Sujet : Re: My reviewers think that halt deciders must report on the behavior of their caller
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory sci.logicDate : 18. Jul 2025, 00:25:09
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <a5759c5a7f464c6bc1fae7040b10bc3327568dda@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
User-Agent : Mozilla Thunderbird
On 7/17/25 9:31 AM, olcott wrote:
On 7/17/2025 2:47 AM, Mikko wrote:
On 2025-07-16 15:15:53 +0000, olcott said:
>
On 7/16/2025 3:55 AM, Mikko wrote:
>
If there were an error in the proof you would quote the erronoeus inference.
>
The error is the requirement that a halt decider
reports on the direct execution of a machine that
is not an input.
>
That was stimpluated before asking the question that the proof answers.
>
No Turing Machine decider can ever report on the
behavior of anything that is not an input encoded
as a finite string.
But it CAN for one that has, and EVERY actual Turing Machine can be, just like any program can.
*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
Ĥ is not a finite string input to Ĥ.embedded_H
⟨Ĥ⟩ ⟨Ĥ⟩ are finite string inputs to Ĥ.embedded_H
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⟩, and
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⟩.
When Ĥ is applied to ⟨Ĥ⟩ and embedded_H is a
simulating partial halt decider
(a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation until
embedded_H sees the repeating pattern and transitions to Ĥ.qn.
And since in you (g) says that embedded_H, that was started in (b) and running in (c), WILL see the same "repeating pattern" that H saw, and abort, and return to the H^ from (b) and that will halt, says that H^ is halting.
Either embedded_H does this and make H^ halting, or H never does this and doesn't halt to answer.
Thus, you demonstrate that H is just wrong.
Sorry, you just admitted that you arguement is a lie.