Re: [Haskell-cafe] A question about "monad laws"

G'day all. Quoting askyle :

If you use this presentation you also need the following law: (a . b) >=> c = (a >=> c) . b

that is, compatibility with ordinary function composition. I like to call this "naturality", since it's instrumental in proving return and bind to be natural transformations.

Define: f >=> g = \x -> f x >>= g fmap f xs = xs >>= return . f Then: fmap f . return = (expand fmap) \x -> (return x >>= return . f) = (defn. of >=>) \x -> (return >=> return . f) x = (left identity) \x -> (return . f) x = return . f Therefore return is natural. Bind (or, equivalently, join) is left as an exercise. Cheers, Andrew Bromage