Sujet : Re: Who knows that DDD correctly simulated by HHH cannot possibly reach its own return instruction final state?
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theoryDate : 05. Aug 2024, 23:59:10
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <dca317e236dd975a3f030ae92ea0aa343833f029@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
User-Agent : Mozilla Thunderbird
On 8/5/24 9:49 AM, olcott wrote:
On 8/5/2024 2:39 AM, Mikko wrote:
On 2024-08-04 18:59:03 +0000, olcott said:
>
On 8/4/2024 1:51 PM, Richard Damon wrote:
On 8/4/24 9:53 AM, olcott wrote:
On 8/4/2024 1:22 AM, Fred. Zwarts wrote:
Op 03.aug.2024 om 18:35 schreef olcott:
>>>> ∞ instructions of DDD correctly emulated by HHH[∞] never
reach their own "return" instruction final state.
>
So you are saying that the infinite one does?
>
>
Dreaming again of HHH that does not abort? Dreams are no substitute for facts.
The HHH that aborts and halts, halts. A tautology.
>
void DDD()
{
HHH(DDD);
return;
}
>
That is the right answer to the wrong question.
I am asking whether or not DDD emulated by HHH
reaches its "return" instruction.
>
But the "DDD emulated by HHH" is the program DDD above,
>
When I say DDD emulated by HHH I mean at any level of
emulation and not and direct execution.
>
If you mean anything other than what the words mean you wihout
a definition in the beginning of the same message then it is
not reasonable to expect anyone to understand what you mean.
Instead people may think that you mean what you say or that
you don't know what you are saying.
>
If you don't understand what the word "emulate" means look it up.
DDD (above) cannot possibly reach its own "return" instruction halt
state when its machine code is correctly emulated by HHH.
Only because an HHH that does so never returns to anybody.
If HHH does return an answer then the behavior of DDD is to halt, but does so after HHH as aborted it PARTIAL (and thus incorrect) emulation of it.
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