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

On 5/31/2024 8:00 AM, Mikko wrote:

On 2024-05-30 13:20:09 +0000, olcott said:
 

On 5/30/2024 2:06 AM, Mikko wrote:

On 2024-05-29 13:13:13 +0000, olcott said:
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On 5/29/2024 3:37 AM, Mikko wrote:

On 2024-05-28 11:34:24 +0000, Richard Damon said:
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On 5/27/24 10:59 PM, olcott wrote:

On 5/27/2024 9:52 PM, Richard Damon wrote:

On 5/27/24 10:41 PM, olcott wrote:

On 5/27/2024 9:23 PM, Richard Damon wrote:

On 5/27/24 10:01 PM, olcott wrote:

On 5/27/2024 8:24 PM, Richard Damon wrote:

On 5/27/24 9:04 PM, olcott wrote:

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I totally do. Can you please write down the
"completely specified state transition/tape operation table."
of this specific (thus uniquely identifiable) machine I would
really like to see it.
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But it was proven that no such machine exists!
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Remember, the proof starts with the hypothetical that such a machine exists. Such a machine WOULD HAVE a completely specified state transition/tape operation table.
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That is not what you said.
 >>>>> There doesn't need to be a unique finite string, but it is a 100%
 >>>>> completely specified state transition/tape operation table.
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"a 100% completely specified state transition/tape operation table"
of a non-existent machine.

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Right, by presuming that you have a Turing Machine, you have a completly specified state transition/tape operation table.
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You may not KNOW what that table is if you don't know what the exact machine is, but you know it exists.

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 >>> But it was proven that no such machine exists!
 > ... but you know it exists.
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 >>> But it was proven that no such machine exists!
 > ... but you know it exists.
>
 >>> But it was proven that no such machine exists!
 > ... but you know it exists.
>
>

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Really, then show that one exists!
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*I am quoting your words. You did contradict yourself*
*I am quoting your words. You did contradict yourself*
*I am quoting your words. You did contradict yourself*
*I am quoting your words. You did contradict yourself*
*I am quoting your words. You did contradict yourself*
*I am quoting your words. You did contradict yourself*
*I am quoting your words. You did contradict yourself*
*I am quoting your words. You did contradict yourself*
*I am quoting your words. You did contradict yourself*
*I am quoting your words. You did contradict yourself*
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Really, where did I say that H exists?
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I said that if a Turing Machine exists, then its transition table does too.
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OK my mistake this time. I did not take into account the full context.
I will go back an read the Linz proof and see if he said anything
about a specific machine.

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Read the DEFINITION of the problem. He talks about "a" machine. Using a singular article means you are working with just one.
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Taking stuff out of context is a common problem with you, when you don't understand something, you just make up what it must mean, and stick to that. That isn't the way to learn.
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None of the proofs ever try to show that there exists one machine that
gets the wrong answer. They are always at least trying to prove that no
machine of the infinite set of machine gets the right answer.
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What I see, is they always start with a prototypical single machine, and show that it gets the answer wrong, and then they use categorical logic to say that we can do this same thing for all of them.

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It is possible to formulate the claim and proof so that H is an universally
quantified variable. But the usual way is apparently equally good for the
target audience.
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*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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That is not a proof structure. That is the counter-hypothesis of Linz' proof.
Also note that both x and y are finite strings.
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The above is what Linz is claiming evaluates to false, he says
there is no such H.

 Yes, and proves the claim.
 

A decider computes the mapping from finite string inputs to
its own accept or reject state.

 An existing decider.
 

A decider does not and cannot compute the mapping from Turing_Machine
inputs to its own accept or reject state.

 An exsiting decider.
 

When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
*Formalizing the Linz Proof structure*
∃H  ∈ Turing_Machines
∀x  ∈ Turing_Machine_Descriptions
∀y  ∈ Finite_Strings
such that H(x,y) = Halts(x,y)
That <is> what Linz is claiming is false.
*Here is the same claim with 100% complete specificity*
such that H(⟨Ĥ⟩, ⟨Ĥ⟩) != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)
*A quick summary of the reasoning provided below*
The LHS is behavior that embedded_H is allowed to report on.
The RHS is behavior that embedded_H NOT is allowed to report on.
The LHS and the RHS specify different behaviors.
Please to not reply here instead reply at the end of my proof
after all of the steps have been presented.
The problem with that claim is the that RHS reports on the behavior
of the Turing Machine that embedded_H is contained within.
No Turing machines are ever allowed to report on the behavior of the
Actual Turing machine that themselves are contained within.
They are not allowed to take actual Turing machines as input they
are only allowed to take finite string Turing machines descriptions
as inputs.
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
When embedded_H <is> a UTM or <is> a halting computation based on a
UTM then the ⟨Ĥ⟩ ⟨Ĥ⟩ input to embedded_H SPECIFIES that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly
simulated by embedded_H cannot possibly reach its own simulated final
state at ⟨Ĥ.qn⟩.
Thus the behavior that embedded_H is allowed to report on is different
than the behavior of Halts(⟨Ĥ⟩, ⟨Ĥ⟩) which is behavior that embedded_H is
NOT allowed to report on.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer