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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: How do simulating termination analyzers work?
Date: Sun, 29 Jun 2025 11:44:15 +0300
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On 2025-06-28 13:17:17 +0000, olcott said:

> On 6/28/2025 6:41 AM, Mikko wrote:
>> On 2025-06-21 17:34:55 +0000, olcott said:
>> 
>>> 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*
>> 
>> Does not follow. HHH and DDD are irrelevant to those proofs.
> 
> void DDD()
> {
>    HHH(DDD);
>    return;
> }
> 
> When I dumbed the original self-referential proof down
> to HHH(DDD) everyone here proved that they did not even
> understand what ordinary recursion is.

That you are dumb does not mean that others don't understand
ordinary recursion.

-- 
Mikko