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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: ChatGPT totally understands exactly how I refuted the conventional halting problem proof technique
Date: Wed, 25 Jun 2025 10:21:05 +0300
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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.

> 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. The definition of
"halting decider" requires that the decider thakes a
description of a Turing machine and a an input to it.
From the construction of Ĥ follows that the domain of Ĥ is
the same as the required domain of a halt decider. As the
proof proves H does not do what a halting decider is
required to do when the input is <Ĥ> <Ĥ>, contradicting
the claim that H is a halting decider.

-- 
Mikko