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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: How do simulating termination analyzers work?
Date: Sun, 22 Jun 2025 11:59:23 +0300
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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*

No, they are not. You have not solved the halting problem and that
(in addition to all proofs) supports the claim that halting problem
is unsolvable.

In order to show that a proof is wrong you need to show an error
in the proof. Even then the conclusion is proven unless you can
show an error in every proof of that conclusion.

-- 
Mikko