Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement--

On 3/9/2024 9:33 AM, immibis wrote:

On 9/03/24 16:04, olcott wrote:

On 3/9/2024 7:11 AM, immibis wrote:

On 8/03/24 20:08, olcott wrote:

On 3/8/2024 12:58 PM, immibis wrote:

On 8/03/24 19:12, olcott wrote:

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Unless it is an extra parameter it has no basis for doing this.
If it is an extra parameter then it is no longer the Linz proof.
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So it's impossible to make a Turing machine that writes 12345 onto its tape unless 12345 is a parameter?

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That is a valid point.
It is impossible to get me to talk about that until
we first have full closure that the Linz H correctly
determines the halt status of the Linz Ĥ ⟨Ĥ⟩ when run
in the Olcott master UTM.
>

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When run in the Olcott master UTM as Olcott machines, the Linz proof does not work because it is designed for Turing machines. A different proof works.

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Because Olcott machines are more powerful than Turing Machines they can correctly determine the halt status of this input: H ⟨Ĥ⟩ ⟨Ĥ⟩ <H>.
>

 Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> always calculates the same answer that H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> calculates.
 

No it does not. Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ has some preceding steps that copy its input
and H ⟨Ĥ⟩ ⟨Ĥ⟩ does not have these steps. Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can see all of these
steps.
It turns out that this single difference is why Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort its simulation and H ⟨Ĥ⟩ ⟨Ĥ⟩ need not abort its simulation.
I only created Olcott machines so that it is easy to see how Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
would detect that it must abort its simulation and transition to Ĥ.Hqn.
Turing machine Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ might still be able to do this thus
nullifying your counter-example.
Your reviews are often quite good. Some of Richard's reviews
were very good and some of them are stuck in circular reasoning
anchored in very persistently false assumptions.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer