Have you tried generating a free theorem for :-> ? (I haven't as I'm writing from my phone) 24.01.2012, в 9:06, Ryan Ingram написал(а):

On Mon, Jan 23, 2012 at 8:05 PM, Daniel Fischer wrote: On Tuesday 24 January 2012, 04:39:03, Ryan Ingram wrote:

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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.

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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

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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 wrote:

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

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On Tuesday 24 January 2012, 04:39:03, Ryan Ingram wrote:

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At the end of that paste, I prove the three Haskell monad laws from

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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

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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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I know a bit of category theory, but I'm trying to look at it from a fundamental perspective; I know that I intend (m :-> n) to mean "natural transformation from functor m to functor n", but the type itself (forall x. m x -> n x) doesn't necessarily enforce that. However, the type of natural transformations comes with a free theorem, for example concat :: [[a]] -> [a] has the free theorem forall f :: a -> b, f strict and total, fmap f . concat = concat . fmap (fmap f) The strictness condition is needed; consider broken_concat :: [[a]] -> [a] broken_concat _ = [undefined] f = const () fmap f (broken_concat []) = fmap f [undefined] = [()] broken_concat (fmap (fmap f) []) = broken_concat [] = [undefined] The 'taming selective strictness' version of the free theorem generator[1] allows removing the totality condition on f, but not the strictness condition. But in the case of concat, you can prove a stronger theorem: forall f :: a -> b, fmap f . concat = concat . fmap (fmap f) My suspicion is that this stronger theorem holds for all strict and total natural transformations, but I don't know how to go about proving that suspicion. I'm a hobbyist mathematician and a professional programmer, not the other way around :) I think it's probably easy to prove that the monoid laws imply that mult' and one' are strict and total.

Thus, you can in principle define plenty of natural transformations which do not have the type f :: forall X. M X -> N X.

Can you suggest one? I don't see how you can get around f needing to act at multiple types since it can occur before and after g's fmap: fmap g . f = f . fmap g M A --fmap_M g--> M B | | f_A f_B | | v v N A --fmap_N g--> N B You can have n = m, of course, but that just means f :: M :-> M. -- ryan [1] http://www-ps.iai.uni-bonn.de/cgi-bin/polyseq.cgi Use this term: /\a. let flipappend =(\xs :: [a]. fix (\rec :: [a] -> [a]. \ys :: [a]. case ys of { [] -> xs; y:zs -> y : rec zs })) in let concat = fix (\rec :: [[a]] -> [a]. \xss :: [[a]]. case xss of { [] -> []_{a}; xs:yss -> flipappend (rec yss) xs}) in concat [2] See http://hpaste.org/56903 Summary: -- both of these types have obvious Functor instances newtype (f :. g) x = O (f (g x)) data Id x = Id x class Functor m => Monad m where one' :: Id :-> m mult' :: (m :. m) :-> m -- instances are required to satisfy monoid laws: -- one' is a left/right identity for mult': -- forall x :: m a -- mult' . O . one' . Id $ x = x = mult' . O . fmap (one' . Id) $ x -- mult' is associative: -- forall x :: m (m (m a))). -- mult' . O . mult' . O $ x = mult' . O . fmap (mult' . O) $ x On Thu, Jan 26, 2012 at 9:30 PM, wren ng thornton wrote:

On 1/23/12 10:39 PM, Ryan Ingram wrote:

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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)?

That is the defining property of natural transformations. To prove it for polymorphic functions in Haskell you'll probably want to leverage parametricity.

I assume you don't know category theory, based on other emails in this thread. But the definition of a natural transformation is that it is a family of morphisms/functions { f_X :: M X -> N X | X an object/type } such that for all g :: a -> b we have that f_b . fmap_m g == fmap_n g . f_a

Thus, you can in principle define plenty of natural transformations which do not have the type f :: forall X. M X -> N X. The only requirement is that the family of morphisms obeys that equation. It's nice however that if a function has that type, then it is guaranteed to satisfy the equation (so long as it doesn't break the rules by playing with strictness or other things that make it so Hask isn't actually a category).

-- Live well, ~wren

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