On 11/12/21 5:53 PM, olcott wrote:
On 11/12/2021 4:36 PM, Richard Damon wrote:
On 11/12/21 5:24 PM, olcott wrote:
On 11/12/2021 4:13 PM, Richard Damon wrote:
On 11/12/21 5:07 PM, olcott wrote:
On 11/12/2021 4:00 PM, Richard Damon wrote:
On 11/12/21 3:11 PM, olcott wrote:I have shown that P always specifies infinite recursion whether or >>>>>> not this infinite recursion is aborted, therefore H(P,P)==0 is
#include <stdint.h>
typedef void (*ptr)();
int H(ptr x, ptr y)
{
x(y);
return 1;
}
// Minimal essence of Linz(1990) Ĥ
// and Strachey(1965) P (see below)
void P(ptr x)
{
H(x, x);
}
int main(void)
{
H(P, P);
}
It is obvious that the direct execution of the above code never >>>>>>>> halts because it is infinitely recursive. It is equally obvious >>>>>>>> that when H performs a correct pure simulation of its input
(instead of directly executing it) that its input never halts. >>>>>>>>
_P()
[00001a5e](01) 55 push ebp
[00001a5f](02) 8bec mov ebp,esp
[00001a61](03) 8b4508 mov eax,[ebp+08]
[00001a64](01) 50 push eax // push P
[00001a65](03) 8b4d08 mov ecx,[ebp+08]
[00001a68](01) 51 push ecx // push P
[00001a69](05) e810000000 call 00001a7e // call H >>>>>>>> [00001a6e](03) 83c408 add esp,+08
[00001a71](01) 5d pop ebp
[00001a72](01) c3 ret
Size in bytes:(0021) [00001a72]
Because there is nothing that H can possibly do to cause or
enable P to reach its final state at 1a72 we correctly conclude >>>>>>>> that the input to H(P,P) never halts.
Wrong. IF H does abort and return 0 then the ACTUAL running of P >>>>>>> will reach that address, and the actual running of P is what
matters.
All you have shown is that it is impossible for H to PROVE that P >>>>>>> will be halting, not that P isn't Halting.
always correct.
No, you haven't.
You logic makes the unsound step of FIRST assuming that H never
aborts its operation, and THEN has H do an abort.
If you DO have a valid proof that P(P) is non-halting when H(P,P)
return 0 then you have just proved you logic system to be
inconsistent as it can also be proved the if H(P,P) returns 0, that
P(P) halts.
A system that can prove a statement and its complement is
inconsestent, and logically worthless.
All rebuttals must take this form:
Find an invocation of H(P,P) at machine address 00001a7e such that >>>>>> the simulation or execution of (the exact byte sequence of) P
reaches its final address of 00001a72.
If no rebuttals exist this conclusively proves that H(P,P)==0 for
every H in the unverse.
WRONG CRITERIA.
Just proves you are looking at POOP.
The REAL halting problems asks what P(P) actually does.
I proven beyond all possible doubt that the real P is infinitely
recursive in my latest example where H directly executes its input
Yes, *IF* H just directly executes its input, then P(P) will be
non-Halting,
The ultimate measure of the halt status of an input is its behavior
when directly executed.
but H(P,P) never returns 0, so it is not a counter example.
The fact that for every possible H that can possibly exist at at
machine address 00001a7e the simulation or execution of (the exact
byte sequence of) P never reaches its final address of 00001a72
conclusively proves that the input to H(P,P) never halts.
The input to H(P,P) never halts therefore when H returns 0 it is
always correct.
No, you have a fundamental error in your logic,
FIRST, as has been explained before, but you just ignorantly ignore,
'inputs' do not have behavior, and as such do not halt or be
non-halting. Halting is a property of COMPUTATIONS, not inputs. Thus
your statement is proved conclusively FALSE because it makes an error in category (Maybe you don't understand these terms, but repeatedly
ignoring them doesn't help your cause).
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