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From: Richard Damon <richard@damon-family.org>
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 21:34:07 -0400
Organization: i2pn2 (i2pn.org)
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On 6/18/25 11:07 AM, olcott wrote:
> 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, that is not true,

The problem is "ever stop runnig" isn't shown to be false by stopping 
the simulation.

EVERY HHH that stops its emulation has FORFETED it ability to say its 
input is non-halting, as it has failed to be a correct simulation, and 
the correct simulation of that input *WILL* halt

> 
> 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.
> 

But you are, as you keep on changing the meaning of "non-halting".

Since it is a property of "THE MACHINE" (not a partial simulation of it) 
all you talk about HHH not reaching the final state is just a lying 
change of topic.

Sorry, you are just proving that you "logic" is based on lying.