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

On 5/29/24 9:55 PM, olcott wrote:

On 5/29/2024 8:25 PM, Richard Damon wrote:

On 5/29/24 9:12 PM, olcott wrote:

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,

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*No that is never the case*

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Of course it is.
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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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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.
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Why do you say that?
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How do you run ALL the machines at once?
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 When the category is examined all at once then there is no need
to look at each individual element.

So, which one or ones gave the correct answer for their input?
Since for EVERY H^ (H^) based on an H that goes to its state qn when it is  applied to (H^) (H^), that H^ (H^) will also go to qn and halt, and thus that H was wrong.

 

Maybe you can think of all of them running INDIVIDUALLY in parrallel, but each machine does what that machine does with the input that THAT machine was given.
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You just don't understand what you are talking about.
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 Existential quantification always looks at all the elements
of an infinite set.
 

Ok, and for EVERY H(H^,H^) that goes to qn creates an H^ (H^) that halts, and thus ALL are wrong, so there doesn't exist ANY that are right.
Actually "Existential" qualifications look for just at least one example, they don't need to find all of them. The issue is that this problem has both an existential and universal qualifications. Since each H needs to handle ALL inputs, that includes the "pathological" input H^ that makes it wrong and thus no H to meet the requirements exist.
You are talking Turing machine and the actual Halting problem so none of this H can't simulate its input to a final state, the ACTUAL QUESTION in you description said "Halts" which is SPECIFICAL about the actual behavior of the actual specific machine given to it.
So, you are CATEGORIALLY incorrect.