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