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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Who here understands that the last paragraph is Necessarily true?
Date: Fri, 19 Jul 2024 15:21:08 -0500
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On 7/19/2024 11:03 AM, joes wrote:
> Am Fri, 19 Jul 2024 09:54:07 -0500 schrieb olcott:
>> On 7/19/2024 1:35 AM, Fred. Zwarts wrote:
>>> Exactly the same input is presented to the direct execution and the
>>> simulation, namely the x86 code of the program.
>>> The semantics of the x86 language does not change in these two cases,
>>> so a correct simulator should interpret the x86 in the same way as the
>>> direct execution.

No it is not.
Although you are the same person when you are hungry before
you eat after you eat you are no longer in the hungry state.
*Same person transitioning from hungry to not-hungry*

Because or actions that HHH does DDD changes its state
from needing to be aborted to not needing to be aborted.

Same program transitioning from needing to be aborted
to not needing to be aborted.

>>      Before HHH(DDD) aborts its emulation the directly executed DDD()
>>      cannot possibly halt.
>>      After HHH(DDD) aborts its emulation the directly executed DDD()
>>      halts.
> What do you mean "after"? The outer DDD called by main? It will halt
> even before HHH has aborted, because it is deterministic and actually
> does halt. It makes no sense to say that something that will, couldn't.
> 

void DDD()
{
   HHH(DDD);
}

int main()
{
   DDD();
}

DDD() is invoked and calls HHH(DDD) that emulates its own
separate DDD instance as a separate process.

Unless HHH(DDD) aborts its emulated DDD:
HHH, emulated DDD and executed DDD never stop running.

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer