Re: My reviewers think that halt deciders must report on the behavior of their caller

On 7/7/2025 9:24 PM, Richard Damon wrote:

On 7/7/25 7:47 PM, olcott wrote:>>

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.
 

Yes.

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.

>
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.
 

Gross ignorance about the reasoning behind the rules
of computation would tell you that.

>
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
*Here is the Linz proof corrected to account for that*
>
*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.
 

Because no Turing machine can take a directly executed
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 been
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 machines
can take directly executing Turing machines (that are not
finite string encodings) as inputs.
Alan had a hard time on this because a directly executed
machine never had to be divided from the simulation of
a finite string encoding before I created the idea of a
simulating halt decider.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer