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Question : 101 of 150
Marks:
+1,
-0
Solution:
We have,
(A∪B∪C)∩(A∩B′∩C′)′∩C′
(A∪B∪C)∩(A∪B′∪C′)′∩C′ [by using distributed law]
=[(A∪B∪C)∩A′]∪[(A∪B∪C) ∩(B∪C)]
=[(B∪C)∪(B∪C)]∩C′
=(B∪C)∩C′
=(B∩C′)∪(C∩C′)
=(B∩C)‌‌‌[∵C∩C′=ϕ and (B∩C′)∪ϕ=B∩C′]
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