On Mon, Jan 23, 2012 at 8:05 PM, Daniel Fischer <daniel.is.fischer@googlemail.com> wrote:

On Tuesday 24 January 2012, 04:39:03, Ryan Ingram wrote:
> At the end of that paste, I prove the three Haskell monad laws from the
> functor laws and "monoid"-ish versions of the monad laws, but my proofs
> all rely on a property of natural transformations that I'm not sure how
> to prove; given
>
> type m :-> n = (forall x. m x -> n x)
> class Functor f where fmap :: forall a b. (a -> b) -> f a -> f b
> -- Functor identity law: fmap id = id
> -- Functor composition law fmap (f . g) = fmap f . fmap g
>
> Given Functors m and n, natural transformation f :: m :-> n, and g :: a
> -> b, how can I prove (f . fmap_m g) = (fmap_n g . f)?

Unless I'm utterly confused, that's (part of) the definition of a natural
transformation (for non-category-theorists).

Alright, let's pretend I know nothing about natural transformations and just have the type declaration

type m :-> n = (forall x. m x -> n x)

And I have

f :: M :-> N

g :: A -> B

instance Functor M -- with proofs of functor laws

instance Functor N -- with proofs of functor laws

How can I prove

fmap g. f :: M A -> N B

f . fmap g :: M A -> N B

I assume I need to make some sort of appeal to the parametricity of M :-> N.

> Is there some
> more fundamental law of natural transformations that I'm not aware of
> that I need to use? Is it possible to write a natural transformation
> in Haskell that violates this law?
>
> -- ryan