CS 2800 – Fall 2008
Homework #1 Solution
Section 1.1
Problem 8
(c) If you miss the ﬁnal exam, then you do not pass the course.
(f) Either you have the ﬂu and miss the ﬁnal exam, or you do not miss the ﬁnal exam and
pass the course.
Problem 10
(a)
r
∧ ¬
q
(b)
p
∧
q
∧
r
(c)
r
→
p
(d)
p
∧ ¬
q
∧
r
(e) (
p
∧
q
)
→
r
(f)
r
↔
(
q
∨
p
)
Problem 28
(c) Truth table:
p
q
p
∨
q
p
⊕
(
p
∨
q
)
T
T
T
F
T
F
T
F
F
T
T
T
F
F
F
F
(f) Truth table:
p
q
¬
q
p
↔
q
p
↔ ¬
q
(
p
↔
q
)
⊕
(
p
↔ ¬
q
)
T
T
F
T
F
T
T
F
T
F
T
T
F
T
F
F
T
T
F
F
T
T
F
T
1
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View Full DocumentProblem 44
No – this is the classic Barber’s Paradox. We will use the male pronoun in what follows,
assuming that we are talking about males shaving their beards here, and assuming that all
men have facial hair. If we restrict ourselves to beards and allow female barbers, then the
barber could be female with no contradiction. If such a barber existed, then who shaves the
barber? If the barber shaved himself, then he would be violating the rule that he shaves
only those people who do not shave themselves. On the other hand, if he does not shave
himself, then the rule says that he must shave himself. Neither is possible, so there can be
no such barber.
Section 1.2
Problem 12
(a) Assume the hypothesis is true. Then,
p
must be false. But (
p
∨
q
) needs to be true and
hence
q
must be true. Alternately, one can make an algebraic argument as follows:
[
¬
p
∧
(
p
∨
q
)]
→
q
≡
[(
¬
p
∧
p
)
∨
(
¬
p
∧
q
)]
→
q
(distributing
∧
over
∨
)
≡
[
F
∨
(
¬
p
∧
q
)]
→
q
(negation law)
≡
(
¬
p
∧
q
)
→
q
(identity law)
≡ ¬
(
¬
p
∧
q
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 Fall '07
 SELMAN
 Logic, Summation, Shaving, Logical connective

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