Hi, Am Donnerstag, den 18.01.2007, 15:05 +0100 schrieb Ulf Norell: > _|_ . id = \x -> _|_ ≠ _|_
Isn’t _|_ = \x -> _|_? Or better stated: For your first _|_ to be used in (.), it has to take an argument, therefore it is \x -> _|_. (.) :: (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) Joachim -- Joachim Breitner e-Mail: [EMAIL PROTECTED] Homepage: http://www.joachim-breitner.de ICQ#: 74513189 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe