Wrong definition of 'bijective'

[BertLisser]

-- Check the length of the intersection of 2 unique lists
-- with both individual lists.
-- If they are equal we have a permutation because we may assume that there are no duplicates in the list
-- and thus have a 1:1 image domain.
propIsPermutation :: [Integer] -> [Integer] -> Bool
propIsPermutation xs ys = isPermutation (nub xs) (nub ys) == bijective
    where bijective = length (nub xs) == length (intersect (nub xs) (nub ys)) && length (nub ys) == length (intersect (nub xs) (nub ys))

For example bijective [1] [2] returns True must be False