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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: How do simulating termination analyzers work?
Date: Wed, 2 Jul 2025 10:04:33 +0300
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On 2025-07-01 13:02:17 +0000, olcott said:

> On 6/30/2025 4:58 AM, Mikko wrote:
>> On 2025-06-29 14:21:55 +0000, olcott said:
>> 
>>> On 6/29/2025 3:44 AM, Mikko wrote:
>>>> 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.
>>>> 
>>> 
>>> Mensa scored me on the top 3% of the population.
>> 
>> Your intelligence, not wisdom.
>> 
>>> This is a little more difficult than ordinary recursion.
>> 
>> Perhaps to your little mind.
>> 
>>> void DDD()
>>> {
>>>    HHH(DDD);
>>>    return;
>>> }
>>> 
>>> _DDD()
>>> [00002192] 55             push ebp
>>> [00002193] 8bec           mov ebp,esp
>>> [00002195] 6892210000     push 00002192  // push DDD
>>> [0000219a] e833f4ffff     call 000015d2  // call HHH
>>> [0000219f] 83c404         add esp,+04
>>> [000021a2] 5d             pop ebp
>>> [000021a3] c3             ret
>>> Size in bytes:(0018) [000021a3]
>>> 
>>> The x86 source code of DDD specifies that this emulated
>>> DDD cannot possibly reach its own emulated "ret" instruction
>>> final halt state when emulated by HHH according to the
>>> semantics of the x86 language.
>> 
>> That defect in HHH is already known and a possible fix has been proposed.
> 
> Four Chatbots all agree that the input to simulating termination
> analyzer HHH(DDD) specifies non-terminating recursive emulation
> even though the directly executed DDD() halts.

It is easier to agree than to think, escpecially as artificial idiots
don't care about truth.

-- 
Mikko