I tried the free theorem generator (
http://www-ps.iai.uni-bonn.de/cgi-bin/free-theorems-webui.cgi) and it
wouldn't let me use generic functors, but playing with [] and Maybe leads
me to believe that the free theorem for :-> is
forall f :: m :-> n, forall g :: a -> b, g strict and total
fmap g . f = f . fmap g
This implies that the monad laws don't necessarily hold in situations like
"\m -> m >>= const Nothing", which seems wrong to me. The counterexamples (
http://www-ps.iai.uni-bonn.de/cgi-bin/exfind.cgi), however, all rely on
"odd" natural transformations like (\_ -> Just undefined). My guess is
that there is a side condition we can put on f that is implied by the
monoid laws which doesn't require g to be strict or total.
-- ryan
On Mon, Jan 23, 2012 at 10:23 PM, Brent Yorgey
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
functor laws and "monoid"-ish versions of the monad laws, but my
On Mon, Jan 23, 2012 at 09:06:52PM -0800, Ryan Ingram wrote: the 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
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