Re: DDD correctly emulated by HHH is correctly rejected as non-halting.

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Sujet : Re: DDD correctly emulated by HHH is correctly rejected as non-halting.
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory
Date : 12. Jul 2024, 03:08:27
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <b61ed323c69ac7f8f7af365e7a7aba5085407d67@i2pn2.org>
References : 1 2 3 4 5 6 7
User-Agent : Mozilla Thunderbird
On 7/11/24 11:05 AM, olcott wrote:
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:
>
void DDD()
{
    HHH(DDD);
}
int main()
{
    HHH(DDD);
}
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.
 
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.
 
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.
 DDD correctly simulated by HHH has provably different
behavior than DDD correctly simulated by HHH1.
 
Which just shows that YOUR definition of "Correctly Simulated" that you are trying to use can't actually be a correct definition of an objective property.

Date Sujet#  Auteur
14 Jul 25 o 

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