Re: A simulating halt decider applied to the The Peter Linz Turing Machine description ⟨Ĥ⟩ -- key details

On 6/3/2024 3:01 AM, Mikko wrote:

On 2024-06-02 13:21:56 +0000, olcott said:
 

On 6/2/2024 2:42 AM, Mikko wrote:

On 2024-06-01 19:26:55 +0000, olcott said:
>

On 6/1/2024 1:52 PM, joes wrote:

Am Sat, 01 Jun 2024 09:37:01 -0500 schrieb olcott:

On 6/1/2024 2:52 AM, Mikko wrote:

On 2024-05-31 15:35:18 +0000, olcott said:

>

*A quick summary of the reasoning provided below*
The LHS is behavior that embedded_H is allowed to report on.

There is no restrictions on what embedded_H is allowed to report on.

>
embedded_H is only allowed to report on the behavior that its finite
string Turing Machine Description specifies to a UTM.
>
embedded_H <is> a UTM except that it stops simulating and reports
non-halting as soon as it correctly recognizes a non-halting behavior
pattern that is specified by its input.

"Except". So it is not an UTM.
>

When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
(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⟩
(g) goto (d)
>
embedded_H is not allowed to be applied to Ĥ ⟨Ĥ⟩ because inputs can
only be finite strings and Ĥ is not a finite string. This means
that embedded_H is not allowed to report on its own actual behavior.

I can't read that notation. What is H^ and what does it look like?
>

>
*Here is the whole Linz proof*
I simplified the Linz notation at the bottom of page 2 of the proof.
https://www.liarparadox.org/Linz_Proof.pdf

>
You are right, that is a sufficient proof. You may change the presentation
but then you must prove that your presentation is equivalent to Linz'.

>
The proof was removed until Joe could understand what Linz was saying
Here is my actual proof.

 Linz' proof is Linz' proof, not yours.
 

When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

 The above does not say what Linz said and hardly anything else, either.
It is not a sentence. If you remove the second last or the last line
or add some conjunction between them and add a point to the end then
you would have a sentence.
 

(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⟩
(g) goto (d)
>
Linz H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ derives a different result than
embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩.

 Irrelevant as embedded_H is not a part of Linz' proof. The part
of Linz' Ĥ that corresponds to your embedded_H does produce the
same result as Linz H does.
 

https://www.liarparadox.org/Linz_Proof.pdf top of page 3
Linz named his embedded_H "Ĥq0" and has input Wm Wm
because this is a second start state in the same machine
Ben agrees that the Linz notation is wrong.

Turns out that you can prove that the result produced by Linz' H is
different from the result produced by linz' H. This is sufficient to
prove that Linz' H does not exist.
 

This is because the in the latter case embedded_H must determine that
⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H cannot possibly stop running
after 1 to ∞ steps of correct simulation. Thus (as we can all see) embedded_H meets its abort simulation criteria.

 Those criteria are not mentioned in Linz' proof and are therefore
irrelevant to it.
 

They are not mentioned because like everyone else Linz rejects
the notion of a simulating halt decider out-of-hand without review.
When you look for "simulating halt decider" or
"simulating termination analyzer" you only find me.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer