For any three sets A, B and C


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) $A \cup(B-C)=(A \cup B)-(A \cup C)$.


(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, $A \cap(B-C)=(A \cap B)-(A \cap C)$

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