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

On 3/9/2024 12:31 PM, Richard Damon wrote:

On 3/9/24 9:27 AM, olcott wrote:

On 3/9/2024 11:08 AM, Richard Damon wrote:

On 3/9/24 7:12 AM, 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.

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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.
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Because if H simulates "long enough" to see the answer, that "long enough" will turn out to be FOREVER, since H^.H will do the same thing (BY DEFINITION)

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It is a verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort the simulation of
its input and H ⟨Ĥ⟩ ⟨Ĥ⟩ need not abort the simulation of its input
and verified facts supersede all theories to the contrary.
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Nope.

 Please switch over to this new post that addresses all of your
objections in advance.
[Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior]