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  • Re: Liar detector: Fred, Richard, Joes and Alan --- Honest Dialogue ?

    From Richard Damon@21:1/5 to olcott on Sat Jul 20 09:45:47 2024
    On 7/20/24 9:28 AM, olcott wrote:
    On 7/20/2024 3:54 AM, Mikko wrote:
    On 2024-07-19 14:39:25 +0000, olcott said:

    On 7/19/2024 3:51 AM, Mikko wrote:
    On 2024-07-18 13:17:22 +0000, olcott said:

    On 7/18/2024 2:40 AM, Mikko wrote:
    On 2024-07-17 13:00:55 +0000, olcott said:

    On 7/17/2024 1:43 AM, Mikko wrote:
    On 2024-07-16 14:21:28 +0000, olcott said:

    When simulated input DDD stops running {if and only if}
    the simulation of this input DDD has been aborted this
    necessitates that input DDD specifies non-halting behavior

    DDD does not stop runnig unless it is completely exeuted.

    _DDD()
    [00002163] 55         push ebp      ; housekeeping
    [00002164] 8bec       mov ebp,esp   ; housekeeping
    [00002166] 6863210000 push 00002163 ; push DDD
    [0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
    [00002170] 83c404     add esp,+04
    [00002173] 5d         pop ebp
    [00002174] c3         ret
    Size in bytes:(0018) [00002174]

    DDD emulated by HHH according to the semantic meaning of
    its x86 instructions never stop running unless aborted.

    You mean HHH's simulation of DDD may not termite before HHH aborts >>>>>> it?

    When we examine the infinite set of every HHH/DDD pair such that:
    HHH₁ one step of DDD₁ is correctly emulated by HHH₁.
    HHH₂ two steps of DDD₂ are correctly emulated by HHH₂.
    HHH₃ three steps of DDD₃ are correctly emulated by HHH₃.
    ...
    HHH∞ The emulation of DDD∞ by HHH∞ never stops running.

    When each DDD of the HHH/DDD pairs above is correctly emulated
    by its corresponding HHH according to the semantic meaning of its
    x86 instructions it CANNOT POSSIBLY reach past its own machine
    address 0000216b, not even by an act of God.

    You apparently mean that no HHHᵢ can simulate the corresponding DDDᵢ to
    its termination?

    No I don't mean that at all that incorrectly allocates the error
    to the emulator.

    Anyway you did not say that some HHHᵢ can simulate the corresponding DDDᵢ
    to its termination. And each DDDᵢ does terminate, whether simulated or
    not.

    *Until you quit lying about this we cannot have an honest dialog*

    int main()
    {
      DDD();
    }

    Calls HHH(DDD) that must abort the emulation of its input
    or {HHH, emulated DDD and executed DDD} never stop running.


    But HHH DOES abort, as you say it correctly answers with non-halting,
    thus you are arguing about a case that you already ruled out.

    DDD halts because HHH aborts emulation of DDD becuase it (incorrectly)
    things that it will not halt. This just shows that HHH is wrong.

    The fact that it halts because HHH wrongly aborts its emulation of it,
    doesn't make it non-halting, or that a correct emulation of DDD will not
    halt.

    Your logic of saying HHH does a correct emulation that gets an incorrect
    answer is just incorrect.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Mikko@21:1/5 to olcott on Sun Jul 21 12:38:37 2024
    On 2024-07-20 13:28:36 +0000, olcott said:

    On 7/20/2024 3:54 AM, Mikko wrote:
    On 2024-07-19 14:39:25 +0000, olcott said:

    On 7/19/2024 3:51 AM, Mikko wrote:
    On 2024-07-18 13:17:22 +0000, olcott said:

    On 7/18/2024 2:40 AM, Mikko wrote:
    On 2024-07-17 13:00:55 +0000, olcott said:

    On 7/17/2024 1:43 AM, Mikko wrote:
    On 2024-07-16 14:21:28 +0000, olcott said:

    When simulated input DDD stops running {if and only if}
    the simulation of this input DDD has been aborted this
    necessitates that input DDD specifies non-halting behavior

    DDD does not stop runnig unless it is completely exeuted.

