ajb-2 wrote:
Define: f >=> g = \x -> f x >>= g
So you're either not taking (>=>) as primitive or you're stating the additional property that there exists (>>=) such that f >=> g === (>>= g) . f (from which you can easily show that (f . g) >=> h === (f >=> h) . g ). A presentation of the monad laws based on (>=>) (I prefer (<=<) since it meshes better with (.) ) should (IMHO) regard (>=>) as primitive and state its needed properties whence one can derive all other formulations (Note that he Identity law then requires the existence of return as another primitive). That done, you define (>>=), fmap and the rest in terms of (<=<) : (>>=) = flip (<=< id) -- I like to put it as (>>= g) = (g <=< id) fmap f = (return . f) <=< id ----- Ariel J. Birnbaum -- View this message in context: http://www.nabble.com/A-question-about-%22monad-laws%22-tp15411587p16045123.... Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.