• Re: The directly executed D(D) does not halt even though it looks like

    From immibis@21:1/5 to olcott on Sat Jan 27 17:45:40 2024
    XPost: sci.logic

    On 1/27/24 17:06, olcott wrote:
    01 int D(ptr x)  // ptr is pointer to int function
    02 {
    03   int Halt_Status = H(x, x);
    04   if (Halt_Status)
    05     HERE: goto HERE;
    06   return Halt_Status;
    07 }
    08
    09 void main()
    10 {
    11   D(D);
    12 }

    *Execution Trace*
    Line 11: main() invokes D(D);
    Line 03: D(D) invokes H(D,D) that simulates D(D)

    *keeps repeating* (unless aborted)
    Line 03: simulated D(D) invokes simulated H(D,D) that simulates D(D)

    *Simulation invariant*
    D correctly simulated by H cannot possibly reach past its own line 03.

    D correctly simulated by H cannot possibly reach its simulated final
    state in 1 to ∞ steps of correct simulation.

    *Every computation that only stops running because some steps of this*
    *same computation had to be aborted to prevent the infinite execution*
    *of this computation is a computation that does not halt*

    Because D specifies that it calls H(D,D) in recursive simulation and H
    must abort this recursive simulation to prevent the infinite execution
    of directly executed D(D) we know that D(D) does not actually halt even though it really looks like it does.

    Within the Turing Machine model of computation only UTM simulations can
    be aborted, direct executions cannot be aborted.


    *Dishonest straw man*
    You are dishonestly pretending that the directly executed D(D) is
    simulated by H.

    If you kill John does it kill his identical twin brother?

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From immibis@21:1/5 to olcott on Sat Jan 27 17:55:29 2024
    XPost: sci.logic

    On 1/27/24 17:50, olcott wrote:
    On 1/27/2024 10:45 AM, immibis wrote:
    On 1/27/24 17:06, olcott wrote:
    01 int D(ptr x)  // ptr is pointer to int function
    02 {
    03   int Halt_Status = H(x, x);
    04   if (Halt_Status)
    05     HERE: goto HERE;
    06   return Halt_Status;
    07 }
    08
    09 void main()
    10 {
    11   D(D);
    12 }

    *Execution Trace*
    Line 11: main() invokes D(D);
    Line 03: D(D) invokes H(D,D) that simulates D(D)

    *keeps repeating* (unless aborted)
    Line 03: simulated D(D) invokes simulated H(D,D) that simulates D(D)

    *Simulation invariant*
    D correctly simulated by H cannot possibly reach past its own line 03.

    D correctly simulated by H cannot possibly reach its simulated final
    state in 1 to ∞ steps of correct simulation.

    *Every computation that only stops running because some steps of this*
    *same computation had to be aborted to prevent the infinite execution*
    *of this computation is a computation that does not halt*

    Because D specifies that it calls H(D,D) in recursive simulation and H
    must abort this recursive simulation to prevent the infinite execution
    of directly executed D(D) we know that D(D) does not actually halt even
    though it really looks like it does.

    Within the Turing Machine model of computation only UTM simulations can
    be aborted, direct executions cannot be aborted.


    *Dishonest straw man*
    You are dishonestly  pretending that the directly executed D(D) is
    simulated by H.

    *Not at all, you merely did not pay close enough attention*

    The directly executed D(D) never halts unless the recursive
    simulation that it specifies in its call to H(D,D) has been
    aborted by H.


    So John never says hello unless his identical twin Jake dies. When Jake
    dies, you think John cannot say hello because John is dead.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to immibis on Sat Jan 27 12:21:03 2024
    XPost: sci.logic

    On 1/27/24 11:45 AM, immibis wrote:
    On 1/27/24 17:06, olcott wrote:
    01 int D(ptr x)  // ptr is pointer to int function
    02 {
    03   int Halt_Status = H(x, x);
    04   if (Halt_Status)
    05     HERE: goto HERE;
    06   return Halt_Status;
    07 }
    08
    09 void main()
    10 {
    11   D(D);
    12 }

    *Execution Trace*
    Line 11: main() invokes D(D);
    Line 03: D(D) invokes H(D,D) that simulates D(D)

    *keeps repeating* (unless aborted)
    Line 03: simulated D(D) invokes simulated H(D,D) that simulates D(D)

    *Simulation invariant*
    D correctly simulated by H cannot possibly reach past its own line 03.

    D correctly simulated by H cannot possibly reach its simulated final
    state in 1 to ∞ steps of correct simulation.

    *Every computation that only stops running because some steps of this*
    *same computation had to be aborted to prevent the infinite execution*
    *of this computation is a computation that does not halt*

    Because D specifies that it calls H(D,D) in recursive simulation and H
    must abort this recursive simulation to prevent the infinite execution
    of directly executed D(D) we know that D(D) does not actually halt even
    though it really looks like it does.

    Within the Turing Machine model of computation only UTM simulations can
    be aborted, direct executions cannot be aborted.


    *Dishonest straw man*
    You are dishonestly  pretending that the directly executed D(D) is
    simulated by H.

    If you kill John does it kill his identical twin brother?

    If H isn't doing that, then how do you expect H to get the right answer,
    since that is the DEFINITION of the question it is being asked?

    "Does the program/input described by the input Halt when run?"

    or, does PO-Computation Theory not include the fact that identical
    copies of programs given identical inputs will return the identical
    answers? That is sort of a fundamental part of Computation Theory.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Sat Jan 27 12:21:01 2024
    XPost: sci.logic

    On 1/27/24 11:06 AM, olcott wrote:
    01 int D(ptr x)  // ptr is pointer to int function
    02 {
    03   int Halt_Status = H(x, x);
    04   if (Halt_Status)
    05     HERE: goto HERE;
    06   return Halt_Status;
    07 }
    08
    09 void main()
    10 {
    11   D(D);
    12 }

    *Execution Trace*
    Line 11: main() invokes D(D);
    Line 03: D(D) invokes H(D,D) that simulates D(D)

    *keeps repeating* (unless aborted)
    Line 03: simulated D(D) invokes simulated H(D,D) that simulates D(D)

    *Simulation invariant*
    D correctly simulated by H cannot possibly reach past its own line 03.

    D correctly simulated by H cannot possibly reach its simulated final
    state in 1 to ∞ steps of correct simulation.

    *Every computation that only stops running because some steps of this*
    *same computation had to be aborted to prevent the infinite execution*
    *of this computation is a computation that does not halt*

    Because D specifies that it calls H(D,D) in recursive simulation and H
    must abort this recursive simulation to prevent the infinite execution
    of directly executed D(D) we know that D(D) does not actually halt even though it really looks like it does.

    Within the Turing Machine model of computation only UTM simulations can
    be aborted, direct executions cannot be aborted.


    And nothing needed to (or could) abort any of the steps of the directly executed D(D) that calls an H(D,D) that aborts its simulation of ITS
    COPY of the program D(D).

    You show your misunderstand by saying a UTM simulation can be aborted,
    as a UTM, by defintion, will not abort its simulation.

    A partial UTMish simulation might be aborted, but such a simulation is
    NOT a "Correct Simulation" per the theory that lets you use a simulation
    as a replacement for direct execution, and so is irrelevent.

    You might be able, in some cases, to use the information from a partial
    UTMish simulation to prove that the ACTUAL UTM simulation of the input
    would never halt, but it turns out you can't in this case.

    If H(D,D) aborts its simulation and returns 0, then it could not have
    correctly determined that the UTM simulation of this input would not
    halt, as it will, as D(D) (which is based on the H that gives the
    claimed right answer) will call its exact copy of this H(D,D) which WILL
    return 0, and this D(D) will Halt, and thus a "Correct Simulation" of
    this input will halt.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From immibis@21:1/5 to olcott on Sat Jan 27 19:08:35 2024
    XPost: sci.logic

    On 1/27/24 17:52, olcott wrote:
    On 1/27/2024 10:50 AM, olcott wrote:

    The directly executed D(D) never *stops running* unless the
    recursive simulation that it specifies in its call to H(D,D)
    has been aborted by H.


    That's a different computation

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From immibis@21:1/5 to olcott on Sat Jan 27 20:19:27 2024
    XPost: sci.logic

    On 1/27/24 20:10, olcott wrote:
    On 1/27/2024 12:08 PM, immibis wrote:
    On 1/27/24 17:52, olcott wrote:
    On 1/27/2024 10:50 AM, olcott wrote:

    The directly executed D(D) never *stops running* unless the
    recursive simulation that it specifies in its call to H(D,D)
    has been aborted by H.


