[000012e8][00102184][00102190] 8bec       mov >>>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00102184][00102190] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>> mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push >>>>>>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][00102180][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08] [000012f1][0010217c][000012e7] >>>>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P >>>>>>>>>>>>>>>>>>>>>>>> [000012f2][00102178][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H

Begin Local Halt Decider Simulation   Execution >>>>>>>>>>>>>>>>>>>>>>>> Trace Stored
at:212244 [000012e7][00212230][00212234] 55 >>>>>>>>>>>>>>>>>>>>>>>> push ebp // enter emulated P
[000012e8][00212230][00212234] 8bec        mov >>>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00212230][00212234] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>   mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push
eax      //
push P [000012ee][0021222c][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08]
[000012f1][00212228][000012e7] 51 push ecx >>>>>>>>>>>>>>>>>>>>>>>> // push P [000012f2][00212224][000012f7] >>>>>>>>>>>>>>>>>>>>>>>> e880feffff call 00001177 // call H >>>>>>>>>>>>>>>>>>>>>>>

So, by what instruction reference manual is a >>>>>>>>>>>>>>>>>>>>>>> call 00001177
followedby the execution of the instruction at >>>>>>>>>>>>>>>>>>>>>>> 000012e7.

Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55          push
ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov >>>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][0025cc58][0025cc5c] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>   mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push
eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08]
[000012f1][0025cc50][000012e7] 51 push ecx >>>>>>>>>>>>>>>>>>>>>>>> // push P [000012f2][0025cc4c][000012f7] >>>>>>>>>>>>>>>>>>>>>>>> e880feffff call 00001177 // call H Local Halt >>>>>>>>>>>>>>>>>>>>>>>> Decider: Infinite Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>>>>>> correctly emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats >>>>>>>>>>>>>>>>>>>>>>>> this process we
know with complete certainty that the correct >>>>>>>>>>>>>>>>>>>>>>>> and complete
emulation of P by H would never reach its final >>>>>>>>>>>>>>>>>>>>>>>> “ret” instruction, thus never halts. >>>>>>>>>>>>>>>>>>>>>>>

Problem, the 7th intruction DOESN't "Just repeat >>>>>>>>>>>>>>>>>>>>>>> the procedure",
because that H always has the option to abort >>>>>>>>>>>>>>>>>>>>>>> its simulation,
just like this onne did, and return to its P and >>>>>>>>>>>>>>>>>>>>>>> see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS >>>>>>>>>>>>>>>>>>>>>> IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to >>>>>>>>>>>>>>>>>>>>>> H(P,P) conclusively proves that the complete and >>>>>>>>>>>>>>>>>>>>>> correct x86 emulation
would never stop running.

You SAY that, but you don't answer the actual >>>>>>>>>>>>>>>>>>>>> questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS >>>>>>>>>>>>>>>>>>>> IS NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT >>>>>>>>>>>>>>>>>>>> THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) >>>>>>>>>>>>>>>>>>>> CONCLUSIVELY PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT >>>>>>>>>>>>>>>>>>>> TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>> instructions of P.
Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>>>>> process we know
with complete certainty that the correct and >>>>>>>>>>>>>>>>>>>> complete emulation of
P by H would never reach its final “ret” >>>>>>>>>>>>>>>>>>>> instruction, thus never
halts.

If P should have halted (i.e. no infinite loop) then >>>>>>>>>>>>>>>>>>> your simulation
detector, S (not H), gets the answer wrong.  You S >>>>>>>>>>>>>>>>>>> is NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>>> NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>>>>> [00001358](01)  50              push eax      //
push P [00001359](03)  8b4d08          mov >>>>>>>>>>>>>>>>>> ecx,[ebp+08] [0000135c](01)  51              push
ecx      // push P [0000135d](05)  e840feffff >>>>>>>>>>>>>>>>>> call 000011a2 // call H [00001362](03)  83c408 >>>>>>>>>>>>>>>>>>    add esp,+08 [00001365](02)  85c0            test
eax,eax [00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369 >>>>>>>>>>>>>>>>>> [0000136b](01)  5d              pop ebp >>>>>>>>>>>>>>>>>> [0000136c](01)  c3              ret >>>>>>>>>>>>>>>>>> Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>> instructions of P. Because
the seventh instruction of P repeats this process we >>>>>>>>>>>>>>>>>> know with
complete certainty that the correct and complete >>>>>>>>>>>>>>>>>> emulation of P by H
would never reach its final “ret” instruction, thus >>>>>>>>>>>>>>>>>> never halts.

We are going around and around and around in circles. >>>>>>>>>>>>>>>>> I will try again:

If you replace the opcodes "EB FE" at 00001369 with >>>>>>>>>>>>>>>>> the opcodes "90 90"
then your H gets the answer wrong: P should have >>>>>>>>>>>>>>>>> halted.

/Flibble

As I already said before this is merely your
cluelessness that when H(P,P) is invoked the correct >>>>>>>>>>>>>>>> x86 emulation of the input to H(P,P) makes and code >>>>>>>>>>>>>>>> after [0000135d] unreachable.

Wrong, because when that H return the value 0, it will >>>>>>>>>>>>>>> get there.

Like I said people that are dumber than a box of rocks >>>>>>>>>>>>>> won't be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to >>>>>>>>>>>>>> H(P,P) cannot possibly reach any instruction beyond >>>>>>>>>>>>>> [0000135d].

So, you are defining that you H(P,P) never returns because >>>>>>>>>>>>> it is caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering >>>>>>>>>>>>> decider you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know >>>>>>>>>> that H performs a correct x86 emulation of its input and then >>>>>>>>>> examine the execution trace.

And a CORRECT emulation of the code will Halt if H(P,P)
returns 0, which it can only do if it does not actually do a >>>>>>>>> correct emulation

The correctly emulated input to H(P,P) never gets past its
machine address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly
return 0 if it correctly emulates its input, but you can't drop
that requirement.

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to
main().

But it also return 0 to the computation P(P), maybe not the copy
that it is simulating, since it aborts that before it get to it,

Finally you are not stupid or deceptive.

If H never returns a value to P then H is not a halting decider; if
H returns a value of 0 to main() when P halts then H is not a
halting decider.

H is not a halting decider; H is a simulation detector, S.

/Flibble

In other words you are saying the infintely nested emulation only
excutes for a few steps and then it stops on its own.

I am saying that your H is not a halting decider; I am saying that your
H is just a simulation detector, S.

/Flibble

You are saying that you simply "don't believe in" simulating halt deciders.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/2022 5:33 PM, Richard Damon wrote:

On 6/9/22 6:18 PM, olcott wrote:

On 6/9/2022 5:15 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:09:08 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 4:54 PM, Richard Damon wrote:

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 6/9/22 11:47 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>> void P(u32 x)

{
      if (H(x, x))
        HERE: goto HERE; >>>>>>>>>>>>>>>>>>>>>>>>       return;
}

int main()
{
      P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax >>>>>>>>>>>>>>>>>>>>>>>> [000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx >>>>>>>>>>>>>>>>>>>>>>>> [000012f2](05)  e880feffff      call 00001177 // >>>>>>>>>>>>>>>>>>>>>>>> call H [000012f7](03)  83c408          add >>>>>>>>>>>>>>>>>>>>>>>> esp,+08 [000012fa](02)  85c0            test
eax,eax [000012fc](02)  7402            jz >>>>>>>>>>>>>>>>>>>>>>>> 00001300 [000012fe](02)  ebfe            jmp
000012fe [00001300](01)  5d              pop ebp
[00001301](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301] >>>>>>>>>>>>>>>>>>>>>>>>
_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 // >>>>>>>>>>>>>>>>>>>>>>>> push P [0000130f](05)  e8d3ffffff      call >>>>>>>>>>>>>>>>>>>>>>>> 000012e7 // call P [00001314](03)  83c404 >>>>>>>>>>>>>>>>>>>>>>>>    add esp,+04 [00001317](02)  33c0 >>>>>>>>>>>>>>>>>>>>>>>> xor eax,eax [00001319](01)  5d              pop
ebp [0000131a](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0020) [0000131a] >>>>>>>>>>>>>>>>>>>>>>>>
     machine   stack     stack     machine
assembly address   address   data      code >>>>>>>>>>>>>>>>>>>>>>>>   language ========  ========  ======== >>>>>>>>>>>>>>>>>>>>>>>> ========= =============
[00001307][00102190][00000000] 55         push >>>>>>>>>>>>>>>>>>>>>>>> ebp [00001308][00102190][00000000] 8bec >>>>>>>>>>>>>>>>>>>>>>>> mov ebp,esp [0000130a][0010218c][000012e7] >>>>>>>>>>>>>>>>>>>>>>>> 68e7120000 push 000012e7 //
push P [0000130f][00102188][00001314] e8d3ffffff >>>>>>>>>>>>>>>>>>>>>>>> call 000012e7
// call P [000012e7][00102184][00102190] 55 push >>>>>>>>>>>>>>>>>>>>>>>> ebp // enter executed P
[000012e8][00102184][00102190] 8bec       mov >>>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00102184][00102190] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>   mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push >>>>>>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][00102180][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08] [000012f1][0010217c][000012e7] >>>>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P >>>>>>>>>>>>>>>>>>>>>>>> [000012f2][00102178][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H

Begin Local Halt Decider Simulation   Execution >>>>>>>>>>>>>>>>>>>>>>>> Trace Stored
at:212244 [000012e7][00212230][00212234] 55 push >>>>>>>>>>>>>>>>>>>>>>>> ebp // enter emulated P
[000012e8][00212230][00212234] 8bec        mov >>>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00212230][00212234] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>    mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push
eax      //
push P [000012ee][0021222c][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>   mov ecx,[ebp+08] [000012f1][00212228][000012e7] >>>>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P >>>>>>>>>>>>>>>>>>>>>>>> [000012f2][00212224][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H

So, by what instruction reference manual is a >>>>>>>>>>>>>>>>>>>>>>> call 00001177
followedby the execution of the instruction at >>>>>>>>>>>>>>>>>>>>>>> 000012e7.

Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55          push
ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov >>>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][0025cc58][0025cc5c] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>    mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push
eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>   mov ecx,[ebp+08] [000012f1][0025cc50][000012e7] >>>>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P >>>>>>>>>>>>>>>>>>>>>>>> [000012f2][0025cc4c][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H Local Halt Decider: >>>>>>>>>>>>>>>>>>>>>>>> Infinite Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>>>>>> correctly emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats >>>>>>>>>>>>>>>>>>>>>>>> this process we
know with complete certainty that the correct >>>>>>>>>>>>>>>>>>>>>>>> and complete
emulation of P by H would never reach its final >>>>>>>>>>>>>>>>>>>>>>>> “ret” instruction, thus never halts. >>>>>>>>>>>>>>>>>>>>>>>

Problem, the 7th intruction DOESN't "Just repeat >>>>>>>>>>>>>>>>>>>>>>> the procedure",
because that H always has the option to abort its >>>>>>>>>>>>>>>>>>>>>>> simulation,
just like this onne did, and return to its P and >>>>>>>>>>>>>>>>>>>>>>> see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS >>>>>>>>>>>>>>>>>>>>>> IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to >>>>>>>>>>>>>>>>>>>>>> H(P,P) conclusively proves that the complete and >>>>>>>>>>>>>>>>>>>>>> correct x86 emulation
would never stop running.

You SAY that, but you don't answer the actual >>>>>>>>>>>>>>>>>>>>> questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>>>>> NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT >>>>>>>>>>>>>>>>>>>> THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) >>>>>>>>>>>>>>>>>>>> CONCLUSIVELY PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT >>>>>>>>>>>>>>>>>>>> TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>> instructions of P.
Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>>>>> process we know
with complete certainty that the correct and >>>>>>>>>>>>>>>>>>>> complete emulation of
P by H would never reach its final “ret” >>>>>>>>>>>>>>>>>>>> instruction, thus never
halts.

If P should have halted (i.e. no infinite loop) then >>>>>>>>>>>>>>>>>>> your simulation
detector, S (not H), gets the answer wrong.  You S is >>>>>>>>>>>>>>>>>>> NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>>> NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>>>>> [00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>>>>>> [0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08 >>>>>>>>>>>>>>>>>> [00001365](02)  85c0            test eax,eax >>>>>>>>>>>>>>>>>> [00001367](02)  7402            jz 0000136b >>>>>>>>>>>>>>>>>> [00001369](02)  ebfe            jmp 00001369 >>>>>>>>>>>>>>>>>> [0000136b](01)  5d              pop ebp >>>>>>>>>>>>>>>>>> [0000136c](01)  c3              ret >>>>>>>>>>>>>>>>>> Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>> instructions of P. Because
the seventh instruction of P repeats this process we >>>>>>>>>>>>>>>>>> know with
complete certainty that the correct and complete >>>>>>>>>>>>>>>>>> emulation of P by H
would never reach its final “ret” instruction, thus >>>>>>>>>>>>>>>>>> never halts.

We are going around and around and around in circles. I >>>>>>>>>>>>>>>>> will try again:

If you replace the opcodes "EB FE" at 00001369 with the >>>>>>>>>>>>>>>>> opcodes "90 90"
then your H gets the answer wrong: P should have halted. >>>>>>>>>>>>>>>>>
/Flibble

As I already said before this is merely your
cluelessness that when H(P,P) is invoked the correct x86 >>>>>>>>>>>>>>>> emulation of the input to H(P,P) makes and code after >>>>>>>>>>>>>>>> [0000135d] unreachable.

Wrong, because when that H return the value 0, it will >>>>>>>>>>>>>>> get there.

Like I said people that are dumber than a box of rocks >>>>>>>>>>>>>> won't be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to >>>>>>>>>>>>>> H(P,P) cannot possibly reach any instruction beyond >>>>>>>>>>>>>> [0000135d].

So, you are defining that you H(P,P) never returns because >>>>>>>>>>>>> it is caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering >>>>>>>>>>>>> decider you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know >>>>>>>>>> that H performs a correct x86 emulation of its input and then >>>>>>>>>> examine the execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) returns >>>>>>>>> 0, which it can only do if it does not actually do a correct >>>>>>>>> emulation

The correctly emulated input to H(P,P) never gets past its
machine address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly
return 0 if it correctly emulates its input, but you can't drop
that requirement.

void P(u32 x)
{
    if (H(x, x))
      HERE: goto HERE;
    return;
}

int main()
{
    Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to
main().

But it also return 0 to the computation P(P), maybe not the copy
that it is simulating, since it aborts that before it get to it,

Finally you are not stupid or deceptive.

If H never returns a value to P then H is not a halting decider; if H
returns a value of 0 to main() when P halts then H is not a halting
decider.

H is not a halting decider; H is a simulation detector, S.

/Flibble

In other words you are saying the infintely nested emulation only
excutes for a few steps and then it stops on its own.

No, the "infinitely nested emulation" halts when the first H(P,P) that
P(P) called aborts it simulation.

When I tell you that an input only halts when it reaches its final state
and I tell you this many hundreds of times because you are a God damned
liar you pretend that I never said this.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/22 6:37 PM, olcott wrote:

On 6/9/2022 5:33 PM, Richard Damon wrote:

On 6/9/22 6:18 PM, olcott wrote:

On 6/9/2022 5:15 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:09:08 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 4:54 PM, Richard Damon wrote:

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote: >>>>>>>>>>>>>>>>>>>> On Thu, 9 Jun 2022 12:15:24 -0500

olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 6/9/22 12:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 6/9/2022 11:34 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 6/9/22 11:47 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> void P(u32 x)

{
      if (H(x, x))
        HERE: goto HERE; >>>>>>>>>>>>>>>>>>>>>>>>>       return;
}

int main()
{
      P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax
[000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx
[000012f2](05)  e880feffff      call 00001177 //
call H [000012f7](03)  83c408          add >>>>>>>>>>>>>>>>>>>>>>>>> esp,+08 [000012fa](02)  85c0            test
eax,eax [000012fc](02)  7402            jz
00001300 [000012fe](02)  ebfe            jmp
000012fe [00001300](01)  5d              pop ebp
[00001301](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301] >>>>>>>>>>>>>>>>>>>>>>>>>
_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 //
push P [0000130f](05)  e8d3ffffff      call >>>>>>>>>>>>>>>>>>>>>>>>> 000012e7 // call P [00001314](03)  83c404 >>>>>>>>>>>>>>>>>>>>>>>>>    add esp,+04 [00001317](02)  33c0 >>>>>>>>>>>>>>>>>>>>>>>>> xor eax,eax [00001319](01)  5d              pop
ebp [0000131a](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0020) [0000131a] >>>>>>>>>>>>>>>>>>>>>>>>>
     machine   stack     stack     machine
assembly address   address   data      code >>>>>>>>>>>>>>>>>>>>>>>>>   language ========  ========  ======== >>>>>>>>>>>>>>>>>>>>>>>>> ========= =============
[00001307][00102190][00000000] 55         push
ebp [00001308][00102190][00000000] 8bec >>>>>>>>>>>>>>>>>>>>>>>>> mov ebp,esp [0000130a][0010218c][000012e7] >>>>>>>>>>>>>>>>>>>>>>>>> 68e7120000 push 000012e7 //
push P [0000130f][00102188][00001314] e8d3ffffff >>>>>>>>>>>>>>>>>>>>>>>>> call 000012e7
// call P [000012e7][00102184][00102190] 55 push >>>>>>>>>>>>>>>>>>>>>>>>> ebp // enter executed P
[000012e8][00102184][00102190] 8bec       mov >>>>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00102184][00102190] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>>   mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push
eax      //
push P [000012ee][00102180][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08] [000012f1][0010217c][000012e7] >>>>>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P >>>>>>>>>>>>>>>>>>>>>>>>> [000012f2][00102178][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H

Begin Local Halt Decider Simulation   Execution >>>>>>>>>>>>>>>>>>>>>>>>> Trace Stored
at:212244 [000012e7][00212230][00212234] 55 push >>>>>>>>>>>>>>>>>>>>>>>>> ebp // enter emulated P
[000012e8][00212230][00212234] 8bec        mov >>>>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00212230][00212234] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>>    mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push
eax      //
push P [000012ee][0021222c][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>>   mov ecx,[ebp+08] [000012f1][00212228][000012e7] >>>>>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P >>>>>>>>>>>>>>>>>>>>>>>>> [000012f2][00212224][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H

So, by what instruction reference manual is a >>>>>>>>>>>>>>>>>>>>>>>> call 00001177
followedby the execution of the instruction at >>>>>>>>>>>>>>>>>>>>>>>> 000012e7.

Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55          push
ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov >>>>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][0025cc58][0025cc5c] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>>    mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push
eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>>   mov ecx,[ebp+08] [000012f1][0025cc50][000012e7] >>>>>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P >>>>>>>>>>>>>>>>>>>>>>>>> [000012f2][0025cc4c][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H Local Halt Decider: >>>>>>>>>>>>>>>>>>>>>>>>> Infinite Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>>>>>>> correctly emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats >>>>>>>>>>>>>>>>>>>>>>>>> this process we
know with complete certainty that the correct >>>>>>>>>>>>>>>>>>>>>>>>> and complete
emulation of P by H would never reach its final >>>>>>>>>>>>>>>>>>>>>>>>> “ret” instruction, thus never halts. >>>>>>>>>>>>>>>>>>>>>>>>

Problem, the 7th intruction DOESN't "Just repeat >>>>>>>>>>>>>>>>>>>>>>>> the procedure",
because that H always has the option to abort its >>>>>>>>>>>>>>>>>>>>>>>> simulation,
just like this onne did, and return to its P and >>>>>>>>>>>>>>>>>>>>>>>> see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS >>>>>>>>>>>>>>>>>>>>>>> IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to >>>>>>>>>>>>>>>>>>>>>>> H(P,P) conclusively proves that the complete and >>>>>>>>>>>>>>>>>>>>>>> correct x86 emulation
would never stop running.

You SAY that, but you don't answer the actual >>>>>>>>>>>>>>>>>>>>>> questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>>>>>> NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT >>>>>>>>>>>>>>>>>>>>> THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) >>>>>>>>>>>>>>>>>>>>> CONCLUSIVELY PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT >>>>>>>>>>>>>>>>>>>>> TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>> instructions of P.
Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>>>>>> process we know
with complete certainty that the correct and >>>>>>>>>>>>>>>>>>>>> complete emulation of
P by H would never reach its final “ret” >>>>>>>>>>>>>>>>>>>>> instruction, thus never
halts.

If P should have halted (i.e. no infinite loop) then >>>>>>>>>>>>>>>>>>>> your simulation
detector, S (not H), gets the answer wrong.  You S is >>>>>>>>>>>>>>>>>>>> NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>>>> NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>>>>>> [00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>>>>>>> [0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08 >>>>>>>>>>>>>>>>>>> [00001365](02)  85c0            test eax,eax >>>>>>>>>>>>>>>>>>> [00001367](02)  7402            jz 0000136b >>>>>>>>>>>>>>>>>>> [00001369](02)  ebfe            jmp 00001369 >>>>>>>>>>>>>>>>>>> [0000136b](01)  5d              pop ebp >>>>>>>>>>>>>>>>>>> [0000136c](01)  c3              ret >>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>>> instructions of P. Because
the seventh instruction of P repeats this process we >>>>>>>>>>>>>>>>>>> know with
complete certainty that the correct and complete >>>>>>>>>>>>>>>>>>> emulation of P by H
would never reach its final “ret” instruction, thus >>>>>>>>>>>>>>>>>>> never halts.

We are going around and around and around in circles. I >>>>>>>>>>>>>>>>>> will try again:

If you replace the opcodes "EB FE" at 00001369 with the >>>>>>>>>>>>>>>>>> opcodes "90 90"
then your H gets the answer wrong: P should have halted. >>>>>>>>>>>>>>>>>>
/Flibble

As I already said before this is merely your >>>>>>>>>>>>>>>>> cluelessness that when H(P,P) is invoked the correct x86 >>>>>>>>>>>>>>>>> emulation of the input to H(P,P) makes and code after >>>>>>>>>>>>>>>>> [0000135d] unreachable.

Wrong, because when that H return the value 0, it will >>>>>>>>>>>>>>>> get there.

Like I said people that are dumber than a box of rocks >>>>>>>>>>>>>>> won't be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to >>>>>>>>>>>>>>> H(P,P) cannot possibly reach any instruction beyond >>>>>>>>>>>>>>> [0000135d].

So, you are defining that you H(P,P) never returns because >>>>>>>>>>>>>> it is caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering >>>>>>>>>>>>>> decider you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know >>>>>>>>>>> that H performs a correct x86 emulation of its input and then >>>>>>>>>>> examine the execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) returns >>>>>>>>>> 0, which it can only do if it does not actually do a correct >>>>>>>>>> emulation

The correctly emulated input to H(P,P) never gets past its
machine address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly
return 0 if it correctly emulates its input, but you can't drop >>>>>>>> that requirement.

void P(u32 x)
{
    if (H(x, x))
      HERE: goto HERE;
    return;
}

int main()
{
    Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to
main().

But it also return 0 to the computation P(P), maybe not the copy
that it is simulating, since it aborts that before it get to it,

Finally you are not stupid or deceptive.

If H never returns a value to P then H is not a halting decider; if H
returns a value of 0 to main() when P halts then H is not a halting
decider.

H is not a halting decider; H is a simulation detector, S.

/Flibble

In other words you are saying the infintely nested emulation only
excutes for a few steps and then it stops on its own.

No, the "infinitely nested emulation" halts when the first H(P,P) that
P(P) called aborts it simulation.

When I tell you that an input only halts when it reaches its final state
and I tell you this many hundreds of times because you are a God damned
liar you pretend that I never said this.

Right, and when H(P,P) returns 0, the P that called it reaches its final state/the ret instrucion.

The key is that the CORRECT simulation doesn't stop running just because
H aborts it simulation.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/22 6:34 PM, olcott wrote:

On 6/9/2022 5:17 PM, Richard Damon wrote:

On 6/9/22 6:09 PM, olcott wrote:

On 6/9/2022 4:54 PM, Richard Damon wrote:

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:15:24 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 6/9/22 12:54 PM, olcott wrote:

On 6/9/2022 11:34 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 6/9/22 11:47 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> void P(u32 x)

{
     if (H(x, x))
       HERE: goto HERE;
     return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>>>> [000012e8](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>>>>> [000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax >>>>>>>>>>>>>>>>>>>>>>> [000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx >>>>>>>>>>>>>>>>>>>>>>> [000012f2](05)  e880feffff      call 00001177 // >>>>>>>>>>>>>>>>>>>>>>> call H
[000012f7](03)  83c408          add esp,+08 >>>>>>>>>>>>>>>>>>>>>>> [000012fa](02)  85c0            test eax,eax
[000012fc](02)  7402            jz 00001300 >>>>>>>>>>>>>>>>>>>>>>> [000012fe](02)  ebfe            jmp 000012fe
[00001300](01)  5d              pop ebp >>>>>>>>>>>>>>>>>>>>>>> [00001301](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301]

_main()
[00001307](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>>>> [00001308](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>>>>> [0000130a](05)  68e7120000      push 000012e7 // >>>>>>>>>>>>>>>>>>>>>>> push P
[0000130f](05)  e8d3ffffff      call 000012e7 // >>>>>>>>>>>>>>>>>>>>>>> call P
[00001314](03)  83c404          add esp,+04 >>>>>>>>>>>>>>>>>>>>>>> [00001317](02)  33c0            xor eax,eax >>>>>>>>>>>>>>>>>>>>>>> [00001319](01)  5d              pop ebp >>>>>>>>>>>>>>>>>>>>>>> [0000131a](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0020) [0000131a]

    machine   stack     stack     machine >>>>>>>>>>>>>>>>>>>>>>> assembly
    address   address   data      code >>>>>>>>>>>>>>>>>>>>>>> language
    ========  ========  ========  ========= >>>>>>>>>>>>>>>>>>>>>>> =============
[00001307][00102190][00000000] 55         push ebp
[00001308][00102190][00000000] 8bec       mov >>>>>>>>>>>>>>>>>>>>>>> ebp,esp
[0000130a][0010218c][000012e7] 68e7120000 push >>>>>>>>>>>>>>>>>>>>>>> 000012e7 //
push P [0000130f][00102188][00001314] e8d3ffffff >>>>>>>>>>>>>>>>>>>>>>> call 000012e7
// call P [000012e7][00102184][00102190] 55 push ebp >>>>>>>>>>>>>>>>>>>>>>>     // enter executed P
[000012e8][00102184][00102190] 8bec       mov >>>>>>>>>>>>>>>>>>>>>>> ebp,esp
[000012ea][00102184][00102190] 8b4508     mov >>>>>>>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push >>>>>>>>>>>>>>>>>>>>>>> eax      //
push P [000012ee][00102180][000012e7] 8b4d08     mov
ecx,[ebp+08] [000012f1][0010217c][000012e7] 51 push >>>>>>>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][00102178][000012f7] >>>>>>>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H

Begin Local Halt Decider Simulation   Execution >>>>>>>>>>>>>>>>>>>>>>> Trace Stored
at:212244 [000012e7][00212230][00212234] 55 push ebp >>>>>>>>>>>>>>>>>>>>>>>     // enter emulated P
[000012e8][00212230][00212234] 8bec        mov >>>>>>>>>>>>>>>>>>>>>>> ebp,esp
[000012ea][00212230][00212234] 8b4508      mov >>>>>>>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push
eax      //
push P [000012ee][0021222c][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>> mov
ecx,[ebp+08] [000012f1][00212228][000012e7] 51 push >>>>>>>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][00212224][000012f7] >>>>>>>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H

So, by what instruction reference manual is a call >>>>>>>>>>>>>>>>>>>>>> 00001177
followedby the execution of the instruction at >>>>>>>>>>>>>>>>>>>>>> 000012e7.

Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55          push
ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov >>>>>>>>>>>>>>>>>>>>>>> ebp,esp
[000012ea][0025cc58][0025cc5c] 8b4508      mov >>>>>>>>>>>>>>>>>>>>>>> eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push
eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>> mov
ecx,[ebp+08] [000012f1][0025cc50][000012e7] 51 push >>>>>>>>>>>>>>>>>>>>>>> ecx      // push P [000012f2][0025cc4c][000012f7] >>>>>>>>>>>>>>>>>>>>>>> e880feffff
call 00001177 // call H Local Halt Decider: >>>>>>>>>>>>>>>>>>>>>>> Infinite Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>>>>> correctly emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats >>>>>>>>>>>>>>>>>>>>>>> this process we
know with complete certainty that the correct and >>>>>>>>>>>>>>>>>>>>>>> complete
emulation of P by H would never reach its final >>>>>>>>>>>>>>>>>>>>>>> “ret”
instruction, thus never halts.

Problem, the 7th intruction DOESN't "Just repeat >>>>>>>>>>>>>>>>>>>>>> the procedure",
because that H always has the option to abort its >>>>>>>>>>>>>>>>>>>>>> simulation,
just like this onne did, and return to its P and >>>>>>>>>>>>>>>>>>>>>> see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS >>>>>>>>>>>>>>>>>>>>> IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to >>>>>>>>>>>>>>>>>>>>> H(P,P)
conclusively proves that the complete and correct >>>>>>>>>>>>>>>>>>>>> x86 emulation
would never stop running.

You SAY that, but you don't answer the actual >>>>>>>>>>>>>>>>>>>> questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>>>> NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT THE >>>>>>>>>>>>>>>>>>> CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) CONCLUSIVELY >>>>>>>>>>>>>>>>>>> PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT >>>>>>>>>>>>>>>>>>> TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>>> instructions of P.
Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>>>> process we know
with complete certainty that the correct and complete >>>>>>>>>>>>>>>>>>> emulation of
P by H would never reach its final “ret” instruction, >>>>>>>>>>>>>>>>>>> thus never
halts.

If P should have halted (i.e. no infinite loop) then >>>>>>>>>>>>>>>>>> your simulation
detector, S (not H), gets the answer wrong.  You S is >>>>>>>>>>>>>>>>>> NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS NO >>>>>>>>>>>>>>>>> ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>>>> [00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>>>>> [0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H >>>>>>>>>>>>>>>>> [00001362](03)  83c408          add esp,+08 >>>>>>>>>>>>>>>>> [00001365](02)  85c0            test eax,eax >>>>>>>>>>>>>>>>> [00001367](02)  7402            jz 0000136b >>>>>>>>>>>>>>>>> [00001369](02)  ebfe            jmp 00001369 >>>>>>>>>>>>>>>>> [0000136b](01)  5d              pop ebp >>>>>>>>>>>>>>>>> [0000136c](01)  c3              ret >>>>>>>>>>>>>>>>> Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven instructions >>>>>>>>>>>>>>>>> of P. Because
the seventh instruction of P repeats this process we >>>>>>>>>>>>>>>>> know with
complete certainty that the correct and complete >>>>>>>>>>>>>>>>> emulation of P by H
would never reach its final “ret” instruction, thus >>>>>>>>>>>>>>>>> never halts.

We are going around and around and around in circles. I >>>>>>>>>>>>>>>> will try again:

If you replace the opcodes "EB FE" at 00001369 with the >>>>>>>>>>>>>>>> opcodes "90 90"
then your H gets the answer wrong: P should have halted. >>>>>>>>>>>>>>>>
/Flibble

As I already said before this is merely your cluelessness >>>>>>>>>>>>>>> that when H(P,P) is invoked the correct x86 emulation of >>>>>>>>>>>>>>> the input to H(P,P) makes and code after [0000135d] >>>>>>>>>>>>>>> unreachable.

Wrong, because when that H return the value 0, it will get >>>>>>>>>>>>>> there.

Like I said people that are dumber than a box of rocks >>>>>>>>>>>>> won't be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to >>>>>>>>>>>>> H(P,P) cannot possibly reach any instruction beyond
[0000135d].

So, you are defining that you H(P,P) never returns because >>>>>>>>>>>> it is caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering >>>>>>>>>>>> decider you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know >>>>>>>>> that H performs a correct x86 emulation of its input and then >>>>>>>>> examine the execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) returns >>>>>>>> 0, which it can only do if it does not actually do a correct
emulation

The correctly emulated input to H(P,P) never gets past its
machine address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly return >>>>>> 0 if it correctly emulates its input, but you can't drop that
requirement.

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to main().

But it also return 0 to the computation P(P), maybe not the copy
that it is simulating, since it aborts that before it get to it,

Finally you are not stupid or deceptive.

So, you agree that your H(P,P) that returns a 0, claiming it to be
correct, also returns that 0 to P(P) and thus P(P) Halts, and thus
H(P,P) is INCORRECT in saying that its input represents a non-halting
computation?

That you are waaaay too stupid to understand that I correctly proved
that P(P) and the correct simulatuon of the input to H(P,P) actually
have different halting behavior is no actual rebuttal at all.

No, you didn't. Wanna try again. You just make that claim without an
actual proof.

The ONLY way thst can happen is if H isn't actually a computation. If I remember your proof, all you did was prove that your H isn't actoally a computation.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/2022 6:09 PM, Richard Damon wrote:

On 6/9/22 6:37 PM, olcott wrote:

On 6/9/2022 5:33 PM, Richard Damon wrote:

On 6/9/22 6:18 PM, olcott wrote:

On 6/9/2022 5:15 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:09:08 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 4:54 PM, Richard Damon wrote:

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote: >>>>>>>>>>>>>>>>>>>>> On Thu, 9 Jun 2022 12:15:24 -0500

olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 6/9/22 12:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 6/9/2022 11:34 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 6/9/22 11:47 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> void P(u32 x)

{
      if (H(x, x))
        HERE: goto HERE; >>>>>>>>>>>>>>>>>>>>>>>>>>       return;
}

int main()
{
      P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax
[000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx
[000012f2](05)  e880feffff      call 00001177 //
call H [000012f7](03)  83c408          add >>>>>>>>>>>>>>>>>>>>>>>>>> esp,+08 [000012fa](02)  85c0            test
eax,eax [000012fc](02)  7402            jz
00001300 [000012fe](02)  ebfe            jmp
000012fe [00001300](01)  5d              pop ebp
[00001301](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301] >>>>>>>>>>>>>>>>>>>>>>>>>>
_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 //
push P [0000130f](05)  e8d3ffffff      call >>>>>>>>>>>>>>>>>>>>>>>>>> 000012e7 // call P [00001314](03)  83c404 >>>>>>>>>>>>>>>>>>>>>>>>>>    add esp,+04 [00001317](02)  33c0 >>>>>>>>>>>>>>>>>>>>>>>>>> xor eax,eax [00001319](01)  5d              pop
ebp [0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a] >>>>>>>>>>>>>>>>>>>>>>>>>>
     machine   stack     stack     machine
assembly address   address   data      code >>>>>>>>>>>>>>>>>>>>>>>>>>   language ========  ========  ======== >>>>>>>>>>>>>>>>>>>>>>>>>> ========= =============
[00001307][00102190][00000000] 55         push
ebp [00001308][00102190][00000000] 8bec >>>>>>>>>>>>>>>>>>>>>>>>>> mov ebp,esp [0000130a][0010218c][000012e7] >>>>>>>>>>>>>>>>>>>>>>>>>> 68e7120000 push 000012e7 // >>>>>>>>>>>>>>>>>>>>>>>>>> push P [0000130f][00102188][00001314] e8d3ffffff >>>>>>>>>>>>>>>>>>>>>>>>>> call 000012e7
// call P [000012e7][00102184][00102190] 55 push >>>>>>>>>>>>>>>>>>>>>>>>>> ebp // enter executed P
[000012e8][00102184][00102190] 8bec       mov >>>>>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00102184][00102190] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>>>   mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push
eax      //
push P [000012ee][00102180][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08] [000012f1][0010217c][000012e7] >>>>>>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P >>>>>>>>>>>>>>>>>>>>>>>>>> [000012f2][00102178][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H

Begin Local Halt Decider Simulation   Execution >>>>>>>>>>>>>>>>>>>>>>>>>> Trace Stored
at:212244 [000012e7][00212230][00212234] 55 push >>>>>>>>>>>>>>>>>>>>>>>>>> ebp // enter emulated P
[000012e8][00212230][00212234] 8bec        mov
ebp,esp [000012ea][00212230][00212234] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>>>    mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push
eax      //
push P [000012ee][0021222c][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>>>   mov ecx,[ebp+08] [000012f1][00212228][000012e7] >>>>>>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P >>>>>>>>>>>>>>>>>>>>>>>>>> [000012f2][00212224][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H

So, by what instruction reference manual is a >>>>>>>>>>>>>>>>>>>>>>>>> call 00001177
followedby the execution of the instruction at >>>>>>>>>>>>>>>>>>>>>>>>> 000012e7.

Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55          push
ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov
ebp,esp [000012ea][0025cc58][0025cc5c] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>>>    mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push
eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>>>   mov ecx,[ebp+08] [000012f1][0025cc50][000012e7] >>>>>>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P >>>>>>>>>>>>>>>>>>>>>>>>>> [000012f2][0025cc4c][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H Local Halt Decider: >>>>>>>>>>>>>>>>>>>>>>>>>> Infinite Recursion
Detected Simulation Stopped >>>>>>>>>>>>>>>>>>>>>>>>>>
It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>>>>>>>> correctly emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats >>>>>>>>>>>>>>>>>>>>>>>>>> this process we
know with complete certainty that the correct >>>>>>>>>>>>>>>>>>>>>>>>>> and complete
emulation of P by H would never reach its final >>>>>>>>>>>>>>>>>>>>>>>>>> “ret” instruction, thus never halts. >>>>>>>>>>>>>>>>>>>>>>>>>

Problem, the 7th intruction DOESN't "Just repeat >>>>>>>>>>>>>>>>>>>>>>>>> the procedure",
because that H always has the option to abort its >>>>>>>>>>>>>>>>>>>>>>>>> simulation,
just like this onne did, and return to its P and >>>>>>>>>>>>>>>>>>>>>>>>> see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS >>>>>>>>>>>>>>>>>>>>>>>> IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to >>>>>>>>>>>>>>>>>>>>>>>> H(P,P) conclusively proves that the complete and >>>>>>>>>>>>>>>>>>>>>>>> correct x86 emulation
would never stop running.

You SAY that, but you don't answer the actual >>>>>>>>>>>>>>>>>>>>>>> questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>>>>>>> NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT >>>>>>>>>>>>>>>>>>>>>> THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) >>>>>>>>>>>>>>>>>>>>>> CONCLUSIVELY PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT >>>>>>>>>>>>>>>>>>>>>> TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>>> instructions of P.
Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>>>>>>> process we know
with complete certainty that the correct and >>>>>>>>>>>>>>>>>>>>>> complete emulation of
P by H would never reach its final “ret” >>>>>>>>>>>>>>>>>>>>>> instruction, thus never
halts.

If P should have halted (i.e. no infinite loop) then >>>>>>>>>>>>>>>>>>>>> your simulation
detector, S (not H), gets the answer wrong.  You S is >>>>>>>>>>>>>>>>>>>>> NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>>>>> NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>>>>>>> [00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08] >>>>>>>>>>>>>>>>>>>> [0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08 >>>>>>>>>>>>>>>>>>>> [00001365](02)  85c0            test eax,eax >>>>>>>>>>>>>>>>>>>> [00001367](02)  7402            jz 0000136b >>>>>>>>>>>>>>>>>>>> [00001369](02)  ebfe            jmp 00001369 >>>>>>>>>>>>>>>>>>>> [0000136b](01)  5d              pop ebp >>>>>>>>>>>>>>>>>>>> [0000136c](01)  c3              ret >>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>> instructions of P. Because
the seventh instruction of P repeats this process we >>>>>>>>>>>>>>>>>>>> know with
complete certainty that the correct and complete >>>>>>>>>>>>>>>>>>>> emulation of P by H
would never reach its final “ret” instruction, thus >>>>>>>>>>>>>>>>>>>> never halts.

We are going around and around and around in circles. I >>>>>>>>>>>>>>>>>>> will try again:

If you replace the opcodes "EB FE" at 00001369 with the >>>>>>>>>>>>>>>>>>> opcodes "90 90"
then your H gets the answer wrong: P should have halted. >>>>>>>>>>>>>>>>>>>
/Flibble

As I already said before this is merely your >>>>>>>>>>>>>>>>>> cluelessness that when H(P,P) is invoked the correct x86 >>>>>>>>>>>>>>>>>> emulation of the input to H(P,P) makes and code after >>>>>>>>>>>>>>>>>> [0000135d] unreachable.

Wrong, because when that H return the value 0, it will >>>>>>>>>>>>>>>>> get there.

Like I said people that are dumber than a box of rocks >>>>>>>>>>>>>>>> won't be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to >>>>>>>>>>>>>>>> H(P,P) cannot possibly reach any instruction beyond >>>>>>>>>>>>>>>> [0000135d].

So, you are defining that you H(P,P) never returns because >>>>>>>>>>>>>>> it is caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering >>>>>>>>>>>>>>> decider you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know >>>>>>>>>>>> that H performs a correct x86 emulation of its input and then >>>>>>>>>>>> examine the execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) returns >>>>>>>>>>> 0, which it can only do if it does not actually do a correct >>>>>>>>>>> emulation

The correctly emulated input to H(P,P) never gets past its >>>>>>>>>> machine address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly
return 0 if it correctly emulates its input, but you can't drop >>>>>>>>> that requirement.

void P(u32 x)
{
    if (H(x, x))
      HERE: goto HERE;
    return;
}

int main()
{
    Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to
main().

But it also return 0 to the computation P(P), maybe not the copy >>>>>>> that it is simulating, since it aborts that before it get to it,

Finally you are not stupid or deceptive.

If H never returns a value to P then H is not a halting decider; if H >>>>> returns a value of 0 to main() when P halts then H is not a halting
decider.

H is not a halting decider; H is a simulation detector, S.

/Flibble

In other words you are saying the infintely nested emulation only
excutes for a few steps and then it stops on its own.

No, the "infinitely nested emulation" halts when the first H(P,P)
that P(P) called aborts it simulation.

When I tell you that an input only halts when it reaches its final
state and I tell you this many hundreds of times because you are a God
damned liar you pretend that I never said this.

Right, and when H(P,P) returns 0, the P that called it reaches its final state/the ret instrucion.

YOU GOD DAMNED LIAR main() calls H(P,P) and H returns to main()

void P(u32 x)
{
if (H(x, x))
HERE: goto HERE;
return;
}

int main()
{
Output("Input_Halts = ", H((u32)P, (u32)P));
}


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/9/22 7:14 PM, olcott wrote:

On 6/9/2022 6:09 PM, Richard Damon wrote:

On 6/9/22 6:37 PM, olcott wrote:

On 6/9/2022 5:33 PM, Richard Damon wrote:

On 6/9/22 6:18 PM, olcott wrote:

On 6/9/2022 5:15 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:09:08 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 4:54 PM, Richard Damon wrote:

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote: >>>>>>>>>>>>>>>>>>>> On Thu, 9 Jun 2022 12:39:32 -0500

olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote: >>>>>>>>>>>>>>>>>>>>>> On Thu, 9 Jun 2022 12:15:24 -0500

olcott <NoOne@NoWhere.com> wrote: >>>>>>>>>>>>>>>>>>>>>>> On 6/9/2022 12:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 6/9/22 12:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 6/9/2022 11:34 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 6/9/22 11:47 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> void P(u32 x)

{
      if (H(x, x))
        HERE: goto HERE; >>>>>>>>>>>>>>>>>>>>>>>>>>>       return;
}

int main()
{
      P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov eax,[ebp+08]
[000012ed](01)  50              push eax
[000012ee](03)  8b4d08          mov ecx,[ebp+08]
[000012f1](01)  51              push ecx
[000012f2](05)  e880feffff      call 00001177 //
call H [000012f7](03)  83c408          add
esp,+08 [000012fa](02)  85c0            test
eax,eax [000012fc](02)  7402            jz
00001300 [000012fe](02)  ebfe            jmp
000012fe [00001300](01)  5d              pop ebp
[00001301](01)  c3              ret >>>>>>>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [00001301] >>>>>>>>>>>>>>>>>>>>>>>>>>>
_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 //
push P [0000130f](05)  e8d3ffffff      call >>>>>>>>>>>>>>>>>>>>>>>>>>> 000012e7 // call P [00001314](03)  83c404 >>>>>>>>>>>>>>>>>>>>>>>>>>>    add esp,+04 [00001317](02)  33c0 >>>>>>>>>>>>>>>>>>>>>>>>>>> xor eax,eax [00001319](01)  5d              pop
ebp [0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a] >>>>>>>>>>>>>>>>>>>>>>>>>>>
     machine   stack     stack     machine
assembly address   address   data      code
  language ========  ========  ======== >>>>>>>>>>>>>>>>>>>>>>>>>>> ========= =============
[00001307][00102190][00000000] 55         push
ebp [00001308][00102190][00000000] 8bec >>>>>>>>>>>>>>>>>>>>>>>>>>> mov ebp,esp [0000130a][0010218c][000012e7] >>>>>>>>>>>>>>>>>>>>>>>>>>> 68e7120000 push 000012e7 // >>>>>>>>>>>>>>>>>>>>>>>>>>> push P [0000130f][00102188][00001314] e8d3ffffff >>>>>>>>>>>>>>>>>>>>>>>>>>> call 000012e7
// call P [000012e7][00102184][00102190] 55 push >>>>>>>>>>>>>>>>>>>>>>>>>>> ebp // enter executed P
[000012e8][00102184][00102190] 8bec       mov >>>>>>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00102184][00102190] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>>>>   mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50         push
eax      //
push P [000012ee][00102180][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08] [000012f1][0010217c][000012e7] >>>>>>>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P >>>>>>>>>>>>>>>>>>>>>>>>>>> [000012f2][00102178][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H

Begin Local Halt Decider Simulation   Execution >>>>>>>>>>>>>>>>>>>>>>>>>>> Trace Stored
at:212244 [000012e7][00212230][00212234] 55 push >>>>>>>>>>>>>>>>>>>>>>>>>>> ebp // enter emulated P
[000012e8][00212230][00212234] 8bec        mov
ebp,esp [000012ea][00212230][00212234] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>>>>    mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50          push
eax      //
push P [000012ee][0021222c][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>>>>   mov ecx,[ebp+08]
[000012f1][00212228][000012e7] >>>>>>>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P >>>>>>>>>>>>>>>>>>>>>>>>>>> [000012f2][00212224][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H

So, by what instruction reference manual is a >>>>>>>>>>>>>>>>>>>>>>>>>> call 00001177
followedby the execution of the instruction at >>>>>>>>>>>>>>>>>>>>>>>>>> 000012e7.

Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55          push
ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec        mov
ebp,esp [000012ea][0025cc58][0025cc5c] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>>>>    mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50          push
eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>>>>   mov ecx,[ebp+08]
[000012f1][0025cc50][000012e7] >>>>>>>>>>>>>>>>>>>>>>>>>>> 51 push ecx      // push P >>>>>>>>>>>>>>>>>>>>>>>>>>> [000012f2][0025cc4c][000012f7] e880feffff >>>>>>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H Local Halt Decider: >>>>>>>>>>>>>>>>>>>>>>>>>>> Infinite Recursion
Detected Simulation Stopped >>>>>>>>>>>>>>>>>>>>>>>>>>>
It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>>>>>>>>> correctly emulates
its input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>>>>>>>> instructions of
P. Because the seventh instruction of P repeats >>>>>>>>>>>>>>>>>>>>>>>>>>> this process we
know with complete certainty that the correct >>>>>>>>>>>>>>>>>>>>>>>>>>> and complete
emulation of P by H would never reach its final >>>>>>>>>>>>>>>>>>>>>>>>>>> “ret” instruction, thus never halts. >>>>>>>>>>>>>>>>>>>>>>>>>>

Problem, the 7th intruction DOESN't "Just repeat >>>>>>>>>>>>>>>>>>>>>>>>>> the procedure",
because that H always has the option to abort its >>>>>>>>>>>>>>>>>>>>>>>>>> simulation,
just like this onne did, and return to its P and >>>>>>>>>>>>>>>>>>>>>>>>>> see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS >>>>>>>>>>>>>>>>>>>>>>>>> IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input to >>>>>>>>>>>>>>>>>>>>>>>>> H(P,P) conclusively proves that the complete and >>>>>>>>>>>>>>>>>>>>>>>>> correct x86 emulation
would never stop running.

You SAY that, but you don't answer the actual >>>>>>>>>>>>>>>>>>>>>>>> questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>>>>>>>> NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT >>>>>>>>>>>>>>>>>>>>>>> THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) >>>>>>>>>>>>>>>>>>>>>>> CONCLUSIVELY PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE INPUT >>>>>>>>>>>>>>>>>>>>>>> TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>>>> instructions of P.
Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>>>>>>>> process we know
with complete certainty that the correct and >>>>>>>>>>>>>>>>>>>>>>> complete emulation of
P by H would never reach its final “ret” >>>>>>>>>>>>>>>>>>>>>>> instruction, thus never
halts.

If P should have halted (i.e. no infinite loop) then >>>>>>>>>>>>>>>>>>>>>> your simulation
detector, S (not H), gets the answer wrong.  You S is >>>>>>>>>>>>>>>>>>>>>> NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS IS >>>>>>>>>>>>>>>>>>>>> NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      //
push P
[00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      //
push P
[0000135d](05)  e840feffff      call 000011a2 // >>>>>>>>>>>>>>>>>>>>> call H
[00001362](03)  83c408          add esp,+08 >>>>>>>>>>>>>>>>>>>>> [00001365](02)  85c0            test eax,eax >>>>>>>>>>>>>>>>>>>>> [00001367](02)  7402            jz 0000136b >>>>>>>>>>>>>>>>>>>>> [00001369](02)  ebfe            jmp 00001369 >>>>>>>>>>>>>>>>>>>>> [0000136b](01)  5d              pop ebp >>>>>>>>>>>>>>>>>>>>> [0000136c](01)  c3              ret >>>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>> instructions of P. Because
the seventh instruction of P repeats this process we >>>>>>>>>>>>>>>>>>>>> know with
complete certainty that the correct and complete >>>>>>>>>>>>>>>>>>>>> emulation of P by H
would never reach its final “ret” instruction, thus >>>>>>>>>>>>>>>>>>>>> never halts.

We are going around and around and around in circles. I >>>>>>>>>>>>>>>>>>>> will try again:

If you replace the opcodes "EB FE" at 00001369 with the >>>>>>>>>>>>>>>>>>>> opcodes "90 90"
then your H gets the answer wrong: P should have >>>>>>>>>>>>>>>>>>>> halted.

/Flibble

As I already said before this is merely your >>>>>>>>>>>>>>>>>>> cluelessness that when H(P,P) is invoked the correct x86 >>>>>>>>>>>>>>>>>>> emulation of the input to H(P,P) makes and code after >>>>>>>>>>>>>>>>>>> [0000135d] unreachable.

Wrong, because when that H return the value 0, it will >>>>>>>>>>>>>>>>>> get there.

Like I said people that are dumber than a box of rocks >>>>>>>>>>>>>>>>> won't be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to >>>>>>>>>>>>>>>>> H(P,P) cannot possibly reach any instruction beyond >>>>>>>>>>>>>>>>> [0000135d].

So, you are defining that you H(P,P) never returns because >>>>>>>>>>>>>>>> it is caught in the infinite rcursion.

Thats fine, just says it can't be the correctly answering >>>>>>>>>>>>>>>> decider you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to know >>>>>>>>>>>>> that H performs a correct x86 emulation of its input and then >>>>>>>>>>>>> examine the execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) returns >>>>>>>>>>>> 0, which it can only do if it does not actually do a correct >>>>>>>>>>>> emulation

The correctly emulated input to H(P,P) never gets past its >>>>>>>>>>> machine address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly >>>>>>>>>> return 0 if it correctly emulates its input, but you can't drop >>>>>>>>>> that requirement.

void P(u32 x)
{
    if (H(x, x))
      HERE: goto HERE;
    return;
}

int main()
{
    Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to
main().

