Olcott admits to being a liar!

On 5/12/24 11:58 AM, olcott wrote:

On 5/12/2024 10:21 AM, Mikko wrote:

On 2024-05-12 11:34:17 +0000, Richard Damon said:
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On 5/12/24 5:19 AM, Mikko wrote:

On 2024-05-11 16:26:30 +0000, olcott said:
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I am working on providing an academic quality definition of this
term.

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The definition in Wikipedia is good enough.
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I think he means, he is working on a definition that redefines the field to allow him to claim what he wants.

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Here one can claim whatever one wants anysay.
In if one wants to present ones claims on some significant forum then
it is better to stick to usual definitions as much as possible.
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Sort of like his new definition of H as an "unconventional" machine that some how both returns an answer but also keeps on running.

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There are systems where that is possible but unsolvable problems are
unsolvable even in those systems.
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 When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
 embedded_H is an actual UTM that has the extra feature of watching
every state transition of its simulated input so that it detects
the non-halting behavior pattern that we can all see.
 Execution trace of  Ĥ applied to ⟨Ĥ⟩
(a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to embedded_H
(b) embedded_H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
 Simulation invariant: ⟨Ĥ⟩ correctly simulated by embedded_H never
reaches its own simulated final state of ⟨Ĥ.qn⟩.
 (1) The directly executed embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to its own
internal state of embedded_H.qn indicating that the correct and
complete simulation of its input never halts. It never reaches Ĥ.qn.
 (2) embedded_H continues to simulate its input after it transitions
to this internal state.

How? Are you admitting that embedded_H isn't actually the equivalent of a Turing Machine?

 (3) Each simulated embedded_H does the same thing.

In other words, you have LIED that embedded_H is just your name for H embedded into H^, and thus your H^ isn't Linz's so it means nothing.

 embedded_H is not a halt decider or a partial halt decider
because all deciders are required to halt. embedded_H is
not even a conventional termination analyzer for this same
reason. None-the-less embedded_H does derive the correct
halt status of its conventional HP input.
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And thus you admit that either H isn't a Halt Decider, or that your H^ isn't the right H^
So, you have just proven that you have wasted your last 20 years on your lies that embedded_H IS the required copy of H as described by Linz's proof.