    That's a different computation

    *D specifies that it calls H(D,D) in recursive simulation*
    A thing is not different than itself.

    A simulation is different than a direct execution.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From immibis@21:1/5 to olcott on Sat Jan 27 21:44:54 2024
    XPost: sci.logic

    On 1/27/24 20:21, olcott wrote:
    On 1/27/2024 1:19 PM, immibis wrote:
    On 1/27/24 20:10, olcott wrote:
    On 1/27/2024 12:08 PM, immibis wrote:
    On 1/27/24 17:52, olcott wrote:
    On 1/27/2024 10:50 AM, olcott wrote:

    The directly executed D(D) never *stops running* unless the
    recursive simulation that it specifies in its call to H(D,D)
    has been aborted by H.


    That's a different computation

    *D specifies that it calls H(D,D) in recursive simulation*
    A thing is not different than itself.

    A simulation is different than a direct execution.

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    That a simulation does not halt if it is not aborted only proves that
    the simulation does not halt.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Sat Jan 27 17:06:16 2024
    XPost: sci.logic

    On 1/27/24 4:31 PM, olcott wrote:
    On 1/27/2024 2:44 PM, immibis wrote:
    On 1/27/24 20:21, olcott wrote:
    On 1/27/2024 1:19 PM, immibis wrote:
    On 1/27/24 20:10, olcott wrote:
    On 1/27/2024 12:08 PM, immibis wrote:
    On 1/27/24 17:52, olcott wrote:
    On 1/27/2024 10:50 AM, olcott wrote:

    The directly executed D(D) never *stops running* unless the
    recursive simulation that it specifies in its call to H(D,D)
    has been aborted by H.


    That's a different computation

    *D specifies that it calls H(D,D) in recursive simulation*
    A thing is not different than itself.

    A simulation is different than a direct execution.

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    That a simulation does not halt if it is not aborted only proves that
    the simulation does not halt.

    *It proves that behavior specified by the input DOES NOT HALT*
    *It proves that behavior specified by the input DOES NOT HALT*
    *It proves that behavior specified by the input DOES NOT HALT*




    NOPE. Not by the REAL definition.

    The CORRECT simulation, as well as the DIRECT EXECTION of the input of D
    built on the H that aborts, shows D(D) calling H(D,D), then that H(D,D) aborting it simulation (since you are claim that it is ok for H to
    abort in this case, and it does) and then returning to D(D) and it halting.

    If you DEFINE H to not abort, then yes, the input D is non-halting, but
    that H can't abort its simulation as it wss defined to be non-aborting.

    So, the ONLY H that is "allowed" to abort its simulation because its
    input is non-halting is also the H that was defined not to do it.

    Any H that tries to take that option, is shown to have an input that
    actually did halt and would have completed the correct simulation if the
    input was given to a real correct simulator. (You can't use this H to do
    that, as it always aborts its input).

    You run into the problem that since D is defined to call the H that you
    claim to get the right answer, that H can't use the results of an H
    simulating a DIFFERENT D based on a DIFFERENT H.

    You are just caught it the trap of not defining programs correctly.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Sat Jan 27 17:46:19 2024
    XPost: sci.logic

    On 1/27/24 5:28 PM, olcott wrote:
    On 1/27/2024 4:06 PM, Richard Damon wrote:
    On 1/27/24 4:31 PM, olcott wrote:
    On 1/27/2024 2:44 PM, immibis wrote:
    On 1/27/24 20:21, olcott wrote:
    On 1/27/2024 1:19 PM, immibis wrote:
    On 1/27/24 20:10, olcott wrote:
    On 1/27/2024 12:08 PM, immibis wrote:
    On 1/27/24 17:52, olcott wrote:
    On 1/27/2024 10:50 AM, olcott wrote:

    The directly executed D(D) never *stops running* unless the
    recursive simulation that it specifies in its call to H(D,D) >>>>>>>>> has been aborted by H.


    That's a different computation

    *D specifies that it calls H(D,D) in recursive simulation*
    A thing is not different than itself.

    A simulation is different than a direct execution.

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    That a simulation does not halt if it is not aborted only proves
    that the simulation does not halt.

    *It proves that behavior specified by the input DOES NOT HALT*
    *It proves that behavior specified by the input DOES NOT HALT*
    *It proves that behavior specified by the input DOES NOT HALT*




    NOPE. Not by the REAL definition.

