Re: D correctly simulated by H cannot possibly halt --- templates and infinite sets

On 5/29/2024 8:02 PM, Richard Damon wrote:

On 5/29/24 8:53 PM, olcott wrote:

On 5/29/2024 7:47 PM, Richard Damon wrote:

On 5/29/24 8:21 PM, olcott wrote:

On 5/29/2024 7:09 PM, Richard Damon wrote:

On 5/29/24 8:01 PM, olcott wrote:

On 5/29/2024 6:47 PM, Richard Damon wrote:

*Formalizing the Linz Proof structure*
∃H  ∈ Turing_Machines
∀x  ∈ Turing_Machines_Descriptions
∀y  ∈ Finite_Strings
such that H(x,y) = Halts(x,y)
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And since NO H, can get right the H^ built to contradict IT, that claim is proven false.
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YOU KEEP TRYING TO GET AWAY WITH CHANGING THE SUBJECT
THE ABOVE FORMALIZATION IS CORRECT
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How?
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The above is the question that Linz asks and the he gets
an answer of no, no such H exists.
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So, you now agree with Linz. Good.
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I said that Linz says that. The point is that the Linz
template examines an infinite set of Turing Machine / input
pairs the same way my H/D template references an infinite set
of C function / input pairs.
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 The difference is, In Linz's formulation, each machine is INDIVIDUALLY EVALUTED with its inputs,

*No that is never the case*
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The entire category of every decider/input pair is examined ALL AT ONCE.
No one is dumb enough to look at each element of an infinite set
one at a time because they know this takes literally forever.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer