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 : 03. Aug 2024, 19:58:02
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <74c4fe66234c5332f4ec6032bc55cc6c5f038aee@i2pn2.org>
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User-Agent : Mozilla Thunderbird
On 8/3/24 2:33 PM, olcott wrote:
On 8/3/2024 1:09 PM, Richard Damon wrote:
On 8/3/24 1:58 PM, olcott wrote:
Every DDD correctly emulated by any HHH for a finite or
infinite number of steps never reaches its own "return"
halt state.
>
>
Nope. And you statment is just a incoherent statement, as no partial simulaitoni for a finite number of steps is "correct".
>
In other words you are trying to get away with saying that
when N instructions are correctly emulated by HHH that none
of these correctly emulated instructions were correctly emulated.
No, I am saying that the result is NOT the final result that the x86 semantics says will happen, because the x86 semantics says it does not stop therme
On top of this you are deceitfully trying to get way
with saying that when and infinite number of instructions
of DDD are correctly emulated by HHH that this emulated
DDD reaches its "return" instruction halt state and halts.
No, because I admitted that when (and only when) HHH is the version of itself that NEVER aborts do we have a DDD that when correctly emulated will never halt.
You seem to forget that because DDD calls HHH, each version of HHH creates a different DDD.
This just shows that you are ignorant of what a program actually is.
Liar Liar swearing your allegiance to the father of
lies. I wonder what your pastor would say about this?
But I don't do that, you do. My words align with the "scriptures" of the field, YOU are the one creating the heretical works.
You don't seem to know what a LIE actually is.