Haskell: Implement list union

Implement the list union. Duplicates, and elements of the first list, are removed from the second list, but if the first list contains duplicates, so will the result.

ListUtil.hs

listUnion :: Eq a => [a] -> [a] -> [a]
listUnion xs [] = xs
listUnion xs (y:ys)
    | (y `elem` xs) = listUnion xs ys
    | otherwise = listUnion (y:xs) ys

*Main> :load ListUtil.hs
[1 of 1] Compiling Main             ( ListUtil.hs, interpreted )
Ok, modules loaded: Main.
*Main> 
*Main> listUnion [2, 3, 3, 4] [3, 4, 5]
[5,2,3,3,4]
*Main> 
*Main> listUnion [2, 3, 3, 4] [3, 4, 5, 2, 3, 2]
[5,2,3,3,4]
*Main> 
*Main> listUnion [2, 3, 3, 4] [3, 4, 5, 2, 3, 2, 3,3, 10, 10, 10]
[10,5,2,3,3,4]
*Main> :t listUnion
listUnion :: Eq a => [a] -> [a] -> [a]

As you observe above output, list2 elements come first, list2 elements come next. In case of union function implementation in Data.List module, result contains list1 followed by list2. Following program gives you output just like the union function in Data.List module.

ListUtil.hs

listUnion :: Eq a => [a] -> [a] -> [a]
listUnion xs [] = xs
listUnion xs (y:ys)
    | (y `elem` xs) = listUnion xs ys
    | otherwise = listUnion (xs++(y:[])) ys

*Main> :load ListUtil.hs
[1 of 1] Compiling Main             ( ListUtil.hs, interpreted )
Ok, modules loaded: Main.
*Main> 
*Main> listUnion [2, 3, 3, 4] [3, 4, 5]
[2,3,3,4,5]
*Main> 
*Main> listUnion [2, 3, 3, 4] [3, 4, 5, 2, 3, 2]
[2,3,3,4,5]
*Main> 
*Main> listUnion [2, 3, 3, 4] [3, 4, 5, 2, 3, 2, 3,3, 10, 10, 10]
[2,3,3,4,5,10]
*Main> 

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