There are two ways to prove this. One way is the "hard" way, where you prove each set is a subset of the other. To do this is tedious, but you can try this if you want. We can do it the easy way and apply the distributive laws repeatedly to the RHS:
(A∩B)∪(A' ∩B' )
= ( A ∪(A'∩B') ) ∩ ( B ∪(A' ∩B' ) )
= ( (A ∩ A' )∪(A∩B' ) ) ∩ ( (B ∩ A' )∪(B∩B' ) )
= ( ∅ ∪(A∩B' ) ) ∩ ( (B ∩ A' )∪∅ )
= (A∪B' )∩(A'∪B)
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There are two ways to prove this. One way is the "hard" way, where you prove each set is a subset of the other. To do this is tedious, but you can try this if you want. We can do it the easy way and apply the distributive laws repeatedly to the RHS:
(A∩B)∪(A' ∩B' )
= ( A ∪(A'∩B') ) ∩ ( B ∪(A' ∩B' ) )
= ( (A ∩ A' )∪(A∩B' ) ) ∩ ( (B ∩ A' )∪(B∩B' ) )
= ( ∅ ∪(A∩B' ) ) ∩ ( (B ∩ A' )∪∅ )
= (A∪B' )∩(A'∪B)