(.:) f g = (f .) . g This can interestingly be generalized(in a way) to get (.::) :: (d -> e) -> (a -> b -> c -> d) -> a -> b -> c -> d (.::) f g = ((

(.:) f g = (f .) . g This can interestingly be generalized(in a way) to get (.::) :: (d -> e) -> (a -> b -> c -> d) -> a -> b -> c -> d (.::) f g = ((f .) .) . g and so on. Since fmap = (.), substitution would get your point free form.