Re: ZFC solution to incorrect questions: reject them

On 3/12/2024 10:49 PM, Richard Damon wrote:

On 3/12/24 7:41 PM, olcott wrote:

On 3/12/2024 9:17 PM, immibis wrote:

On 13/03/24 02:47, olcott wrote:

On 3/12/2024 8:05 PM, immibis wrote:

On 13/03/24 01:18, olcott wrote:

On 3/12/2024 7:10 PM, immibis wrote:

So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?

Exactly one element of Q differs by writing a 1 instead of a 0.

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That's part of δ but this mistake doesn't matter.
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It wasn't clear whether you were talking about a Turing machine that was somehow identical but gave a different return value, or one that was not identical. Now you have explained it is not identical.
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They are identical except for their return value that is specified
in a single state that is different.
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*This means that they implement the exact same algorithm*

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OK. Well, one of them gets the right answer and one of them gets the wrong answer. What is the confusion?

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The Linz Ĥ.H machine gets the wrong answer on its own
machine description no matter how its Linz H is defined.
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This means that it gets the wrong answer on YES and the
wrong answer on NO.
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 Not quite. It always gets the wrong answer, but only one of them for each quesiton.
 

They all gets the wrong answer on a whole class of questions because
epistemological antinomies are not rejected as semantically invalid input.

For EACH SEPARATE definition of H, and thus H^, we have a different question.
 

They are all the same epistemological antinomy category of question.

Note, the machine H^ isn't DEFINED to just get H^ as an input.
 H^ is defined to get as an input, the description of ANY Turing Machine, and to ask H what that machine applied to its description will do, and then it does the opposite.
 Thus, for every different H we go to test, we get a DIFFERENT H^ machine. and when we look at the question to H (or H^.H) about the description (H^) (H^),
 If H (H^) (H^) goes to qn, then H^ (H^) goes to qn too and halts, so the correct answer would have been to go to qy.
 If H (H^) (H^) goes to qy, then H^ (H^) goes to qy too, and loops, so the correct answer would have been to go to qn.
 So, each case HAS a correct answer, just not the one that H (or H^.H) goes to,
 So yes, which ever one it goes to (and a given machine will only go to one with this input) will be wrong, but the other one would have been right, and an H* machine that answer the opposite of H would have been correct.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer