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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: HHH(DDD) is correct to reject its input as non-halting ---
 EVIDENCE THAT I AM CORRECT
Date: Tue, 17 Jun 2025 11:33:08 +0200
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In-Reply-To: <102pikk$1odus$4@dont-email.me>

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.