Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --closure yet?--

On 3/8/2024 10:18 AM, immibis wrote:

On 8/03/24 17:15, olcott wrote:

On 3/8/2024 10:09 AM, immibis wrote:

On 8/03/24 07:30, olcott wrote:

Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> immediately detects that is about to simulate a
copy of itself with a copy of its own input thus immediately
detects recursive simulation just like H(D,D).
>
H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> immediately detects that is NOT about to simulate a
copy of itself with a copy of its own input thus immediately
rejects recursive simulation just like H1(D,D).

>
And what about Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> ?
>

>
Let's first determine that Olcott machines refute
the uncomputability of the Peter Linz ⟨Ĥ⟩ ⟨Ĥ⟩.

 Peter Linz Ĥ ⟨Ĥ⟩ is a Turing machine. The exact same proof does not work for Olcott machines because they are different. A different proof proves that Olcott machines cannot compute the halting problem. I have already told you this proof (but not in great detail).

I am working on the computability of the halting problem
(the exact same TMD / input pairs) by a slightly augmented
notion of Turing machines as elaborated below:
Olcott machines are entirely comprised of a UTM + TMD and one
extra step that any UTM could perform, append the TMD to the end
of its own tape.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> ⊢* Ĥ.Hqn
therefore
H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> ⊢* H.qy
Olcott machines take Turing Machine Descriptions.
Thus an Olcott machine does compute halting for
the Linz H ⟨Ĥ⟩ ⟨Ĥ⟩.

>
Because my cancer came back in less than 24 months that
means I don't have time for endless divergence.
>

Then why do you keep doing it?

You are the one diverging here.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer