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On 3/24/2025 12:35 PM, dbush wrote:Why should that be proven?On 3/24/2025 12:44 PM, olcott wrote:The whole point of this post is to prove thatOn 3/24/2025 10:14 AM, dbush wrote:The HHH you implemented is computing *a* computable function, but it's not computing the halting function:On 3/24/2025 11:03 AM, olcott wrote:given an input of the function domain it canOn 3/24/2025 6:23 AM, Richard Damon wrote:Which includes the machine code of DDD, the machine code of HHH, and the machine code of everything it calls down to the OS level.On 3/23/25 11:09 PM, olcott wrote:DDD is a semantically and syntactically correct finiteIt is impossible for HHH compute the function from the directWHy isn't DDD made into the correct finite string?i
execution of DDD because DDD is not the finite string input
basis from which all computations must begin.
https://en.wikipedia.org/wiki/Computable_function
stirng of the x86 machine language.
Which is another way of saying that HHH can't determine that DDD halts when executed directly.That seems to be your own fault.DDD emulated by HHH directly causes recursive emulation
The problem has always been that you want to use the wrong string for DDD by excluding the code for HHH from it.
because it calls HHH(DDD) to emulate itself again. HHH
complies until HHH determines that this cycle cannot
possibly reach the final halt state of DDD.
return the corresponding output.
https://en.wikipedia.org/wiki/Computable_function
Computable functions are only allowed to compute the
mapping from their input finite strings to an output.
no Turing machine ever reports on the behavior
of the direct execution of another Turing machine.
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