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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: How do simulating termination analyzers work?
Date: Thu, 19 Jun 2025 10:09:09 -0500
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On 6/19/2025 4:09 AM, Fred. Zwarts wrote:
> Op 19.jun.2025 om 08:59 schreef olcott:
>> On 6/19/2025 1:17 AM, Mikko wrote:
>>> On 2025-06-18 13:46:16 +0000, olcott said:
>>>
>>>> On 6/18/2025 5:12 AM, Fred. Zwarts wrote:
>>>>> Op 18.jun.2025 om 03:54 schreef olcott:
>>>>>> On 6/17/2025 8:19 PM, Richard Damon wrote:
>>>>>>> On 6/17/25 4:34 PM, olcott wrote:
>>>>>>>> void Infinite_Recursion()
>>>>>>>> {
>>>>>>>>    Infinite_Recursion();
>>>>>>>>    return;
>>>>>>>> }
>>>>>>>>
>>>>>>>> void Infinite_Loop()
>>>>>>>> {
>>>>>>>>    HERE: goto HERE;
>>>>>>>>    return;
>>>>>>>> }
>>>>>>>>
>>>>>>>> void DDD()
>>>>>>>> {
>>>>>>>>    HHH(DDD);
>>>>>>>>    return;
>>>>>>>> }
>>>>>>>>
>>>>>>>> When it is understood that HHH does simulate itself
>>>>>>>> simulating DDD then any first year CS student knows
>>>>>>>> that when each of the above are correctly simulated
>>>>>>>> by HHH that none of them ever stop running unless aborted.
>>>>>>>
>>>>>>> WHich means that the code for HHH is part of the input, and thus 
>>>>>>> there is just ONE HHH in existance at this time.
>>>>>>>
>>>>>>> Since that code aborts its simulation to return the answer that 
>>>>>>> you claim, you are just lying that it did a correct simulation 
>>>>>>> (which in this context means complete)
>>>>>>>
>>>>>>
>>>>>> *none of them ever stop running unless aborted*
>>>>>
>>>>> All of them do abort and their simulation does not need an abort.
>>>>>
>>>>
>>>> *It is not given that any of them abort*
>>>
>>> It is known a priori that HHH either does or does not abort. 
>>
>> Very good.
>>
>>> If HHH does
>>> not abort it does not terminate the simulation of DDD and therefore 
>>
>> DDD never stops running.
> 
> because HHH never stops running and therefore this HHH
> 

So you agree that when HHH never aborts that none
of the above three functions ever stop running?

>>
>>> does
>>> not report correctly. If HHH does abort it reports that DDD does not
>>> halt, which is incorrect as in that case DDD does halt. HHH is correct
>>> about DDD only if it does abort its simulation and reports "halts".
>>> But you HHH does not do that.
>>>
>>
>>
> 
> So, both the aborting and the non-aborting HHH do not provide a correct 
> report.

I am not yet talking about any reports.
I am only talking about:
(a) stops running
(b) never stop running.

-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer