I think you were having a similar problem in an earlier thread. Others have given answers, but I'll try to drive the point home further. In a pattern match like this: f a = case a of [Rule x y] -> ... the pattern is *NOT* matched against each element of the list, but against the list as a whole. Say we have an expression like: let f [x] = True l = [1,2,3] in f l In the third line we attempt to match the list [1,2,3] against the pattern [x]. This cannot possibly work -- [x] can only match a list with a single element! I hope it's more clear now -- and that I answered the right question at all! There's more detail below if you're interested. {-# nitty-gritty on #-} In more detail, recall that [x] is nothing but syntactic sugar for x:[]. That is, the (:) constructor applied to x and []. [1,2,3] in turn stands for 1:2:3:[]. Pattern matching then proceeds: Match 1:2:3:[] against x:[] (:) matched -> match 1 against x; 2:3:[] against [] Match 1 against x x is a variable, bind it to 1 Match 2:3:[] against [] Match failed! {-# nitty-gritty off #-} Have fun =) Jealous 'cause I never had any homework in Haskell, -- Ariel J. Birnbaum