• Re: ChatGPT refutes the key rebuttal of my work --- CORRECTION

    From Richard Damon@21:1/5 to olcott on Thu Oct 17 21:13:19 2024
    On 10/17/24 12:06 PM, olcott wrote:
    On 10/17/2024 5:07 AM, joes wrote:
    Am Wed, 16 Oct 2024 15:06:01 -0500 schrieb olcott:
    On 10/16/2024 3:02 PM, joes wrote:
    Am Wed, 16 Oct 2024 14:39:37 -0500 schrieb olcott:
    On 10/16/2024 2:33 PM, joes wrote:
    Am Wed, 16 Oct 2024 13:59:58 -0500 schrieb olcott:
    On 10/16/2024 1:47 PM, joes wrote:
    Am Wed, 16 Oct 2024 13:35:01 -0500 schrieb olcott:
    On 10/16/2024 1:06 PM, joes wrote:
    Am Wed, 16 Oct 2024 12:46:01 -0500 schrieb olcott:
    On 10/16/2024 12:27 PM, joes wrote:
    Am Wed, 16 Oct 2024 10:39:21 -0500 schrieb olcott:
    On 10/16/2024 9:45 AM, joes wrote:
    Am Wed, 16 Oct 2024 09:11:22 -0500 schrieb olcott: >>>>>>>>>>>>>>> On 10/16/2024 9:01 AM, joes wrote:
    Am Wed, 16 Oct 2024 08:31:43 -0500 schrieb olcott: >>>>>>>>>>>>>>>>> On 10/16/2024 1:33 AM, joes wrote:

    Terminating C functions must reach their "return" >>>>>>>>>>>>>>>>> statement.
    Which DDD does.
    THIS IS ALSO THE INDUSTRY STANDARD DEFINITION It is >>>>>>>>>>>>>>> stipulated that *correct_x86_emulation* means that a finite >>>>>>>>>>>>>>> string of x86 instructions is emulated according to the >>>>>>>>>>>>>>> semantics of the x86 language beginning with the first bytes >>>>>>>>>>>>>>> of this string.
    You are not simulating the given program, but a version that >>>>>>>>>>>>>> differs in the abort check.
    HHH is correctly emulating (not simulating) the x86 language >>>>>>>>>>>>> finite string of DDD including emulating the finite string of >>>>>>>>>>>>> itself emulating the finite string of DDD up until the point >>>>>>>>>>>>> where the emulated emulated DDD would call HHH(DDD) again. >>>>>>>>>>>> Whereupon the simulated HHH would abort, if it weren't >>>>>>>>>>>> unnecessarily aborted.
    If the first HHH to meet its abort criteria does not act on this >>>>>>>>>>> criteria then none of them do.
    And if the first one does, all of them do.
    In theory this seems true when ignoring or failing to comprehend >>>>>>>>> key details.
    In practice you programmed H impurely.
    Which totally does not matter to the slightest degree when you have >>>>>>> the discipline to stay within the precisely designated scope of the >>>>>>> exact words that I am saying.
    When HHH is an x86 emulation based termination analyzer then each >>>>>>> DDD *correctly_emulated_by* any HHH that it calls cannot possibly >>>>>>> return no matter what this HHH does.
    Exactly, because your nested HHHs do not abort.
    In other words you continue to fail to understand that unless the
    first one aborts then none of them can possibly abort because they all >>>>> have the exact same code.
    Then HHH should report itself as halting, when they would all abort.
    They would not all abort when you pay close attention to ALL of the
    details. It is utterly impossible for any of them besides the outermost
    one to abort because it aborts before any of the rest of them see their
    abort criteria has been met.
    Only because they aren't simulated. But they are still terminating

    programs. An infinite loop is still non-halting even if I never run it.


    This is just over your head.

    _DDD()
    [00002172] 55         push ebp      ; housekeeping
    [00002173] 8bec       mov ebp,esp   ; housekeeping
    [00002175] 6872210000 push 00002172 ; push DDD
    [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
    [0000217f] 83c404     add esp,+04
    [00002182] 5d         pop ebp
    [00002183] c3         ret
    Size in bytes:(0018) [00002183]

    When DDD is correctly emulated by HHH according
    to the semantics of the x86 language DDD cannot
    possibly reach its own machine address [00002183]
    no matter what HHH does.



