Re: Working out the details of the steps of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> ⊢* Ĥ.Hqn

On 9/03/24 17:53, olcott wrote:

On 3/9/2024 10:21 AM, immibis wrote:

On 9/03/24 17:00, olcott wrote:

On 3/9/2024 1:05 AM, Richard Damon wrote:

On 3/8/24 10:40 PM, olcott wrote:

On 3/9/2024 12:25 AM, Richard Damon wrote:

On 3/8/24 9:55 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
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Expecting Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to correctly report on the behavior of
Ĥ ⟨Ĥ⟩ is a little nuts because Ĥ contradicts Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩.

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But that is the job it signed up for when it tried to call itself a Halt Decider.
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There is no correct answer that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly
provide that corresponds to the behavior of this Ĥ ⟨Ĥ⟩.
Therefore it must have a basis for its wrong answer.
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WHY?
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What is the actual grounds for that statement?

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Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot just sit around scratching its head it
must do something. As soon as we have the basis for what
it will do, then H ⟨Ĥ⟩ ⟨Ĥ⟩ has its basis to correctly
decide halting.

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H ⟨A⟩ ⟨B⟩ cannot just sit around scratching its head, it must do something. As soon as we have the basis for what H will do, then we have a basis to construct an Ĥ where H ⟨Ĥ⟩ ⟨Ĥ⟩ is wrong.
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 Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can fool itself yet cannot possibly fool an external H ⟨Ĥ⟩ ⟨Ĥ⟩.
 

If H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to state qn then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ also transitions to state H.qn. If it does not do this then you did not follow the instructions properly when you wrote Ĥ.