It was I that he quoted, and now I am totally flummoxed: thm1 :: (a -> a) -> a thm1 f = let x = f x in x
thm1 (const 1) 1
I *thought* that the theorem ((a => a) => a) is not derivable (after all, 0^(0^0) = 0^1 = 0), but it appears somehow that thm1 is a proof of its type. Help, I just unlearned everything I ever thought I new about the C-H correspondence! jerzy.karczmarczuk@info.unicaen.fr wrote:
Tim Newsham quotes somebody /I didn't follow this thread!/:
I assume you mean then that it is a valid proof because it halts for *some* argument? Suppose I have: thm1 :: (a -> a) -> a thm1 f = let x = f x in x There is no f for which (thm1 f) halts (for the simple reason that _|_ is the only value in every type), so thm1 is not a valid theorem. ================= Since, as I said, I didn't follow you, it would be indecent of me to try to be clever now, but the statement above (that there is no f for which thm1 halts) is false *IN HASKELL*. Try f = const 1. Unless I missed some important point elsewhere...
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