Answer:
a) From A ∩ A' = ∅, we have;
A ∩ (A' ∪ B) = A ∩ B
b) From A ∩ (A' ∩ B') = (A ∩ A') ∩ B' and A ∩ A' = ∅, we have;
A ∩ (A ∪ B)' = ∅
Step-by-step explanation:
a) By distributive law of sets, we have;
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
From the complementary law of sets, we have;
A ∩ A' = ∅
Therefore, for A ∩ (A' ∪ B) = A ∩ B, we have
A ∩ (A' ∪ B) = (A ∩ A') ∪ (A ∩ B) (distributive law of sets)
A ∩ A' = ∅ (complementary law of sets)
Therefore;
(A ∩ A') ∪ (A ∩ B) = ∅ ∪ (A ∩ B) = (A ∩ B) (Addition to zero identity property)
∴ A ∩ (A' ∪ B) = A ∩ B
b) By De Morgan's law
(A ∪ B)' = A' ∩ B'
Therefore, A ∩ (A ∪ B)' = A ∩ (A' ∩ B')
By associative law of sets, we have;
A ∩ (A' ∩ B') = (A ∩ A') ∩ B'
A ∩ A' = ∅ (complementary law of sets)
Therefore, (A ∩ A') ∩ B' = ∅ ∩ B' = ∅
Which gives;
A ∩ (A ∪ B)' = ∅.
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