C-01-Q056mediumsingle_mcqDe Morgan's law: (A∪B)′(A \cup B)'(A∪B)′ = ?aA′∪B′A' \cup B'A′∪B′bA′∩B′A' \cap B'A′∩B′cA∩BA \cap BA∩BdA′∖B′A' \setminus B'A′∖B′