Hi everyone, I use and love Haskell, but I just have this nagging concern, that maybe someone can help me reason about. If I'm missing something completely obvious here and making the wrong assumptions, please be gentle. :) I'm reading the third (bind associativity) law for monads in this form: m >>= (\x -> k x >>= h) = (m >>= k) >>= h Now considering the definition of liftM2: liftM2 f m1 m2 = m1 >>= (\x1 -> m2 >>= (\x2 -> return (f x1 x2))) Isn't this liftM2 definition in the same form as the LHS of the third law equation, with (\x2 -> return (f x1 x2)) being the h function? Comparing this definition with the third law equation, the equation doesn't work because on the RHS equivalent; the x1 argument would be lost. So, why wasn't I finding a mention of a qualification that states that the third law only applies as long as the function in the h position doesn't reference arguments bound from previous 'binds'? It took going all the way back to Philip Wadler's 1992 paper, 'Monads for functional programming' to find reassurance: "The scope of variable x includes h on the left but excludes h on the right, so this law is valid only when x does not appear free in h." I'm also thinking of the Maybe monad, where Nothing >>= \x -> Just (x + 1) >>= \y -> return (y + 2) evaluates to Nothing after the first monadic bind and doesn't evaluate the rest of the expression. However, Nothing >>= Just . (+ 1) >>= return . (+ 2) should evaluate through the first and second monadic bind, evaluating to Nothing each time, of course. For the Maybe monad, both expressions give the same result, but if there were another monad defined like, data MadMaybe a = Nothing | Perhaps | Just a instance Monad MadMaybe where (Just x) >>= k = k x Nothing >>= _ = Perhaps Perhaps >>= _ = Nothing - then the two previous expressions run in the MadMaybe monad would evaluate to different values. Since the first of the previous expressions evaluates like the LHS of the third law equation above and the second expression evaluates like the RHS, the expressions should be equivalent, but they are not. Does this put into question the third monad law's relevance to Haskell monads, or is it that MadMaybe shouldn't be made a monad because it violates the third law? Thanks for any insight. Jon