Prove and
.
Proof












The other inclusion is identical: if











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, .∎