Hi, Am Montag, den 08.12.2008, 15:59 -0600 schrieb Nathan Bloomfield:
Slightly off topic, but the A^B notation for hom-sets also makes the natural isomorphism we call currying expressable as A^(BxC) = (A^B)^C.
So A^(B+C) = A^B × A^C ? Oh, right, I guess that’s actually true: uncurry either :: (a -> c, b -> c) -> (Either a b -> c) (\f -> (f . Left, f . Right)) :: (Either a b -> c) -> (a -> c, b -> c) It’s always nice to see that I havn’t learned the elementary power calculation rules for nothing :-) Greetings, Joachim PS: For those who prefer Control.Arrow to points: (.Left) &&& (.Right) :: (Either a b -> c) -> (a -> c, b -> c) (found by trial and error :-)) -- Joachim "nomeata" Breitner mail: mail@joachim-breitner.de | ICQ# 74513189 | GPG-Key: 4743206C JID: nomeata@joachim-breitner.de | http://www.joachim-breitner.de/ Debian Developer: nomeata@debian.org