But it also return 0 to the computation P(P), maybe not the copy >>>>>>>> that it is simulating, since it aborts that before it get to it, >>>>>>>

Finally you are not stupid or deceptive.

If H never returns a value to P then H is not a halting decider; if H >>>>>> returns a value of 0 to main() when P halts then H is not a halting >>>>>> decider.

H is not a halting decider; H is a simulation detector, S.

/Flibble

In other words you are saying the infintely nested emulation only
excutes for a few steps and then it stops on its own.

No, the "infinitely nested emulation" halts when the first H(P,P)
that P(P) called aborts it simulation.

When I tell you that an input only halts when it reaches its final
state and I tell you this many hundreds of times because you are a
God damned liar you pretend that I never said this.

Right, and when H(P,P) returns 0, the P that called it reaches its
final state/the ret instrucion.

YOU GOD DAMNED LIAR main() calls H(P,P) and H returns to main()

void P(u32 x)
{
  if (H(x, x))
    HERE: goto HERE;
  return;
}

int main()
{
  Output("Input_Halts = ", H((u32)P, (u32)P));
}

Right, and P calls H which returns the 0 to P and P halts.

If you deny the existance of the program P, then you invalidate your
whole proof.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On Thu, 9 Jun 2022 17:35:05 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 5:27 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:18:38 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 5:15 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:09:08 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 4:54 PM, Richard Damon wrote:

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote: >>>>>>>>>>>>>>>>>>> On Thu, 9 Jun 2022 12:15:24 -0500

olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>> On 6/9/22 12:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>> On 6/9/2022 11:34 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 6/9/22 11:47 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>> void P(u32 x)

{
     if (H(x, x))
       HERE: goto HERE; >>>>>>>>>>>>>>>>>>>>>>>>      return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov >>>>>>>>>>>>>>>>>>>>>>>> eax,[ebp+08] [000012ed](01)  50 >>>>>>>>>>>>>>>>>>>>>>>> push eax [000012ee](03)  8b4d08          mov
ecx,[ebp+08] [000012f1](01)  51 >>>>>>>>>>>>>>>>>>>>>>>> push ecx [000012f2](05)  e880feffff      call >>>>>>>>>>>>>>>>>>>>>>>> 00001177 // call H [000012f7](03)  83c408 >>>>>>>>>>>>>>>>>>>>>>>>     add esp,+08 [000012fa](02)  85c0 >>>>>>>>>>>>>>>>>>>>>>>> test eax,eax [000012fc](02)  7402 >>>>>>>>>>>>>>>>>>>>>>>> jz 00001300 [000012fe](02)  ebfe >>>>>>>>>>>>>>>>>>>>>>>> jmp 000012fe [00001300](01)  5d >>>>>>>>>>>>>>>>>>>>>>>> pop ebp [00001301](01)  c3              ret
Size in bytes:(0027) [00001301] >>>>>>>>>>>>>>>>>>>>>>>>
_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 >>>>>>>>>>>>>>>>>>>>>>>> // push P [0000130f](05)  e8d3ffffff >>>>>>>>>>>>>>>>>>>>>>>> call 000012e7 // call P [00001314](03) >>>>>>>>>>>>>>>>>>>>>>>> 83c404 add esp,+04 [00001317](02)  33c0 >>>>>>>>>>>>>>>>>>>>>>>> xor eax,eax [00001319](01)  5d >>>>>>>>>>>>>>>>>>>>>>>> pop ebp [0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a] >>>>>>>>>>>>>>>>>>>>>>>>
    machine   stack     stack     machine
assembly address   address   data      code >>>>>>>>>>>>>>>>>>>>>>>> language ========  ========  ======== >>>>>>>>>>>>>>>>>>>>>>>> ========= =============
[00001307][00102190][00000000] 55 >>>>>>>>>>>>>>>>>>>>>>>> push ebp [00001308][00102190][00000000] 8bec >>>>>>>>>>>>>>>>>>>>>>>> mov ebp,esp [0000130a][0010218c][000012e7] >>>>>>>>>>>>>>>>>>>>>>>> 68e7120000 push 000012e7 //
push P [0000130f][00102188][00001314] >>>>>>>>>>>>>>>>>>>>>>>> e8d3ffffff call 000012e7
// call P [000012e7][00102184][00102190] 55 >>>>>>>>>>>>>>>>>>>>>>>> push ebp // enter executed P >>>>>>>>>>>>>>>>>>>>>>>> [000012e8][00102184][00102190] 8bec       mov >>>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00102184][00102190] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>> mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50 >>>>>>>>>>>>>>>>>>>>>>>> push eax      //
push P [000012ee][00102180][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08]
[000012f1][0010217c][000012e7] 51 push ecx >>>>>>>>>>>>>>>>>>>>>>>>    // push P [000012f2][00102178][000012f7] >>>>>>>>>>>>>>>>>>>>>>>> e880feffff call 00001177 // call H >>>>>>>>>>>>>>>>>>>>>>>>
Begin Local Halt Decider Simulation >>>>>>>>>>>>>>>>>>>>>>>> Execution Trace Stored
at:212244 [000012e7][00212230][00212234] 55 >>>>>>>>>>>>>>>>>>>>>>>> push ebp // enter emulated P >>>>>>>>>>>>>>>>>>>>>>>> [000012e8][00212230][00212234] 8bec >>>>>>>>>>>>>>>>>>>>>>>> mov ebp,esp [000012ea][00212230][00212234] >>>>>>>>>>>>>>>>>>>>>>>> 8b4508 mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50 >>>>>>>>>>>>>>>>>>>>>>>> push eax      //
push P [000012ee][0021222c][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08]
[000012f1][00212228][000012e7] 51 push ecx >>>>>>>>>>>>>>>>>>>>>>>> // push P [000012f2][00212224][000012f7] >>>>>>>>>>>>>>>>>>>>>>>> e880feffff call 00001177 // call H >>>>>>>>>>>>>>>>>>>>>>>

So, by what instruction reference manual is a >>>>>>>>>>>>>>>>>>>>>>> call 00001177
followedby the execution of the instruction at >>>>>>>>>>>>>>>>>>>>>>> 000012e7.

Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55 >>>>>>>>>>>>>>>>>>>>>>>> push ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec >>>>>>>>>>>>>>>>>>>>>>>> mov ebp,esp [000012ea][0025cc58][0025cc5c] >>>>>>>>>>>>>>>>>>>>>>>> 8b4508 mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50 >>>>>>>>>>>>>>>>>>>>>>>> push eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08]
[000012f1][0025cc50][000012e7] 51 push ecx >>>>>>>>>>>>>>>>>>>>>>>> // push P [000012f2][0025cc4c][000012f7] >>>>>>>>>>>>>>>>>>>>>>>> e880feffff call 00001177 // call H Local Halt >>>>>>>>>>>>>>>>>>>>>>>> Decider: Infinite Recursion
Detected Simulation Stopped

It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>>>>>> correctly emulates
its input that it must emulate the first >>>>>>>>>>>>>>>>>>>>>>>> seven instructions of
P. Because the seventh instruction of P >>>>>>>>>>>>>>>>>>>>>>>> repeats this process we
know with complete certainty that the correct >>>>>>>>>>>>>>>>>>>>>>>> and complete
emulation of P by H would never reach its >>>>>>>>>>>>>>>>>>>>>>>> final “ret” instruction, thus never halts. >>>>>>>>>>>>>>>>>>>>>>>

Problem, the 7th intruction DOESN't "Just >>>>>>>>>>>>>>>>>>>>>>> repeat the procedure",
because that H always has the option to abort >>>>>>>>>>>>>>>>>>>>>>> its simulation,
just like this onne did, and return to its P >>>>>>>>>>>>>>>>>>>>>>> and see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND >>>>>>>>>>>>>>>>>>>>>> THIS IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input >>>>>>>>>>>>>>>>>>>>>> to H(P,P) conclusively proves that the >>>>>>>>>>>>>>>>>>>>>> complete and correct x86 emulation >>>>>>>>>>>>>>>>>>>>>> would never stop running.

You SAY that, but you don't answer the actual >>>>>>>>>>>>>>>>>>>>> questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS >>>>>>>>>>>>>>>>>>>> IS NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT >>>>>>>>>>>>>>>>>>>> THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) >>>>>>>>>>>>>>>>>>>> CONCLUSIVELY PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE >>>>>>>>>>>>>>>>>>>> INPUT TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>> correctly emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>> instructions of P.
Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>>>>> process we know
with complete certainty that the correct and >>>>>>>>>>>>>>>>>>>> complete emulation of
P by H would never reach its final “ret” >>>>>>>>>>>>>>>>>>>> instruction, thus never
halts.

If P should have halted (i.e. no infinite loop) >>>>>>>>>>>>>>>>>>> then your simulation
detector, S (not H), gets the answer wrong.  You S >>>>>>>>>>>>>>>>>>> is NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS >>>>>>>>>>>>>>>>>> IS NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>>>>> [00001358](01)  50              push eax      //
push P [00001359](03)  8b4d08          mov >>>>>>>>>>>>>>>>>> ecx,[ebp+08] [0000135c](01)  51              push
ecx      // push P [0000135d](05)  e840feffff >>>>>>>>>>>>>>>>>> call 000011a2 // call H [00001362](03)  83c408 >>>>>>>>>>>>>>>>>>    add esp,+08 [00001365](02)  85c0
test eax,eax [00001367](02)  7402            jz
0000136b [00001369](02)  ebfe            jmp >>>>>>>>>>>>>>>>>> 00001369 [0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret >>>>>>>>>>>>>>>>>> Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>> instructions of P. Because
the seventh instruction of P repeats this process >>>>>>>>>>>>>>>>>> we know with
complete certainty that the correct and complete >>>>>>>>>>>>>>>>>> emulation of P by H
would never reach its final “ret” instruction, thus >>>>>>>>>>>>>>>>>> never halts.

We are going around and around and around in >>>>>>>>>>>>>>>>> circles. I will try again:

If you replace the opcodes "EB FE" at 00001369 with >>>>>>>>>>>>>>>>> the opcodes "90 90"
then your H gets the answer wrong: P should have >>>>>>>>>>>>>>>>> halted.

/Flibble

As I already said before this is merely your >>>>>>>>>>>>>>>> cluelessness that when H(P,P) is invoked the correct >>>>>>>>>>>>>>>> x86 emulation of the input to H(P,P) makes and code >>>>>>>>>>>>>>>> after [0000135d] unreachable.

Wrong, because when that H return the value 0, it will >>>>>>>>>>>>>>> get there.

Like I said people that are dumber than a box of rocks >>>>>>>>>>>>>> won't be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to >>>>>>>>>>>>>> H(P,P) cannot possibly reach any instruction beyond >>>>>>>>>>>>>> [0000135d].

So, you are defining that you H(P,P) never returns >>>>>>>>>>>>> because it is caught in the infinite rcursion.

Thats fine, just says it can't be the correctly
answering decider you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to
know that H performs a correct x86 emulation of its input >>>>>>>>>> and then examine the execution trace.

And a CORRECT emulation of the code will Halt if H(P,P)
returns 0, which it can only do if it does not actually do a >>>>>>>>> correct emulation

The correctly emulated input to H(P,P) never gets past its
machine address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly
return 0 if it correctly emulates its input, but you can't
drop that requirement.

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to
main().

But it also return 0 to the computation P(P), maybe not the copy
that it is simulating, since it aborts that before it get to
it,

Finally you are not stupid or deceptive.

If H never returns a value to P then H is not a halting decider;
if H returns a value of 0 to main() when P halts then H is not a
halting decider.

H is not a halting decider; H is a simulation detector, S.

/Flibble

In other words you are saying the infintely nested emulation only
excutes for a few steps and then it stops on its own.

I am saying that your H is not a halting decider; I am saying that
your H is just a simulation detector, S.

/Flibble

You are saying that you simply "don't believe in" simulating halt
deciders.

You haven't got a simulating halt decider, what you have got is a
simulation detector.

/Flibble

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 6/10/2022 6:11 AM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:35:05 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 5:27 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:18:38 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 5:15 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:09:08 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 4:54 PM, Richard Damon wrote:

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote:

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote:

On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 12:39:32 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote: >>>>>>>>>>>>>>>>>>>>> On Thu, 9 Jun 2022 12:15:24 -0500

olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>> On 6/9/22 12:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 6/9/2022 11:34 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 6/9/22 11:47 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> void P(u32 x)

{
     if (H(x, x))
       HERE: goto HERE; >>>>>>>>>>>>>>>>>>>>>>>>>>      return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov ebp,esp
[000012ea](03)  8b4508          mov >>>>>>>>>>>>>>>>>>>>>>>>>> eax,[ebp+08] [000012ed](01)  50 >>>>>>>>>>>>>>>>>>>>>>>>>> push eax [000012ee](03)  8b4d08          mov
ecx,[ebp+08] [000012f1](01)  51 >>>>>>>>>>>>>>>>>>>>>>>>>> push ecx [000012f2](05)  e880feffff      call >>>>>>>>>>>>>>>>>>>>>>>>>> 00001177 // call H [000012f7](03)  83c408 >>>>>>>>>>>>>>>>>>>>>>>>>>     add esp,+08 [000012fa](02)  85c0 >>>>>>>>>>>>>>>>>>>>>>>>>> test eax,eax [000012fc](02)  7402 >>>>>>>>>>>>>>>>>>>>>>>>>> jz 00001300 [000012fe](02)  ebfe >>>>>>>>>>>>>>>>>>>>>>>>>> jmp 000012fe [00001300](01)  5d >>>>>>>>>>>>>>>>>>>>>>>>>> pop ebp [00001301](01)  c3              ret
Size in bytes:(0027) [00001301] >>>>>>>>>>>>>>>>>>>>>>>>>>
_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov ebp,esp
[0000130a](05)  68e7120000      push 000012e7 >>>>>>>>>>>>>>>>>>>>>>>>>> // push P [0000130f](05)  e8d3ffffff >>>>>>>>>>>>>>>>>>>>>>>>>> call 000012e7 // call P [00001314](03) >>>>>>>>>>>>>>>>>>>>>>>>>> 83c404 add esp,+04 [00001317](02)  33c0 >>>>>>>>>>>>>>>>>>>>>>>>>> xor eax,eax [00001319](01)  5d >>>>>>>>>>>>>>>>>>>>>>>>>> pop ebp [0000131a](01)  c3              ret
Size in bytes:(0020) [0000131a] >>>>>>>>>>>>>>>>>>>>>>>>>>
    machine   stack     stack     machine
assembly address   address   data      code >>>>>>>>>>>>>>>>>>>>>>>>>> language ========  ========  ======== >>>>>>>>>>>>>>>>>>>>>>>>>> ========= =============
[00001307][00102190][00000000] 55 >>>>>>>>>>>>>>>>>>>>>>>>>> push ebp [00001308][00102190][00000000] 8bec >>>>>>>>>>>>>>>>>>>>>>>>>> mov ebp,esp [0000130a][0010218c][000012e7] >>>>>>>>>>>>>>>>>>>>>>>>>> 68e7120000 push 000012e7 // >>>>>>>>>>>>>>>>>>>>>>>>>> push P [0000130f][00102188][00001314] >>>>>>>>>>>>>>>>>>>>>>>>>> e8d3ffffff call 000012e7
// call P [000012e7][00102184][00102190] 55 >>>>>>>>>>>>>>>>>>>>>>>>>> push ebp // enter executed P >>>>>>>>>>>>>>>>>>>>>>>>>> [000012e8][00102184][00102190] 8bec       mov >>>>>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea][00102184][00102190] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>>> mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50 >>>>>>>>>>>>>>>>>>>>>>>>>> push eax      //
push P [000012ee][00102180][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08]
[000012f1][0010217c][000012e7] 51 push ecx >>>>>>>>>>>>>>>>>>>>>>>>>>    // push P [000012f2][00102178][000012f7] >>>>>>>>>>>>>>>>>>>>>>>>>> e880feffff call 00001177 // call H >>>>>>>>>>>>>>>>>>>>>>>>>>
Begin Local Halt Decider Simulation >>>>>>>>>>>>>>>>>>>>>>>>>> Execution Trace Stored
at:212244 [000012e7][00212230][00212234] 55 >>>>>>>>>>>>>>>>>>>>>>>>>> push ebp // enter emulated P >>>>>>>>>>>>>>>>>>>>>>>>>> [000012e8][00212230][00212234] 8bec >>>>>>>>>>>>>>>>>>>>>>>>>> mov ebp,esp [000012ea][00212230][00212234] >>>>>>>>>>>>>>>>>>>>>>>>>> 8b4508 mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50 >>>>>>>>>>>>>>>>>>>>>>>>>> push eax      //
push P [000012ee][0021222c][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08]
[000012f1][00212228][000012e7] 51 push ecx >>>>>>>>>>>>>>>>>>>>>>>>>> // push P [000012f2][00212224][000012f7] >>>>>>>>>>>>>>>>>>>>>>>>>> e880feffff call 00001177 // call H >>>>>>>>>>>>>>>>>>>>>>>>>

So, by what instruction reference manual is a >>>>>>>>>>>>>>>>>>>>>>>>> call 00001177
followedby the execution of the instruction at >>>>>>>>>>>>>>>>>>>>>>>>> 000012e7.

Your "CPU" is broken, or emulation incorrect. >>>>>>>>>>>>>>>>>>>>>>>>>
FAIL.

[000012e7][0025cc58][0025cc5c] 55 >>>>>>>>>>>>>>>>>>>>>>>>>> push ebp      //
enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec >>>>>>>>>>>>>>>>>>>>>>>>>> mov ebp,esp [000012ea][0025cc58][0025cc5c] >>>>>>>>>>>>>>>>>>>>>>>>>> 8b4508 mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50 >>>>>>>>>>>>>>>>>>>>>>>>>> push eax      //
push P [000012ee][0025cc54][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08]
[000012f1][0025cc50][000012e7] 51 push ecx >>>>>>>>>>>>>>>>>>>>>>>>>> // push P [000012f2][0025cc4c][000012f7] >>>>>>>>>>>>>>>>>>>>>>>>>> e880feffff call 00001177 // call H Local Halt >>>>>>>>>>>>>>>>>>>>>>>>>> Decider: Infinite Recursion >>>>>>>>>>>>>>>>>>>>>>>>>> Detected Simulation Stopped >>>>>>>>>>>>>>>>>>>>>>>>>>
It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>>>>>>>> correctly emulates
its input that it must emulate the first >>>>>>>>>>>>>>>>>>>>>>>>>> seven instructions of
P. Because the seventh instruction of P >>>>>>>>>>>>>>>>>>>>>>>>>> repeats this process we
know with complete certainty that the correct >>>>>>>>>>>>>>>>>>>>>>>>>> and complete
emulation of P by H would never reach its >>>>>>>>>>>>>>>>>>>>>>>>>> final “ret” instruction, thus never halts. >>>>>>>>>>>>>>>>>>>>>>>>>

Problem, the 7th intruction DOESN't "Just >>>>>>>>>>>>>>>>>>>>>>>>> repeat the procedure",
because that H always has the option to abort >>>>>>>>>>>>>>>>>>>>>>>>> its simulation,
just like this onne did, and return to its P >>>>>>>>>>>>>>>>>>>>>>>>> and see it halt.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND >>>>>>>>>>>>>>>>>>>>>>>> THIS IS NO ACTUAL
REBUTTAL AT ALL:

The partial correct x86 emulation of the input >>>>>>>>>>>>>>>>>>>>>>>> to H(P,P) conclusively proves that the >>>>>>>>>>>>>>>>>>>>>>>> complete and correct x86 emulation >>>>>>>>>>>>>>>>>>>>>>>> would never stop running.

You SAY that, but you don't answer the actual >>>>>>>>>>>>>>>>>>>>>>> questions about HOW.

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS >>>>>>>>>>>>>>>>>>>>>> IS NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE THAT >>>>>>>>>>>>>>>>>>>>>> THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) >>>>>>>>>>>>>>>>>>>>>> CONCLUSIVELY PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE >>>>>>>>>>>>>>>>>>>>>> INPUT TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>>>> correctly emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>>> instructions of P.
Because the seventh instruction of P repeats this >>>>>>>>>>>>>>>>>>>>>> process we know
with complete certainty that the correct and >>>>>>>>>>>>>>>>>>>>>> complete emulation of
P by H would never reach its final “ret” >>>>>>>>>>>>>>>>>>>>>> instruction, thus never
halts.

If P should have halted (i.e. no infinite loop) >>>>>>>>>>>>>>>>>>>>> then your simulation
detector, S (not H), gets the answer wrong.  You S >>>>>>>>>>>>>>>>>>>>> is NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND THIS >>>>>>>>>>>>>>>>>>>> IS NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp >>>>>>>>>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08] >>>>>>>>>>>>>>>>>>>> [00001358](01)  50              push eax      //
push P [00001359](03)  8b4d08          mov >>>>>>>>>>>>>>>>>>>> ecx,[ebp+08] [0000135c](01)  51              push
ecx      // push P [0000135d](05)  e840feffff >>>>>>>>>>>>>>>>>>>> call 000011a2 // call H [00001362](03)  83c408 >>>>>>>>>>>>>>>>>>>>    add esp,+08 [00001365](02)  85c0 >>>>>>>>>>>>>>>>>>>> test eax,eax [00001367](02)  7402            jz
0000136b [00001369](02)  ebfe            jmp >>>>>>>>>>>>>>>>>>>> 00001369 [0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret >>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) correctly >>>>>>>>>>>>>>>>>>>> emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>> instructions of P. Because
the seventh instruction of P repeats this process >>>>>>>>>>>>>>>>>>>> we know with
complete certainty that the correct and complete >>>>>>>>>>>>>>>>>>>> emulation of P by H
would never reach its final “ret” instruction, thus >>>>>>>>>>>>>>>>>>>> never halts.

We are going around and around and around in >>>>>>>>>>>>>>>>>>> circles. I will try again:

If you replace the opcodes "EB FE" at 00001369 with >>>>>>>>>>>>>>>>>>> the opcodes "90 90"
then your H gets the answer wrong: P should have >>>>>>>>>>>>>>>>>>> halted.

/Flibble

As I already said before this is merely your >>>>>>>>>>>>>>>>>> cluelessness that when H(P,P) is invoked the correct >>>>>>>>>>>>>>>>>> x86 emulation of the input to H(P,P) makes and code >>>>>>>>>>>>>>>>>> after [0000135d] unreachable.

Wrong, because when that H return the value 0, it will >>>>>>>>>>>>>>>>> get there.

Like I said people that are dumber than a box of rocks >>>>>>>>>>>>>>>> won't be able to correctly understand this.

When H(P,P) is invoked the correctly emulated input to >>>>>>>>>>>>>>>> H(P,P) cannot possibly reach any instruction beyond >>>>>>>>>>>>>>>> [0000135d].

So, you are defining that you H(P,P) never returns >>>>>>>>>>>>>>> because it is caught in the infinite rcursion.

Thats fine, just says it can't be the correctly
answering decider you claim it to be.

I have corrected you on this too many times.

How. You need to define what H(P,P) actually does.

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to >>>>>>>>>>>> know that H performs a correct x86 emulation of its input >>>>>>>>>>>> and then examine the execution trace.

And a CORRECT emulation of the code will Halt if H(P,P)
returns 0, which it can only do if it does not actually do a >>>>>>>>>>> correct emulation

The correctly emulated input to H(P,P) never gets past its >>>>>>>>>> machine address [0000135d].

Only if H actually doesn't return 0. Yes, H can't correctly
return 0 if it correctly emulates its input, but you can't
drop that requirement.

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to
main().

But it also return 0 to the computation P(P), maybe not the copy >>>>>>> that it is simulating, since it aborts that before it get to
it,

Finally you are not stupid or deceptive.

If H never returns a value to P then H is not a halting decider;
if H returns a value of 0 to main() when P halts then H is not a
halting decider.

H is not a halting decider; H is a simulation detector, S.

/Flibble

In other words you are saying the infintely nested emulation only
excutes for a few steps and then it stops on its own.

I am saying that your H is not a halting decider; I am saying that
your H is just a simulation detector, S.

/Flibble

You are saying that you simply "don't believe in" simulating halt
deciders.

You haven't got a simulating halt decider, what you have got is a
simulation detector.

/Flibble

It is an easily verified fact that H does correctly decide the halt
status of its input.

void Infinite_Loop()
{
HERE: goto HERE;
}

int main()
{
Output("Input_Halts = ", H0(Infinite_Loop));
}

_Infinite_Loop()
[00001342](01) 55 push ebp
[00001343](02) 8bec mov ebp,esp
[00001345](02) ebfe jmp 00001345
[00001347](01) 5d pop ebp
[00001348](01) c3 ret
Size in bytes:(0007) [00001348]

_main()
[00001372](01) 55 push ebp
[00001373](02) 8bec mov ebp,esp
[00001375](05) 6842130000 push 00001342
[0000137a](05) e833fdffff call 000010b2
[0000137f](03) 83c404 add esp,+04
[00001382](01) 50 push eax
[00001383](05) 6823040000 push 00000423
[00001388](05) e8e5f0ffff call 00000472
[0000138d](03) 83c408 add esp,+08
[00001390](02) 33c0 xor eax,eax
[00001392](01) 5d pop ebp
[00001393](01) c3 ret
Size in bytes:(0034) [00001393]

machine stack stack machine assembly
address address data code language
======== ======== ======== ========= ============= [00001372][0010228f][00000000] 55 push ebp [00001373][0010228f][00000000] 8bec mov ebp,esp [00001375][0010228b][00001342] 6842130000 push 00001342 // push
_Infinite_Loop
[0000137a][00102287][0000137f] e833fdffff call 000010b2 // call H0

Begin Local Halt Decider Simulation Execution Trace Stored at:212343 [00001342][00212333][00212337] 55 push ebp [00001343][00212333][00212337] 8bec mov ebp,esp [00001345][00212333][00212337] ebfe jmp 00001345 [00001345][00212333][00212337] ebfe jmp 00001345
Local Halt Decider: Infinite Loop Detected Simulation Stopped

[0000137f][0010228f][00000000] 83c404 add esp,+04 [00001382][0010228b][00000000] 50 push eax [00001383][00102287][00000423] 6823040000 push 00000423 [00001388][00102287][00000423] e8e5f0ffff call 00000472
Input_Halts = 0
[0000138d][0010228f][00000000] 83c408 add esp,+08 [00001390][0010228f][00000000] 33c0 xor eax,eax [00001392][00102293][00100000] 5d pop ebp [00001393][00102297][00000004] c3 ret
Number of Instructions Executed(680)

int Sum(int X, int Y)
{
return X + Y;
}

int main()
{
Output("Input_Halts = ", H2(Sum, 3, 4));
}

_Sum()
[000012b6](01) 55 push ebp
[000012b7](02) 8bec mov ebp,esp
[000012b9](03) 8b4508 mov eax,[ebp+08]
[000012bc](03) 03450c add eax,[ebp+0c]
[000012bf](01) 5d pop ebp
[000012c0](01) c3 ret
Size in bytes:(0011) [000012c0]

_main()
[00001316](01) 55 push ebp
[00001317](02) 8bec mov ebp,esp
[00001319](02) 6a04 push +04
[0000131b](02) 6a03 push +03
[0000131d](05) 68b6120000 push 000012b6
[00001322](05) e85ffaffff call 00000d86
[00001327](03) 83c40c add esp,+0c
[0000132a](01) 50 push eax
[0000132b](05) 6807040000 push 00000407
[00001330](05) e821f1ffff call 00000456
[00001335](03) 83c408 add esp,+08
[00001338](02) 33c0 xor eax,eax
[0000133a](01) 5d pop ebp
[0000133b](01) c3 ret
Size in bytes:(0038) [0000133b]

machine stack stack machine assembly
address address data code language
======== ======== ======== ========= ============= ...[00001316][001021d7][00000000] 55 push ebp ...[00001317][001021d7][00000000] 8bec mov ebp,esp ...[00001319][001021d3][00000004] 6a04 push +04 ...[0000131b][001021cf][00000003] 6a03 push +03 ...[0000131d][001021cb][000012b6] 68b6120000 push 000012b6 ...[00001322][001021c7][00001327] e85ffaffff call 00000d86

Begin Local Halt Decider Simulation Execution Trace Stored at:21228b ...[000012b6][00212273][00212277] 55 push ebp ...[000012b7][00212273][00212277] 8bec mov ebp,esp ...[000012b9][00212273][00212277] 8b4508 mov eax,[ebp+08] ...[000012bc][00212273][00212277] 03450c add eax,[ebp+0c] ...[000012bf][00212277][00000e54] 5d pop ebp ...[000012c0][0021227b][00000003] c3 ret ...[00001327][001021d7][00000000] 83c40c add esp,+0c ...[0000132a][001021d3][00000001] 50 push eax ...[0000132b][001021cf][00000407] 6807040000 push 00000407 ---[00001330][001021cf][00000407] e821f1ffff call 00000456
Input_Halts = 1
...[00001335][001021d7][00000000] 83c408 add esp,+08 ...[00001338][001021d7][00000000] 33c0 xor eax,eax ...[0000133a][001021db][00100000] 5d pop ebp ...[0000133b][001021df][00000004] c3 ret
Number of Instructions Executed(660)



--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On Fri, 10 Jun 2022 08:27:28 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/10/2022 8:23 AM, Mr Flibble wrote:

On Fri, 10 Jun 2022 08:05:57 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/10/2022 6:11 AM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:35:05 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 5:27 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:18:38 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 5:15 PM, Mr Flibble wrote:

On Thu, 9 Jun 2022 17:09:08 -0500
olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 4:54 PM, Richard Damon wrote:

On 6/9/22 5:29 PM, olcott wrote:

On 6/9/2022 4:13 PM, Richard Damon wrote:

On 6/9/22 5:00 PM, olcott wrote:

On 6/9/2022 3:25 PM, Richard Damon wrote:

On 6/9/22 4:13 PM, olcott wrote:

On 6/9/2022 2:57 PM, Richard Damon wrote:

On 6/9/22 3:52 PM, olcott wrote:

On 6/9/2022 2:38 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>

On 6/9/22 3:26 PM, olcott wrote:

On 6/9/2022 1:12 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 6/9/22 1:55 PM, olcott wrote:

On 6/9/2022 12:46 PM, Mr Flibble wrote: >>>>>>>>>>>>>>>>>>>>> On Thu, 9 Jun 2022 12:39:32 -0500

olcott <NoOne@NoWhere.com> wrote:

On 6/9/2022 12:28 PM, Mr Flibble wrote: >>>>>>>>>>>>>>>>>>>>>>> On Thu, 9 Jun 2022 12:15:24 -0500 >>>>>>>>>>>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 6/9/2022 12:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 6/9/22 12:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 6/9/2022 11:34 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>

On 6/9/22 11:47 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> void P(u32 x)

{
     if (H(x, x)) >>>>>>>>>>>>>>>>>>>>>>>>>>>>        HERE: goto HERE; >>>>>>>>>>>>>>>>>>>>>>>>>>>>      return;
}

int main()
{
     P(P);
}

_P()
[000012e7](01)  55              push ebp
[000012e8](02)  8bec            mov >>>>>>>>>>>>>>>>>>>>>>>>>>>> ebp,esp [000012ea](03)  8b4508 >>>>>>>>>>>>>>>>>>>>>>>>>>>> mov eax,[ebp+08] [000012ed](01)  50 >>>>>>>>>>>>>>>>>>>>>>>>>>>> push eax [000012ee](03)  8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08] [000012f1](01)  51 >>>>>>>>>>>>>>>>>>>>>>>>>>>> push ecx [000012f2](05)  e880feffff >>>>>>>>>>>>>>>>>>>>>>>>>>>> call 00001177 // call H [000012f7](03) >>>>>>>>>>>>>>>>>>>>>>>>>>>> 83c408 add esp,+08 [000012fa](02)  85c0 >>>>>>>>>>>>>>>>>>>>>>>>>>>> test eax,eax [000012fc](02)  7402 >>>>>>>>>>>>>>>>>>>>>>>>>>>> jz 00001300 [000012fe](02)  ebfe >>>>>>>>>>>>>>>>>>>>>>>>>>>> jmp 000012fe [00001300](01)  5d >>>>>>>>>>>>>>>>>>>>>>>>>>>> pop ebp [00001301](01)  c3 >>>>>>>>>>>>>>>>>>>>>>>>>>>> ret Size in bytes:(0027) [00001301] >>>>>>>>>>>>>>>>>>>>>>>>>>>>
_main()
[00001307](01)  55              push ebp
[00001308](02)  8bec            mov >>>>>>>>>>>>>>>>>>>>>>>>>>>> ebp,esp [0000130a](05)  68e7120000 >>>>>>>>>>>>>>>>>>>>>>>>>>>> push 000012e7 // push P [0000130f](05) >>>>>>>>>>>>>>>>>>>>>>>>>>>> e8d3ffffff call 000012e7 // call P >>>>>>>>>>>>>>>>>>>>>>>>>>>> [00001314](03) 83c404 add esp,+04 >>>>>>>>>>>>>>>>>>>>>>>>>>>> [00001317](02)  33c0 xor eax,eax >>>>>>>>>>>>>>>>>>>>>>>>>>>> [00001319](01)  5d pop ebp [0000131a](01) >>>>>>>>>>>>>>>>>>>>>>>>>>>> c3              ret Size in bytes:(0020)
[0000131a]

    machine   stack     stack >>>>>>>>>>>>>>>>>>>>>>>>>>>> machine assembly address   address   data >>>>>>>>>>>>>>>>>>>>>>>>>>>>     code language ========  ======== >>>>>>>>>>>>>>>>>>>>>>>>>>>> ======== ========= ============= >>>>>>>>>>>>>>>>>>>>>>>>>>>> [00001307][00102190][00000000] 55 >>>>>>>>>>>>>>>>>>>>>>>>>>>> push ebp [00001308][00102190][00000000] >>>>>>>>>>>>>>>>>>>>>>>>>>>> 8bec mov ebp,esp
[0000130a][0010218c][000012e7] 68e7120000 >>>>>>>>>>>>>>>>>>>>>>>>>>>> push 000012e7 // push P >>>>>>>>>>>>>>>>>>>>>>>>>>>> [0000130f][00102188][00001314] e8d3ffffff >>>>>>>>>>>>>>>>>>>>>>>>>>>> call 000012e7 // call P >>>>>>>>>>>>>>>>>>>>>>>>>>>> [000012e7][00102184][00102190] 55 push >>>>>>>>>>>>>>>>>>>>>>>>>>>> ebp // enter executed P >>>>>>>>>>>>>>>>>>>>>>>>>>>> [000012e8][00102184][00102190] 8bec >>>>>>>>>>>>>>>>>>>>>>>>>>>> mov ebp,esp
[000012ea][00102184][00102190] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>>>>> mov eax,[ebp+08]
[000012ed][00102180][000012e7] 50 push >>>>>>>>>>>>>>>>>>>>>>>>>>>> eax // push P
[000012ee][00102180][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08]
[000012f1][0010217c][000012e7] 51 push >>>>>>>>>>>>>>>>>>>>>>>>>>>> ecx // push P
[000012f2][00102178][000012f7] >>>>>>>>>>>>>>>>>>>>>>>>>>>> e880feffff call 00001177 // call H >>>>>>>>>>>>>>>>>>>>>>>>>>>>
Begin Local Halt Decider Simulation >>>>>>>>>>>>>>>>>>>>>>>>>>>> Execution Trace Stored >>>>>>>>>>>>>>>>>>>>>>>>>>>> at:212244 [000012e7][00212230][00212234] >>>>>>>>>>>>>>>>>>>>>>>>>>>> 55 push ebp // enter emulated P >>>>>>>>>>>>>>>>>>>>>>>>>>>> [000012e8][00212230][00212234] 8bec >>>>>>>>>>>>>>>>>>>>>>>>>>>> mov ebp,esp
[000012ea][00212230][00212234] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>>>>> mov eax,[ebp+08]
[000012ed][0021222c][000012e7] 50 push >>>>>>>>>>>>>>>>>>>>>>>>>>>> eax      // push P >>>>>>>>>>>>>>>>>>>>>>>>>>>> [000012ee][0021222c][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08]
[000012f1][00212228][000012e7] 51 push >>>>>>>>>>>>>>>>>>>>>>>>>>>> ecx // push P
[000012f2][00212224][000012f7] >>>>>>>>>>>>>>>>>>>>>>>>>>>> e880feffff call 00001177 // call H >>>>>>>>>>>>>>>>>>>>>>>>>>>

So, by what instruction reference manual >>>>>>>>>>>>>>>>>>>>>>>>>>> is a call 00001177
followedby the execution of the >>>>>>>>>>>>>>>>>>>>>>>>>>> instruction at 000012e7. >>>>>>>>>>>>>>>>>>>>>>>>>>>
Your "CPU" is broken, or emulation >>>>>>>>>>>>>>>>>>>>>>>>>>> incorrect.

FAIL.

[000012e7][0025cc58][0025cc5c] 55 >>>>>>>>>>>>>>>>>>>>>>>>>>>> push ebp      // >>>>>>>>>>>>>>>>>>>>>>>>>>>> enter emulated P
[000012e8][0025cc58][0025cc5c] 8bec >>>>>>>>>>>>>>>>>>>>>>>>>>>> mov ebp,esp
[000012ea][0025cc58][0025cc5c] 8b4508 >>>>>>>>>>>>>>>>>>>>>>>>>>>> mov eax,[ebp+08]
[000012ed][0025cc54][000012e7] 50 push >>>>>>>>>>>>>>>>>>>>>>>>>>>> eax      // push P >>>>>>>>>>>>>>>>>>>>>>>>>>>> [000012ee][0025cc54][000012e7] 8b4d08 >>>>>>>>>>>>>>>>>>>>>>>>>>>> mov ecx,[ebp+08]
[000012f1][0025cc50][000012e7] 51 push >>>>>>>>>>>>>>>>>>>>>>>>>>>> ecx // push P
[000012f2][0025cc4c][000012f7] >>>>>>>>>>>>>>>>>>>>>>>>>>>> e880feffff call 00001177 // call H Local >>>>>>>>>>>>>>>>>>>>>>>>>>>> Halt Decider: Infinite Recursion >>>>>>>>>>>>>>>>>>>>>>>>>>>> Detected Simulation Stopped >>>>>>>>>>>>>>>>>>>>>>>>>>>>
It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>>>>>>>>>> correctly emulates
its input that it must emulate the first >>>>>>>>>>>>>>>>>>>>>>>>>>>> seven instructions of >>>>>>>>>>>>>>>>>>>>>>>>>>>> P. Because the seventh instruction of P >>>>>>>>>>>>>>>>>>>>>>>>>>>> repeats this process we >>>>>>>>>>>>>>>>>>>>>>>>>>>> know with complete certainty that the >>>>>>>>>>>>>>>>>>>>>>>>>>>> correct and complete >>>>>>>>>>>>>>>>>>>>>>>>>>>> emulation of P by H would never reach its >>>>>>>>>>>>>>>>>>>>>>>>>>>> final “ret” instruction, thus never >>>>>>>>>>>>>>>>>>>>>>>>>>>> halts.

Problem, the 7th intruction DOESN't "Just >>>>>>>>>>>>>>>>>>>>>>>>>>> repeat the procedure", >>>>>>>>>>>>>>>>>>>>>>>>>>> because that H always has the option to >>>>>>>>>>>>>>>>>>>>>>>>>>> abort its simulation,
just like this onne did, and return to >>>>>>>>>>>>>>>>>>>>>>>>>>> its P and see it halt. >>>>>>>>>>>>>>>>>>>>>>>>>> THAT YOU ARE SIMPLY TOO STUPID TO >>>>>>>>>>>>>>>>>>>>>>>>>> UNDERSTAND THIS IS NO ACTUAL >>>>>>>>>>>>>>>>>>>>>>>>>> REBUTTAL AT ALL:

The partial correct x86 emulation of the >>>>>>>>>>>>>>>>>>>>>>>>>> input to H(P,P) conclusively proves that >>>>>>>>>>>>>>>>>>>>>>>>>> the complete and correct x86 emulation >>>>>>>>>>>>>>>>>>>>>>>>>> would never stop running. >>>>>>>>>>>>>>>>>>>>>>>>>>

You SAY that, but you don't answer the >>>>>>>>>>>>>>>>>>>>>>>>> actual questions about HOW. >>>>>>>>>>>>>>>>>>>>>>>>

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND >>>>>>>>>>>>>>>>>>>>>>>> THIS IS NO EVIDENCE
WHAT-SO-EVER THAT I DID NOT COMPLETELY PROVE >>>>>>>>>>>>>>>>>>>>>>>> THAT THE CORRECT
PARTIAL EMULATION OF THE INPUT TO H(P,P) >>>>>>>>>>>>>>>>>>>>>>>> CONCLUSIVELY PROVES THAT
THE CORRECT AND COMPLETE X86 EMULATION OF THE >>>>>>>>>>>>>>>>>>>>>>>> INPUT TO H(P,P)
WOULD NEVER STOP RUNNING.

It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>>>>>> correctly emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>>>>> instructions of P.
Because the seventh instruction of P repeats >>>>>>>>>>>>>>>>>>>>>>>> this process we know
with complete certainty that the correct and >>>>>>>>>>>>>>>>>>>>>>>> complete emulation of
P by H would never reach its final “ret” >>>>>>>>>>>>>>>>>>>>>>>> instruction, thus never
halts.

If P should have halted (i.e. no infinite >>>>>>>>>>>>>>>>>>>>>>> loop) then your simulation
detector, S (not H), gets the answer wrong. >>>>>>>>>>>>>>>>>>>>>>> You S is NOT a halting
decider.

/Flibble

THAT YOU ARE SIMPLY TOO STUPID TO UNDERSTAND >>>>>>>>>>>>>>>>>>>>>> THIS IS NO ACTUAL
REBUTTAL AT ALL.

_P()
[00001352](01)  55              push ebp >>>>>>>>>>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov >>>>>>>>>>>>>>>>>>>>>> eax,[ebp+08] [00001358](01)  50 >>>>>>>>>>>>>>>>>>>>>> push eax      // push P [00001359](03)  8b4d08 >>>>>>>>>>>>>>>>>>>>>>          mov ecx,[ebp+08] [0000135c](01)  51 >>>>>>>>>>>>>>>>>>>>>>            push ecx      // push P >>>>>>>>>>>>>>>>>>>>>> [0000135d](05)  e840feffff call 000011a2 // >>>>>>>>>>>>>>>>>>>>>> call H [00001362](03)  83c408 add esp,+08 >>>>>>>>>>>>>>>>>>>>>> [00001365](02)  85c0 test eax,eax >>>>>>>>>>>>>>>>>>>>>> [00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp >>>>>>>>>>>>>>>>>>>>>> [0000136c](01)  c3              ret Size in
bytes:(0027) [0000136c]

It is completely obvious that when H(P,P) >>>>>>>>>>>>>>>>>>>>>> correctly emulates its
input that it must emulate the first seven >>>>>>>>>>>>>>>>>>>>>> instructions of P. Because
the seventh instruction of P repeats this >>>>>>>>>>>>>>>>>>>>>> process we know with
complete certainty that the correct and >>>>>>>>>>>>>>>>>>>>>> complete emulation of P by H
would never reach its final “ret” instruction, >>>>>>>>>>>>>>>>>>>>>> thus never halts.

We are going around and around and around in >>>>>>>>>>>>>>>>>>>>> circles. I will try again:

If you replace the opcodes "EB FE" at 00001369 >>>>>>>>>>>>>>>>>>>>> with the opcodes "90 90"
then your H gets the answer wrong: P should have >>>>>>>>>>>>>>>>>>>>> halted.

/Flibble

As I already said before this is merely your >>>>>>>>>>>>>>>>>>>> cluelessness that when H(P,P) is invoked the >>>>>>>>>>>>>>>>>>>> correct x86 emulation of the input to H(P,P) >>>>>>>>>>>>>>>>>>>> makes and code after [0000135d] unreachable. >>>>>>>>>>>>>>>>>>>

Wrong, because when that H return the value 0, it >>>>>>>>>>>>>>>>>>> will get there.

Like I said people that are dumber than a box of >>>>>>>>>>>>>>>>>> rocks won't be able to correctly understand this. >>>>>>>>>>>>>>>>>>
When H(P,P) is invoked the correctly emulated input >>>>>>>>>>>>>>>>>> to H(P,P) cannot possibly reach any instruction >>>>>>>>>>>>>>>>>> beyond [0000135d].

So, you are defining that you H(P,P) never returns >>>>>>>>>>>>>>>>> because it is caught in the infinite rcursion. >>>>>>>>>>>>>>>>>
Thats fine, just says it can't be the correctly >>>>>>>>>>>>>>>>> answering decider you claim it to be.

I have corrected you on this too many times. >>>>>>>>>>>>>>>>

How. You need to define what H(P,P) actually does. >>>>>>>>>>>>>>

I have explained that too many times.

To understand that H(P,P)==0 is correct we only need to >>>>>>>>>>>>>> know that H performs a correct x86 emulation of its >>>>>>>>>>>>>> input and then examine the execution trace.

And a CORRECT emulation of the code will Halt if H(P,P) >>>>>>>>>>>>> returns 0, which it can only do if it does not actually >>>>>>>>>>>>> do a correct emulation

The correctly emulated input to H(P,P) never gets past >>>>>>>>>>>> its machine address [0000135d].

Only if H actually doesn't return 0. Yes, H can't
correctly return 0 if it correctly emulates its input, >>>>>>>>>>> but you can't drop that requirement.

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
   return;
}

int main()
{
   Output("Input_Halts = ", H((u32)P, (u32)P));
}

When H returns 0 it does not returns 0 to P it returns 0 to >>>>>>>>>> main().

But it also return 0 to the computation P(P), maybe not the >>>>>>>>> copy that it is simulating, since it aborts that before it >>>>>>>>> get to it,

Finally you are not stupid or deceptive.

If H never returns a value to P then H is not a halting
decider; if H returns a value of 0 to main() when P halts
then H is not a halting decider.

H is not a halting decider; H is a simulation detector, S.

/Flibble

In other words you are saying the infintely nested emulation
only excutes for a few steps and then it stops on its own.

I am saying that your H is not a halting decider; I am saying
that your H is just a simulation detector, S.

/Flibble

You are saying that you simply "don't believe in" simulating halt
deciders.

You haven't got a simulating halt decider, what you have got is a
simulation detector.

/Flibble

It is an easily verified fact that H does correctly decide the halt
status of its input.

void Infinite_Loop()
{
HERE: goto HERE;
}

int main()
{
Output("Input_Halts = ", H0(Infinite_Loop));
}

_Infinite_Loop()
[00001342](01) 55 push ebp
[00001343](02) 8bec mov ebp,esp
[00001345](02) ebfe jmp 00001345
[00001347](01) 5d pop ebp
[00001348](01) c3 ret
Size in bytes:(0007) [00001348]

_main()
[00001372](01) 55 push ebp
[00001373](02) 8bec mov ebp,esp
[00001375](05) 6842130000 push 00001342
[0000137a](05) e833fdffff call 000010b2
[0000137f](03) 83c404 add esp,+04
[00001382](01) 50 push eax
[00001383](05) 6823040000 push 00000423
[00001388](05) e8e5f0ffff call 00000472
[0000138d](03) 83c408 add esp,+08
[00001390](02) 33c0 xor eax,eax
[00001392](01) 5d pop ebp
[00001393](01) c3 ret
Size in bytes:(0034) [00001393]

machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
[00001372][0010228f][00000000] 55 push ebp
[00001373][0010228f][00000000] 8bec mov ebp,esp
[00001375][0010228b][00001342] 6842130000 push 00001342 // push
_Infinite_Loop
[0000137a][00102287][0000137f] e833fdffff call 000010b2 // call H0

Begin Local Halt Decider Simulation Execution Trace Stored
at:212343 [00001342][00212333][00212337] 55 push ebp
[00001343][00212333][00212337] 8bec mov ebp,esp
[00001345][00212333][00212337] ebfe jmp 00001345
[00001345][00212333][00212337] ebfe jmp 00001345
Local Halt Decider: Infinite Loop Detected Simulation Stopped

[0000137f][0010228f][00000000] 83c404 add esp,+04
[00001382][0010228b][00000000] 50 push eax
[00001383][00102287][00000423] 6823040000 push 00000423
[00001388][00102287][00000423] e8e5f0ffff call 00000472
Input_Halts = 0
[0000138d][0010228f][00000000] 83c408 add esp,+08
[00001390][0010228f][00000000] 33c0 xor eax,eax
[00001392][00102293][00100000] 5d pop ebp
[00001393][00102297][00000004] c3 ret
Number of Instructions Executed(680)

int Sum(int X, int Y)
{
return X + Y;
}

int main()
{
Output("Input_Halts = ", H2(Sum, 3, 4));
}

_Sum()
[000012b6](01) 55 push ebp
[000012b7](02) 8bec mov ebp,esp
[000012b9](03) 8b4508 mov eax,[ebp+08]
[000012bc](03) 03450c add eax,[ebp+0c]
[000012bf](01) 5d pop ebp
[000012c0](01) c3 ret
Size in bytes:(0011) [000012c0]

_main()
[00001316](01) 55 push ebp
[00001317](02) 8bec mov ebp,esp
[00001319](02) 6a04 push +04
[0000131b](02) 6a03 push +03
[0000131d](05) 68b6120000 push 000012b6
[00001322](05) e85ffaffff call 00000d86
[00001327](03) 83c40c add esp,+0c
[0000132a](01) 50 push eax
[0000132b](05) 6807040000 push 00000407
[00001330](05) e821f1ffff call 00000456
[00001335](03) 83c408 add esp,+08
[00001338](02) 33c0 xor eax,eax
[0000133a](01) 5d pop ebp
[0000133b](01) c3 ret
Size in bytes:(0038) [0000133b]

machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
...[00001316][001021d7][00000000] 55 push ebp
...[00001317][001021d7][00000000] 8bec mov ebp,esp
...[00001319][001021d3][00000004] 6a04 push +04
...[0000131b][001021cf][00000003] 6a03 push +03
...[0000131d][001021cb][000012b6] 68b6120000 push 000012b6
...[00001322][001021c7][00001327] e85ffaffff call 00000d86

Begin Local Halt Decider Simulation Execution Trace Stored
at:21228b ...[000012b6][00212273][00212277] 55 push
ebp ...[000012b7][00212273][00212277] 8bec mov ebp,esp
...[000012b9][00212273][00212277] 8b4508 mov eax,[ebp+08]
...[000012bc][00212273][00212277] 03450c add eax,[ebp+0c]
...[000012bf][00212277][00000e54] 5d pop ebp
...[000012c0][0021227b][00000003] c3 ret
...[00001327][001021d7][00000000] 83c40c add esp,+0c
...[0000132a][001021d3][00000001] 50 push eax
...[0000132b][001021cf][00000407] 6807040000 push 00000407
---[00001330][001021cf][00000407] e821f1ffff call 00000456
Input_Halts = 1
...[00001335][001021d7][00000000] 83c408 add esp,+08
...[00001338][001021d7][00000000] 33c0 xor eax,eax
...[0000133a][001021db][00100000] 5d pop ebp
...[0000133b][001021df][00000004] c3 ret
Number of Instructions Executed(660)

All you are showing here is that your H can decide the halt status
of two trivial cases however you have not shown that H can decide
the halt status of non-trivial cases.

You said:
"You haven't got a simulating halt decider"
Apologize and we can move on.

I posted four paragraphs but you ignored three of them. Lets try again.

All you are showing here is that your H can decide the halt status of
two trivial cases however you have not shown that H can decide the halt
status of non-trivial cases.

Publish the source code of your halt decider so we can see what classes
of programs your H can decide.

I am interested in deciding the halt status of programs containing
non-trivial branching logic predicated on arbitrary program input.

Crucially if your H cannot return an answer for an arbitrary
non-halting program in finite time then it isn't a halt decider.

/Flibble

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)