Question 15

Maths Question: 

$\text{prove that }(X-Y)\cup (Y-X)=(X\cup Y)\cap (X'\cup Y')$

Maths Solution: 

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

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