C-01-Q057mediumsingle_mcqDe Morgan's law: (A∩B)′(A \cap B)'(A∩B)′ = ?aA′∩B′A' \cap B'A′∩B′bA′∪B′A' \cup B'A′∪B′cA∖BA \setminus BA∖Bd(A∪B)′(A \cup B)'(A∪B)′