Question

For any two sets A and B, prove the following:
(i) $A\cap \left(A\text{'}\cup B\right)=A\cap B$
(ii) $A-\left(A-B\right)=A\cap B$
(iii) $A\cap \left(A\cup B\right)\text{'}=\varphi$
(iv) $A-B=A\Delta \left(A\cap B\right)$.

Answer 1

(i)
$$\begin{gathered} \mathrm{LHS}=A\cap \left(A\text{'}\cup B\right) \\ =\left(A\cap A\text{'}\right)\cup \left(A\cap B\right) \\ =\left(\varphi \right)\cup \left(A\cap B\right) \\ =A\cap B=\mathrm{RHS} \end{gathered}$$
Hence proved.

(ii)
$$\begin{gathered} \mathrm{LHS}=A-\left(A-B\right) \\ =A-\left(A\cap B\text{'}\right) \\ =A\cap \left(A\cap B\text{'}\right)\text{'} \\ =A\cap \left\{A\text{'}\cup \left(B\text{'}\right)\text{'}\right\} \\ =A\cap \left(A\text{'}\cup B\right) \\ =\left(A\cap A\text{'}\right)\cup \left(A\cup B\right) \\ =\left(\varnothing \right)\cup \left(A\cup B\right) \\ =\left(A\cup B\right)=\mathrm{RHS} \end{gathered}$$
Hence proved.

(iii)
$$\begin{gathered} \mathrm{LHS}=A\cap \left(A\cup B\right)\text{'} \\ =A\cap \left(A\text{'}\cap B\text{'}\right) \\ =\left(A\cap A\text{'}\right)\cap \left(A\cap B\text{'}\right) \\ =\left(\varphi \right)\cap \left(A\cap B\text{'}\right) \\ =\varphi =\mathrm{RHS}\left[\varphi \cap A=\varphi \right] \end{gathered}$$
Hence proved.

(iv)
$$\begin{gathered} \mathrm{LHS}=A\Delta \left(A\cap B\right) \\ =\left\{A-\left(A\cap B\right)\right\}\cup \left\{\left(A\cap B\right)-A\right\} \\ =\left\{A\cap \left(A\cap B\right)\text{'}\right\}\cup \left\{\left(A\cap B\right)\cap A\text{'}\right\} \\ =\left\{A\cap \left(A\text{'}\cup B\text{'}\right)\right\}\cup \left\{\left(A\cap B\right)\cap A\text{'}\right\} \\ =\left\{\left(A\cap A\text{'}\right)\cup \left(A\cap B\text{'}\right)\right\}\cup \left\{\left(A\cap A\text{'}\right)\cap \left(B\cap A\text{'}\right)\right\} \\ =\left\{\left(\varphi \right)\cup \left(A\cap B\text{'}\right)\right\}\cup \left\{\left(\varphi \right)\cap \left(B\cap A\text{'}\right)\right\} \\ =\left(A\cap B\text{'}\right)\cup \left(\varphi \right) \\ =\left(A\cap B\text{'}\right) \\ =A-B=\mathrm{RHS} \end{gathered}$$.

Hence proved.