Question:
For any three sets A, B and C
(a) $A \cap(B-C)=(A \cap B)-(A \cap C)$
(b) $A \cap(B-C)=(A \cap B)-C$
(c) $A \cup(B-C)=(A \cup B) \cap\left(A \cup C^{\prime}\right)$
(d) $A \cup(B-C)=(A \cup B)-(A \cup C)$.
Solution:
(a) $A \cap(B-C)=(A \cap B)-(A \cap C)$
Let $x$ be any arbitrary element of $A \cap(B-C)$.
Thus, we have,
$x \in[A \cap(B-C)] \Rightarrow x \in A$ 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 $x \notin C)$
$\Rightarrow x(A \cap B)$ and $x \notin(A \cap C)$
$\Rightarrow x \in[(A \cap B)-(A \cap C)]$
$\Rightarrow A \cap(B-C) \subseteq(A \cap B)-(A \cap C)$
Similarly, $(A \cap B)-(A \cap C) \subseteq A \cap(B-C)$
Hence, $A \cap(B-C)=(A \cap B)-(A \cap C)$