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!
[EMAIL PROTECTED] 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...
Jerzy Karczmarczuk
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