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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: Wed, 18 Jun 2025 12:08:48 +0300
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On 2025-06-17 13:33:05 +0000, olcott said:
> On 6/17/2025 5:09 AM, Mikko wrote:
>> On 2025-06-16 17:01:08 +0000, olcott said:
>>
>>> 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.
>>
>> Tnat means that you think the program
>>
>> int HHH((void *)x(void)) {
>> return 1;
>> }
>>
>> when called with HHH(DDD) would return 0
>>
>> void DDD()
>> {
>> HHH(DDD);
>> return;
>> }
>>
>> to indicate that DDD does not halt?
>>
>
> Your HHH is not s simulating termination analyzer as required:
Your "every HHH that could possibly exist" does not mention simulation.
--
Mikko