    _DDD()
    [00002163] 55         push ebp      ; housekeeping
    [00002164] 8bec       mov ebp,esp   ; housekeeping
    [00002166] 6863210000 push 00002163 ; push DDD
    [0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
    [00002170] 83c404     add esp,+04
    [00002173] 5d         pop ebp
    [00002174] c3         ret
    Size in bytes:(0018) [00002174]

    DDD emulated by HHH according to the semantic meaning of
    its x86 instructions never stop running unless aborted.

    You mean HHH's simulation of DDD may not termite before HHH aborts it? >>>>>
    When we examine the infinite set of every HHH/DDD pair such that:
    HHH₁ one step of DDD₁ is correctly emulated by HHH₁.
    HHH₂ two steps of DDD₂ are correctly emulated by HHH₂.
    HHH₃ three steps of DDD₃ are correctly emulated by HHH₃.
    ...
    HHH∞ The emulation of DDD∞ by HHH∞ never stops running.

    When each DDD of the HHH/DDD pairs above is correctly emulated
    by its corresponding HHH according to the semantic meaning of its
    x86 instructions it CANNOT POSSIBLY reach past its own machine
    address 0000216b, not even by an act of God.

    You apparently mean that no HHHᵢ can simulate the corresponding DDDᵢ to
    its termination?

    No I don't mean that at all that incorrectly allocates the error
    to the emulator.

    Anyway you did not say that some HHHᵢ can simulate the corresponding DDDᵢ
    to its termination. And each DDDᵢ does terminate, whether simulated or not.

    *Until you quit lying about this we cannot have an honest dialog*

    I don't believe that you can have a honest dialog, at least not without
    a chairman who wants to and can keep the dialog honest.

    --
    Mikko

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Mikko@21:1/5 to olcott on Mon Jul 22 11:01:20 2024
    On 2024-07-21 13:50:17 +0000, olcott said:

    On 7/21/2024 4:38 AM, Mikko wrote:
    On 2024-07-20 13:28:36 +0000, olcott said:

    On 7/20/2024 3:54 AM, Mikko wrote:
    On 2024-07-19 14:39:25 +0000, olcott said:

    On 7/19/2024 3:51 AM, Mikko wrote:

    You apparently mean that no HHHᵢ can simulate the corresponding DDDᵢ to
    its termination?

    No I don't mean that at all that incorrectly allocates the error
    to the emulator.

    Anyway you did not say that some HHHᵢ can simulate the corresponding DDDᵢ
    to its termination. And each DDDᵢ does terminate, whether simulated or not.

    *Until you quit lying about this we cannot have an honest dialog*

    I don't believe that you can have a honest dialog, at least not without
    a chairman who wants to and can keep the dialog honest.


    void DDD()
    {
    HHH(DDD);
    return;
    }

    When N steps of DDD are emulated by pure function HHH according
    to the semantics of the x86 language then N steps are emulated correctly.

    The subscripts to HHH and DDD pairs are each element of
    the set of positive integers ℤ+

    When we examine the infinite set of every HHH/DDD pair
    such that:

    HHH₁ one step of DDD₁ is correctly emulated by HHH₁.
    HHH₂ two steps of DDD₂ are correctly emulated by HHH₂.
    HHH₃ three steps of DDD₃ are correctly emulated by HHH₃.
    ...
    HHHₙ n steps of DDDₙ are correctly emulated by HHHₙ.

    Then DDD correctly simulated by any pure function HHH cannot
    possibly reach its own return instruction and halt, therefore
    every HHH is correct to reject its DDD as non-halting.

    That does not follow. It is never correct to reject a halting comoputation
    as non-halting.

    --
    Mikko

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