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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Anyone with sufficient knowledge of C knows that DD specifies non-terminating behavior to HHH
Date: Wed, 19 Feb 2025 11:01:26 +0200
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On 2025-02-18 11:26:25 +0000, olcott said:

> On 2/18/2025 3:24 AM, Mikko wrote:
>> On 2025-02-17 09:05:42 +0000, Fred. Zwarts said:
>> 
>>> Op 16.feb.2025 om 23:51 schreef olcott:
>>>> On 2/16/2025 4:30 PM, joes wrote:
>>>>> Am Sun, 16 Feb 2025 15:58:14 -0600 schrieb olcott:
>>>>>> On 2/16/2025 2:02 PM, joes wrote:
>>>>>>> Am Sun, 16 Feb 2025 13:24:14 -0600 schrieb olcott:
>>>>>>>> On 2/16/2025 10:35 AM, joes wrote:
>>>>>>>>> Am Sun, 16 Feb 2025 06:51:12 -0600 schrieb olcott:
>>>>>>>>>> On 2/15/2025 2:49 AM, Mikko wrote:
>>>>>>>>>>> On 2025-02-14 12:40:04 +0000, olcott said:
>>>>>>>>>>>> On 2/14/2025 2:58 AM, Mikko wrote:
>>>>>>>>>>>>> On 2025-02-14 00:07:23 +0000, olcott said:
>>>>>>>>>>>>>> On 2/13/2025 3:20 AM, Mikko wrote:
>>>>>>>>>>>>>>> On 2025-02-13 04:21:34 +0000, olcott said:
>>>>>>>>>>>>>>>> On 2/12/2025 4:04 AM, Mikko wrote:
>>>>>>>>>>>>>>>>> On 2025-02-11 14:41:38 +0000, olcott said:
>>>>> 
>>>>>>>>>>>> DD  correctly simulated by HHH cannot possibly terminate normally.
>>>>>>>>>>> 
>>>>>>>>>>> That claim has already shown to be false. Nothing above shows that
>>>>>>>>>>> HHH does not return 0. If it does DD also returns 0.
>>>>>>>>>> When we are referring to the above DD simulated by HHH and not
>>>>>>>>>> trying to get away with changing the subject to some other DD
>>>>>>>>>> somewhere else
>>>>>>>>> such as one that calls a non-aborting version of HHH
>>>>>>>>> 
>>>>>>>>>> then anyone with sufficient knowledge of C programming knows that no
>>>>>>>>>> instance of DD shown above simulated by any corresponding instance
>>>>>>>>>> of HHH can possibly terminate normally.
>>>>>>>>> Well, then that corresponding (by what?) HHH isn’t a decider.
>>>>>>>> I am focusing on the isomorphic notion of a termination analyzer.
>>>>>>> (There are other deciders that are not termination analysers.)
>>>>>>> 
>>>>>>>> A simulating termination analyzer correctly rejects any input that
>>>>>>>> must be aborted to prevent its own non-termination.
>>>>>>> Yes, in particular itself is not such an input, because we *know* that
>>>>>>> it halts, because it is a decider. You can’t have your cake and eat it
>>>>>>> too.
>>>>>> I am not even using the confusing term "halts".
>>>>>> Instead I am using in its place "terminates normally".
>>>>>> DD correctly simulated by HHH cannot possibly terminate normally.
>>>>> What’s confusing about „halts”? I find it clearer as it does not imply
>>>>> an ambiguous „abnormal termination”. How does HHH simulate DD
>>>>> terminating abnormally, then? Why doesn’t it terminate abnormally
>>>>> itself?
>>>>> You can substitute the term: the input DD to HHH does not need to be
>>>>> aborted, because the simulated decider terminates.
>>>>> 
>>>> 
>>>> typedef void (*ptr)();
>>>> int HHH(ptr P);
>>>> 
>>>> int DD()
>>>> {
>>>>   int Halt_Status = HHH(DD);
>>>>   if (Halt_Status)
>>>>     HERE: goto HERE;
>>>>   return Halt_Status;
>>>> }
>>>> 
>>>> int main()
>>>> {
>>>>   HHH(DD);
>>>> }
>>>> 
>>>> Every simulated input that must be aborted to
>>>> prevent the non-termination of HHH is stipulated
>>>> to be correctly rejected by HHH as non-terminating.
>>>> 
>>> A very strange and invalid stipulation.
>> 
>> It merely means that the words do not have their ordinary meaning.
>> 
> 
> Unless HHH(DD) aborts its simulation of DD itself cannot possibly 
> terminate normally.

That cannot be determined without examination of HHH, which is not in the
scope of OP.

> Every expert in the C programming language can see this.

They can't when they can't see HHH and even then it is not obvious,
so the claim on the subject line is false.


-- 
Mikko