I'll assume the usual definition of set difference, .
Let . Then
and
. If
, then
and
. This means
and
, so it follows that
. Hence
.
Now let . Then
and
. By definition of set difference,
and
. Since
, we have
, and so
. Hence
.
The two sets are subsets of one another, so they must be equal.
The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices .
Proof of one direction for example:
Let . Then
and
, which in turn means
for all
. This means
, and
, and so on, where
, for all
. This means
, and
, and so on, so
. Hence
.
