Question
For any two sets A and B, prove that $\text{A}\cup(\text{B}-\text{A})=\text{A}\cup\text{B}$

Answer

Let $\text{x}\in \text{A} \cup \text{(B} - \text{A)}\Rightarrow\text{x}\in\text{A or x}\in\text{(B} - \text{A)}$ $\Rightarrow \text{x} \in \text{A or x} \in \text{B and x}\not\in \text{A}$ $\Rightarrow \text{x}\in \text{B}$ $\Rightarrow \text{x}\in \text{A}\cup\text{B}$ $[\because \text{B}\subset\text{(A}\cup\text{B})]$ This is true for all $\text{x}\in\text{A}\cup\text{(B} - \text{A)}$ $\therefore \text{A}\cup\text{(B} - \text{A)}\subset\text{(A}\cup \text{B})....\text{(i)}$ Conversely, Let, $\text{x}\in \text{(A}\cup\text{B)}$ $\Rightarrow\text{x} \in \text{A or x} \in\text{B}$ $\Rightarrow \text{x}\in \text{A or x} \in \text{(B – A)}$ $[\because\text{B}\subset \text{(B – A)]}$ $\therefore\text{(A}\cup\text{B}) \subset \text{A}\cup \text{(B}- \text{A)}.....\text{(ii)}$ From (i) and (ii), we get $\text{A}\cup\text{(B} - \text{A) = (A}\cup\text{B}).$

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