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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: How do simulating termination analyzers work?
Date: Sat, 21 Jun 2025 12:52:54 +0300
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On 2025-06-20 13:59:02 +0000, olcott said:

> On 6/20/2025 4:20 AM, Fred. Zwarts wrote:
>> Op 19.jun.2025 om 17:17 schreef olcott:
>>> On 6/19/2025 4:21 AM, Fred. Zwarts wrote:
>>>> Op 18.jun.2025 om 15:46 schreef olcott:
>>>>> 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*
>>>>> 
>>>> 
>>>> At least it is true for all aborting ones, such as the one you 
>>>> presented in Halt7.c.
>>> 
>>> My claim is that each of the above functions correctly
>>> simulated by any termination analyzer HHH that can possibly
>>> exist will never stop running unless aborted by HHH.
>>> Can you affirm or correctly refute this?
> 
>> Yes, I confirmed many times that we can confirm this vacuous claim, 
>> because no such HHH exists. All of them fail to do a correct simulation 
>> up to the point where they can see whether the input specifies a 
>> halting program.
> 
> if DDD correctly simulated by any simulating termination
> analyzer HHH never aborts its simulation of DDD then

that HHH is not interesting.

-- 
Mikko