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

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.
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)
Linz H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ derives a different result than
embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩.
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.
The input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H correctly simulated by adapted UTM
embedded_H remains stuck in recursive simulation until embedded_H
recognizes this and stops simulating its input.
Introduction to the Theory of Computation, by Michael Sipser
https://www.amazon.com/Introduction-Theory-Computation-Michael-Sipser/dp/113318779X/
On 10/13/2022 11:29:23 AM
MIT Professor Michael Sipser agreed that these verbatim words are correct
(He has neither reviewed nor agreed to anything else in this paper)
<Professor Sipser agreed>
If simulating halt decider H correctly simulates its input D until H correctly determines that its simulated D would never stop running unless aborted then
H can abort its simulation of D and correctly report that D specifies a non-halting sequence of configurations.
</Professor Sipser agreed>
Linz H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ can see that embedded_H has already aborted
its simulation, thus it never reaches its own abort criteria.
Thus embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ meets its abort status criteria and
H ⟨Ĥ⟩ ⟨Ĥ⟩ NEVER meets its abort status criteria, thus proving
that these are two different sequences of configurations.
It is only because everyone since 1936 has rejected simulation
OUT-OF-HAND without review that no one ever noticed this before.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer