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

On 3/9/24 9:39 AM, olcott wrote:

On 3/9/2024 11:12 AM, Richard Damon wrote:

On 3/9/24 8:41 AM, olcott wrote:

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.

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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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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.

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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.

>
How?

 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

Comment are SPECIFICATION, not actual behavior until existance of a conforming H is proven

 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
 When every H reports halting on the basis of its own POV then
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ rejects its input and H ⟨Ĥ⟩ ⟨Ĥ⟩ accepts its input.

Except it is supposed to report on the OBJECTIVE requirement of what the machine does, so you are just admitting to LYING and all your work is VOID for the actual Halting Problem.

 This causes Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to incorrectly report on the behavior of
Ĥ ⟨Ĥ⟩ and causes H ⟨Ĥ⟩ ⟨Ĥ⟩ to correctly report on the behavior of Ĥ ⟨Ĥ⟩

Only in the case of your LIE, so you aren't saying anything about the actual Halting Problem

 When we can see that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort the simulation of its
input and H ⟨Ĥ⟩ ⟨Ĥ⟩ need not abort the simulation of its input
then the theory that identical machines on the same input must
have identical behavior is refuted or the fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
is embedded within Ĥ makes Ĥ.H not an identical machine to H.

It "Must" to meet specifications. Not that is WILL do that.
You don't get that until you prove that an H that meed the specifications exists.
YOu are just proving you don't understand how logic works.

 I am going with the latter because the former seems self-evidently
correct. Identical machine on identical input must have the same
behavior. That Ĥ.H is embedded within Ĥ makes Ĥ.H a different
machine than H even though it has the exact same states in its
sub-machine portion.
 

You can't say you "refuted" something because it doesn't allow you to show something you want to show.
That just shows you are nothing but a pathetic ignorant pathological lying idiot that shouldn't be trusted to know anything about anything.