Consider the following type class for free monoids.
class FreeMonoid f where
inj :: a -> f a
univ :: Monoid m => (a -> m) -> f a -> m
inj injects a value into the free monoid, and univ is the universal property of free monoids.
Instances of this class should satisfy the following laws.
- Identity:
univ f . inj = f - Free empty:
univ f mempty = mempty - Free append:
univ f (m <> n) = univ f m <> univ f n
Note that if f is an instance of FreeMonoid then (f a) must be an instance of Monoid. Otherwise, the last two laws don't make sense. So, how do I specify this constraint? Here's what I tried.
class Monoid (f a) => FreeMonoid f where
inj :: a -> f a
univ :: Monoid m => (a -> m) -> f a -> m
Not having this constraint makes it inconvenient to use this class. For example, consider the following function.
mapFreeMonoid :: (FreeMonoid f, Monoid (f b)) => (a -> b) -> f a -> f b
mapFreeMonoid f = univ (inj . f)
Since f is an instance of FreeMonoid, we shouldn't have to specify the Monoid (f b) constraint. Ideally, we should be able to define the above function as follows.
mapFreeMonoid :: FreeMonoid f => (a -> b) -> f a -> f b
mapFreeMonoid f = univ (inj . f)
You can try experimenting with the
QuantifiedConstraintsextension.Your code then compiles without the additional constraint.