Re: I have just proven the error of all of the halting problem proofs --- Mackenzie

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