Chapter 3
3.1 Exercises
1. “February 2, 2009, was a Monday” is a
declarative sentence that is true and,
therefore, is considered a statement.
2. “The ZIP code for Oscar, LA, is 70762” is a
declarative sentence that is true and,
therefore, is considered a statement.
3. “Listen my children and you shall hear of
the midnight ride of Paul Revere” is not a
declarative sentence and does not have the
property of being true or false. Hence, it is
not considered a statement.
4. “Did you yield to oncoming traffic?” is a
question, not a declarative sentence and
therefore, it is not a statement.
5. “5 + 9  14 and 4  1 = 12” is a declarative
sentence that is false and, therefore, is
considered a statement.
6. “5 + 9  12 or 4  2 = 5” is a declarative
sentence that is true and, hence is considered
a statement.
7. “Some numbers are positive” is a
declarative sentence that is true and,
therefore, is a statement.
8. “Grover Cleveland was president of the
United States in 1885 and 1897” is a
declarative sentence that is true and is,
therefore, a statement.
9. “Accidents are the main cause of deaths of
children under the age of 7” is a declarative
sentence that has the property of being true
or false and, therefore, is considered to be a
statement.
10. “It is projected that in the United States
between 2010 and 2020, there will be over
500,000 job openings per year for
elementary school teachers, with median
annual salaries of about $51,000” is a
declarative sentence that has the property of
being either true or false and, therefore, is
considered to be a statement.
12. “Behave yourself and sit down” is a
command, not a declarative sentence and,
therefore, is not considered a statement.
13. “Kevin ‘Catfish’ McCarthy once took a
prolonged continuous shower for 340 hours,
40 minutes” is a declarative sentence that
has the property of being either true or false
and, therefore, is considered to be a
statement.
14. One gallon of milk weighs more than 3
pounds” is a declarative sentence that has
the property of being true and, therefore, is
considered to be a statement.
15. “I read the Detroit Free Press, and I read the
Sacramento Bee” is a compound statement
because it consists of two simple statements
combined by the connective “and.”
16. “My brother got married in Copenhagen” is
not a compound statement because only one
assertion is being made.
17. “Tomorrow is Saturday” is a simple
statement because only one assertion is
being made.
18. “Jing is younger than 18 years of age, and
so is her friend Shu-fen” is a compound
statement because it consists of two simple
statements combined by the connective
“and.”
19. “Jay’s wife loves Ben and Jerry’s ice
cream” is not compound because only one
assertion is being made.
20. “The sign on the back of the car read
‘Canada or bust!’” is not compound because
only one assertion is being made.
21. “If Lorri sells her quota, then Michelle will
be happy” is a compound statement because
it consists of two simple statements
combined by the connective “if...then.”
22. “If Bobby is a politician, then Mitch is a
crook” is a compound statement because it
consists of two simple statements combined
by the connective “if...then.”
11. “Where are you going tomorrow?” is a
question, not a declarative sentence and,
therefore, is not considered a statement.
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3.1: STATEMENTS AND QUANTIFIERS 49
23. The negation of “Her aunt’s name is
Hermione” is “Her aunt’s name is not
Hermione.”
24. A negation of “No rain fell in southern
California today” is “Some rain fell in
southern California today.”
25. A negation of “Some books are longer than
this book” is “No book is longer than this
book.”
26. A negation of “All students present will get
another chance” is “At least one student
present will not get another chance.”
27. A negation of “No computer repairman can
play blackjack” is “At least one computer
repairman can play blackjack.”
28. A negation of “Some people have all the
luck” is “No people have all the luck.”
29. A negation of “Everybody loves somebody
sometime” is “Someone does not love
somebody sometime.”
39. A translation for “~p” is “She does not have
green eyes.”
40. A translation for “~q” is “He is not 60 years
old.”
41. A translation for “p  q” is “She has green
eyes and he is 60 years old.”
42. A translation for “p  q” is “She has green
eyes or he is 60 years old.
43. A translation for “~p  q” is “She does not
have green eyes or he is 60 years old.”
44. A translation for “p  ~ q” is “She has
green eyes and he is not 60 years old.”
45. A translation for “~p  ~q” is “She does
not have green eyes or he is not 60 years
old.”
46. A translation for “~p  ~q” is “She does
not have green eyes and he is not 60 years
old.”
30. A negation of “Everyone needs a friend” is
“Someone does not need a friend.”
47. A translation for “~(~p  q)” is “It is not
the case that she does not have green eyes
and he is 60 years old.”
31. A negation of "The trash needs to be
collected" is "The trash does not need to be
collected."
48. A translation for “~(p  ~q)” is “It is not
the case that she has green eyes or he is not
60 years old.”
32. A negation of "Every architect who wants a
job can find one" is "Not every architect
who wants a job can find one."
49. “Tyler collects DVDs and Josh is not an art
major” may be symbolized as p  ~q.
33. A negation of “x > 12” (without using a
slash sign) would be “x  12.”
50. “Tyler does not collect DVDs or Josh is not
an art major” may be symbolized as
~p  ~q.
34. A negation for “x < 6” (without using a
slash sign) would be “x  6.”
51. “Tyler does not collect DVDs or Josh is an
art major” may be symbolized as ~p  q.
35. A negation for “x  5” would be “x < 5.”
52. “Josh is an art major and Tyler does not
collect DVDs” may be symbolized as
q  ~p.
36. A negation for “x  19” would be “x > 19.”
37. Writing exercise; answers will vary.
38. Writing exercise; answers will vary.
Let p represent the statement “She has green
eyes,” and let q represent “He is 60 years old.”
Translate each symbolic compound statement into
words.
53. “Neither Tyler collects DVDs nor Josh is an
art major” may be symbolized as ~(p  q)
or equivalently, ~p  ~q.
54. “Either Josh is an art major or Tyler collects
DVDs, and it is not the case that both Josh is
an art major and Tyler collects DVDs” may
be symbolized as (q  p)  [~(q  p)].
55. Writing exercise; answers will vary.
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CHAPTER 3 INTRODUCTION TO LOGIC
56. Writing exercise; answers will vary.
Refer to the sketches labeled A, B, and C in the
text, and identify the sketch (or sketches) that is
(are) satisfied by the given statement involving a
quantifier.
57. The condition that “all pictures have
frames” is satisfied by group C.
58. The condition that “No picture has a frame”
is satisfied by group B.
59. The condition that “At least one picture does
not have a frame” is met by groups A and B.
60. The condition that “Not every picture has a
frame” is satisfied by groups A and B.
Observe that this statement is equivalent to
“At least one picture does not have a
frame.”
69. Since rational numbers are real numbers, the
statement “All rational numbers are real
numbers” is true.
70. Since irrational numbers are real numbers,
the statement “All irrational numbers are
real numbers” is true.
1
is a rational number but not an
2
integer, the statement “Some rational
numbers are not integers” is true.
71. Since
72. Since whole numbers are rational, the
statement “Some whole numbers are not
rational numbers” is false.
73. The number 0 is a whole number but not
positive. Thus, the statement “Each whole
number is a positive number” is false.
1
is a rational number. Thus,
2
the statement “Each rational number is a
positive number” is false.
61. The condition that “At least one picture has
a frame” is satisfied by groups A and C.
74. The number 
62. The condition that “No picture does not
have a frame” is satisfied by group C.
Observe that this statement is equivalent to
“All pictures have a frame.”
75. Writing exercise; answers will vary.
63. The condition that “all pictures do not have
frames” is satisfied by group B. Observe
that this statement is equivalent to “No
pictures have a frame.”
64. The condition that “Not every picture does
not have a frame” is satisfied by groups A
and C. Observe that this statement is
equivalent to “At least one picture does have
a frame.”
65. Since all whole numbers are integers, the
statement “Every whole number is an
integer” is true.
66. Since not every integer is a whole number,
the statement “Every integer is a whole
number” is false.
67. Since all natural numbers are integers, the
statement “There exists a natural number
that is not an integer” is false.
68. Since 3 is an integer but not a natural
number, the statement “There exists an
integer that is not a natural number” is true.
76. Writing exercise; answers will vary.
77. We might write the statement “There is no
one here who has not made mistakes before”
using the word “every” as “Every person
here has made mistakes before.”
78. A. The statement “For some real number x,
x  0” is true. Let x = 10, for example.
This real number satisfies the statement.
B. The statement “For all real numbers x,
x3  0 ” is false. For example, let
x = 2. Observe that ( 2)3  8 ò 0.
C. The statement “For all real numbers x
less than 0, x 2 is also less than 0” is
false. Let x = 4, which is less than 0,
but x 2  ( 4) 2  16, which is not less
than 0.
D. The statement “For some real number x,
x 2  0 ” is false. If you square any real
number (negative, positive, or 0), the
result will be  0.
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3.2: TRUTH TABLES AND EQUIVALENT STATEMENTS 51
79. The statement “Some cats have fleas” is
symbolized  c  f  , and its negation “No
cat has fleas” could be stated “Every cat
does not have fleas.” This negation is
expressed symbolically as  c  ~ f  .
80. In part (b) of Example 5, the statement
“Some cats do not have fleas can be
expressed symbolically as  c  ~ f  . The
negation is  c  f  . In part (c) of Example
5, the statement “No cats have fleas” can be
expressed symbolically as  c  ~ f  , and
its negation is expressed as  c  f  .
3.2 Exercises
1. If q is false, then (p  ~q)  q must be
false, since both conjuncts (parts of the
conjunction) must be true for the compound
statement to be true.
2. If q is true, then q  (q  ~p) is true, since
only one disjunct (parts of the disjunctive
statement) need be true for the compound
statement to be true.
3. If p  q is true, and p is true, then q must
also be true in order for the conjunctive
statement to be true. Observe that both
conjuncts must be true for a conjunctive
statement to be true.
4. If p  q is false, and p is false, then q must
also be false. Observe that both disjuncts
must be false for a disjunctive statement to
be false.
5. In order for p   q  ~ q  to be true, either p
must be true, or  q  ~ q  must be true. But
 q ~ q 
cannot be true, so p must be true.
6. The statement p  ~  q  r  is a
conjunction. In order for a conjunction to be
true, both component statements must be
true. Therefore, p must be true, and q  r
must be false. But a disjunction is false only
if both components are false, so both q and r
are false. In conclusion, p must be true, and
q and r must both be false.
7. If ~  p  q  is true, both components
(disjuncts) must be false. Thus, the
disjunction itself is false making its negation
true.
8. If ~  p  q  is false, then both p and q must
be true. This will assure that the conjunction
itself is true making its negation false.
In exercises 920, p represents a false statement
and q represents a true statement.
9. Since p = F, ~p = ~F = T. That is, replace p
by F and determine the truth of ~F.
10. Since q = T, ~q = ~T = F. Thus, ~q is false.
11. Since p is false and q is true, we may
consider the “or” statement as
F  T
T,
by logical definition of an “or” statement.
That is p  q is true.
12. Since p is false and q is true, we may
consider the “and” statement as
F  T
F,
by the logical definition of an “and”
statement. That is, p  q is false.
13. With the given truth values for p and q, we
may consider p  ~q as
F  ~T
F  F
F,
by the logical definition of “  .”
14. With the given truth values for p and q, we
may consider ~p  q as
~F  T
T  T
T,
by the logical definition of “  .” Thus, the
compound statement is true.
15. With the given truth values for p and q, we
may consider ~p  ~q as
~F  ~T
T  F
T.
Thus, the compound statement is true.
16. Replacing p and q with the given truth
values, we have
F  ~T
F  F
F.
Thus, the compound statement p  ~q is
false.
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CHAPTER 3 INTRODUCTION TO LOGIC
17. Replacing p and q with the given truth
values, we have
~(F  ~T)
~(F  F)
~F
T.
Thus, the compound statement ~(p  ~q) is
true.
24. Replacing p, q and r with the given truth
values, we have
(F  ~F)  T
(F  T)  T
T  T
T.
Thus, the compound statement
(q  ~r)  p is true.
18. Replacing p and q with the given truth
values, we have
~(~F  ~T)
~(T  F)
~(T)
F.
Thus, the compound statement ~(~p  ~q)
is false.
25. Replacing p, q and r with the given truth
values, we have
T  (F  F)
T  F
F.
Thus, the compound statement p  (q  r)
is false.
19. Replacing p and q with the given truth
values, we have
~[~F  (~T  F)]
~[T  (F  F)]
~[T  F]
~F
T.
Thus, the compound statement
~[~p  (~q  p)] is true.
26. Replacing p, q and r with the given truth
values, we have
(~T  F)  ~F
(F  F)  T
F T
T.
Thus, the compound statement
(~p  q)  ~r is true.
20. Replacing p and q with the given truth
values, we have
~[(~F  ~T)  ~T]
~[(T  F)  F]
~[F  F]
~F
T.
Thus, the compound statement
~[(~p  ~q)  ~q] is true.
21. The statement 6  2 is a disjunction since it
means “6 > 2” or “6 = 2.”
22. The statement “8  3” is true because 8 > 3
and only one disjunct need be true.
The statement “5  5” is true because 5 = 5
and only one disjunct need be true.
In exercises 2332, p represents a true statement,
and q and r represent false statements.
23. Replacing p, q and r with the given truth
values, we have
(T  F)  ~F
F  T
T.
Thus, the compound statement
(p  r)  ~q is true.
27. Replacing p, q and r with the given truth
values, we have
~(T  F)  (F  ~F)
~F  (F  T)
T  T
T.
Thus, the compound statement
~(p  q)  (r  ~q) is true.
28. Replacing p, q and r with the given truth
values, we have
(~F  ~F)  (~F  F)
(T  T)  (T  F)
T  F
T.
Thus, the compound statement
(~r  ~q)  (~r  q) is true.
29. Replacing p, q and r with the given truth
values, we have
~[(~T  F)  F]
~[(F  F)  F]
~[F  F]
~F
T.
Thus, the compound statement
~[(~p  q)  r] is true.
30. Replacing p, q and r with the given truth
values, we have
~[F  (~F  ~T)]
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3.2: TRUTH TABLES AND EQUIVALENT STATEMENTS 53
~[F  (T  F)]
~[F  F]
~F
T.
Thus, the compound statement
~[r  (~q  ~p)] is true.
31. Replacing p, q and r with the given truth
values, we have
~[~F  (F  ~T)]
~[T  F]
~T
F.
Thus, the compound statement
~[~q  (r  ~p)] is false.
32. Replacing p, q and r with the given truth
values, we have
~ (T  F)  ~ (T  F)
~T  ~F
FT
F.
Thus, the compound statement
~ (p  q)  ~ (p  q) is false.
Let p represent the statement “16 < 8,” which is
false, let q represent “5  4,” which is false and
let r represent “17  17,” which is true.
[E.g. p = F, q = F and r = T.]
33. Replacing p and r with the given truth
values, we have
F  T
F.
The compound statement p  r is false.
34. Replacing p and q with the observed truth
values, we have
F  ~F
F  T
T.
The compound statement p  ~q is true.
35. Replacing q and r with the observed truth
values, we have
~F  ~T
T  F
T.
The compound statement ~q  ~r is true.
36. Replacing p and r with the observed truth
values, we have
~F  ~T
T  F
F.
The compound statement ~p  ~r is false.
(F  F)  T
F  T
T.
The compound statement (p  q)  r is true.
38. Replacing p, q and r with the observed truth
values, we have
~F  (~T  ~F)
T  (F  T)
T  T
T.
The compound statement ~p  (~r  ~q) is
true.
39. Replacing p, q and r with the observed truth
values, we have
(~T  F)  ~F
(F  F)  T
F  T
T.
The compound statement (~r  q)  ~p is
true.
40. Replacing p, q and r with the observed truth
values, we have
~(F  ~F)  ~T
~(F  T)  F
~T  F
F  F
F.
The compound statement ~(p  ~q)  ~r is
false.
41. Since there are two simple statements (p and
r), we have 22  4 combinations of truth
values, or rows in the truth table, to
examine.
42. Since there are three simple statements (p, r,
and s), we have 23  8 combinations of
truth values, or rows in the truth table, to
examine.
43. Since there are four simple statements (p, q,
r, and s), we have 24  16 combinations of
truth values, or rows in the truth table, to
examine.
44. Since there are five simple statements (p, q,
r, s, and t), we have 25  32 combinations
of truth values, or rows in the truth table, to
examine.
37. Replacing p, q and r with the observed truth
values, we have
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CHAPTER 3 INTRODUCTION TO LOGIC
45. Since there are seven simple statements (p,
52. p  ~q
7
q, r, s, t, u, and v), we have 2  128
combinations of truth values, or rows in the
truth table, to examine.
46. Since there are eight simple statements (p, q,
r, s, m, n, u, and v), we have
47. If the truth table for a certain compound
statement has 64 rows, then there must be
six distinct component statements
(26  64).
48. It is not possible for a truth table of a
compound statement to have exactly 54
rows, because 54 is not a natural-number
power of 2.
49. ~p  q
p
q
~p
p  ~q
T
F
T
F
T
T
F
T
F
F
F
F
T
T
q
T
T
8
2  256 combinations of truth values, or
rows in the truth table, to examine.
~q
p
53. (q  ~p)  ~q
p
q
~p
~q
(q  ~p)
(q  ~p)  ~q
T
T
F
F
T
T
T
F
F
T
F
T
F
T
T
F
T
T
F
F
T
T
T
T
54. (p  ~q)  p
p  ~q
~q
(p  ~q)  p
p
q
~p  q
T
T
F
F
F
F
T
T
T
T
T
F
F
T
T
F
F
F
F
T
F
F
F
F
T
T
T
F
F
T
F
F
F
F
T
F
55. (p  ~q)  (p  q)
50. ~p  ~q
p
q
(p

~q)

(p

q)
p
q
~p
~q
~p  ~q
T
T
T
T
F
T
T
T
T
T
T
F
F
F
T
F
T
T
T
F
T
F
F
T
F
F
T
T
F
T
F
F
F
F
F
F
T
F
T
T
F
T
F
F
F
T
T
F
F
F
F
1
2
1
3
1
2
1
F
F
T
T
T
56. (~p  ~q)  (~p  q)
51. ~(p  q)
p
q
p  q
~(p  q)
p
q
(~p
T
T
T
F
T
T
F
F
F

~q)

(~p

q)
F
F
T
F
T
T
F
T
F
F
F
F
T
F
F
T
T
F
T
F
T
F
T
T
F
F
T
T
T
T
F
F
F
T
F
F
T
T
T
T
T
T
F
1
2
1
3
1
2
1
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3.2: TRUTH TABLES AND EQUIVALENT STATEMENTS 55
57. (~p  q)  r
59. (~p  ~q)  (~r  ~p)
q)

r
p
q
r
F
T
F
T
T
T
T
F
F
T
F
F
T
T
F
F
F
F
F
T
T
F
T
F
F
F
F
F
F
T
F
F
T
T
T
T
T
T
F
T
T
F
T
T
T
F
F
F
F
F
T
T
F
F
F
T
F
F
F
T
F
F
F
F
1
2
1
3
1
p
q
r
T
T
T
F
T
T
F
F
T
F
T
F
T
F
F
F
T
F

(~p
58. r  (p  ~q)

~q)


