Patrick_(orthonormal) comments on Variable Question Fallacies - Less Wrong
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More to the point: (P or ~P) isn't a theorem, it's an axiom. It is (so far as we can tell) consistent with our other axioms and absolutely necessary for many important theorems (any proof by contradictionâ and there are some theorems like Brouwer's Fixed Point Theorem which, IIRC, don't seem to be provable any other way), so we accept a few counterintuitive but consistent consequences like (G or ~G) as the price of doing business. (The Axiom of Choice with the Banach-Tarski Paradox is the same way.)
OK, I've said enough on that tangent.