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On 3/8/2024 10:33 PM, Richard Damon wrote:But the algorith at H^.H IS an instanc eof the decide,r, and must get the answer right.On 3/8/24 8:20 PM, olcott wrote:When the input data gets the wrong answer this allows theOn 3/8/2024 9:41 PM, Richard Damon wrote:>On 3/8/24 7:06 PM, olcott wrote:>Ĥ.H computes the mapping from ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> to Ĥ.Hqn>
therefore
H computes the mapping from ⟨Ĥ⟩ ⟨Ĥ⟩ <H> to H.qy
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I still don't know the detailed steps of how
H computes the mapping from ⟨H⟩ ⟨H⟩ <H> to H.qy
YET WE CAN SEE THAT IT IS CORRECT
No, you see that you have a set of results that makes
Every H needs a criteria to use whether it is contradicted
or not. When it is not contradicted this same criteria
correctly decides halting for its input.
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When H is contradicted it provides a consistent way
that H can provide the wrong answer.
But if you admit that it will get a wrong answer, how can you claim it is a correct Halt Decider, which means it always gets the Right answer.
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actual decider to get the right answer.
And thus you admit that H is WRONG, as all copies of it are the ACTUAL DECIDER.You are just admitting that you are lying about refuting the proof that we can not make an always correct Halt Decider.When the input data gets the wrong answer this allows the
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You don't get to claim to be right, when the machine gives the wrong answer.
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You already know that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must be aborted by Ĥ.H
or it will never halt.
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When every H has the same way to say that, then when it is
contradicted it has something to say and when it is not
contradicted this is the correct halt status of its input.
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No, if ANY copy of H gives the wrong answer to some input, then H is not a Halt Decider.
actual decider to get the right answer.
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Just saying "something" when stuck, doesn't make it correct.
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