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Old 11-13-2008   #7 (permalink)
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Re: Halting Undecidability proof faulty?

Quote:
Originally Posted by Buffy View Post
Yes you do. The theorem makes no assumption of that the Halting Problem Program must use itself as input.
But that's not a requirement for the proof! The core of the proof is simply that the f() algorithm is different from h() .

For your argument to disprove the theorem, you need to demonstrate why Halting Problem Programs are somehow different than any other algorithm in a fundamental way.

That the operation is "recursive" is of no consequence: an algorithm is an algorithm. There's no special effect on *any* algorithm simply because it takes itself as input. That's quite non-sensical.

In other words:

Yes, it's not only trivial because you've simply assumed that there is such an algorithm, it's more than insufficient in showing any sort of weakness in the proof.

Why are you making assumptions about what a halting machine would do? Isn't that the very charge that you are laying against the proof?

Why should your argument--for which you are providing no evidence or logic at all--be taken at face value when that is the very weakness you're--quite incorrectly--accusing the theorem of doing?

Human speech is like a cracked kettle on which we tap crude rhythms for bears to dance to, while we long to make music that will melt the stars,
Buffy
In a proof by contradiction, you don't have to explicitly state that you are making an assumption. Any logical assumption you make, whether you recognize it or not, can be the cause of the contradiction.

The proof assumes that the halting machine can operate on itself, by operating the machine on itself during the course of the proof and assuming it behaves as it normally would.

It isn't an issue of disproving the "proof". "proofs" make assertions, and this assertion either fails due to a logical oversight, or if you simply define the halting machine in the proof to be able to operate on itself then it proves nothing more than that there can be no halting machine that operates normally on itself. There could still be one that doesn't work on itself.

The fact that my philosophical knowledge alerts me to the fact that it's definition most likely demands that it not be able to operate on itself is not required for the above reasoning.

Last edited by Kriminal99; 11-13-2008 at 09:50 AM..
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