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On 7/7/2025 9:24 PM, Richard Damon wrote:Then evdrything in reality is outside the domain of Turing Machines.On 7/7/25 7:47 PM, olcott wrote:>>Yes.That Turing machines cannot take directly executing Turing>
Machines as inputs entails that these directly executed
machines are outside of the domain of every Turing machine
based halt decider.
But they can take the finite-stringt encoding of those machines.
>
I guess you idea of Turing Machine is so limited that you think they can't do arithmatic, as you can't actually put a "Number" as the input, only the finite-string encoding of a number, which puts it outside the domain of them.No one here has any understanding of the philosophy of
>
computation. They can only memorize the rules and have
no idea about the reasoning behind these rules.
Gross ignorance about the reasoning behind the rules>>
That you cannot understand that is a truism is only your
own lack of understanding.
But it isn't a truism, it is just a stupid lie that ignores that almost everything done with programs is via an "encoding" for the input.
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of computation would tell you that.
Because no Turing machine can take a directly executed>>
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
*Here is the Linz proof corrected to account for that*
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*adapted from bottom of page 319*
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches
its simulated final halt state of ⟨Ĥ.qn⟩
>
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly
reach its simulated final halt state of ⟨Ĥ.qn⟩
>
>
Which is just an admission of your lying strawman, as the question is NOT about the (partial) simulation done by your H / embedded_H, but about the direct execution of the input H^ (H^) as that is what the input to H is encoding.
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Turing machine as an input, directly executed Turing
machines have always been outside of the domain of every
Turing machine based decider.
"the direct execution of the input H^ (H^)" has always beenBut it hasn't been, and thus you are just wrong.
out-of-scope for every Turing machine based halt decider.
That no one bothered to notice this ever before
*DOES NOT MAKE ME WRONG*
An actual rebuttal requires proving that Turing machinesBut the Turing Machine that is directly executed CAN be represented by a finite string,.
can take directly executing Turing machines (that are not
finite string encodings) as inputs.
Alan had a hard time on this because a directly executedBecause that isn't a valid operation, just your insanity.
machine never had to be divided from the simulation of
a finite string encoding before I created the idea of a
simulating halt decider.
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