typeclass challenge: having both variadic arguments and results

While writing some Arbitrary instances, I implemented a couple of functions with the following quite mechanical pattern:

type A = Arbitrary -- to cut down on the size of the annotations below
shrink1 :: (A a          ) => (a           -> r) -> (a           -> [r])
shrink2 :: (A a, A b     ) => (a -> b      -> r) -> (a -> b      -> [r])
shrink3 :: (A a, A b, A c) => (a -> b -> c -> r) -> (a -> b -> c -> [r])

shrink1 f a     = [f a'     | a' <- shrink a]
shrink2 f a b   = [f a' b   | a' <- shrink a] ++ [f a b'   | b' <- shrink b]
shrink3 f a b c = [f a' b c | a' <- shrink a] ++ [f a b' c | b' <- shrink b] ++ [f a b c' | c' <- shrink c]

I wrote out these functions by hand up to shrink7, and that seems to be sufficient for my needs. But I can't help but wonder: can this reasonably be automated? Bonus points for a solution that:

allows for shrink0 f = []
generates all the shrinkers
has loads of typeclass hackery, I love that
skips the scary extensions like incoherent/undecidable/overlapping instances
lets me have my cake and eat it, too: doesn't require me to uncurry f when passing it in or curry the application shrinkX f when applying it to a, b, and c

Solution

This compiles, I hope it works:

{-# LANGUAGE TypeFamilies #-}
import Test.QuickCheck

class Shrink t where
  type Inp t :: *
  shrinkn :: Inp t -> t
  (++*) :: [Inp t] -> t -> t

instance Shrink [r] where
  type Inp [r] = r
  shrinkn _ = []
  (++*) = (++) 

instance (Arbitrary a, Shrink s) => Shrink (a -> s) where
  type Inp (a -> s) = a -> Inp s
  shrinkn f a = [ f a' | a' <- shrink a ] ++* shrinkn (f a)
  l ++* f = \b -> map ($ b) l ++* f b

(++*) is only for implementing shrinkn.

Sorry for the relative lack of typeclass hackery. The [r] provides a nice stop condition for the type recursion, so hackery isn't needed.