Re: A simulating halt decider applied to the The Peter Linz Turing Machine description ⟨Ĥ⟩

On 5/26/2024 10:30 PM, Richard Damon wrote:

On 5/26/24 11:17 PM, olcott wrote:

On 5/26/2024 10:05 PM, Richard Damon wrote:

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:

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i.e. when we assume it is something it isn't, i.e we LIE to ourselves.
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If you assume embedded_H is something it isn't,

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Not at all.
*It looks like you may be utterly clueless about what-if scenarios*

You can only ask what-ifs about things that are possible.
 

>
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

 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^).
 

*Great this is a step of progress*
This conclusively proves that ⟨Ĥ⟩ will not reach ⟨Ĥ.qn⟩ is less than
an infinite number of steps. A decider is not allowed to simulate
an infinite number of steps.

That just proves that you are just a pathological liar.

*You have a clownish lack of professionalism that academicians*
*coming into these posts for the first would find quite pathetic*
Several have called you out on this already.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer