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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH is correctly rejected as
 non-halting.
Date: Fri, 12 Jul 2024 13:46:21 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Fri, 12 Jul 2024 07:34:59 -0500 schrieb olcott:

> On 7/12/2024 3:08 AM, joes wrote:
>> Am Thu, 11 Jul 2024 15:56:09 -0500 schrieb olcott:
>>> On 7/11/2024 3:19 PM, joes wrote:
>>>> Am Thu, 11 Jul 2024 10:05:58 -0500 schrieb olcott:
>>>>> On 7/11/2024 9:25 AM, joes wrote:
>>>>>> Am Thu, 11 Jul 2024 09:10:24 -0500 schrieb olcott:
>>>>>>> On 7/11/2024 1:25 AM, Mikko wrote:
>>>>>>>> On 2024-07-10 17:53:38 +0000, olcott said:
>>>>>>>>> On 7/10/2024 12:45 PM, Fred. Zwarts wrote:
>>>>>>>>>> Op 10.jul.2024 om 17:03 schreef olcott:
>> 
>>>>>>> When DDD is correctly emulated by HHH according to the semantics
>>>>>>> of the x86 programming language HHH must abort its emulation of
>>>>>>> DDD or both HHH and DDD never halt.
>>>>>> If the recursive call to HHH from DDD halts, the outer HHH doesn't
>>>>>> need to abort.
>>>> Do you mean that HHH doesn't halt?

>>>>>> DDD depends totally on HHH; it halts exactly when HHH does.
>>>>>> Which it does, because it aborts.
>>>> What does HHH do after it aborts?

>>>>> DDD correctly simulated by HHH has provably different behavior than
>>>>> DDD correctly simulated by HHH1.
>>>> Which means that HHH is not doing the simulation correctly.
>>> When HHH simulates DDD according to the semantics of the x86 language
>>> then HHH is simulating correctly. When people disagree with the
>>> semantics of the x86 language THEY ARE WRONG !!!
>> Aborting is not a correct simulation.
>> Please answer the other questions above.
> Aborting is what a simulating termination analyzer must do for any input
> that cannot possibly otherwise stop running.
Yes, which makes it not a simulator.
Why does DDD not halt?

-- 
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.