Question

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

Answer 1

$(A\cap B)\cup (A-B)$

= $(A\cap B)\cup (A\cap B^{\prime })$

= $A\cap (B\cup B^{\prime })$ (By distributive law)

= $A\cap U=A$

Hence $A=(A\cap B)\cup (A-B)$

Also $A\cup (B-A)$

= $A\cup (B\cap A^{\prime })$

= $(A\cup B)\cap (A\cup A^{\prime })$ (By distributive law)

= $(A\cup B)\cap U$

= $A\cup B$

Hence $A\cup (B-A)=A\cup B.$