Hi Jeremy,
I have seen it said that all Monads are Applicative Functors because you can just do:
instance (Monad f, Applicative f) => Applicative (ReaderT r f) where pure = return (<*>) = ap
However, that instance for ReaderT does not exhibit the properties I was hoping for.
OK, let's calculate. Here are the necessary definitions for the reader monad (not the monad transformer). m >>= k = \ x -> k (m x) x u <*> v = \ x -> (u x) (v x) return a = pure a = \ x -> a So, <*> is the S combinator and pure is the K combinator. u >>= \ f -> v >>= \ a -> return (f a) = { definition of >>= } \ x -> (\ f -> v >>= \ a -> return (f a)) (u x) x = { beta } \ x -> (v >>= \ a -> return ((u x) a)) x = { definition of >>= } \ x -> (\ a -> return ((u x) a)) (v x) x = { beta } \ x -> return ((u x) (v x)) x = { definition of return } \ x -> (u x) (v x) = { definition of <*> } u <*> v Yes, both definitions are, in fact, equal. So, what went wrong in your program? Observe that the first instance declaration can be simplified to instance (Monad f) => Applicative (ReaderT r f) where pure = return (<*>) = ap which is suspicious. Hope this helps, Ralf