(A∪B) c (A∪B)c

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Let,

$x ∈ (A ∪ B)^c⇔ x \notin A ∪ B ⇔ x \notin A \text{ and }x \notin B ⇔ x ∈ A^c$

and $x ∈ B^c ⇔ x ∈ A^c ∩ B^c$

Hence, $(A\cup B)^c⇔ (A^c \cap B^c)$

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Question ID: mtid-5-stid-8-sqid-3009-qpid-1708