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

On 6/3/2024 3:07 AM, Mikko wrote:

On 2024-06-02 13:07:25 +0000, olcott said:
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On 6/2/2024 2:36 AM, Mikko wrote:

On 2024-06-01 14:37:01 +0000, olcott said:
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On 6/1/2024 2:52 AM, Mikko wrote:

On 2024-05-31 15:35:18 +0000, olcott said:

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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

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Of those two lines one is false.
As embedded_H is a copy of H both lines imply that H is not a halt decider.
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*Formalizing the Linz Proof structure*
∃H  ∈ Turing_Machines
∀x  ∈ Turing_Machine_Descriptions
∀y  ∈ Finite_Strings
such that H(x,y) = Halts(x,y)

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As already noted, the above is not a part of a proof structure.
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Unless and until you provide reasoning to back that up it counts
as if you said nothing about it.

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If there are no more questions about the details of the reasoning
we may assume that the presiented reasoning is sufficieant.
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The above <is> the structure of his proof your empty assertion utterly bereft of any supporting (EAUBoaSR)) reasoning counts for zilch.

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Those how know what "structure" means can see that it is not a structure.
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Linz claims that of every Turing Machine there are none that solve the halting problem.

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And proves the claim.
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∃!H  ∈ Turing_Machines  (What Richard was saying)
would say that there does not exist exactly one Turing Machine that
solves the halting problem thus fails if there are more than one.

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Irrelevant as that is not what Linz' says.
And you should not lie about Rchard.
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 Richard was confused bout this. Linz words were confusing
unless one had read many other proofs one might be confused.