On Mon, Jan 23, 2012 at 09:06:52PM -0800, Ryan Ingram wrote: > On Mon, Jan 23, 2012 at 8:05 PM, Daniel Fischer < > daniel.is.fisc...@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.
This is in fact precisely the "free theorem" you get from the parametricity of f. Parametricity means that f must act "uniformly" for all x -- which is an intuitive way of saying that f really is a natural transformation. -Brent _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe