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On 8/3/2024 3:45 PM, Richard Damon wrote:Why not? That *IS* the definition of correct emulation as defined by the concept of a UTM, which is the only definition of emulation that lets you get the actual full behavior of the program given.On 8/3/24 4:14 PM, olcott wrote:*No you damned liar it does not mean that*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.
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.
It means that when 0 to infinity steps of DDD areWhere do you get that from?
correctly emulated by its corresponding HHH not a
single DDD ever reaches its own halt state of "return".
This means that every HHH can take a wild guess thatNope.
its DDD does not halt and it would be correct because
we exhaustively covered every damn one of them.
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