    Halt deciders were always required to report on the behavior
    that their input specifies and D specifies recursive simulation
    to H. *Alternative understandings have always been misconceptions*


    No, D specifies FINITE recursion to an H that aborts its simulation.

    FINITE recursion is not non-halting.

    Yes, if you define your H to not abort, then D specifies non-halting
    behavior to a "decider" that fails to answer, and thus isn't a decider.

    If H aborts its simulation and returns a value, D specifies behavior
    based on that return value.

    Your H makes an error by not understanding that the H it is seeing
    called is EXACTLY like it, so if this H aborts, so does the H being called.

    REALITY IS REALITY, you can't look at a fantasy that isn't what actually happens.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Sat Jan 27 18:40:52 2024
    XPost: sci.logic

    On 1/27/24 5:56 PM, olcott wrote:
    On 1/27/2024 4:46 PM, Richard Damon wrote:
    On 1/27/24 5:28 PM, olcott wrote:
    On 1/27/2024 4:06 PM, Richard Damon wrote:
    On 1/27/24 4:31 PM, olcott wrote:
    On 1/27/2024 2:44 PM, immibis wrote:
    On 1/27/24 20:21, olcott wrote:
    On 1/27/2024 1:19 PM, immibis wrote:
    On 1/27/24 20:10, olcott wrote:
    On 1/27/2024 12:08 PM, immibis wrote:
    On 1/27/24 17:52, olcott wrote:
    On 1/27/2024 10:50 AM, olcott wrote:

    The directly executed D(D) never *stops running* unless the >>>>>>>>>>> recursive simulation that it specifies in its call to H(D,D) >>>>>>>>>>> has been aborted by H.


    That's a different computation

    *D specifies that it calls H(D,D) in recursive simulation*
    A thing is not different than itself.

    A simulation is different than a direct execution.

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    That a simulation does not halt if it is not aborted only proves
    that the simulation does not halt.

    *It proves that behavior specified by the input DOES NOT HALT*
    *It proves that behavior specified by the input DOES NOT HALT*
    *It proves that behavior specified by the input DOES NOT HALT*




    NOPE. Not by the REAL definition.

    Halt deciders were always required to report on the behavior
    that their input specifies and D specifies recursive simulation
    to H. *Alternative understandings have always been misconceptions*


    No, D specifies FINITE recursion to an H that aborts its simulation.

    FINITE recursion is not non-halting.

    The key fact that H(D,D) only halts when H aborts its simulation
    of D conclusively proves that D DOES NOT HALT. Aborted simulations
    do not count as D halting.


    Nope.

    H aborted a DIFFERENT instnace of D, The D that calls H halts because of
    this. That D was not being "simulated" and thus wasn't aborted.

    This shows that H was INCORRECT in its decision to abort.

    Yes, H aborting its simulation doesn't, by itself, say the input is
    Halting, as the simulation didn't reach a final state. It didn't also
    show that an unbounded number of steps of simulation by a correct
    simulation wouldn't halt (rememver, D calls the H that does what the H
    that you claim gave the right aswer does, which is abort it simulation).

    The fact that this correct simulation isn't done by H doesn't affect the
    fact that the correct simulation shows that the direct execution halts.

    You keep on confusing different things thinking they are the same,

    This could be a sign of mental problems, or could just be signs of
    stupidity.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Sat Jan 27 19:15:35 2024
    XPost: sci.logic

    On 1/27/24 6:51 PM, olcott wrote:
    On 1/27/2024 5:40 PM, Richard Damon wrote:
    On 1/27/24 5:56 PM, olcott wrote:
    On 1/27/2024 4:46 PM, Richard Damon wrote:
    On 1/27/24 5:28 PM, olcott wrote:
    On 1/27/2024 4:06 PM, Richard Damon wrote:
    On 1/27/24 4:31 PM, olcott wrote:
    On 1/27/2024 2:44 PM, immibis wrote:
    On 1/27/24 20:21, olcott wrote:
    On 1/27/2024 1:19 PM, immibis wrote:
    On 1/27/24 20:10, olcott wrote:
    On 1/27/2024 12:08 PM, immibis wrote:
    On 1/27/24 17:52, olcott wrote:
    On 1/27/2024 10:50 AM, olcott wrote:

    The directly executed D(D) never *stops running* unless the >>>>>>>>>>>>> recursive simulation that it specifies in its call to H(D,D) >>>>>>>>>>>>> has been aborted by H.


    That's a different computation

    *D specifies that it calls H(D,D) in recursive simulation* >>>>>>>>>>> A thing is not different than itself.

    A simulation is different than a direct execution.

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    That a simulation does not halt if it is not aborted only proves >>>>>>>> that the simulation does not halt.

    *It proves that behavior specified by the input DOES NOT HALT*
    *It proves that behavior specified by the input DOES NOT HALT*
    *It proves that behavior specified by the input DOES NOT HALT*




    NOPE. Not by the REAL definition.

    Halt deciders were always required to report on the behavior
    that their input specifies and D specifies recursive simulation
    to H. *Alternative understandings have always been misconceptions*


    No, D specifies FINITE recursion to an H that aborts its simulation.

    FINITE recursion is not non-halting.

    The key fact that H(D,D) only halts when H aborts its simulation
    of D conclusively proves that D DOES NOT HALT. Aborted simulations
    do not count as D halting.


    Nope.

    H aborted a DIFFERENT instnace of D,
    I don't believe that you actually believe that the levels
    of recursive simulation in the same recursive simulation
    chain are separate instances.

    That would be the same as believing that infinite recursion
    is not infinite because each call has a different stack frame.


    Nope.

    The the fact that H aborts its simulation has ZERO errect on any caller
    of it. Thus H(D,D) aborting its simulation of D(D) does not affect the
    direct execution of D(D) that calls that H.

    Please show how it does.

    Note, your thought of equivalence of simulation to execution only apply
    to UNCONDITIONAL simulation, the fact that an outer layer of simulation
    can see in to the behavior of the inner layers is different than with
    calls, where the other level are "inert" while the inner layers are running.

    If an outer layer aborts, that makes all the inner layers just
    disappear, but doesn't affect any layers outside of it, except by what
    it returns to them.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Sat Jan 27 20:33:23 2024
    XPost: sci.logic

    On 1/27/24 7:35 PM, olcott wrote:
    On 1/27/2024 6:15 PM, Richard Damon wrote:
    On 1/27/24 6:51 PM, olcott wrote:
    On 1/27/2024 5:40 PM, Richard Damon wrote:
    On 1/27/24 5:56 PM, olcott wrote:
    On 1/27/2024 4:46 PM, Richard Damon wrote:
    On 1/27/24 5:28 PM, olcott wrote:
    On 1/27/2024 4:06 PM, Richard Damon wrote:
    On 1/27/24 4:31 PM, olcott wrote:
    On 1/27/2024 2:44 PM, immibis wrote:
    On 1/27/24 20:21, olcott wrote:
    On 1/27/2024 1:19 PM, immibis wrote:
    On 1/27/24 20:10, olcott wrote:
    On 1/27/2024 12:08 PM, immibis wrote:
    On 1/27/24 17:52, olcott wrote:
    On 1/27/2024 10:50 AM, olcott wrote:

    The directly executed D(D) never *stops running* unless the >>>>>>>>>>>>>>> recursive simulation that it specifies in its call to H(D,D) >>>>>>>>>>>>>>> has been aborted by H.


    That's a different computation

    *D specifies that it calls H(D,D) in recursive simulation* >>>>>>>>>>>>> A thing is not different than itself.

    A simulation is different than a direct execution.

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    That a simulation does not halt if it is not aborted only
    proves that the simulation does not halt.

    *It proves that behavior specified by the input DOES NOT HALT* >>>>>>>>> *It proves that behavior specified by the input DOES NOT HALT* >>>>>>>>> *It proves that behavior specified by the input DOES NOT HALT* >>>>>>>>>



    NOPE. Not by the REAL definition.

    Halt deciders were always required to report on the behavior
    that their input specifies and D specifies recursive simulation
    to H. *Alternative understandings have always been misconceptions* >>>>>>>

    No, D specifies FINITE recursion to an H that aborts its simulation. >>>>>>
    FINITE recursion is not non-halting.

    The key fact that H(D,D) only halts when H aborts its simulation
    of D conclusively proves that D DOES NOT HALT. Aborted simulations
    do not count as D halting.


    Nope.

    H aborted a DIFFERENT instnace of D,
    I don't believe that you actually believe that the levels
    of recursive simulation in the same recursive simulation
    chain are separate instances.

    That would be the same as believing that infinite recursion
    is not infinite because each call has a different stack frame.


    Nope.

    The the fact that H aborts its simulation has ZERO errect on any
    caller of it. Thus H(D,D) aborting its simulation of D(D) does not
    affect the direct execution of D(D) that calls that H.

    So in other words you are claiming to be too stupid to understand
    that unless H every H that can possibly exist aborts its simulated D
    that neither H nor directly executed D would ever stop running?


    No, I am claiming that YOU are too stupid to understand that programs do
    what they are programmed to do.

    Thus any H that was programmed to abort, is given a D that calls a copy
    of that H, and that input, if correctly simulated (which H doesn't do
    since it aborts) will reach a terminal state,

    The only ground for H to correctly abort is if it can prove that a
    correct simulation of its input would be non-halting.

    Since a correct simulation of its input Halted, H does not have the
    needed facts to make its aborting correct.

    IF you want to try to justify based on "ITS" correct simulation, then NO
    H can correctly abort, as no H that aborts mets the requirement of doing
    a correct simulation.

    Your inability to see this just proves you are too stupid (perhaps intentionally made so) to understand the basics of Programs.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From immibis@21:1/5 to olcott on Sun Jan 28 13:17:52 2024
    XPost: sci.logic

    On 1/28/24 01:35, olcott wrote:
    On 1/27/2024 6:15 PM, Richard Damon wrote:
    On 1/27/24 6:51 PM, olcott wrote:
    On 1/27/2024 5:40 PM, Richard Damon wrote:
    On 1/27/24 5:56 PM, olcott wrote:
    On 1/27/2024 4:46 PM, Richard Damon wrote:
    On 1/27/24 5:28 PM, olcott wrote:
    On 1/27/2024 4:06 PM, Richard Damon wrote:
    On 1/27/24 4:31 PM, olcott wrote:
    On 1/27/2024 2:44 PM, immibis wrote:
    On 1/27/24 20:21, olcott wrote:
    On 1/27/2024 1:19 PM, immibis wrote:
    On 1/27/24 20:10, olcott wrote:
    On 1/27/2024 12:08 PM, immibis wrote:
    On 1/27/24 17:52, olcott wrote:
    On 1/27/2024 10:50 AM, olcott wrote:

    The directly executed D(D) never *stops running* unless the >>>>>>>>>>>>>>> recursive simulation that it specifies in its call to H(D,D) >>>>>>>>>>>>>>> has been aborted by H.


    That's a different computation

    *D specifies that it calls H(D,D) in recursive simulation* >>>>>>>>>>>>> A thing is not different than itself.

    A simulation is different than a direct execution.

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    *The direct execution of D(D) specifies*
    *that it calls H(D,D) in recursive simulation*

    That a simulation does not halt if it is not aborted only
    proves that the simulation does not halt.

    *It proves that behavior specified by the input DOES NOT HALT* >>>>>>>>> *It proves that behavior specified by the input DOES NOT HALT* >>>>>>>>> *It proves that behavior specified by the input DOES NOT HALT* >>>>>>>>>



    NOPE. Not by the REAL definition.

    Halt deciders were always required to report on the behavior
    that their input specifies and D specifies recursive simulation
    to H. *Alternative understandings have always been misconceptions* >>>>>>>

    No, D specifies FINITE recursion to an H that aborts its simulation. >>>>>>
    FINITE recursion is not non-halting.

    The key fact that H(D,D) only halts when H aborts its simulation
    of D conclusively proves that D DOES NOT HALT. Aborted simulations
    do not count as D halting.


    Nope.

    H aborted a DIFFERENT instnace of D,
    I don't believe that you actually believe that the levels
    of recursive simulation in the same recursive simulation
    chain are separate instances.

    That would be the same as believing that infinite recursion
    is not infinite because each call has a different stack frame.


    Nope.

    The the fact that H aborts its simulation has ZERO errect on any
    caller of it. Thus H(D,D) aborting its simulation of D(D) does not
    affect the direct execution of D(D) that calls that H.

    So in other words you are claiming to be too stupid to understand
    that unless H every H that can possibly exist aborts its simulated D
    that neither H nor directly executed D would ever stop running?

    Every H that can possibly exist is a hippopotamous because no H can
    possibly exist.

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