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