Dan Weston writes:
... 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!
No, the world is not so cruel. This is not a theorem, you cannot really derive it by constructing a polymorphic function, I cheated, because your statement was negligent. (const 1) is not a function of type (a->a), the only function *really* of that type is id. People, use Djinn! The slave of the Lamp of Alladenartson will send you to Walhalla if you ask: what ? (a->a)->a Jerzy Karczmarczuk