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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: How do simulating termination analyzers work? ---Truth Maker
 Maximalism FULL_TRACE
Date: Sat, 12 Jul 2025 10:08:21 +0200
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Op 11.jul.2025 om 23:23 schreef olcott:
> On 7/11/2025 4:07 AM, Fred. Zwarts wrote:
>> Op 10.jul.2025 om 16:30 schreef olcott:
>>> On 7/10/2025 6:24 AM, Fred. Zwarts wrote:
>>>> Op 09.jul.2025 om 14:54 schreef olcott:>>
>>>>> One of these "errors" was that HHH cannot simulate itself at all.
>>>>
>>>> As usual you twist the words of your reviewers.
>>>> The claim was that no HHH can simulate itself correctly *up to the 
>>>> end*.
>>>>
>>>>> https://liarparadox.org/HHH(DDD)_Full_Trace.pdf
>>>>> Is the full execution trace of
>>>>>
>>>>> executed HHH simulates DDD that calls emulated HHH(DDD)
>>>>> that simulates DDD that calls emulated emulated HHH(DDD)
>>>>
>>>> And it also proves my claim that HHH did not simulate itself 
>>>> correctly *up to the end*.
>>>>
>>>
>>> Exactly what is your professional programming experience?
>>> I have 20 years in C++ and became a professional programmer
>>> in 1986.
>>>
>>
>> Irrelevant, even when it is more than your experience.
> 
> void DDD()
> {
>    HHH(DDD);
>    return;
> }
> 
> Anyone that cannot see that DDD simulated by HHH
> cannot possibly reach its own "return" instruction
> (a) Has woefully deficient knowledge
> (b) Is a liar.

No need to repeat it. Everyone sees this failure of HHH to simulate it 
correctly up to the end.
> 
>> Even a proof of a first year student can be a correct proof.
>> Apparently you ran out of counter arguments and try the authority card.
> 
> I want to see if you have the capacity to understand.
> COBOL experience counts as zero programming experience
> relative to anything about recursion.

Irrelevant. We know that when it turns out that I have more experience, 
you will find another irrelevant reason to ignore the facts.

> 
>> So, we see that HHH cannot possibly simulate itself up to the end.
> 
> Because DDD unconditionally calls HHH(DDD).
> Do you know what unconditionally means?

Yes I do. It means that the halting of DDD depends on the halting of HHH 
and HHH has conditional branch instructions, which make that it returns 
to DDD, so that DDD reaches its final halt state, if not disturbed by a 
premature abort.
Your misconception is that you think that the behaviour of DDD does not 
depend on the behaviour of HHH.