Using properties of sets show that : a) A ∩ (A’ U B) = A ∩ B b) A ∩ (A U B )’ = Ф

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)' = ∅.

20%

step-by-step explanation:

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(8,15)

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