~p)
F
F
F
F
F
F
F
F
T
T
T
F
F
T
F
F
F
F
F
F
T
T
T
T
F
T
T
F
F
T
F
T
T
T
F
T
F
F
T
T
T
T
F
F
T
T
F
F
F
T
T
T
T
F
T
T
T
T
T
T
T
T
1
2
1
3
1
2
1
(~p
(~r
60. (~r  ~p)  (~p  ~q)
p
q
r
r

(p

~q)
p
q
r
(~r

~p)

(~p

~q)
T
T
T
T
T
T
F
F
T
T
T
F
F
F
F
F
F
F
T
T
F
F
F
T
F
F
T
T
F
T
T
F
F
F
F
F
T
F
T
T
T
T
T
T
T
F
T
F
F
F
F
F
T
T
T
F
F
F
T
T
T
T
T
F
F
T
T
F
T
F
T
T
F
T
T
T
T
F
F
F
F
T
T
F
T
T
T
T
T
F
F
T
F
F
F
F
F
F
F
T
F
T
T
T
T
T
T
F
F
F
T
T
T
F
F
T
F
F
T
F
T
T
T
T
T
T
F
F
F
F
F
F
F
T
F
F
F
T
T
T
T
T
T
T
1
3
1
2
1
1
2
1
3
1
2
1
Copyright © 2016 Pearson Education, Inc.
56
CHAPTER 3 INTRODUCTION TO LOGIC
61. ~(~p  ~q)  (~r  ~s)
~q)

(~r
F
F
T
F
F
F
F
F
T
F
T
T
F
F
T
T
T
F
F
F
F
T
T
T
T
T
F
F
T
T
F
F
F
T
F
F
T
T
F
T
T
F
T
T
T
T
F
F
T
T
T
T
T
F
F
T
F
F
F
T
F
F
T
F
T
T
T
T
F
F
T
T
T
F
T
T
F
F
T
T
T
T
T
F
F
F
F
T
T
F
T
T
T
T
T
T
F
T
T
T
T
T
T
2
1
4
2
3
2
~

p
q
r
s
(~p
T
T
T
T
T
F
T
T
T
F
T
F
T
T
F
T
T
F
T
T
F
F
T
T
F
T
T
T
F
T
F
T
F
F
T
T
F
T
F
F
F
T
F
F
T
T
T
T
T
F
T
T
F
T
F
T
F
T
F
T
F
F
F
F
T
T
F
T
T
F
F
T
F
F
T
T
F
F
F
T
F
T
T
F
F
F
F
F
T
3
1

~s)
62. (~r  s)  (~p  q)
p
q
r
s
(~r

s)

(~p

q)
T
T
T
T
F
T
T
F
F
F
T
T
T
T
F
F
F
F
F
F
F
T
T
T
F
T
T
T
T
F
F
F
T
T
T
F
F
T
T
F
F
F
F
T
T
F
T
T
F
T
T
F
F
F
F
T
F
T
F
F
F
F
F
F
F
F
T
F
F
T
T
T
T
F
F
F
F
T
F
F
F
T
T
F
F
F
F
F
F
T
T
T
F
T
T
T
T
T
T
F
T
T
F
F
F
F
F
T
T
T
F
T
F
T
T
T
T
T
T
T
T
F
T
F
F
T
T
F
T
T
T
T
F
F
T
T
F
T
T
F
T
F
F
F
F
T
F
F
F
F
F
T
F
F
F
F
F
T
T
T
T
F
T
F
F
F
F
F
F
T
T
F
F
T
F
F
1
2
1
3
1
2
1
Copyright © 2016 Pearson Education, Inc.
3.2: TRUTH TABLES AND EQUIVALENT STATEMENTS 57
63. “You can pay me now or you can pay me
later” has the symbolic form (p  q). The
negation, ~(p  q), is equivalent, by one of
De Morgan’s laws, to (~p  ~q). The
corresponding word statement is “You can’t
pay me now and you can’t pay me later.”
64. “I am not going or she is going” has the
symbolic form ~p  q. Its negation,
~(~p  q), is equivalent, by De Morgan’s,
to p  ~q. The word translation for the
negation is “I am going and she is not
going.”
65. “It is summer and there is no snow” has the
symbolic form p  ~q. The negation,
~(p  ~q), is equivalent by De Morgan’s to
~p  q. The word translation for the
negation is “It is not summer or there is
snow.”
1
is a positive number and 9 is less than
2
zero” is of the form p  q. The negation,
~(p  q), is equivalent, by De Morgan’s to
~p  ~q. The word translation for the
1
negation is “ is not a positive number or
2
9  zero.” (Note that the inequality “” is
equivalent to “not less than.”)
66. “
67. “I said yes but she said no” is of the form
p  q. The negation, ~(p  q), is
equivalent, by De Morgan’s, to ~p  ~q.
The word translation for the negation is “I
did not say yes or she did not say no.”
(Note: The connective “but” is equivalent to
that of “and.”)
68. “Dan tried to sell the software, but he was
unable to do so” is of the form
p  q. The negation, ~(p  q), is
equivalent, by De Morgan’s, to ~p  ~q.
The word translation for the negation is Dan
did not try to sell the software, or he was
able to do so.”
69. “6  1 = 5 and 9 + 13  7” is of the form
p  ~q. The negation, ~(p  ~q), is
equivalent, by De Morgan’s, to ~p  q. The
translation for the negation is “6  1  5 or
9 + 13 = 7.”
negation is “8  10 and 5 = 2.” (Note that
the inequality “” is equivalent to “  ”).
71. “Prancer or Vixen will lead Santa’s sleigh
next Christmas” is of the form p  q. The
negation, ~(p  q), is equivalent, by De
Morgan’s, to ~p  ~q. A translation for the
negation is “Neither Prancer nor Vixen will
lead Santa’s sleigh next Christmas.”
72. “The lawyer and the client appeared in
court” is of the form p  q. The negation,
~(p  q), is equivalent, by De Morgan’s, to
~p  ~q. The word translation for the
negation is “The lawyer did not appear in
court or the client did not appear in court.”
73. “For every real number x, x < 14 or x > 6” is
true since for any real number at least one of
the component statements is true.
74. “For every real number x, x > 9 or x < 9” is
false since for the real number x = 9, both
component statements are false.
75. “There exists an integer n such that n > 0
and n < 0” is false since any integer that
makes one of the component statements true
will make the other false.
76. “For some integer n, n  3 and n  3” is true
since both component statements are true for
the integer n = 3.
77. p  q
p
q
p  q
T
T
F
T
F
T
F
T
T
F
F
F
Observe that it is only the first line in the
truth table that changes for “exclusive
disjunction” since the component statements
cannot both be true at the same time.
78. The phrase “and/or” represents inclusive
disjunction since both or either component
statement(s) may be true.
70. “8 < 10 or 5  2” is of the form p  ~q. The
negation, ~(p  ~q), is equivalent, by De
Morgan’s, to ~p  q, A translation for the
Copyright © 2016 Pearson Education, Inc.
58
CHAPTER 3 INTRODUCTION TO LOGIC
79. “3 + 1 = 4  2 + 5 = 7” is false since both
component statements are true.
equivalent, and it can be said that “OR
distributes over AND.”
80. “3 + 1 = 4  2 + 5 = 10” is true since the
first component statement is true and the
second is false.
(b)
p
q
r
p

