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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: ChatGPT totally understands exactly how I refuted the
 conventional halting problem proof technique
Date: Sat, 28 Jun 2025 07:56:04 -0500
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On 6/28/2025 6:38 AM, Mikko wrote:
> On 2025-06-27 15:22:50 +0000, olcott said:
> 
>> On 6/26/2025 5:00 AM, Mikko wrote:
>>> On 2025-06-25 15:26:28 +0000, olcott said:
>>>
>>>> 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.
>>>
>>> That is not the main point.
>>
>> It is the *only* reason why
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>> is incorrectly construed as being incorrect.
> 
> It is neither correct nor incorrect. There are no requirements about Ĥ.
> 

The above shows that Ĥ.embedded_H decided not halting.
This is either correct or incorrect depending on the
criterion measure.

If Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ is to report on the behavior
that its inputs specify then transitioning to Ĥ.qn
is correct.

When it is understood that the directly executing
Ĥ applied to ⟨Ĥ⟩ is not in the domain of Ĥ.embedded_H
then the behavior of Ĥ applied to ⟨Ĥ⟩ does not contradict
the reporting of non-halting.

>>> The main point is that Ĥ ⟨Ĥ⟩ halts if
>>> iH ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to Ĥ.qn but not if H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to
>>> Ĥ.qy. Anything said about embedded_H is merely an intermediate
>>> step in the proof of the main point if not totally irrelevant.
>>>
>>>> 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.
>>>
>>> Irrelevant. It can and does take the same input as H and from that
>>> computes the same as H. That is all that is needed for the proof.
>>>
>>>>>> 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 a halt decider requires that a halt decider
>>> correctly predicts whether a direct execution halts
>>
>> That has always been incorrect because no Turing machine
>> can ever take any directly executing Turing machine as
>> its input all of these directly executed Turing machines
>> are outside of the domain of any partial halt decider.
> 
> No, it is not incorrect. It is what the words mean.
> 

Likewise with the definition of a circle as having four
equal length sides. The requirement that a halt decider
report on the behavior of the direct execution of a machine
is contradicted by the fact that no Turing Machine can take
a directly executing machine as its input.

Just because no one has ever noticed this before does not
mean that I am wrong. Partial halt deciders are only held
accountable for *inputs* in their domain. Their own directly
executed selves are not *inputs* in their domain.

>> Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>> is correct because ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by
>> Ĥ.embedded_H cannot possibly reach its own simulated
>> final halt state ⟨Ĥ.qn⟩.
> 
> Correct or incorrect does not apply to Ĥ as there are no requirements.
> 

The requirement is that Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly
determine the halt status that its input specifies.

*Here is the whole Linz proof*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer