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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: Tue, 10 Jun 2025 12:40:49 -0500
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On 6/10/2025 6:24 AM, Richard Damon wrote:
> On 6/9/25 8:47 PM, olcott wrote:
>> 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?
> 
> Who said correctly simulated by HHH?
> 
> You need to make a decision on which part of your arguement is a lie?
> 
> Is HHH a program? if not, your whole argument is a category error. (you 
> have actually admitted this error)
> 

In other words you are stupidly making the counter-factual
statement that termination analyzers cannot operate on C
functions. It is a verified fact that they do operate
on C functions so shut-the-f_ck-up about this.

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