Practicing Success
Let A and B be two sets. The $(A∪B)'∪(A' ∩ B) =$ |
$A'$ A $B'$ none of these |
$A'$ |
We have, $(A∪B)'∪(A' ∩ B)$ $=(A' ∩ B')∪(A' ∩ B)$ $=(A'∪A')∩(A'∪B)∩(B'∪A')∩(B'∪B)$ [By dist. of ∪ over ∩] $=A' ∩(A'∪B)∩(B'∪A')∩U$ $=A' ∩(A'∪B)∩(B'∪A')$ $=A' ∩\{(A∩B')'∪(A∩B)\}'$ $=A' ∩\{(A∩B')∩(A∩B)\}'=A'∩A'=A'$ |