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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Everyone on this forum besides Keith has been a damned liar about
 this point
Date: Mon, 9 Jun 2025 19:47:12 -0500
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On 6/9/2025 7:26 PM, Richard Damon wrote:
> On 6/9/25 10:43 AM, olcott wrote:
>> On 6/9/2025 5:31 AM, Fred. Zwarts wrote:
>>> Op 09.jun.2025 om 06:15 schreef olcott:
>>>> On 6/8/2025 10:42 PM, dbush wrote:
>>>>> On 6/8/2025 11:39 PM, olcott wrote:
>>>>>> On 6/8/2025 10:32 PM, dbush wrote:
>>>>>>> On 6/8/2025 11:16 PM, olcott wrote:
>>>>>>>> On 6/8/2025 10:08 PM, dbush wrote:
>>>>>>>>> On 6/8/2025 10:50 PM, olcott wrote:
>>>>>>>>>> void DDD()
>>>>>>>>>> {
>>>>>>>>>>    HHH(DDD);
>>>>>>>>>>    return;
>>>>>>>>>> }
>>>>>>>>>>
>>>>>>>>>> The *input* to simulating termination analyzer HHH(DDD)
>>>>>>>>>
>>>>>>>>> No it's not, as halt deciders / termination analyzers work with 
>>>>>>>>> algorithms,
>>>>>>>>
>>>>>>>> That is stupidly counter-factual.
>>>>>>>>
>>>>>>>
>>>>>>> That you think that shows that 
>>>>>>
>>>>>> My understanding is deeper than yours.
>>>>>> No decider ever takes any algorithm as its input.
>>>>>
>>>>> But they take a description/specification of an algorithm,
>>>>
>>>> There you go.
>>>>
>>>>> which is what is meant in this context. 
>>>>
>>>> It turns out that this detail makes a big difference.
>>>>
>>>>> And because your HHH does not work with the description/ 
>>>>> specification of an algorithm, by your own admission, you're not 
>>>>> working on the halting problem.
>>>>>
>>>>
>>>> HHH(DDD) takes a finite string of x86 instructions
>>>> that specify that HHH simulates itself simulating DDD.
>>>
>>> And HHH fails to see the specification of the x86 instructions. It 
>>> aborts before it can see how the program ends.
>>>
>>
>> This is merely a lack of sufficient technical competence
>> on your part. It is a verified fact that unless the outer
>> HHH aborts its simulation of DDD that DDD simulated by HHH
>> the directly executed DDD() and the directly executed HHH()
>> would never stop running. That you cannot directly see this
>> is merely your own lack of sufficient technical competence.
> 
> And it is a verified fact that you just ignore that if HHH does in fact 
> abort its simulation of DDD and return 0, then the behavior of the 
> input, PER THE ACTUAL DEFINITIONS, is to Halt, and thus HHH is just 
> incorrect.
> 

void DDD()
{
   HHH(DDD);
   return;
}

How the f-ck does DDD correctly simulated by HHH
reach its own "return" statement final halt state?

_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

How the f-ck does DDD correctly emulated by HHH
reach its own "ret" instruction final halt state?

That you have dodged this question hundreds of times
proves that you are a liar.

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