Question

For any two sets A and B, prove that

(i) $\left(A\cup B\right)-B=A-B$
(ii) $A-\left(A\cap B\right)=A-B$
(iii) $A-\left(A-B\right)=A\cap B$
(iv) $A\cup \left(B-A\right)=A\cup B$ [NCERT EXEMPLAR]
(v) $\left(A-B\right)\cup \left(A\cap B\right)=A$ [NCERT EXEMPLAR]

Answer 1

(i)
$$\begin{gathered} \left(A\cup B\right)-B=\left(A\cup B\right)\cap B\text{'}\left(X-Y=X\cap Y\text{'}\right) \\ =\left(A\cap B\text{'}\right)\cup \left(B\cap B\text{'}\right)\left(\mathrm{Distributive}\mathrm{law}\right) \\ =\left(A\cap B\text{'}\right)\cup \varphi \\ =A\cap B\text{'} \\ =A-B \end{gathered}$$

(ii)
$$\begin{gathered} A-\left(A\cap B\right)=A\cap \left(A\cap B\right)\text{'}\left(X-Y=X\cap Y\text{'}\right) \\ =A\cap \left(A\text{'}\cup B\text{'}\right)\left(\mathrm{De}\mathrm{Morgan}\mathrm{law}\right) \\ =\left(A\cap A\text{'}\right)\cup \left(A\cap B\text{'}\right) \\ =\varphi \cup \left(A\cap B\text{'}\right) \\ =A\cap B\text{'} \\ =A-B \end{gathered}$$

(iii)
$$\begin{gathered} A-\left(A-B\right)=A-\left(A\cap B\text{'}\right)\left(X-Y=X\cap Y\text{'}\right) \\ =A\cap \left(A\cap B\text{'}\right)\text{'} \\ =A\cap \left[A\text{'}\cup \left(B\text{'}\right)\text{'}\right]\left(\mathrm{De}\mathrm{Morgan}\mathrm{law}\right) \\ =A\cap \left(A\text{'}\cup B\right) \\ =\left(A\cap A\text{'}\right)\cup \left(A\cap B\right)\left(\mathrm{Distributive}\mathrm{law}\right) \\ =\varphi \cup \left(A\cap B\right) \\ =A\cap B \end{gathered}$$

(iv)
$$\begin{gathered} A\cup \left(B-A\right)=A\cup \left(B\cap A\text{'}\right)\left(X-Y=X\cap Y\text{'}\right) \\ =\left(A\cup B\right)\cap \left(A\cup A\text{'}\right)\left(\mathrm{Distributive}\mathrm{law}\right) \\ =\left(A\cup B\right)\cap \mathit{\cup }\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathit{\cup }\mathit{}\mathrm{is}\mathrm{the}\mathrm{universal}\mathrm{set}\right) \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=A\cup B \end{gathered}$$

(v)
$$\begin{gathered} \left(A-B\right)\cup \left(A\cap B\right)=\left(A\cap B\text{'}\right)\cup \left(A\cap B\right)\left(X-Y=X\cap Y\text{'}\right) \\ =A\cap \left(B\text{'}\cup B\right)\left(\mathrm{Distributive}\mathrm{law}\right) \\ =A\cap \mathit{\cup } \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=A \end{gathered}$$