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On 6/21/24 2:22 PM, olcott wrote:In this case the definition of the x86 language specifiesOn 6/21/2024 1:00 PM, Richard Damon wrote:But strings don't "have" behavior, or even "specify" behavior by themselves, the behavior comes from applying the string to the DEFINITION of the problem.On 6/21/24 1:55 PM, olcott wrote:>On 6/21/2024 12:40 PM, Richard Damon wrote:>On 6/21/24 1:22 PM, olcott wrote:>>>
When there is no mapping from the finite string x86 machine
language input to H(D,D) to the behavior of D(D) then
H(D,D) IS NOT being asked about the behavior of D(D).
But there *IS* a mapping, it just isn't a COMPUTABLE MAPPING.
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If there is a mapping yet not a computable mapping then
the actual halt decider cannot even see the question that
the textbooks expect it to see.
But a decider doesn't "See" the question. it just computes the result it was programmed to give.
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It must be the behavior that the input finite string actually specifies.
It cannot be the behavior that the programmer imagines that it specifies.
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