(q

r)
T
T
T
T
T
T
T
T
81. “3 + 1 = 6  2 + 5 = 7” is true since the
first component statement is false and the
second is true.
T
T
F
T
T
T
T
F
T
F
T
T
T
F
T
T
T
F
F
T
F
F
F
F
82. “3 + 1 = 12  2 + 5 = 9” is false since both
component statements are false.
F
T
T
F
F
T
T
T
F
T
F
F
F
T
T
F
83. The lady is behind Door 2. Reasoning:
Suppose that the sign on Door 1 is true.
Then the sign on Door 2 would also be true,
but this is impossible. So the sign on door 2
must be true, and the sign on door 1 must be
false. Because the sign on Door 1 says the
lady is in Room 1 and this is false, the lady
must be behind Door 2.
F
F
T
F
F
F
T
T
F
F
F
F
F
F
F
F
1
3
1
2
1
84. (a)
p

(q

r)
T
T
T
T
T
T
F
T
T
T
F
F
F
T
T
T
F
F
T
T
F
F
T
T
F
F
F
F
T
T
F
T
T
T
T
F
T
F
F
F
T
F
F
F
F
T
F
F
F
F
T
F
F
F
F
F
F
F
F
1
3
1
2
1
p
q
r
T
T
T
T
T
p
q
r
(p

q)

(p

r)
T
T
T
T
T
T
T
T
T
T
T
T
F
T
T
T
T
T
F
F
T
F
T
T
F
F
T
T
T
T
T
F
F
T
F
F
F
T
F
F
F
T
T
F
F
T
F
F
F
T
F
T
F
F
F
T
F
F
F
F
F
F
T
F
F
F
F
F
F
T
F
F
F
F
F
F
F
F
F
F
1
2
1
3
1
2
1
Since the final column (step 3) of the truth
tables are identical, the statements are
equivalent, and it can be said that “AND
distributes over OR.”
(c) Writing exercise; answers will vary.
p
q
r
(p

q)

(p

r)
T
T
T
T
T
T
T
T
T
T
T
T
F
T
T
T
T
T
T
F
T
F
T
T
T
F
T
T
T
T
T
F
F
T
T
F
T
T
T
F
F
T
T
F
T
T
T
F
T
T
F
T
F
F
T
T
F
F
F
F
F
F
T
F
F
F
F
F
T
T
F
F
F
F
F
F
F
F
F
F
1
2
1
3
1
2
1
Since the final column (step 3) of the truth
tables are identical, the statements are
85. The statement ~  p  q   ~ p  ~ q can
be expressed verbally as “The negation of a
conjunction is equivalent to the disjunction
of the negations.
3.3 Exercises
1. The statement “You can do it if you just
believe” becomes “If you just believe, then
you can do it.”
2. The statement “It must be bad for you if it’s
sweet” becomes “If it’s sweet, then it must
be bad for you.”
Copyright © 2016 Pearson Education, Inc.
3.3: THE CONDITIONAL AND CIRCUITS 59
3. The statement “Every even integer divisible
by 5 is divisible by 10” becomes “If it is an
even integer divisible by 5, then it is
divisible by 10.”
4. The statement “No perfect square integers
have units digit 2, 3, 7, or 8” becomes “If an
integer has units digit 2, 3, 7, or 8, then it’s
not a perfect square.”
5. The statement “No grizzly bears live in
California” becomes “If it is a grizzly bear,
then it doesn’t live in California.”
6. The statement “No guinea pigs get lonely”
becomes “If it is a guinea pig, then it doesn’t
get lonely.”
7. The statement “Surfers can’t stay away from
the beach” becomes “If they are surfers,
then they can’t stay away from the beach.”
8. The statement “Running Bear loves Little
White Dove” becomes “If he is Running
Bear, then he loves Little White Dove.”
9. The statement “If the antecedent of a
conditional statement is false, the
conditional statement is true” is true, since a
false antecedent will always yield a true
conditional statement.
logically equivalent. To decide if the above
is true, examine the corresponding truth
tables for each individual statement.
p
q
p

