Pls answer q21 b part...

Dear student

We have, 1) A - (B - C) = A - (B \cap C') [∵ B - C = B \cap C'] = A \cap (B \cap C')' [∵ X - Y = X \cap Y'] = A \cap (B' \cup C) [∵ (B \cap C')' = B' \cup (C')' = B' \cup C = (A \cap B') \cup (A \cap C) = (A - B) \cup (A \cap C) 2) A \cap (B - C) = (A \cap B) - (A \cap C) Let x be any arbitary element of A \cap (B - C) . Then, \Rightarrow x \in A \cap (B - C) \Rightarrow x \in A and x \in (B - C) \Rightarrow x \in A and (x \in B and x \notin C) \Rightarrow (x \in A and x \in B) and (x \in A and x \notin C) \Rightarrow x \in (A \cap B) and x \notin (A \cap C) \Rightarrow x \in (A \cap B) - (A \cap C) ∴ A \cap (B - C) \subseteq (A \cap B) - (A \cap C) Similarly, (A \cap B) - (A \cap C) \subseteq A \cap (B - C) Hence, A \cap (B - C) = (A \cap B) - (A \cap C)

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