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

On 6/2/2024 2:36 AM, Mikko wrote:

On 2024-06-01 14:37:01 +0000, olcott said:
 

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.

 If there are no more questions about the details of the reasoning
we may assume that the presiented reasoning is sufficieant.
 

The above <is> the structure of his proof your empty assertion utterly bereft of any supporting (EAUBoaSR)) reasoning counts for zilch.
Linz claims that of every Turing Machine there are none that solve the halting problem.
∃!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.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer