Practicing Success
For any three sets A, B and C the set $(A∪B∪C)∩(A∩B'∩C')'∩C'$ is equal to |
$B∩C'$ $B'∩C'$ $B∩C$ $A∩B∩C$ |
$B∩C'$ |
We have, $(A∩B'∩C')'=A'∪B∪C$ $∴(A∪B∪C)∩(A∩B'∩C')'$ $=(A∪B∪C)∩(A'∪B∪C)$ $=(A∩A')∪(B∪C)$ [By distributivity of ∪ over ∩] $=\phi∪(B∪C)=B∪C$ Hence, $(A∪B∪C)∩(A∩B'∩C')'∪C'$ $=(B∪C)∩C'=(B∩C')∪(C∩C')=(B∩C')$ |