ajb-2 wrote:
If you wanted to prove that bind is natural, you would need to define bind, no?
Yup: bind f = f <=< id -- whence you can say (>>=) = flip bind My point is that (as far as I can see) you cannot prove the properties of bind by only assuming identity and associativity for (<=<). You need some way to relate (<=<) to normal composition. To be more explicit: Given: 1. m :: * -> * 2. return :: a -> m a 3. (<=<) :: (b -> m c) -> (a -> m b) -> (a -> m c) such that: 1. return <=< f === f <=< return === f 2. (f <=< g) <=< h === f <=< (g <=< h) Define: bind f = f <=< id (>>=) = flip bind fmap f = bind (return . f) join = bind id Show the monad laws hold, either for (return,bind), (return,(>>=)) or (fmap,return,join) (in the latter case, there's also showing that fmap is a functor and return and join are natural transformations). If someone can show it can be done without also assuming: 3. (f <=< g) . h === f <=< (g . h) I'll stand corrected. ----- Ariel J. Birnbaum -- View this message in context: http://www.nabble.com/A-question-about-%22monad-laws%22-tp15411587p16065960.... Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.