We have
(A ∩(B ∪ C)′)′ = {∵ (A ∩ B)′ = A′ ∪ B′}
= (A′ ∪((B ∪ C)′)′)
= (A′ ∪(B ∪ C))
A′ = {6, 7, 8, 9}, B = {2, 4, 6, 7} C = {2, 7, 8, 9}
= ({6, 7, 8, 9} ∪ ({2, 4, 6, 7} ∪ {2, 7, 8, 9}))
= {6, 7, 8, 9}
So, (A ∩ (B ∪ C)′)′ = {6, 7, 8, 9}.
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