On 18/01/07, Joachim Breitner <[EMAIL PROTECTED]> wrote:
(.) :: (b -> c) -> (a -> b) -> a -> c
id :: a -> a
therefore b = a
therefore _|_ :: a -> c
(This is mostly rough guesswork, I might be totally wrong)
That much is right, but remember that just because _|_ has type a -> c
doesn't mean it takes a parameter. Bottom can take any type, and I
don't think _|_ == \x -> _|_.
--
-David House, [EMAIL PROTECTED]
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