• Re: Rebutting the Sipser Halting Problem Proof --- damned liar

    From Fred. Zwarts@21:1/5 to All on Mon Sep 16 15:36:51 2024
    Op 16.sep.2024 om 14:09 schreef olcott:
    On 9/16/2024 6:21 AM, Fred. Zwarts wrote:
    Op 15.sep.2024 om 16:23 schreef olcott:

    Rebutting the Sipser Halting Problem Proof
    D(D) correctly reports its own halt status

    https://www.researchgate.net/
    publication/364302709_Rebutting_the_Sipser_Halting_Problem_Proof


    We can see that the first seven instructions of D emulated by H
    precisely match the first seven instructions of the x86 source-code
    of D. This conclusively proves that these instructions were emulated
    correctly.

    Yes H makes a good start, but fails to complete the simulation,
    because of a bug in the code to recognise an infinite 'recursion'.

    Then if you are not a damned liar you can see this
    next part that you dishonestly erased.

    D()
    [0000218e] 55             push ebp         ; begin D [0000218f] 8bec           mov ebp,esp
    [00002191] 8b4508         mov eax,[ebp+08]
    [00002194] 50             push eax         ; push param [00002195] 8b4d08         mov ecx,[ebp+08]
    [00002198] 51             push ecx         ; push param [00002199] e8a0f2ffff     call 0000143e    ; call H

    After D calls H(D,  D) we can see that H correctly emulates itself
    emulating D because again we see that the first seven instructions of D emulated by the emulated H precisely match the first seven instructions
    of the x86 source-code of D. This conclusively proves that these
    instructions were emulated correctly.

    Yes, nobody denies that H made a good start, but it failed with the
    seventh instruction, where it did not correctly simulate the call
    instruction, which should be followed by the simulation of instructions
    within H. Therefore, it misses many conditional branch instructions.
    Then, because of a bug in the recognition of an infinite recursion, it
    stops the simulation after two cycles. That makes that it also misses
    that in the next cycle the simulated H would also (incorrectly) see the 'special condition', after which it would also abort and return to DDD,
    which would halt.

    In conclusion, after a good start, H failed to reach the end of the
    simulation of a halting program.
    That the semantics of the x86 language of the string w describes a
    halting program, is proved beyond any doubt, by overwhelming evidence by
    direct execution and many correct simulations. These semantics allow
    only one behaviour for the program described by this string. But H fails
    to show this behaviour.
    H cannot possibly simulate itself correctly up to the end.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Mon Sep 16 18:52:13 2024
    On 9/16/24 8:09 AM, olcott wrote:
    On 9/16/2024 6:21 AM, Fred. Zwarts wrote:
    Op 15.sep.2024 om 16:23 schreef olcott:

    Rebutting the Sipser Halting Problem Proof
    D(D) correctly reports its own halt status

    https://www.researchgate.net/
    publication/364302709_Rebutting_the_Sipser_Halting_Problem_Proof


    We can see that the first seven instructions of D emulated by H
    precisely match the first seven instructions of the x86 source-code
    of D. This conclusively proves that these instructions were emulated
    correctly.

    Yes H makes a good start, but fails to complete the simulation,
    because of a bug in the code to recognise an infinite 'recursion'.

    Then if you are not a damned liar you can see this
    next part that you dishonestly erased.

    D()
    [0000218e] 55             push ebp         ; begin D [0000218f] 8bec           mov ebp,esp
    [00002191] 8b4508         mov eax,[ebp+08]
    [00002194] 50             push eax         ; push param [00002195] 8b4d08         mov ecx,[ebp+08]
    [00002198] 51             push ecx         ; push param [00002199] e8a0f2ffff     call 0000143e    ; call H

    After D calls H(D,  D) we can see that H correctly emulates itself
    emulating D because again we see that the first seven instructions of D emulated by the emulated H precisely match the first seven instructions
    of the x86 source-code of D. This conclusively proves that these
    instructions were emulated correctly.



    But the first seven instruction of D are never executed again, so that
    CAN'T be a part of a correct simulation of D, including the part where
    we look at the H that D calls simulatng D, because that should show the
    actions of H, not the results of the simulation.

    At the very least, it needs to say "Based on the CONDITIONAL simulation
    by H".

    Sorry, you are just proving how STUPID and IGNORANT you are of the
    meaning of the words, that or you are just TOTALLY DISHONEST and lie
    about what you are doing (or both).

    So, YOU are the "DAMNED LIAR", beeause the words you say are just UNTRUE.

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