q
~p

q
T
T
T
T
T
F
T
T
T
F
T
F
F
F
F
F
F
T
F
T
T
T
T
T
F
F
F
T
F
T
T
F
1
2
1
1
2
1
This statement is true because the truth
values (columns 2) for each compound
statement are the same, showing that the
statements are equivalent.
15. “Given that ~p is true and q is false, the
conditional p  q is true” is a true statement
since the antecedent, p, must be false.
16. “Given that ~p is false and q is false, the
conditional p  q is true” is a false
statement since the antecedent, p, is true and
the consequent, q is false.
17. Writing exercise; answers will vary.
18. Writing exercise; answers will vary.
10. The statement “If the consequent of a
conditional statement is true, the conditional
statement is true” is true, since a true
consequent is always associated with a true
conditional statement (i.e., it doesn’t matter
what the truth value of the antecedent is if
the consequent itself is true).
19. “T  (7 < 3)” is a false statement, since the
antecedent is true and the consequent is
false.
11. The statement “If q is true, then
 p   q  r    q is true” is true, since
21. “F  (5  5)” is a true statement, since a
false antecedent always yields a true
conditional statement.
with a true consequent the conditional
statement is always true (even though the
antecedent may be false).
12. The statement “If p is true, the ~p  (q 
r) is true” is true since the antecedent, ~p, is
false.
13. The negation of “If pigs fly, I’ll believe it”
is “If pigs don’t fly, I won’t believe it.” This
statement is false. The negation is “Pigs fly
and I won’t believe it.”
20. “F  (4  8)” is a true statement, since a
false antecedent always yields a conditional
statement which is true.
22. “(8  8)  F” is a false statement, since the
antecedent is true and the consequent is
false.
23. “ (52  25)  (8  8  16) ” is a true
statement, since a false antecedent always
yields a true conditional statement.
24. “(5 = 12  7)  (9 > 0)” is true, since the
antecedent and the consequent are both true.
14. The statements “If it flies, then it’s a bird”
and “It does not fly or it’s a bird” are
Copyright © 2016 Pearson Education, Inc.
60
CHAPTER 3 INTRODUCTION TO LOGIC
Let s represent the statement “She sings for a
living,” let p represent the statement “he fixes
cars,” and let m represent “they collect classics.”
25. “~m  p” expressed in words, becomes “If
they do not collect classics, then he fixes
cars.”
26. “p  ~m” expressed in words, becomes “If
he fixes cars, then they do not collect
classics.”
27. “s  (m  p)” expressed in words,
becomes “If she sings for a living, then they
collect classics and he fixes cars.”
28. “(s  p)  m” expressed in words,
becomes “If she sings for a living and he
fixes cars, then they collect classics.”
29. “~p  (~m  s)” expressed in words,
becomes “If he does not fix cars, then they
do not collect classics or she sings for a
living.”
30. “(~s  ~m)  ~p” expressed in words,
becomes “If she does not sing for a living or
they do not collect classics, then he does not
fix cars.”
Let b represent the statement “I take my ball,” let s
represent the statement “it is sunny” and let p
represent “the park is open.”
31. The statement “If I take my ball, then the
park is open,” can be symbolized as
“b  p.”
32. The statement “If I do not take my ball, then
it is not sunny” can be symbolized as
“~b  ~s.”
33. The statement “The park is open, and if it is
sunny then I do not take my ball” can be
symbolized as “p  (s  ~b).”
34. The statement “I take my ball, or if the park
is open, then it is sunny” can be symbolized
as “b  (p  s).”
35. The statement “It is sunny if the park is
open” can be symbolized as “p  s.”
36. The statement “I’ll take my ball if it is not
sunny” can be symbolized as “~s  b.”
Assume that p and r are false, and q is true.
37. Replacing r and q with the given truth
values, we have
~F  T
TT
T.
Thus, the compound statement ~r  q is
true.
38. Replacing p and q with the given truth
values, we have
TF
F.
Thus, the compound statement q  p is
false.
39. Replacing p and q with the given truth
values, we have
FT
T.
Thus, the compound statement p  q is
true.
40. Replacing r and p with the given truth
values, we have
~F  F
TF
F.
Thus, the compound statement ~r  p is
false.
41. Replacing p, r and q with the given truth
values, we have
~F  (T  F)
TF
F.
Thus, the compound statement
~p  (q  r) is false.
42. Replacing p, r and q with the given truth
values, we have
(~F  F)  F
(T  F)  F
TF
F.
Thus, the compound statement
(~r  p)  p is false.
43. Replacing p, r and q with the given truth
values, we have
~T  (F  F)
FF
T.
Thus, the compound statement
~q  (p  r) is true.
Copyright © 2016 Pearson Education, Inc.
3.3: THE CONDITIONAL AND CIRCUITS 61
51. (~p  q)  p
44. Replacing p, r and q with the given truth
values, we have
(~F  ~T)  (F  ~F)
(T  F)  (F  T)
FF
T.
Thus, the compound statement
(~p  ~q)  (p  ~r) is true.
45. Replacing p, r and q with the given truth
values, we have
(F  ~T)  (~F  ~F)
(F  F)  (T  T)
TT
T.
Thus, the compound statement
(p  ~q)  (~p  ~r) is true.
p
q
(~p

q)

p
T
T
F
T
T
T
T
T
F
F
T
F
T
T
F
T
T
T
T
F
F
F
F
T
F
F
T
F
1
2
1
3
2
52. (p  q)  (p  q)
46. Replacing p, r and q with the given truth
values, we have
[(F  ~T)  (F  F)]  F
[(F  F)  T]  F
T  T  F
T  F
F
Thus, the compound statement
(p  ~q)  (p  r)  r is false.
(p

q)

(p

q)
T
T
T
T
T
T
T
T
F
T
F
F
T
T
T
F
F
T
F
F
T
T
F
T
T
F
F
F
F
F
T
F
F
F
1
2
1
3
1
2
1
p
q
T
T
Since this statement is always true (column
3), it s a tautology.
53. (p  q)  (q  p)
p
q
(p

q)

(q

p)
T
T
T
T
T
T
T
T
T
47. Writing exercise; answers will vary.
T
F
T
T
F
T
F
T
T
48. Answers will vary. One example is
p  [(~q  r)  p]. Observe that any “If...,
then...” statement with antecedent p, q or r
will work, since all that is needed to make
the statement true is a false antecedent.
F
T
F
T
T
T
T
T
F
F
F
F
F
F
T
F
F
F
1
2
1
3
1
2
1
49. ~q  p
p
q
~q

p
T
T
F
T
T
T
F
T
T
T
F
T
F
T
F
F
F
T
F
F
1
2
1
50. (~q  ~p)  ~q
(~q

T
F
T
F
F
F
F
T
F
F
T
T
F
T
F
T
T
F
F
F
F
T
T
T
T
T
1
2
1
3
2
p
q
T
T
~p)

Since this statement is always true (column
3), it is a tautology.
54. (~p  ~q)  (p  q)
~q)

(p

q)
T
F
T
T
T
T
T
T
F
T
F
F
F
F
T
F
F
T
T
T
T
F
F
F
F
1
2
1
3
1
2
1
(~p
p
q
T
T
F
T
F
F
F
T
T
F
F
~q
Copyright © 2016 Pearson Education, Inc.

62
CHAPTER 3 INTRODUCTION TO LOGIC
55. [(r  p)  ~q]  p
p
q
r
[(r

p)

~q]

p
T
T
T
T
T
T
F
F
T
T
T
T
F
F
T
T
F
F
T
T
T
F
T
T
T
T
T
T
T
T
T
F
F
F
T
T
T
T
T
T
F
T
T
T
T
F
F
F
T
F
F
T
F
F
F
F
F
F
T
F
F
F
T
T
T
F
T
T
F
F
F
F
F
F
F
F
F
T
T
F
1
2
1
3
2
4
3
56. [(r  p)  (p  q)]  p
p
q
r
[(r

p)

(p  q)]

p
T
T
T
T
T
T
T
T
T
T
T
T
F
F
F
T
F
T
T
T
T
F
T
T
T
T
F
F
T
T
T
F
F
F
F
T
F
F
T
T
F
T
T
T
F
F
F
F
T
F
F
T
F
F
F
F
F
F
T
F
F
F
T
T
F
F
F
F
T
F
F
F
F
F
F
F
F
F
T
F
1
2
1
3
2
4
3
Since this statement is always true (column 4), it is a tautology.
Copyright © 2016 Pearson Education, Inc.
3.3: THE CONDITIONAL AND CIRCUITS 63
57. (~r  s)  (p  ~q)
p
q
r
s
(~r

s)

(p

~q)
T
T
T
T
F
T
T
T
T
F
F
T
T
T
F
F
T
F
T
T
F
F
T
T
F
T
T
T
T
T
T
F
F
T
T
F
F
T
F
F
F
T
F
F
T
F
T
T
F
T
T
T
T
T
T
T
F
T
F
F
T
F
T
T
T
T
T
F
F
T
T
T
T
T
T
T
T
T
F
F
F
T
F
F
T
T
T
T
F
T
T
T
F
T
T
T
F
T
F
F
T
T
F
F
T
F
T
F
T
F
F
T
F
T
T
T
T
T
F
T
F
F
T
F
F
T
F
F
T
F
T
F
F
F
T
T
F
T
T
T
F
T
T
F
F
T
F
F
T
F
T
F
T
T
F
F
F
T
T
T
T
T
F
T
T
F
F
F
F
T
F
F
T
F
T
T
1
2
1
3
1
2
1
58. (~p  ~q)  (s  r)
p
q
r
s
(~p

~q)

(s

r)
T
T
T
T
F
F
F
T
T
T
T
T
T
T
F
F
F
F
T
F
T
T
T
T
F
T
F
F
F
T
T
F
F
T
T
F
F
F
F
F
T
F
T
F
T
F
T
T
F
F
T
T
T
T
T
T
F
T
F
F
F
T
T
F
T
T
T
F
F
T
F
F
T
T
T
F
F
T
F
F
F
F
F
T
T
F
T
F
F
T
T
T
T
F
F
T
T
T
T
F
T
T
F
T
F
F
T
F
T
T
F
T
F
T
T
F
F
T
T
F
F
F
T
F
F
T
F
F
T
F
T
F
F
F
T
T
T
T
T
T
T
T
T
F
F
T
F
T
T
T
T
F
T
T
F
F
F
T
T
T
T
F
T
F
F
F
F
F
F
T
T
T
T
F
T
F
1
2
1
3
1
2
1
Copyright © 2016 Pearson Education, Inc.
64
CHAPTER 3 INTRODUCTION TO LOGIC
59. The statement is not a tautology if only one
F appears in the final column of a truth
table, since a tautology requires all T’s in
the final column.
60. Consider the negation of the statement.
61. The negation of “If that is an authentic
Coach bag, I’ll be surprised” is “That is an
authentic Coach bag and I am not
surprised.”
62. The negation of “If Muley Jones hits that
note, he will shatter glass” is “Muley Jones
hits that note and he will not shatter glass.”
Florida.” An equivalent statement would be
“The person is not a resident of Pensacola or
is a resident of Florida.”
72. An equivalent conditional statement to “All
women were once girls” is “If you are a
woman, then you were once a girl.” An
equivalent statement would be “The person
is not a woman or was once a girl.”
73. The statements p  q and ~p  q are
equivalent if they have the same truth tables.
63. The negation of “If the bullfighter doesn’t
get going, he’s going to get gored” is “The
bullfighter doesn’t get going and he doesn’t
get gored.”
64. The negation of “If you don’t say ‘I do,’
then you’ll regret it for the rest of your life”
is “Don’t say ‘I do’ and you will not regret
it for the rest of your life.”
65. The negation of “If you want to be happy
for the rest of your life, never make a pretty
woman your wife” is “You want to be
happy for the rest of your life and you make
a pretty woman your wife.”
74.
67. An equivalent statement to “If you give your
plants tender, loving care, they will flourish”
is “You do not give your plants tender,
loving care or they flourish.”
69. An equivalent statement to “If she doesn’t,
he will” is “She does or he will.”
70. An equivalent statement to “If I say ‘black’,
she says ‘white’” is “I do not say ‘black’ or
she says ‘white.’”
71. An equivalent conditional statement to “All
residents of Pensacola are residents of
Florida” is “If you are a resident of
Pensacola, then you are a resident of

q
T
F
T
T
F
F
F
F
T
T
T
T
T
F
T
F
T
T
F
1
2
1
1
2
1

q
T
T
T
F
T
F
F
T
F
F
F
q
T
T
Since the truth values in the final columns
for each statement are the same, the
statements are equivalent.
66. The negation of “If I had a hammer, I’d
hammer in the morning” is “I had a hammer,
and I wouldn’t hammer in the morning.”
68. An equivalent statement to “If you scratch
my back, I’ll scratch yours” is “You do not
scratch my back or I will scratch yours.”
~p
p
p
p
q
~
(p

q)
p

~q
T
T
F
T
T
T
T
F
F
T
F
T
T
F
F
T
T
T
F
T
F
F
T
T
F
F
F
F
F
F
F
T
F
F
F
T
3
1
2
1
1
2
1
Since the truth values in the final columns
for each statement are the same, the
statements are equivalent.
75.
p
q
p

q
~q

~p
T
T
T
T
T
F
T
F
T
F
T
F
F
T
F
F
F
T
F
T
T
F
T
T
F
F
F
T
F
T
T
T
1
2
1
1
2
1
Since the truth values in the final columns
for each statement are the same, the
statements are equivalent.
Copyright © 2016 Pearson Education, Inc.
3.3: THE CONDITIONAL AND CIRCUITS 65
76.
p

q
q

p
T
T
T
T
T
T
T
F
T
F
F
F
T
T
F
T
F
T
T
T
F
F
F
F
F
T
F
F
T
F
1
2
1
1
2
1
p
q
T
T
Since the truth values in the final columns for each statement are not the same, the statements are not
equivalent.
77.
p
q
p

~q
~p

~q
T
T
T
F
F
F
F
F
T
F
T
T
T
F
T
T
F
T
F
T
F
T
T
F
F
F
F
T
T
T
T
T
1
2
1
1
2
1
Since the truth values in the final columns for each statement are the same, the statements are equivalent.
78.
p
q
~p

q
~p

q
T
T
F
F
T
F
T
T
T
F
F
F
F
F
T
F
F
T
T
T
T
T
T
T
F
F
T
F
F
T
F
F
1
2
1
1
2
1
Since the truth values in the final columns for each statement are not the same, the statements are not
equivalent.
79.
p
q
q

