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, .∎