Hey, a few notes: 1. The (==) function in the second equation of the Maybe instance of (==) is not complete yet, since you need to match on instances of the Maybe type and "Just :: a -> Maybe a". Your idea "Just a == Just a" goes in the right direction, but please note that you can't bind two variable names ("a") in the same equation. You need to give each "boxed value" a different name. I'm sure you can work it out from there. 2.  The right hand side of "(x:xs) == (y:ys)" is not implicitly recursive, but rather explicitly! Since we apply the function (==) to the smaller lists "xs" and "ys" until we arrive at a base case.  Greetings, Tobias ----- Nachricht von trent shipley <trent.shipley@gmail.com> ---------      Datum: Sat, 15 Sep 2018 01:23:47 -0700        Von: trent shipley <trent.shipley@gmail.com> Antwort an: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell <beginners@haskell.org>    Betreff: [Haskell-beginners] Hutton Ex 8.9.7         An: Haskell Beginners <beginners@haskell.org>

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   

 

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