Section 1.2
11. [BB] The third and sixth
columns
of
the truth table show
that
(p
t
q)
'¢=}
((,q)
t
(,p)).
12. The third and sixth
columns
of
the truth table
show that
(p
~
q)
'¢=}
((p
t
q)
/\
(q
t
p)).
13. [BB] The third and fifth columns
0
the truth table show that
(p
t
q)
'¢=}
((,p)
V
q).
2.
(a) We construct a truth table.
Since
p
V
[,(p
/\
q)]
is true
for all values
of
p
and
q,
this statement is a
tautology.
p
q
T
T
T
F
F
T
F
F
f
p
T
T
F
F
17
p
q
p+q
'q
'p
(,q)
+
(,p)
T
T
T
F
F
T
T
F
F
T
F
F
F
T
T
F
T
T
F
F
T
T
T
T
p
.....
q
p+q
q+p
(p
+
q)
1\
(q
+
p)
T
T
T
T
F
F
T
F
F
T
F
F
T
T
T
T
p
q
p+q
,p
(,p)
V
q
T
T
T
F
T
T
F
F
F
F
F
T
T
T
T
F
F
T
T
T
q
p/\q
,(p
/\
q)
P
V
[,(p
/\
q)]
T
T
F
T
F
F
T
T
T
F
T
T
F
F
T
T
(b)
By
one
of
the laws
of
DeMorgan, the negation is
(,p)
/\
(p
/\
q).
By
associativity, this is logically
equivalent to
[(
,p)
/\
p]/\
q
'¢=}
0 /\
q
'¢=}
O.
So
the negation is a contradiction.
3.
(a) [BB] Using one
of
the laws
of
De
Morgan and one distributive property, we obtain
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 Summer '10
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 Graph Theory

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