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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Proof that DDD specifies non-halting behavior --- point by point
Date: Fri, 16 Aug 2024 11:53:52 +0300
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On 2024-08-15 15:25:07 +0000, olcott said:

> On 8/15/2024 5:22 AM, Mikko wrote:
>> On 2024-08-14 13:06:27 +0000, olcott said:
>> 
>>> On 8/14/2024 3:17 AM, Mikko wrote:
>>>> On 2024-08-14 00:52:36 +0000, olcott said:
>>>> 
>>>>> void DDD()
>>>>> {
>>>>>    HHH(DDD);
>>>>>    return;
>>>>> }
>>>> 
>>>> In order to prove that the above specifies a non-halting behavour
>>>> you must prove that HHH(DDD) does not terminate.
>>> 
>>> Wrong.
>> 
>> At least the proof that DDD does not terminate also proves as an
>> intermedate result or an obvious corollary that HHH does not halt.
>> 
>> Non-halting means that an infinite number of instructions can be
>> executed without halting. That means that at least one instruction
>> is executed infinitely many times as there are only finitely many
>> instructions. But not instrunctions of DDD outside HHH is executed
>> infinitely many times.
>> 
> 
> Wrong. Non-halting only means that when DDD is emulated
> according to the semantics of the x86 language and this
> emulation is unlimited that DDD would never reach its
> own "return" instruction.

If what I said is wrong then what you said is wrong, too,
as you say what I said.

-- 
Mikko