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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: HHH(DD) does correctly reject its input as non-halting ---
 VERIFIED FACT +++
Date: Thu, 19 Jun 2025 11:35:43 -0500
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On 6/19/2025 3:32 AM, Mikko wrote:
> On 2025-06-18 15:32:50 +0000, olcott said:
> 
>> On 6/18/2025 4:16 AM, Mikko wrote:
>>> On 2025-06-17 14:45:42 +0000, olcott said:
>>>
>>>> On 6/17/2025 3:42 AM, Mikko wrote:
>>>>> On 2025-06-16 19:35:52 +0000, olcott said:
>>>>>
>>>>>> On 6/16/2025 5:36 AM, Mikko wrote:
>>>>>>> On 2025-06-15 13:49:51 +0000, olcott said:
>>>>>>>
>>>>>>>> On 6/15/2025 3:24 AM, Fred. Zwarts wrote:
>>>>>>>>> Op 14.jun.2025 om 15:38 schreef olcott:
>>>>>>>>>> On 6/14/2025 4:10 AM, Fred. Zwarts wrote:
>>>>>>>>>>> Op 13.jun.2025 om 17:53 schreef olcott:
>>>>>>>>>>>> On 6/13/2025 5:51 AM, Mikko wrote:
>>>>>>>>>>>>> On 2025-06-12 15:30:05 +0000, olcott said:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> int DD()
>>>>>>>>>>>>>> {
>>>>>>>>>>>>>>    int Halt_Status = HHH(DD);
>>>>>>>>>>>>>>    if (Halt_Status)
>>>>>>>>>>>>>>      HERE: goto HERE;
>>>>>>>>>>>>>>    return Halt_Status;
>>>>>>>>>>>>>> }
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> It is a verified fact that DD() *is* one of the forms
>>>>>>>>>>>>>> of the counter-example input as such an input would
>>>>>>>>>>>>>> be encoded in C. Christopher Strachey wrote his in CPL.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> // rec routine P
>>>>>>>>>>>>>> //   §L :if T[P] go to L
>>>>>>>>>>>>>> //     Return §
>>>>>>>>>>>>>> // https://academic.oup.com/comjnl/article/7/4/313/354243
>>>>>>>>>>>>>> void Strachey_P()
>>>>>>>>>>>>>> {
>>>>>>>>>>>>>>    L: if (HHH(Strachey_P)) goto L;
>>>>>>>>>>>>>>    return;
>>>>>>>>>>>>>> }
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> https://academic.oup.com/comjnl/article- 
>>>>>>>>>>>>>> abstract/7/4/313/354243? redirectedFrom=fulltext
>>>>>>>>>>>>>
>>>>>>>>>>>>> Strachey only informally presents the idea of the proof. 
>>>>>>>>>>>>> Formalism
>>>>>>>>>>>>> and details needed in a rigorous proof is not shown.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> void DDD()
>>>>>>>>>>>> {
>>>>>>>>>>>>    HHH(DDD);
>>>>>>>>>>>>    return;
>>>>>>>>>>>> }
>>>>>>>>>>>>
>>>>>>>>>>>> _DDD()
>>>>>>>>>>>> [00002192] 55             push ebp
>>>>>>>>>>>> [00002193] 8bec           mov ebp,esp
>>>>>>>>>>>> [00002195] 6892210000     push 00002192
>>>>>>>>>>>> [0000219a] e833f4ffff     call 000015d2  // call HHH
>>>>>>>>>>>> [0000219f] 83c404         add esp,+04
>>>>>>>>>>>> [000021a2] 5d             pop ebp
>>>>>>>>>>>> [000021a3] c3             ret
>>>>>>>>>>>> Size in bytes:(0018) [000021a3]
>>>>>>>>>>>>
>>>>>>>>>>>> Exactly how would DDD correctly emulated by HHH
>>>>>>>>>>>> reach its own "ret" instruction final halt state?
>>>>>>>>>>>
>>>>>>>>>>> Indeed, HHH fails where other world-class simulators have no 
>>>>>>>>>>> problem to simulate the program specified in the input.
>>>>>>>>>>
>>>>>>>>>> So you still don't understand what recursive simulation is?
>>>>>>>>>
>>>>>>>>> It seems I understand it better than you do. You seem to think 
>>>>>>>>> that every recursion is a infinite recursion. As soon as you 
>>>>>>>>> see a recursion, you think it has been proven that it is an 
>>>>>>>>> infinite recursion, even if the code specifies an abort and halt.
>>>>>>>>
>>>>>>>> <MIT Professor Sipser agreed to ONLY these verbatim words 
>>>>>>>> 10/13/2022>
>>>>>>>>      If simulating halt decider H correctly simulates its
>>>>>>>>      input D until H correctly determines that its simulated D
>>>>>>>>      would never stop running unless aborted then
>>>>>>>>
>>>>>>>>      H can abort its simulation of D and correctly report that D
>>>>>>>>      specifies a non-halting sequence of configurations.
>>>>>>>> </MIT Professor Sipser agreed to ONLY these verbatim words 
>>>>>>>> 10/13/2022>
>>>>>>>>
>>>>>>>> It is an easily verified fact that the input to HHH(DDD) and
>>>>>>>> the input to HHH(DD) meets the above self-evidently true criteria.
>>>>>>>
>>>>>>> No, they don't meet the second cireterion. HHH does not correctly
>>>>>>> determine that its input would never stop running unless aborted.
>>>>>>> Perhaps you may deceive with someting like equivocation someone to
>>>>>>> believe it does but in reality it does not.
>>>>>>
>>>>>> No one has ever even attempted to show the details
>>>>>> of how this is not correct:
>>>>>>
>>>>>> void DDD()
>>>>>> {
>>>>>>    HHH(DDD);
>>>>>>    return;
>>>>>> }
>>>>>>
>>>>>> When one or more instructions of DDD are correctly
>>>>>> simulated by ANY simulating termination analyzer HHH
>>>>>> then this correctly simulated DDD never reaches its
>>>>>> simulated "return" statement final halt state.
>>>>>
>>>>> I have. I have shown that there is a simulating termination
>>>>> analyser that has the name HHH and that simulates until it
>>>>> finds either a call to HHH or termination. If it finds HHH
>>>>> it continues simulation after the call. If it finds a return
>>>>> from the input runction it returns 1. Your "any simulating
>>>>> termination analyzer HHH" does not exclude my HHH. Therefore
>>>>> your claim is false.
>>>>>
>>>>
>>>> void Infinite_Loop()
>>>> {
>>>>    HERE: goto HERE;
>>>>    return;
>>>> }
>>>>
>>>> void Infinite_Recursion()
>>>> {
>>>>    Infinite_Recursion();
>>>> }
>>>>
>>>> void DDD()
>>>> {
>>>>    HHH(DDD);
>>>>    return;
>>>> }
>>>>
>>>> When it is understood that HHH does simulate itself
>>>
>>> Can you prove that my HHH does simulate itself?
>>>
>>
>> *That is a given* for this thought experiment.
> 
> Above is only required that HHH partially simulates the behaviour
> specified by its input. 

counter-factual

> That the simulated part includes all of
> or even a part of HHH is not required.
> 

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 one has ever been able to refute this in three years*

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