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

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Sujet : Re: A simulating halt decider applied to the The Peter Linz Turing Machine description ⟨Ĥ⟩
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory sci.logic
Date : 27. May 2024, 16:25:56
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v3255k$2pkb$2@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33
User-Agent : Mozilla Thunderbird
On 5/27/2024 8:27 AM, Richard Damon wrote:
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
>
  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^).
>
>
*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,
*Sure it does, you just like to deny verified facts*
When ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H cannot possibly reach its
own simulated final state of ⟨Ĥ.qn⟩ and halt in an infinite number of
simulated steps
then it can't possibly reach its own simulated final state of ⟨Ĥ.qn⟩ and halt in less than an infinite (AKA a finite) number of simulated steps.
*When infinity is not enough then less than infinity is also not enough*
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

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