kdrag.solvers.kb.multiset.sub

kdrag.solvers.kb.multiset.sub(xs, ys)

Difference two multisets. Return None if the second is not a submultiset of the first

>>> sub([("a", 1), ("b", 2)], [("a", 1), ("c", 3)]) is None
True
>>> sub([("a", 1), ("b", 2)], [("a", 1), ("b", 2)])
[]