type Couple a = forall z. (a -> z) -> z

consCouple :: a -> Couple a
consCouple = a -> p -> p a

type Test = forall x. x -> x

t :: Test
t = x -> x

c :: Couple Test
c = consCouple (t :: Test)

main :: IO ()
main = do{
  print "test"
}

Couldn't match type `x0 -> x0' with `forall x. x -> x'
Expected type: Test
  Actual type: x0 -> x0
In the first argument of `consCouple', namely `(t :: Test)'
In the expression: consCouple (t :: Test)
In an equation for `c': c = consCouple (t :: Test)