(A∪B)
c
(A∪B)c
Let,
x∈(A∪B)c⇔x∉A∪B⇔x∉A and x∉B⇔x∈Acx ∈ (A ∪ B)^c⇔ x \notin A ∪ B ⇔ x \notin A \text{ and }x \notin B ⇔ x ∈ A^cx∈(A∪B)c⇔x∈/A∪B⇔x∈/A and x∈/B⇔x∈Ac
andx∈Bc⇔x∈Ac∩Bcx ∈ B^c ⇔ x ∈ A^c ∩ B^cx∈Bc⇔x∈Ac∩Bc
Hence, (A∪B)c⇔(Ac∩Bc)(A\cup B)^c⇔ (A^c \cap B^c)(A∪B)c⇔(Ac∩Bc)
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