Prove that and .

* Proof (). * Let

be an arbitrary element of

. This means that

and

. Further, since

we know

or

. But then, this implies

or

depending on whether

or

, respectively. Then, since

or

, we have

. Thus,

.

For the opposite inclusion, we let

be an arbitrary element of

. This means that

or

. If

then

and

, while if

we have

and

. Thus, we have

no matter what, and either

or

. Since

or

, we know

. Since we already had that

no matter what, we now have

. Therefore,

.

Hence,

.∎

* Proof (). * Let be an arbitrary element of . This means or . If then and . Hence, . Otherwise, if then and . Hence, and ; therefore, . Therefore, .

For the reverse inclusion, let be an arbitrary element of . This means and . This implies that or and (since if then the fact that and means must be in both and ). If , then . On the other hand, if and , then . Hence, . Therefore, .

Therefore, .∎