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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: HHH(DDD) is correct to reject its input as non-halting ---
 EVIDENCE THAT I AM CORRECT
Date: Wed, 18 Jun 2025 10:07:14 -0500
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On 6/17/2025 8:27 PM, Richard Damon wrote:
> On 6/17/25 10:46 AM, olcott wrote:
>> On 6/17/2025 4:33 AM, Fred. Zwarts wrote:
>>> Op 16.jun.2025 om 19:01 schreef olcott:
>>>> On 6/16/2025 6:37 AM, Mikko wrote:
>>>>> On 2025-06-16 00:57:42 +0000, olcott said:
>>>>>
>>>>>> On 6/15/2025 6:44 PM, Richard Damon wrote:
>>>>>>> On 6/15/25 4:10 PM, olcott wrote:
>>>>>>>> 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.
>>>>>>>
>>>>>>> And it seems you don't understand that the problem is that while, 
>>>>>>> yes, if HHH does infact do a correct simulation, it will not 
>>>>>>> reach a final state, that fact only applie *IF* HHH does that, 
>>>>>>> and all the other HHHs which differ see different inputs.
>>>>>>>
>>>>>>
>>>>>> *I should have said*
>>>>>
>>>>> No, that is not how you should have said.
>>>>>
>>>>>> When one or more instructions of DDD are correctly
>>>>>> simulated by ANY simulating termination analyzer HHH
>>>>>> then DDD never reaches its simulated "return" statement
>>>>>> final halt state.
>>>>>
>>>>> How does ANY simulating termination analyzer HHH differ form some
>>>>> other simulating termination alalyzer?
>>>>>
>>>>
>>>> I changed the evaluation from the HHH that I have coded
>>>> to every HHH that could possibly exist.
>>>>
>>>
>>> And even a beginner can see that they all fail to reach the end of 
>>> the simulation, even though the input is a pointer to code that 
>>> includes the code to abort and halt.
>>
>> 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.
>>
>>
> 
> No, they understand that a pattern seen is a halting program (since you 
> admit that DDD halts when run directly) can't be a pattern that proves 
> the program is non-halting.
> 

You changed the subject from THIS EXACT POINT
*none of them ever stop running unless aborted*
(a) YES that is true
(b) No that is not true

Here are the exact steps of how X stops running
without every being aborted.

> It seems you think that you can proves false statements.
> 
> In other words, you logic lies.

I am not the one that perpetually changes the subject
to avoid addressing the actual point.

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