Re: ChatGPT totally understands exactly how I refuted the conventional halting problem proof technique

On 6/25/2025 2:21 AM, Mikko wrote:

On 2025-06-24 21:41:37 +0000, olcott said:
 

On 6/24/2025 4:07 PM, joes wrote:

Am Tue, 24 Jun 2025 13:06:22 -0500 schrieb olcott:

On 6/24/2025 12:57 PM, joes wrote:

Am Tue, 24 Jun 2025 12:46:01 -0500 schrieb olcott:
>

It is an easily verified fact that no *input* to any partial halt
decider (PHD) can possibly do the opposite of what its corresponding
PHD decides. In all of the years of all of these proofs no such
*input* was ever presented.

>
You should clarify that you don't even think programs can be passed as
input.
>

It is common knowledge the no Turing Machine can take another directly
executed Turing Machine as an input.

So common that nobody would suggest such. You are the king of strawmen.

>
*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
>
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
   if Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
   if Ĥ applied to ⟨Ĥ⟩ does not halt
>
Ĥ applied to ⟨Ĥ⟩ does not have embedded_H reporting on
the behavior specified by its input ⟨Ĥ⟩ ⟨Ĥ⟩ it has embedded_H
reporting on its own behavior.

 As made clear in the source text, embedded_H does the same as
H when given the same input. The only difference is that if
that same behaviour reaches its qy state then H halts there
but Ĥ runs forever in a tight loop.
 

*You are not getting the main point*
The fact that Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to ⟨Ĥ.qn⟩ is
not contradicted by the fact that Ĥ.embedded_H itself halts.
Because Ĥ.embedded_H cannot possibly take any directly
executing TM as its input that makes the behavior of
Ĥ applied to ⟨Ĥ⟩ outside of the domain of Ĥ.embedded_H.

Since Turing Machines cannot take directly executing
Turing Machines as inputs this means that the directly
executed Ĥ applied to ⟨Ĥ⟩ is not in the domain of
Ĥ.embedded_H, *thus no contradiction is ever formed*

 False. That Turing Machines cannot take directly executing
Turing Machnes as inputs is irrelevant.

Directly executing TM's are not in the domain of any
halt decider.

The definition of
"halting decider" requires that the decider thakes a
description of a Turing machine and a an input to it.

Yes.

 From the construction of Ĥ follows that the domain of Ĥ is
the same as the required domain of a halt decider. As the

Maybe, IDK. What I do know is that
Ĥ applied to ⟨Ĥ⟩ outside of the domain of Ĥ.embedded_H.

proof proves H does not do what a halting decider is
required to do

It is required to take a directly executing TM as input.
   and
It is not allowed to take a directly executing TM as input.

when the input is <Ĥ> <Ĥ>, contradicting
the claim that H is a halting decider.
 

*It never has been doing any such thing*
When Ĥ.embedded_H is a simulating partial halt decider
then Ĥ.embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ specifies recursive
simulation that cannot possibly reach its own simulated
final halt state of ⟨Ĥ.qn⟩.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer