ajb-2 wrote:
One thing that may help is that if you can prove that fmap is sane: fmap (f . g) = fmap f . fmap g
I get stuck after expanding the rhs into: ((return . f) <=< id) . ((return . g) <=< id) ajb-2 wrote:
then the naturality of return is precisely its free theorem, and ditto for bind.
Care to develop? IIRC the free theorems have a certain parametericity requirement (which probably holds in all interesting cases, but still sounds like an additional assumption). I'm not too familiar with these, so a link would be appreciated =) ajb-2 wrote:
So perhaps this law: (f <=< g) . h === f <=< (g . h) is actually the fmap law in disguise?
Could be. Maybe the fmap law is this one in disguise ;) ----- Ariel J. Birnbaum -- View this message in context: http://www.nabble.com/A-question-about-%22monad-laws%22-tp15411587p16069396.... Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.