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From: olcott <polcott333@gmail.com>
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On 6/1/2024 2:52 AM, Mikko wrote:
> On 2024-05-31 15:35:18 +0000, olcott said:

>>
>> When Ĥ is applied to ⟨Ĥ⟩
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> 
> Of those two lines one is false.
> As embedded_H is a copy of H both lines imply that H is not a halt decider.
> 
>> *Formalizing the Linz Proof structure*
>> ∃H  ∈ Turing_Machines
>> ∀x  ∈ Turing_Machine_Descriptions
>> ∀y  ∈ Finite_Strings
>> such that H(x,y) = Halts(x,y)
> 
> As already noted, the above is not a part of a proof structure.
> 

Unless and until you provide reasoning to back that up it counts
as if you said nothing about it.

>> That <is> what Linz is claiming is false.
>> *Here is the same claim with 100% complete specificity*
>> such that H(⟨Ĥ⟩, ⟨Ĥ⟩) != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)
> 
> That does not make sense. Every H such that H(⟨Ĥ⟩, ⟨Ĥ⟩) != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)
> is uninteresting.
> 

*Unless it is proven that there is a fatal flaw in the proof*

>> *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.

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.

embedded_H only allowed to report on the behavior specified by its
finite string input. That behavior never stops running for 1 to ∞ steps
of ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H.

> The only reauirement is that embedded_H has the same transition
> rules as H. Therefore embedded_H reports the same as H, whether
> allowed or not.
> 

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 embedded_H meets its
abort simulation criteria.

The former case of Linz H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ can see that embedded_H
has already aborted its simulation, thus it never reaches its own
abort criteria.

It is only because everyone since 1936 has rejected simulation
OUT-OF-HAND without review that no one ever noticed this before.

>> The RHS is behavior that embedded_H NOT is allowed to report on.
>> The LHS and the RHS specify different behaviors.
> 
> You have not shown anything with behaviours as LHS and RHS.
> 
>> Please to not reply here instead reply at the end of my proof
>> after all of the steps have been presented.
> 
> Not a reasonable request. Correctness of a step of proof does not
> depend on what follows. If one step is erroneous the rest is
> irrelevant.
> 

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer