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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: How do simulating termination analyzers work?
Date: Sat, 21 Jun 2025 12:34:55 -0500
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On 6/21/2025 4:52 AM, Mikko wrote:
> 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.
> 

*then the HP proofs are proved to be wrong*

int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}


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