(2) All the set identities are valid.
(2).1 Suppose
x
∈
(
AB
)
0
. Then
x
is not in both of
A
and
B
, i.e. either
x
is not in
A
or
x
is
not in
B
. So either
x
∈
A
0
or
x
∈
B
0
. Thus
x
∈
A
0
S
B
0
.
On the other hand, suppose
x
∈
A
0
S
B
0
. Then either
x
∈
A
0
or
x
∈
B
0
. If
x
∈
A
0
,
x
is
not in
A
and so
x
cannot be in
AB
. If
x
∈
B
0
, then
x
is not in
B
and so
x
cannot be
in
AB
. Hence
x
∈
(
AB
)
0
.
So we conclude (
AB
)
0
=
A
0
S
B
0
.
(2).2 Suppose
x
∈
A
T
(
B
S
C
). Then
x
∈
A
and
x
∈
B
S
C
. As
x
∈
B
S
C
, either
x
∈
B
or
x
∈
C
. If
x
∈
B
, then
x
∈
AB
and if
x
∈
C
, then
x
∈
AC
. So, either
x
∈
AB
or
x
∈
AC
. Hence
x
∈
(
AB
)
S
(
AC
).
On the other hand, suppose
x
∈
(
AB
)
S
(
AC
). Then either
x
∈
AB
or
x
∈
AC
. If
x
∈
AB
, then
x
∈
A
and
x
∈
B
, and in that case we can say
x
∈
A
T
(
B
S
C
) as
B
⊆
(
B
S
C
). Similarly in the other case, if
x
∈
AC
, then
x
∈
A
and
x
∈
C
, and we
can say that
x
∈
A
T
(
B
S
C
), as
C
⊆
(
B
S
C
).
So we conclude
A
T
(
B
S
C
) = (
AB
)
S
(
AC
).
(2).3 Suppose
x
∈
A
S
(
BC
). Then either
x
∈
A
or
x
∈
BC
. If
x
∈
A
, then
x
∈
(
A
S
B
)
and
x
∈
(
A
S
C
), as
A
is subset of both
A
S
B
and
A
S
C
, and hence in this case
x
∈
(
A
S
B
)
T
(
A
S
C
). In the other case, if
x
∈
BC
, then
x
∈
B
and
x
∈
C
. As
B
i s
a subset of
A
S
B
and
C
is a subset of
A
S
C
, we see that
x
∈
A
S
B
and
x
∈
A
S
C
.
Hence in this case also
x
∈
(
A
S
B
)
T
(
A
S
C
).
On the other hand, suppose
x
∈
(
A
S
B
)
T
(
A
S
C
). Then
x
∈
A
S
B
and
x
∈
A
S
C
.
Now if
x
∈
A
, we see
x
∈
A
S
(
BC
), as
A
is a subset of
A
S
(
BC
). Lastly if
x
∈
A
c
,
then
x
∈
A
S
B
will imply
x
∈
B
and
x
∈
A
S
C
will imply
x
∈
C
.
Hence,
x
∈
BC
⊆
A
S
(
BC
).