Re: ZFC solution to incorrect questions: reject them --Gödel--

On 3/12/24 1:11 PM, olcott wrote:

On 3/12/2024 2:40 PM, Richard Damon wrote:

On 3/12/24 12:02 PM, olcott wrote:

On 3/12/2024 1:31 PM, immibis wrote:

On 12/03/24 19:12, olcott wrote:

∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions  |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
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There is some input TMD to every H such that
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

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And it can be a different TMD to each H.
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When we disallow decider/input pairs that are incorrect
questions where both YES and NO are the wrong answer

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Once we understand that either YES or NO is the right answer, the whole rebuttal is tossed out as invalid and incorrect.
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Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩

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No, because a given H will only go to one of the answers. THAT will be wrong, and the other one right.
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 ∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions  |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
 Not exactly. A pair of otherwise identical machines that
(that are contained within the above specified set)
only differ by return value will both be wrong on the
same pathological input.

You mean a pair of DIFFERENT machines. Any difference is different.
And no, they won't be fooled by the SAME pathological input.
Lets call the machine Ha and Hb, and Hb will ALWAYS generate the opposite answer for A for all inputs.
We build the pathological input Ha^ built on it calling Ha and doing the opposite.
If Ha (Ha^) (Ha^) goes to qn, then Ha^ (Ha^) goes to qn and Halts, so Ha was wrong, but since Hb says the opposite, Hb (Ha^) (Ha^) by that definition goes to qy, and is right,
If Ha (Ha^) (Ha^) goes to qy, then Ha^ (Ha^) goes to qy and loops, so Ha was wrong, but since Hb says the opposite, Hb (Ha^) (Ha^) goes to qn and is right again.
Note, you said SAME INPUT, which is (Ha^) (Ha^) in both cases,
Changing the decider for the same input doesn't change the input to the new decider, because that isn't how the input works.
So, that infact PROVES that some decider can get the answer rights.
Note though, the clause doesn't say that there is one input that makes ALL deciders wrong, it says that for all decider, each one has a (potentially different) input that it gets wrong.
So, for Hb, we can make an Hb^ that Hb (Hb^) (Hb^) will get wrong, but Ha will get right.

 When you say that the opposite answer is correct you
are sneaking outside of the above specified set.

Nope. You don't know what that says.

 

Remember, you above statement was built on the ASSUMPTION that a correct H existed, and thus the contradiction you see just says that no such H exists, not that the original question was incorrect.
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