Re: Who knows that DDD correctly simulated by HHH cannot possibly reach its own return instruction final state? BUT ONLY that DDD

On 8/3/24 4:14 PM, olcott wrote:

On 8/3/2024 3:00 PM, Richard Damon wrote:

On 8/3/24 3:06 PM, olcott wrote:

On 8/3/2024 1:58 PM, Richard Damon wrote:

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.
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Nope. And you statment is just a incoherent statement, as no partial simulaitoni for a finite number of steps is "correct".
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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.

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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
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The x86 semantics says that DDD correctly emulated by HHH
never reaches its own halt state of "return" in any finite
or infinite number of steps.
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But only if HHH DOES correct emulation that never aborts.
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 The x86 semantics says that
*DDD correctly emulated by HHH*

...

*DDD correctly emulated by HHH*
never reaches its own halt state of
"return" in any finite
or infinite number of steps.

Yes, but only for an HHH that corectly emulates its input, which means it never aborts, and only for the DDD that calls THAT HHH.
So, the ONLY DDD you can say by that information to be non-halting is the DDD that calls an HHH that actually does that,
Thus, ANY HHH that is given an input DDD that is built on itself, can not use that fact, becuase it isn't true, becuase that HHH doesn't correctly emulate its input.
It seems, you don't understand that a program is all of itself, and you can't judge the book by just its cover.
Sorry, but it seems you are just too stupid to understand these simple facts.