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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: Thu, 11 Jul 2024 20:19:14 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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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:
>> 
>>>>>> Unneeded complexity. It is equivalent to:
>>>>>>         int main()
>>>>>>         {
>>>>>>           return HHH(main);
>>>>>>         }
>>>>> Every time any HHH correctly emulates DDD it calls the x86utm
>>>>> operating system to create a separate process context with its own
>>>>> memory virtual registers and stack, thus each recursively emulated
>>>>> DDD is a different instance.
>>>> However, each of those instances has the same sequence of
>>>> instructions that the x86 language specifies the same operational
>>>> meaning.
>>> *That is counter-factual*
>> Contradicting yourself? "Counterfactual" usually means "if it were
>> different".
>> 
>>> 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.
> Sure and when squares are round you can measure the radius of a square.
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.
> Halting means reaching its own last instruction and terminating
> normally.
What does HHH do after it aborts?

>>> When DDD is correctly emulated by HHH1 according to the semantics of
>>> the x86 programming language HHH1 need not abort its emulation of DDD
>>> because HHH has already done this.
>> Where does HHH figure into this? It is not the simulator here.
>>> The behavior of DDD emulated by HHH1 is identical to the behavior of
>>> the directly executed DDD().
>> At last!
> HHH must abort its simulation. HHH1 does not need to do that because HHH
> has already done this.
No, HHH1 doesn't need to because DDD is just a regular program to it,
not constructed to be unsimulatable.

> 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.

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