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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory,sci.logic
Subject: Re: Liar detector: Fred, Richard, Joes and Alan --- Ben's agreement
Date: Mon, 8 Jul 2024 18:16:28 +0200
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Op 08.jul.2024 om 18:07 schreef olcott:
> On 7/8/2024 11:04 AM, Fred. Zwarts wrote:
>> Op 08.jul.2024 om 17:04 schreef olcott:
>>> On 7/8/2024 9:25 AM, Fred. Zwarts wrote:
>>>> Op 07.jul.2024 om 15:46 schreef olcott:
>>>>>
>>>>> Correctly is measured by the semantics of the x86 language.
>>>>> This specifies that when DDD is correctly simulated by HHH
>>>>> calls emulated HHH(DDD) that this call cannot return.
>>>>
>>>> Yes. This shows that the simulation is incorrect.
>>>>
>>>>>
>>>>> You smash a bottle on the ground. No matter how much you
>>>>> want the bottle to hold water it will not hold water.
>>>>
>>>> Similarly, HHH cannot possibly simulate itself correctly, no matter 
>>>> how much you want it to be correct, 
>>>
>>> Where correct is understood to be what-ever-the-Hell that the
>>> machine code of DDD specifies within the semantics of the x86
>>> language then:
>>>
>>> When DDD is correctly simulated by any pure function x86 emulator
>>> HHH that aborts its emulation at some point calls HHH(DDD) then
>>> it is correctly understood that this call cannot possibly return.
>>> The proof of this is something like mathematical induction.
>>>
>>> When DDD is correctly emulated by any HHH that aborts
>>> its emulation after N repetitions:
>>> (1) DDD is correctly emulated by HHH
>>
>> But only the first part is simulated, not the full input. The 
>> simulation must simulate the full input. It will will only become 
>> correct if also the other behaviour of HHH is simulated. But HHH is 
>> unable to simulate itself up to that point.
>> That is what the x86 code specifies.
>>
>>> (2) that calls an emulated HHH(DDD) that
>>> (3) emulates another DDD... goto (2) or abort
>>
>> And when it aborts, it is one cycle to soon.
> 
> Try to show how infinity is one cycle too soon.
>
You believe that two equals infinity. I do not. A program that aborts 
after two repetitions and returns, does not repeat and infinite number 
of times and does not need to be aborted for the simulation to halt.

void Finite_Recursion (int N) {
   if (N > 0) Finite_Recursion (N - 1);
}

This does not repeat infinitely.
Similarly, a program that aborts the recursive simulation after two 
cycles, does not repeat infinitely.
Your dreams of an infinite recursion seem to influence your judgement 
when there is only an HHH that aborts after two repetitions.
TWO IS NOT INFINITELY!