It seems you've already figured this out, but here's a quick counterexample:
{-# LANGUAGE ExistentialQuantification, RankNTypes #-} module Box where data Box = forall a. B a
--mapBox :: forall a b. (a -> b) -> Box -> Box -- incorrect type mapBox f (B x) = B (f x)
then:
boxedInt :: Box boxedInt = B (1 :: Int)
f :: [Int] -> Int f = sum
mf :: Box -> Box mapBox f -- this is well-typed according to the specified type of "mapBox" But, mf boxedInt :: Box mf boxedInt = mapBox f boxedInt = mapBox sum (B (1 :: Int)) = B (sum (1 :: Int)) which is not well-typed. The least specific type for MapBox is
mapBox :: forall b. (forall a. (a -> b)) -> Box -> Box
There are non-bottom functions of this type, for example: const (1 :: Int) :: forall a. a -> Int With this type,
ok :: Box ok = mapBox (const 1) (B "hello")
is well-typed.
-- ryan
On 11/30/07, Pablo Nogueira
A question about existential quantification:
Given the existential type:
data Box = forall a. B a
in other words:
-- data Box = B (exists a.a) -- B :: (exists a.a) -> Box
I cannot type-check the function:
mapBox :: forall a b. (a -> b) -> Box -> Box -- :: forall a b. (a -> b) -> (exists a.a) -> (exists a.a) mapBox f (B x) = B (f x)
Nor can I type check:
mapBox :: forall a. (a -> a) -> Box -> Box -- :: forall a. (a -> a) -> (exists a.a) -> (exists a.a) mapBox f (B x) = B (f x)
The compiler tells me that in both functions, when it encounters the expression |B (f x)|, it cannot unify the universally quantified |a| (which generates a rigid type var) with the existentially quatified |a| (which generates a different rigid type var) -- or so I interpret the error message.
However, at first sight |f| is polymorphic so it could be applied to any value, included the value hidden in |Box|.
Of course, this works:
mapBox :: (forall a b. a -> b) -> Box -> Box mapBox f (B x) = B (f x)
Because it's a tautology: given a proof of a -> b for any a or b I can prove Box -> Box. However the only value of type forall a b. a -> b is bottom.
This also type-checks:
mapBox :: (forall a. a -> a) -> Box -> Box mapBox f (B x) = B (f x)
When trying to give an explanation about why one works and the other doesn't, I see that, logically, we have:
forall a. P(a) => Q implies (forall a. P(a)) => Q when a does not occur in Q.
The proof in our scenario is trivial:
p :: (forall a. (a -> a) -> (Box -> Box)) -> (forall a. a -> a) -> Box -> Box p mapBox f b = mapBox f b
However, the converse is not true.
Yet, could somebody tell me the logical reason for the type-checking error? In other words, how does the unification failure translate logically. Should the first mapBox actually type-check?
Isn't the code for mapBox :: forall a. (a -> a) -> Box -> Box encoding the proof:
Assume forall a. a -> a Assume exists a.a unpack the existential, x :: a = T for some T apply f to x, we get (f x) :: a pack into existential, B (f x) :: exists a.a Discharge first assumption Discharge second assumption
The error must be in step 3. Sorry if this is obvious, it's not to me right now. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe