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From: olcott <polcott2@gmail.com>
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Date: Sat, 9 Mar 2024 15:55:22 -0600
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On 3/9/2024 3:17 PM, Richard Damon wrote:
> On 3/9/24 10:33 AM, olcott wrote:
>> *Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior*
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
> 
> Specifications, not actual behavior until the existance of such an H is 
> shown.
> 
> IF taken as actual behavior, then it is conditional on such an H existing.
> 
>>
>> Execution trace of Ĥ applied to ⟨Ĥ⟩
>> (a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
>> (b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
>> (c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
>> *This proves that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort its simulation*
> 
> It NEEDS to in order to meet its specification
> 
> It DOESN'T unless its algorithm says it does,
> 
> If it just fails to answer, then it has failed to be a correct Halt 
> Decider.
> 
> The fact that you reach this conflict in actions, is the reason Halt 
> Deciding is uncomputable.
> 
>>
>> *This is a verified fact*
>> When simulating halt deciders always report on the behavior of
>> their simulated input from their own POV then when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
>> transitions to Ĥ.Hqn it is correct from its own POV.
> 
> In other words, you are admitting to changing the question, and thus 
> LYING that you are working on the actual original problem.
> 
>>
>> *This is a verified fact*
>> When that occurs then H ⟨Ĥ⟩ ⟨Ĥ⟩ would transition to H.qy from
>> its own POV.
> 
> Which just means you are LYING that this apply to an actual Halt Decider 
> per the Halting Theory,
> 
>>
>> When Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ report on the basis of their own
>> POV then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ reports incorrectly about the behavior of
>> Ĥ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ reports the behavior of Ĥ ⟨Ĥ⟩ correctly.
> 
> And thus you are admitting that the H in H^.H is WRONG and thus H is not 
> a correct Halt Decider, because it gets some cases wrong.

It is a verified fact that when H ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ meet this 
criteria that H gets the right answer and Ĥ.H gets the wrong answer.

What it not a verified fact is whether or not Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can meet
this criteria as a Turing machine.

If it cannot meet this criteria as a Turing machine then it is
still that case that Ĥ ⟨Ĥ⟩ either halts or fails to halt.

It may fail to halt by looping without ever transitioning to
Ĥ.Hqy or Ĥ.Hqn. I see no reason why H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot see this.

I generally agree that a pair of identical machines
must have the same behavior on the same input.

This may not apply when these machines having identical
states and identical inputs:

(a) Are out-of-sync by a whole execution trace or

(b) When one of the machines is embedded within another machine
that would cause this embedded machine to have recursive
simulation that the non-embedded machine cannot possibly have.

*I think that the actual difference is the latter case because*
*we have the exact same issue when the infinite loop is removed*

Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly get stuck in recursive simulation and
H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly get stuck in recursive simulation.

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