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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: How do simulating termination analyzers work? ---Truth Maker
 Maximalism FULL_TRACE
Date: Fri, 11 Jul 2025 19:35:55 -0400
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On 7/11/25 5:23 PM, olcott wrote:
> 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.

Only because your HHH doesn't do a correct and complete simualtion,

The correct and complete simulation of ANY verision of DDD paired with 
an HHH that returns an answer will halt, and thus any HHH that reports 
that DDD doesn't halt is just wrong.

> 
>> 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.

Wrong. COBOL programs can use recursion. I guess you don't understand 
what you are talking about.

> 
>> 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?

Right, but HHH only conditionally simulated DDD, and that includes the 
HHH the DDD calls, and thus the actual recursion is conditional, and finite.

> 
>> This failure of HHH is 
> 
> entirely your own misconception.

Something beyound your underwstanding.


> 
>> an important fact in your code, because it shows that simulation is 
>> not the right tool for all inputs.
> 
>