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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: Proof that DDD specifies non-halting behavior --- point by point
Date: Fri, 16 Aug 2024 13:21:06 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Fri, 16 Aug 2024 07:42:22 -0500 schrieb olcott:
> On 8/16/2024 3:53 AM, Mikko wrote:
>> 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:

>>>>>> In order to prove that the above specifies a non-halting behavour
>>>>>> you must prove that HHH(DDD) does not terminate.
>>>>>
>>>> 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.
>> 
> *You are getting the computer science incorrectly*
> On 8/2/2024 11:32 PM, Jeff Barnett wrote:
>  > ...In some formulations, there are specific states
>  >    defined as "halting states" and the machine only halts if either
>  >    the start state is a halt state...
>  > ...these and many other definitions all have
>  >    equivalent computing prowess...
> The "return" instruction is the halt state of DDD.
It's kind of amazing to see you quote unrelated stuff.
How does HHH not simulate DDD returning, while not executing the same
instructions infinitely?

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.