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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: How do simulating termination analyzers work?
Date: Fri, 20 Jun 2025 11:45:56 +0300
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On 2025-06-19 09:09:34 +0000, Fred. Zwarts said:

> 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
> 
>> 
>>> 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.

My HHH, if given DDD for input, does abort and does give the correct report
but gives the worng report if given DD.

-- 
Mikko