[Haskell-cafe] Rank-2 polymorphism and overloading

Thomas van Noort thomas at cs.ru.nl
Mon Apr 26 13:52:23 EDT 2010

Hello all,

I'm having difficulties understanding rank-2 polymorphism in combination 
with overloading. Consider the following contrived definition:

f :: (forall a . Eq a => a -> a -> Bool) -> Bool
f eq = eq True True

Then, we pass f both an overloaded function and a regular polymorphic 

x :: forall a . Eq => a -> a -> Bool
x = \x y -> x == y

y :: forall a . a -> a -> Bool
y = \x y -> True

g :: (Bool, Bool)
g = (f x, f y)

Could someone explain to me, or point me to some reading material, why g 
is correctly typed?

I understand that x's type is what f expects, but why does y's 
polymorphic type fulfill the overloaded type of f's argument? I can 
imagine that it is justified since f's argument type is more restrictive 
than y's type. However, it requires y to throw away the provided 
dictionary under the hood, which seems counter intuitive to me.


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