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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic,comp.ai.philosophy
Subject: Re: I have just proven the error of all of the halting problem proofs
 --- Mackenzie
Date: Sat, 26 Jul 2025 17:49:48 -0500
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On 7/26/2025 2:58 PM, olcott wrote:
> On 7/26/2025 2:52 PM, Mr Flibble wrote:
>> On Sat, 26 Jul 2025 14:26:27 -0500, olcott wrote:
>>
>>> On 7/26/2025 1:30 PM, Alan Mackenzie wrote:
>>>> In comp.theory olcott <polcott333@gmail.com> wrote:
>>>>
>>>>> The error of all of the halting problem proofs is that they require a
>>>>> Turing machine halt decider to report on the behavior of a directly
>>>>> executed Turing machine.
>>>>
>>>>> It is common knowledge that no Turing machine decider can take another
>>>>> directly executing Turing machine as an input, thus the above
>>>>> requirement is not precisely correct.
>>>>
>>>>> When we correct the error of this incorrect requirement it becomes a
>>>>> Turing machine decider indirectly reports on the behavior of a
>>>>> directly executing Turing machine through the proxy of a finite string
>>>>> description of this machine.
>>>>
>>>>> Now I have proven and corrected the error of all of the halting
>>>>> problem proofs.
>>>>
>>>> No you haven't, the subject matter is too far beyond your intellectual
>>>> capacity.
>>>>
>>>>
>>> It only seems to you that I lack understanding because you are so sure
>>> that I must be wrong that you make sure to totally ignore the subtle
>>> nuances of meaning that proves I am correct.
>>>
>>> No Turing machine based (at least partial) halt decider can possibly
>>> *directly* report on the behavior of any directly executing Turing
>>> machine.  The best that any of them can possibly do is indirectly report
>>> on this behavior through the proxy of a finite string machine
>>> description.
>>
>> Partial decidability is not a hard problem.
>>
>> /Flibble
> 
> My point is that all of the halting problem proofs
> are wrong when they require a Turing machine decider
> H to report on the behavior of machine M on input i
> because machine M is not in the domain of any Turing
> machine decider. Only finite strings such as ⟨M⟩ the
> Turing machine description of machine M are its
> domain.
> 

Definition of Turing Machine Ĥ
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
   if Ĥ applied to ⟨Ĥ⟩ halts, and        // incorrect requirement
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
   if Ĥ applied to ⟨Ĥ⟩ does not halt.    // incorrect requirement

(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⟩ ...

The fact that the correctly simulated input
specifies recursive simulation prevents the
simulated ⟨Ĥ⟩ from ever reaching its simulated
final halt state of ⟨Ĥ.qn⟩, thus specifies non-termination.

This is not contradicted by the fact that
Ĥ applied to ⟨Ĥ⟩ halts because Ĥ is outside of
the domain of every Turing machine computed function.

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