    But it is also a fact that the HHH that tries to correctly emulate that
    DDD in a way that shows the final behavior of that input, doesn't ever
    answer.

    Since your HHH does abort its emulation, to give an answer, it doesn't
    do a proper "correct emulation" by the definition that allows it to say something about the final behavior of DDD, so that fact isn't applicable.

    But, the actually correct and complete emulation of that DDD (that
    calls that HHH that does abort and return and thus not show what you
    claim) does reach the return statement, showing that the actually
    correct emulation reaches that point, and thus HHH is wrong.

    It also shows that you are nothing but a stupid liar that doesn't know
    what he is talking about.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From joes@21:1/5 to All on Fri Oct 18 08:13:43 2024
    Am Thu, 17 Oct 2024 11:06:59 -0500 schrieb olcott:
    On 10/17/2024 5:07 AM, joes wrote:
    Am Wed, 16 Oct 2024 15:06:01 -0500 schrieb olcott:
    On 10/16/2024 3:02 PM, joes wrote:
    Am Wed, 16 Oct 2024 14:39:37 -0500 schrieb olcott:
    On 10/16/2024 2:33 PM, joes wrote:
    Am Wed, 16 Oct 2024 13:59:58 -0500 schrieb olcott:
    On 10/16/2024 1:47 PM, joes wrote:
    Am Wed, 16 Oct 2024 13:35:01 -0500 schrieb olcott:
    On 10/16/2024 1:06 PM, joes wrote:
    Am Wed, 16 Oct 2024 12:46:01 -0500 schrieb olcott:
    On 10/16/2024 12:27 PM, joes wrote:
    Am Wed, 16 Oct 2024 10:39:21 -0500 schrieb olcott:
    On 10/16/2024 9:45 AM, joes wrote:
    Am Wed, 16 Oct 2024 09:11:22 -0500 schrieb olcott: >>>>>>>>>>>>>>> On 10/16/2024 9:01 AM, joes wrote:
    Am Wed, 16 Oct 2024 08:31:43 -0500 schrieb olcott: >>>>>>>>>>>>>>>>> On 10/16/2024 1:33 AM, joes wrote:

    Terminating C functions must reach their "return" >>>>>>>>>>>>>>>>> statement.
    Which DDD does.
    THIS IS ALSO THE INDUSTRY STANDARD DEFINITION It is >>>>>>>>>>>>>>> stipulated that *correct_x86_emulation* means that a >>>>>>>>>>>>>>> finite string of x86 instructions is emulated according to >>>>>>>>>>>>>>> the semantics of the x86 language beginning with the first >>>>>>>>>>>>>>> bytes of this string.
    You are not simulating the given program, but a version >>>>>>>>>>>>>> that differs in the abort check.
    HHH is correctly emulating (not simulating) the x86 language >>>>>>>>>>>>> finite string of DDD including emulating the finite string >>>>>>>>>>>>> of itself emulating the finite string of DDD up until the >>>>>>>>>>>>> point where the emulated emulated DDD would call HHH(DDD) >>>>>>>>>>>>> again.
    Whereupon the simulated HHH would abort, if it weren't >>>>>>>>>>>> unnecessarily aborted.
    If the first HHH to meet its abort criteria does not act on >>>>>>>>>>> this criteria then none of them do.
    And if the first one does, all of them do.
    In theory this seems true when ignoring or failing to comprehend >>>>>>>>> key details.
    In practice you programmed H impurely.
    Which totally does not matter to the slightest degree when you
    have the discipline to stay within the precisely designated scope >>>>>>> of the exact words that I am saying.
    When HHH is an x86 emulation based termination analyzer then each >>>>>>> DDD *correctly_emulated_by* any HHH that it calls cannot possibly >>>>>>> return no matter what this HHH does.
    Exactly, because your nested HHHs do not abort.
    In other words you continue to fail to understand that unless the
    first one aborts then none of them can possibly abort because they
    all have the exact same code.
    Then HHH should report itself as halting, when they would all abort.
    They would not all abort when you pay close attention to ALL of the
    details. It is utterly impossible for any of them besides the
    outermost one to abort because it aborts before any of the rest of
    them see their abort criteria has been met.
    Only because they aren't simulated. But they are still terminating
    programs. An infinite loop is still non-halting even if I never run it. [topic change ignored]

    --
    Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
    It is not guaranteed that n+1 exists for every n.

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