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

Op 09.jul.2025 om 15:02 schreef olcott:

On 7/9/2025 3:58 AM, Fred. Zwarts wrote:

Op 08.jul.2025 om 17:17 schreef olcott:

On 7/8/2025 6:18 AM, Richard Damon wrote:

On 7/7/25 10:52 PM, olcott wrote:

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.
>

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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.
>

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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.
>

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That you cannot understand that is a truism is only your
own lack of understanding.

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

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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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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.

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Then evdrything in reality is outside the domain of Turing Machines.
>

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"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*

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But it hasn't been, and thus you are just wrong.
>

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An actual rebuttal requires proving that Turing machines
can take directly executing Turing machines (that are not
finite string encodings) as inputs.

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But the Turing Machine that is directly executed CAN be represented by a finite string,.
>

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In the case of pathological self-reference the
representation has different behavior than the
directly executed machine and halt deciders only
report on the actual behavior actually specified
by their inputs.
>

As usual a false claim, without evidence.

 *If you have enough knowledge then that is self evident*

You always think your dreams are self-evident. You seem to be the only one.

All Turing machine deciders only compute the mapping
from their actual inputs. This entails that they never
compute any mapping from non-inputs.

At least one thing you understand.

 

The evidence is that the input includes the code to abort and halt,

 abort and stop running
*IS NOT THE SAME THING AS*
abort and halt

Another claim without evidence.