Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior

On 3/9/24 2:46 PM, olcott wrote:

On 3/9/2024 12:33 PM, olcott wrote:

*Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior*
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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
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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*
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*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.
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*This is a verified fact*
When that occurs then H ⟨Ĥ⟩ ⟨Ĥ⟩ would transition to H.qy from
its own POV.
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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.
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*Verified facts*
(a) It is a verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort the
simulation of its input to prevent its own infinite execution.
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(b) It is a verified fact when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can see that it
must abort its simulation then it would transition to Ĥ.Hqn
to reject this input as non-halting from its own POV.
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(c) It is a verified fact when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot see that it
must abort its simulation then it would transition to Ĥ.Hqy
and loop.
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(d) (b) gives H ⟨Ĥ⟩ ⟨Ĥ⟩ the basis to transition to H.qy.
(e) (c) gives H ⟨Ĥ⟩ ⟨Ĥ⟩ the basis to transition to H.qn.
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Because Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ seem to be identical machines
on identical input that have different behavior we must
somehow explain how they are not identical machines with
identical inputs.

 Because Linz H only contradicts itself that means that it
does not contradict Linz H.

No, Linz H^ contradicts Linz H, not H^, as Linz H^ isn't a decider.

 Thus the only reason why anyone is saying that Linz H cannot get
the right answer is because they assume that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and Linz
H ⟨Ĥ⟩ ⟨Ĥ⟩ are the same computation when I proved otherwise:

Then you did something wrong.

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

But it CAN, as has been shown.
If H waits, so does H^.H, and so does ever version down the line leading us to the infinite depth recursion.
H can get stuck in that because it never sees the way out, or it would never happen.