Prove and .

*Proof*If is any element in then, by definition of union, we have or . But, if is in or , then it is in or , and by definition of union, this means . Therefore, .

The other inclusion is identical: if is any element of , then we know or . But, or implies that is in or ; and hence, . Therefore, .

Hence, .∎

* Proof * If is any element in , then we know by definition of intersection that and . Hence, and , and so, . Therefore, .

The reverse inclusion is again identical: if is any element of , then we know and . Hence, and . This implies . Hence, .

So, .∎