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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: HHH(DDD) is correct to reject its input as non-halting --- EVIDENCE THAT I AM CORRECT
Date: Sat, 21 Jun 2025 13:00:39 +0300
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On 2025-06-20 17:11:30 +0000, olcott said:

> On 6/20/2025 3:35 AM, Mikko wrote:
>> On 2025-06-19 15:25:45 +0000, olcott said:
>> 
>>> On 6/19/2025 3:21 AM, Mikko wrote:
>>>> On 2025-06-18 15:07:14 +0000, olcott said:
>>>> 
>>>>> 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
>>>> 
>>>> No, he did not. The paragraph responded to was about first year CS
>>>> students and what know, and so is the response.
>>> 
>>> My claim is that each of the above functions correctly
>>> simulated by any termination analyzer HHH that can possibly
>>> exist will never stop running unless aborted by HHH.
>>> Can you affirm or correctly refute this?
>> 
>> Now you are changed the topic.
> 
> That is what I said (less clearly) all along.

No, you accused that it was someone else. But that does not matter
anymore as you now admit that you did it.

-- 
Mikko