With both `augment` and `foldr` from GHC.Base: augment :: (forall b. (a -> b -> b) -> b -> b) -> [a] -> [a] augment g = g (:) foldr :: (a -> b -> b) -> b -> [a] -> b ... The function: \xs -> augment (g xs) where g xs c z = foldr c z xs -- pointfree: g == flip (flip . foldr) typechecks (it is in fact just (++)) λ> let g xs c z = foldr c z xs λ> :t \xs -> augment (g xs) \xs -> augment (g xs) :: [a] -> [a] -> [a] Is it possible to explicitly type annotate some `g'` so that the function: (\xs -> (augment . g') xs) :: [a] -> [a] -> [a] is also just (++). That is, basically `g'` is the same as g, but with an explicit type signature that makes the type checker accept the eta-reduced composition? A completely naive attempt, with no type annotations yields: λ> let g xs c z = foldr c z xs λ> :t g g :: [a] -> (a -> b -> b) -> b -> b λ> :t (augment . g) <interactive>:1:2: error: • Couldn't match type: forall b. (a1 -> b -> b) -> b -> b with: (a -> b0 -> b0) -> b0 -> b0 Expected: ((a -> b0 -> b0) -> b0 -> b0) -> [a1] -> [a1] Actual: (forall b. (a1 -> b -> b) -> b -> b) -> [a1] -> [a1] • In the first argument of ‘(.)’, namely ‘augment’ In the expression: augment . g I haven't found any variant type signatures for "g" that make this go. I think this may require impredicative polymorphism, and so can't be done until perhaps QuickLook lands (don't know whether that'll be enough). -- Viktor.