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On 5/26/24 11:17 PM, olcott wrote:*Great this is a step of progress*On 5/26/2024 10:05 PM, Richard Damon wrote:You can only ask what-ifs about things that are possible.On 5/26/24 10:43 PM, olcott wrote:>On 5/26/2024 9:06 PM, olcott wrote:>When Ĥ is applied to ⟨Ĥ⟩>
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
Ĥ copies its own Turing machine description: ⟨Ĥ⟩
then invokes embedded_H that simulates ⟨Ĥ⟩ with ⟨Ĥ⟩ as input.
>
It is an easily verified fact that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by
embedded_H cannot possibly reach its own simulated final state of
⟨Ĥ.qn⟩ in any finite sequence of steps.
*To other reviewers that are not dishonest*
The complete proof of the above statement is that when we hypothesize
that embedded_H is a UTM we can see that:
i.e. when we assume it is something it isn't, i.e we LIE to ourselves.
>
If you assume embedded_H is something it isn't,
Not at all.
*It looks like you may be utterly clueless about what-if scenarios*
>So, If your H was a UTM, and H^ built on that, then embedded_H would be a UTM and H^ (H^) would be non-halting as would H (H^) (H^).
What-if embedded_H was a UTM would ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated
by embedded_H reach its own simulated final state of ⟨Ĥ.qn⟩ ?
(a) YES
(b) NO
(c) DISHONEST HONEST ATTEMPT TO CHANGE THE SUBJECT
That just proves that you are just a pathological liar.*You have a clownish lack of professionalism that academicians*
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