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

On 5/26/24 11:47 PM, olcott wrote:

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
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  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.

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*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.

First, it doesn't "Prove" it, as we are only having an informal discussion, you could go through the effort to actually prove it, and you would see that this result ONLY occurs if H (and embedded_H) are ACTUALLY UTMs, and as shown, such versions of H are not deciders and thus fail.
Note, (H^) doesn't have a property of reaching any state. That is a property of whatever processes the input.
It should be noted that if your H is of the form that you actually use, that which will abort the simulation and go to qn, then we can say that (H^) (H^) when given to an actual correct simulator (aka a real UTM), then it WILL reach that final state of H^,qn.
The fact that the PARTIAL simulation by any of those H doesn't reach there for the particular H^ built on them doesn't actually mean anything.

 

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.
 

And you have an even more clownish lask of understanding of what you talk about.
I think academicians will be more turned off by your utter disregard for the truth and the meaning of the words will understand my demeaner.
I will point out that YOU started it, and you have even stated you have been wrong about such statements at times, because you don't actually care if you have proof of it, but that you presume others are wrong until they can convince you otherwise. THAT is not a valid posture.