[1] funk f x = f (funk f) x f :: a x :: b funk f x :: c therefore funk :: a -> b -> c RHS f (funk f) x :: c f (funk f) :: d -> c x :: d f :: e -> d -> c funk :: h -> e f :: h unification f :: a = h = (e -> d -> c) x b = d No. x :: b = d (a typo?) Paul: What's wrong with x being of type b and of type d? Could you perhaps explain the error please? Don't forget also that funk :: a -> b -> c = h -> e, which means that e = b -> c Paul: is that something to do with partial application? (funk f) is a partially applied function, correct? Again an explanation would be appreciated. therefore funk :: ((h -> e) -> b -> c) -> b -> c No. I don't understand where you've got this expression from. It's funk :: a -> b -> c = (e -> d -> c) -> b -> c = ((b -> c) -> b -> c) - > b -> c According to GHCi: Prelude> let funk f x = f (funk f) x Prelude> :t funk funk :: ((t1 -> t) -> t1 -> t) -> t1 -> t which is about the same. Thanks Paul