{- starter program for Assignment 8, CSE 341, Winter 2003
   simplifying set expressions in Haskell
-}


union :: Eq a => [a] -> [a] -> [a]
-- write your union function here
-- the type says that it takes a list of a's, which must be a type
-- that supports equality tests
--
-- the following definition of union is wrong -- it's just to get Haskell
-- to accept this program
union x y = []


-- also write declarations and definitions for intersect and difference


-- here is a partial definition of the SetExpr data type
-- you just need to add elements for Intersect and Difference
data SetExpr = Set [Integer] 
               | Var String 
               | Union SetExpr SetExpr
               deriving (Eq,Show,Read)


-- here is the type declaration for set_simplify, and an (incorrect)
-- definition (all this version does is simplify an expression to itself)
set_simplify :: SetExpr -> SetExpr
set_simplify s = s



-- here are some functions to run test cases on set_simplify 
-- (you'll need to add additional tests)

t1 :: SetExpr 
t1 = set_simplify (Set [1,3]) -- doesn't simplify any further

t2 :: SetExpr 
t2 = set_simplify (Union (Var "x") (Var "x")) -- should simplify to Var "x"

t3 :: SetExpr 
t3 = set_simplify (Union (Var "x") (Var "y")) -- doesn't simplify any further