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  • Re: Anyone that disagrees with this is not telling the truth ---V2

    From Richard Damon@21:1/5 to olcott on Sun Aug 18 13:30:20 2024
    On 8/18/24 8:32 AM, olcott wrote:
    On 8/17/2024 7:29 AM, olcott wrote:
    void DDD()
    {
       HHH(DDD);
    }

    _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]

    *It is a basic fact that DDD emulated by HHH according to*
    *the semantics of the x86 language cannot possibly stop*
    *running unless aborted* (out of memory error excluded)


    X = DDD emulated by HHH according to the semantics of the x86 language
    Y = HHH never aborts its emulation of DDD
    Z = DDD never stops running

    (X ∧ Y) ↔ Z


    With that input, you can't have those propositions, and HHH can't
    emulated DDD by the semanitc of the x86 language, since it doesn't HAVE
    the instructions it needs to emulate.

    The problem is when you include the memory that has HHH as part of the
    input, you loose the ability to say that all the DDDs are the same.

    Sorry, you are still backed into the corner.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Mikko@21:1/5 to olcott on Mon Aug 19 10:36:09 2024
    On 2024-08-18 12:32:54 +0000, olcott said:

    On 8/17/2024 7:29 AM, olcott wrote:
    void DDD()
    {
      HHH(DDD);
    }

    _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]

    *It is a basic fact that DDD emulated by HHH according to*
    *the semantics of the x86 language cannot possibly stop*
    *running unless aborted* (out of memory error excluded)


    X = DDD emulated by HHH according to the semantics of the x86 language
    Y = HHH never aborts its emulation of DDD
    Z = DDD never stops running

    (X ∧ Y) ↔ Z

    What was wrong in the original version?

    --
    Mikko

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Mikko@21:1/5 to olcott on Tue Aug 20 12:10:43 2024
    On 2024-08-19 11:54:52 +0000, olcott said:

    On 8/19/2024 2:36 AM, Mikko wrote:
    On 2024-08-18 12:32:54 +0000, olcott said:

    On 8/17/2024 7:29 AM, olcott wrote:
    void DDD()
    {
      HHH(DDD);
    }

    _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]

    *It is a basic fact that DDD emulated by HHH according to*
    *the semantics of the x86 language cannot possibly stop*
    *running unless aborted* (out of memory error excluded)


    X = DDD emulated by HHH according to the semantics of the x86 language
    Y = HHH never aborts its emulation of DDD
    Z = DDD never stops running

    (X ∧ Y) ↔ Z

    What was wrong in the original version?


    I appended the above formula

    How does that formula contradict the original version?

    --
    Mikko

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