Show that for any sets A and B,
$A=(A\cap B)\cup (A-B)$ and $A\cup (B-A)=(A\cup B)$

(i) $A=(A\cap B)\cup (A-B)$

Consider $RHS=(A\cap B)\cup (A-B)$
$=(A\cap B)\cup (A\cap B^{\prime })$      (by def of difference of sets, $A-B=A\cap B^{\prime }$)
$=A\cap (B\cup B^{\prime })$     (by distributive )
$=A\cap U$       ($\because A\cup A^{\prime }=U$)
$=A$
$=LHS$
Hence, $A=(A\cap B)\cup (A-B)$ 

(ii) $A\cup (B-A)=A\cup B$
Consider, $A\cup (B-A)$
$=A\cup (B\cap A^{\prime })$      (by def of difference of sets, $A-B=A\cap B^{\prime }$)
$=(A\cup B)\cap (A\cup A^{\prime })$   (by distributive property)
$=(A\cup B)\cap U$        ($\because A\cup A^{\prime }=U$)
$=A\cup B$