$\begin{align}  & \text{From the L}\text{.H}\text{.S} \\ & \text{Let }a\in Z-(X\cup Y) \\ & a\in Z\cap (X\cup Y)' \\ & a\in Z\text{ and }a\in (X\cup Y)' \\ & a\in Z\text{ and }a\notin (X\cup Y) \\ & a\in Z\text{ and }a\in X'\text{ and }a\in Y' \\ & a\in Z\text{ and }a\in X'\text{ and }a\in Z\text{ and }a\in Y' \\ & a\in (Z\cap X')\text{ and }a\in (Z\cap Y') \\ & a\in (Z-X)\text{ and }a\in (Z-Y) \\ & a\in (Z-X)\cap (Z-Y) \\ & (Z-X)\cap (Z-Y)\subseteq Z-(X\cup Y)----(i) \\ & \text{From the R}\text{.H}\text{.S} \\ & \text{Let }b\in (Z-X)\cap (Z-Y) \\ & b\in (Z\cap X')\cap (Z\cap Y') \\ & b\in Z\text{ and }b\in X'\text{ and }b\in Z\text{ and }b\in Y' \\ & b\in Z\text{ and }b\in X'\text{ and }b\in Y' \\ & b\in Z\text{ and }b\notin X\text{ or }b\notin Y \\ & b\in Z\text{ and }b\notin (X\cup Y) \\ & b\in Z\text{ and }b\in (X\cup Y)' \\ & b\in Z\cap (X\cup Y)' \\ & b\in Z-(X\cup Y) \\ & Z-(X\cup Y)\subseteq (Z-X)\cap (Z-Y)---(ii) \\ & \therefore Z-(X\cup Y)=(Z-X)\cap (Z-Y) \\\end{align}$