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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic,comp.ai.philosophy
Subject: Re: HHH(DDD) is correct to reject its input as non-halting ---
 EVIDENCE THAT I AM CORRECT
Date: Sun, 22 Jun 2025 14:27:57 -0500
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On 6/22/2025 11:01 AM, Fred. Zwarts wrote:
> Op 20.jun.2025 om 16:53 schreef olcott:
>> On 6/20/2025 4:42 AM, Fred. Zwarts wrote:
>>> Op 19.jun.2025 om 17:23 schreef olcott:
>>>> On 6/19/2025 3:55 AM, Fred. Zwarts wrote:
>>>>> Op 18.jun.2025 om 17:41 schreef olcott:
>>>>>> On 6/18/2025 4:36 AM, Fred. Zwarts wrote:
>>>>>>> Op 17.jun.2025 om 16:36 schreef olcott:
>>>>>>>> On 6/17/2025 4:28 AM, Fred. Zwarts wrote:
>>>>>>>>> Op 17.jun.2025 om 00:26 schreef olcott:
>>>>>>>>>> On 6/16/2025 3:53 AM, Fred. Zwarts wrote:
>>>>>>>>>>> Op 15.jun.2025 om 22:10 schreef olcott:
>>>>>>>>>>>> void DDD()
>>>>>>>>>>>> {
>>>>>>>>>>>>    HHH(DDD);
>>>>>>>>>>>>    return;
>>>>>>>>>>>> }
>>>>>>>>>>>>
>>>>>>>>>>>> When I challenge anyone to show the details of exactly
>>>>>>>>>>>> how DDD correctly simulated by ANY simulating termination
>>>>>>>>>>>> analyzer HHH can possibly reach its own simulated "return"
>>>>>>>>>>>> statement final halt state they ignore this challenge.
>>>>>>>>>>>
>>>>>>>>>>> It seems very difficult for you to read.
>>>>>>>>>>> We clearly stated that the challenge is improper.
>>>>>>>>>>
>>>>>>>>>> Are you too stupid to understand that dogmatic
>>>>>>>>>> assertions that are utterly bereft of any supporting
>>>>>>>>>> reasoning DO NOT COUNT AS REBUTTALS ???
>>>>>>>>>
>>>>>>>>> No, you are too stupid to realise that challenging for a recipe 
>>>>>>>>> to draw a square circle does not count as a proof that square 
>>>>>>>>> circles exist.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Claiming that I made a mistake with no ability to
>>>>>>>>>> show this mistake is DISHONEST.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Indeed, but irrelevant,
>>>>>>>>
>>>>>>>> That alternative is that you are dishonest.
>>>>>>>> When you claim that I am wrong and have
>>>>>>>> no ability to show how and where I am wrong
>>>>>>>> this would seem to make you a liar.
>>>>>>>>
>>>>>>>> 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.
>>>>>>>
>>>>>>> Indeed, HHH fails to reach the end of the simulation, even though 
>>>>>>> the end is only one cycle further from the point where it gave up 
>>>>>>> the simulation.
>>>>>>>
>>>>>>
>>>>>> That is counter-factual and over-your-head.
>>>>>>
>>>>>
>>>>> No evidence presented for this claim. Dreaming again?
>>>>> Even a beginner understands that when HHH has code to abort and 
>>>>> halt, the simulated HHH runs one cycle behind the simulating HHH, 
>>>>> so that when the simulating HHH aborts, the simulated HHH is only 
>>>>> one cycle away from the same point.
>>>>
>>>> Proving that you do not understand what unreachable code is.
>>>> First year CS students and EE majors may not understand this.
>>>> All CS graduates would understand this.
>>>
>>> That you do not understand what I write makes it difficult for you to 
>>> learn from your errors.
>>> It is not that difficult. Try again and pay full attention to it.
>>> Even a beginner understands that when HHH has code to abort and halt,
>>> the simulated HHH runs one cycle behind the simulating HHH, so that 
>>> when the simulating HHH aborts, the simulated HHH is only one cycle 
>>> away from the same point.
>>
>> Yes this is factual.
>>
>> *This is only ordinary computer programming with*
>> *no theory of computation computer science required*
>>
>> Every simulated HHH remains one cycle behind its simulator
>> no matter how deep the recursive simulations go. This means
>> that the outermost directly executed HHH reaches its abort
>> criteria first.
> 
> And it fails to see that the simulated HHH would reach exactly the same 
> abort criteria one cycle later.
> In this way, it misses the fact that it is simulating an HHH that would 
> abort and halt.
> 

void Infinite_Loop()
{
   HERE: goto HERE;
   printf("Fred Zwarts can't understand this is never reached\n");
}


>>
>> This means that none of simulated HHH have reached their
>> abort criteria. This means that their own abort code is
>> unreachable at the point where the outermost HHH would
>> abort.
> 
> They do not reach its, even if the abort criteria are reachable, because 
> the simulator halts its simulation too soon. That is not the behaviour 
> of the program specified in the input, but erroneous behaviour of the 
> simulator.
> 


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