Subsumption in polymorphic types

Question

In ‘Practical type inference for arbitrary-rank types’, the authors talk about subsumption:

I try to test things in GHCi as I read, but even though g k2 is meant to typecheck, it doesn't when I try with GHC 7.8.3:

λ> :set -XRankNTypes λ> let g :: ((forall b. [b] -> [b]) -> Int) -> Int; g = undefined λ> let k1 :: (forall a. a -> a) -> Int; k1 = undefined λ> let k2 :: ([Int] -> [Int]) -> Int; k2 = undefined λ> :t g k1  <interactive>:1:3: Warning:     Couldn't match type ‘a’ with ‘[a]’       ‘a’ is a rigid type variable bound by           the type forall a1. a1 -> a1 at <interactive>:1:3     Expected type: (forall b. [b] -> [b]) -> Int       Actual type: (forall a. a -> a) -> Int     In the first argument of ‘g’, namely ‘k1’     In the expression: g k1 g k1 :: Int λ> :t g k2  <interactive>:1:3: Warning:     Couldn't match type ‘[Int] -> [Int]’ with ‘forall b. [b] -> [b]’     Expected type: (forall b. [b] -> [b]) -> Int       Actual type: ([Int] -> [Int]) -> Int     In the first argument of ‘g’, namely ‘k2’     In the expression: g k2 g k2 :: Int 

I haven't really gotten to the point where I understand the paper, yet, but still, I worry that I have misunderstood something. Should this typecheck? Are my Haskell types wrong?

Answer 1

The typechecker doesn't know when to apply the subsumption rule.

You can tell it when with the following function.

Prelude> let u :: ((f a -> f a) -> c) -> ((forall b. f b -> f b) -> c); u f n = f n 

This says, given a function from a transformation for a specific type, we can make a function from a natural transformation forall b. f b -> f b.

We can then try it successfully on the second example.

Prelude> :t g (u k2) g (u k2) :: Int 

The first example now gives a more informative error as well.

Prelude> :t g (u k1)     Couldn't match type `forall a. a -> a' with `[a0] -> [a0]'     Expected type: ([a0] -> [a0]) -> Int       Actual type: (forall a. a -> a) -> Int     In the first argument of `u', namely `k1'     In the first argument of `g', namely `(u k1)' 

I don't know if we can write a more general version of u; we'd need a constraint-level notion of less polymorphic to write something like let s :: (a :<: b) => (a -> c) -> (b -> c); s f x = f x