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On 10/6/2024 1:52 PM, Richard Damon wrote:Really? What instruction actually emulated had a different result than when it was executed.On 10/6/24 2:32 PM, olcott wrote:The execution trace proves that the executed DDD hasOn 10/6/2024 1:13 PM, Richard Damon wrote:>On 10/6/24 2:03 PM, olcott wrote:>On 10/6/2024 12:29 PM, Richard Damon wrote:>On 10/6/24 1:07 PM, olcott wrote:>On 10/6/2024 11:59 AM, Richard Damon wrote:>On 10/6/24 8:39 AM, olcott wrote:>>>
DDD emulated by each corresponding HHH that can possibly
exist never returns. Each of these HHH emulators that does
return 0 correctly reports the above non-halting behavior.
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No, the DDD return (if the HHH(DDD) gives an answer), just after the HHH that emulated them gave up.
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DDD emulated by each corresponding HHH that can possibly
exist never returns.
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DDD emulated by each corresponding HHH
DDD emulated by each corresponding HHH
DDD emulated by each corresponding HHH
DDD emulated by each corresponding HHH
Which, as you have been told but seems to be above your head means that the execution of DDD,
gets to ignore the fact that DDD was defined to
have a pathological relationship with HHH that
HHH cannot ignore.
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No, that isn't ignoring it, but taking into account that since HHH is defined to be a specific program, it has specific behavior.
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The behavior of the executed DDD after the emulated
DDD has already been aborted is different than the
behavior of the emulated DDD that must be aborted.
Nope, it is the exact same code on the exact same data, and thus does the exact same behavior.
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different behavior that need not be aborted because
emulated DDD must be an is aborted.
No one can be stupid enough to think thatNo, but the fact that it DOES abort it simulation means its simulation doesn't show the actual behavior of the input, and means that an actual correct simulation will halt.
NEED NOT BE ABORTED is exactly the same as NEED NOT BE ABORTED
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