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

On 3/9/2024 9:36 AM, immibis wrote:

On 9/03/24 16:12, olcott wrote:

On 3/9/2024 7:47 AM, immibis wrote:

On 8/03/24 22:34, olcott wrote:

And since H^ can "lie" to that embedded H^.H about what its description is, that H can't tell that it is part of an H^ computation that is simulating an H^ computation.

>
That subject must be postponed until after the Olcott refutation
of the exact Linz proof is either fully accepted by three people
or actual errors or gaps are found that cannot be addressed or
corrected.

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It's accepted that the Linz proof doen't work on Olcott machines because the Linz proof is designed for Turing machines. But you can't refute the Linz-immibis proof designed for Olcott machines, where H is lied to about its own description.

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I am not sure what Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do except halt or fail to halt
and H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> could see that.
>

 I am not sure what Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do except for exactly the same thing that H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do.

It is easily proven that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort the simulation of
its input and H ⟨Ĥ⟩ ⟨Ĥ⟩ need not abort the simulation of its input.
Everyone has agreed to this yet no one besides me has ever been
able to account for this. It is like they are saying yes this
is a true fact yet in theory it is not true because Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
and H ⟨Ĥ⟩ ⟨Ĥ⟩ have the same inputs their behavior must be the same.
It is a verified fact that their behavior is not the same and
verified facts supersede theories to the contrary.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer