I couldn't get close on my own. From: https://github.com/pankajgodbole/hutton/blob/master/exercises.hs {- 7. Complete the following instance declarations: instance Eq a => Eq (Maybe a) where ... instance Eq a => Eq [a] where ... -} -- suggested answer instance Eq a => Eq (Maybe a) where -- Defines the (==) operation. Nothing == Nothing = True Just == Just = True -- why isn't this Just a == Just a ? -- My guess is that a and Just a are different types and can't be == in Haskell _ == _ = False instance Eq a => Eq [a] where -- Defines the (==) operation. [] == [] = True [x] == [y] = x == y (x:xs) == (y:ys) = x==y && xs==ys -- I assume this is implicitly recursive. _ == _ = False