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34
Solutions to Exercises
(b) First, we prove
A
U
(B
n
G)
~
(A
U
B)
n
(A
U
G).
So let
x
E
Au (B
nO).
Then
x
E
A
or
x
E
B
n
G.
If
x
E
A,
then
x
E
Au B
and
x
E
AUG,
so
x
E
(AU B)
n
(AUG).
If
x
E
Bn G,
then
x
E
B
and
x
E
G
so
x
E
AUB
and
x
E
AU G;
that
is,
x
E
(A
U
B)
n
(A
U
G).
In either case,
x
E
(A
U
B)
n
(A
U
G)
giving the desired inclusion.
Second, we prove
(A
U
B)
n
(A
U
G)
~
Au (B
n
G).
So let
x
E
(AUB)
n
(AUG).
Thus,
x
E
AUB
and
x
E
AuG.
If
x
E
A,
then
x
E
Au (BnG).
If
x
¢:.
A,
then we must have
x
E
B
and
x
E
G;
that is,
x
E
B
n
G,
so
x
E
Au (B
n
G).
In either
case,
x
E
A
U
(B
n
G)
giving the desired inclusion and equality.
21. We use the fact that
(XC)C
=
X for any set X.
Let X
=
Nand
Y
=
BC.
Then
A
=
XC and
B
=
y
c
,
so
(AnB)C
=
[XC
nycr
=
[(XUy)C]C
(by
the first law of De Morgan)
=
X
U
Y
=
A
C
U
BC,
as required.
22. [BB] Using the fact that
X,
Y
= X
nYc,
we have
(A, B)
,
G
=
(A nBC)
n
G
C
=
An (B
C
n
G
C
)
=
An (B
U
G)C
=
A
,
(B
U
G).
23. We use the laws of De Morgan and the facts that
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 Summer '10
 any
 Graph Theory

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