DeMorgan's Laws : 1. (A∩B)′=A′∪B′
2. (A∪B)′=A′∩B′
Let us assume (2) is false, and x∈(A∪B)′⟹x∈/(A∪B)
⟹x∈/Aandx∈/B⟹x∈A′andx∈B′
x∈A′∩B′ .
As x was a general element; (A∪B)′=A′∩B′ , which is a contradiction.
Hence, our assumption was false and (2) is true.
Similarly, (1) can be proved as well.
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