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

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:

>
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*

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^).
If your H wasn't actually a UTM, and H^ was built on that H, then when you assume embedded_H is a UTM and give it that (H^) (H^) then, since the hypothetical doesn't change the input, then this hypothetical embecced_H (which isn't the embedded_H that H^ uses) applied to the input (H^) (H^) would reach a final state.
You get this sort of paradox when you assume things that are not true.
So, you need to clarify your Hypothetical to see which case it is that you are actually trying to look at.
Are you STIPULATING that you H is actually that UTM that you are hypotosizing your embedded_H to be.
Or, are you just giving the input to embedded_H, which isn't actually a UTM, to a UTM to see what it will do.
The first shows that you H fails to be a decider.
The second shows that you answer is wrong.
Note, in both cases we need H^ to be built on the copy of the ACTUAL H that you are claiming is correct, not a hypothetical embedded_H that defers from that H.
If you hypothosize about changing H^, then the answer you get it just not applicable to the original problem, as different inputs can have different answers.

 *We can't bounce all around this point like a manic person on meth*
*We can't bounce all around this point like a manic person on meth*
*We can't bounce all around this point like a manic person on meth*
 

And you can't keep on assuming falsehoods and claim to get valid answers.
That just proves that you are just a pathological liar.