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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: How do simulating termination analyzers work?
Date: Wed, 25 Jun 2025 09:43:29 +0300
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On 2025-06-24 14:57:47 +0000, olcott said:

> On 6/24/2025 3:52 AM, Mikko wrote:
>> On 2025-06-23 16:48:20 +0000, olcott said:
>> 
>>> On 6/23/2025 2:08 AM, Mikko wrote:
>>>> On 2025-06-22 16:37:37 +0000, olcott said:
>>>> 
>>>>> On 6/22/2025 3:47 AM, Mikko wrote:
>>>>>> On 2025-06-21 17:33:10 +0000, olcott said:
>>>>>> 
>>>>>>> On 6/21/2025 4:46 AM, Mikko wrote:
>>>>>>>> On 2025-06-20 17:12:30 +0000, olcott said:
>>>>>>>> 
>>>>>>>>> On 6/20/2025 3:45 AM, Mikko wrote:
>>>>>>>>>> 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.
>>>>>>>>> 
>>>>>>>>> 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?
>>>>>>>> 
>>>>>>>> No, because that is not well claimed. You have used "HHH" in at least
>>>>>>>> two different meanings and it is not clear which meaning is intended.
>>>>>>> 
>>>>>>> *clearer words*
>>>>>>> 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?
>>>>>> 
>>>>>> Not sufficiently clearer than the previous attempt.
>>>>> 
>>>>> Since you did not say exactly what seems unclear to
>>>>> you I am taking this as a dishonest dodge away from the point.
>>>> 
>>>> You are lying. I did say. Your second attermpt does not clarify what
>>>> I did say was unclear. You didn't say what it did clarify, so
>>>> apparently nothing. You just claimed that an exact copy is clearer.
>>> 
>>> 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?
>>> 
>>> All of those words are perfectly clear to me.
>> 
>> Isn't my HHH as a suffifient answer? If not, ask again when you
>> have clarified all points I or someone else has identified as
>> ambiguous.
> 
> You are playing head games.

Indeed I am playing your games. But if they were head gaems you would win.

-- 
Mikko