Question

For any three sets A, B and C
(a) $A\cap \left(B-C\right)=\left(A\cap B\right)-\left(A\cap C\right)$
(b) $A\cap \left(B-C\right)=\left(A\cap B\right)-C$
(c) $A\cup \left(B-C\right)=\left(A\cup B\right)\cap \left(A\cup C\text{'}\right)$
(d) $A\cup \left(B-C\right)=\left(A\cup B\right)-\left(A\cup C\right).$

Answer 1

(a) $A\cap \left(B-C\right)=\left(A\cap B\right)-\left(A\cap C\right)$

Let x be any arbitrary element of A$\cap \left(B-C\right)$.
Thus, we have,
x$\in \left[A\cap \left(B-C\right)\right]$$\Rightarrow$ x $\in A$ and x $\in \left(B-C\right)$
$$\begin{gathered} \Rightarrow x\in Aand\left(x\in Bandx\notin C\right) \\ \Rightarrow \left(x\in Aandx\in B\right)and\left(x\in Aandx\notin C\right) \\ \Rightarrow x\left(A\cap B\right)andx\notin \left(A\cap C\right) \\ \Rightarrow x\in \left[\left(A\cap B\right)-\left(A\cap C\right)\right] \\ \Rightarrow A\cap \left(B-C\right)\subseteq \left(A\cap B\right)-\left(A\cap C\right) \\ \mathrm{Similarly},\left(A\cap B\right)-\left(A\cap C\right)\subseteq A\cap \left(B-C\right) \\ \mathrm{Hence},A\cap \left(B-C\right)=\left(A\cap B\right)-\left(A\cap C\right) \end{gathered}$$