~p
p

~q
T
T
T
F
F
T
F
F
T
F
F
T
F
T
T
T
F
T
T
T
T
F
T
F
F
F
F
T
T
F
T
T
1
3
2
1
3
2
Since the truth values in the final columns for each statement are the same, the statements are equivalent.
Copyright © 2016 Pearson Education, Inc.
66
CHAPTER 3 INTRODUCTION TO LOGIC

q)

r
T
T
T
T
T
T
T
T
T
T
F
T
T
T
F
T
T
T
F
T
T
F
T
F
F
F
T
T
T
F
F
F
T
T
T
T
F
F
F
T
F
F
F
3
1
2
1
~(
80. p
q
r
T
T
T
F
T
T
F
F
T
F
T
F
T
F
F
F
T
T
F
T
F
F
F
F
F
p
q)

r
T
T
T
T
T
T
T
F
T
F
T
T
T
T
F
T
F
T
F
T
T
T
T
T
F
F
T
T
T
F
T
T
F
F
F
T
T
F
F
F
F
F
F
F
4
1
1
2
1
3
1
(p

Since the truth values in the final columns for each statement are the same, the statements are equivalent.
81. In the diagram, two series circuits are shown, which correspond to p  q and
p  ~q. These circuits, in turn, form a parallel circuit. Thus, the logical statement is
(p  q)  (p  ~q).
One pair of equivalent statements listed in the text includes
(p  q)  (p  ~q)  p  (q  ~q).
Since (q  ~q) is always true, p  (q  ~q) simplifies to p  T  p.
82. In the diagram, a parallel circuit is shown, which corresponds to r  q. This circuit, in turn, is in series
with p. Thus, the logical statement is p  (r  q).
83. In the diagram, a series circuit is shown, which corresponds to ~q  r. This circuit, in turn, forms a
parallel circuit with p. Thus, the logical statement is p  (~q  r).
84. The diagram shows p in series with a parallel circuit consisting of two smaller parallel circuits involving
q and ~p, and ~p and ~q, respectively. Thus, the logical statement is p   ~ p  q    ~ p  ~ q   .
This is equivalent to p   ~ p  q  ~ p  ~ q  , which is equivalent to p  T , which equivalent to p.
85. In the diagram, a parallel circuit corresponds to p  q. This circuit is parallel to ~p. Thus, the total
circuit corresponds to the logical statement ~p  (p  q).
This statement in turn, is equivalent to
(~p  p)  (~p  q).
Since ~p  p is always true, we have
T  (~p  q)  T.
86. The diagram shows two parallel circuits,
~p  q and ~p  ~q which are parallel to each other. Thus, the total circuit can be represented as (~p 
q)  (~p  ~q).
This circuit can be simplified using the following equivalences:
(~p  q)  (~p  ~q)  ~p  q  ~p  ~q
 ~p  q  ~q
 ~p  (q  ~q)
 ~p  T
 T.
87. The logical statement, p  (q  ~p), can be represented by the following circuit.
The statement, p  (q  ~p), simplifies to p  q as follows:
Copyright © 2016 Pearson Education, Inc.
3.3: THE CONDITIONAL AND CIRCUITS 67
p  (q  ~p)  (p  q)  (p  ~p)
 (p  q)  F
 p  q.
88. The logical statement, (~p  ~q)  ~r, can
be represented by the following circuit.
89. The logical statement, (p  q)  (~p  ~q),
can be represented by the following circuit.
The statement, (p  q )  (~p  ~q),
simplifies to F as follows:
(p  q)  (~p  ~q)
 [p  (~p  ~q)]  [q  (~p  ~q)]
 [p  ~p  ~q]  [q  ~q  ~p]
 [F  ~q]  [F  ~p]
 F  F
 F.
90. The logical statement,
(~q  ~p)  (~p  q), can be represented
by the following circuit.
The statement, (~q  ~p)  (~p  q),
simplifies to ~p  q as follows:
(~q  ~p)  (~p  q)
 [~q  (~p  q)]  [~p  (~p  q)]
 [~q  ~p  q]  [~p  ~p  q]
 [(~q  q)  ~p]  [(~p  ~p)  q]
 (T  ~p)  (~p  q)
 T  (~p  q)
 ~p  q.
91. The logical statement, [(p  q)  r]  ~p,
can be represented by the following circuit.
The statement, [(p  q)  r]  ~p,
simplifies to (r  ~p)  q as follows:
[(p  q)  r]  ~p
 [(p  r)  (q  r)]  ~p
 [(p  r)  ~p]  [(q  r)  ~p]
 [p  r  ~p]  [(q  r)  ~p]
 [(p  ~p)  r]  [(r  ~p)  q]
 (F  r)  [(r  ~p)  q]
 F  [(r  ~p)  q]
 (r  ~p)  q or q  (r  ~p).
92. The logical statement,
[(~p  ~r)  ~q]  (~p  r), can be
represented by the following circuit.
The statement,
[(~p  ~r)  ~q]  (~p  r), can simplify
to (~p  r)  ~q in the following manner.
Both [(~p  ~r)  ~q] and (~p  r) must
be true. But if (~p  r) is true, then (~p 
~r) is false. If (~p  ~r) is false, then ~q
must be true for the original disjunction to
be true. Thus,
[(~p  ~r)  ~q]  (~p  r)
 (F  ~q)  (~p  r)
 ~q  (~p  r)
 (~p  r)  ~q or ~q  (~p  r).
93. The logical statement, ~q  (~p  q), can
be represented by the following circuit.
The statement, ~q  (~p  q), simplifies to
p  q as follows:
~q  (~p  q)  ~q  (p  q)
 q  (p  q)
 q  p  q
 p  q  q
 p  (q  q)
 p  q.
94. The logical statement, ~p  (~p  ~q), can
be represented by the following circuit.
The statement ~p  (~p  ~q), simplifies
to T as follows:
~p  (~p  ~q)  p  (~p  ~q)
 p  ~p  ~q
 (p  ~p)  ~q
 T  ~q
 T.
95. Referring to Figures 5 and 6 of Example 6
in the text:
Cost per year of the circuit in Figure 5
= number of switches  $.06  24 hr
 365 days
= (4)  (.06)  24  365
= $2102.40.
Cost per year of the circuit in Figure 6
= number of switches  $.06  24 hr
 365 days
= (3)  (.06)  24  365
= $1576.80.
Copyright © 2016 Pearson Education, Inc.
68
CHAPTER 3 INTRODUCTION TO LOGIC
Thus, the savings is $2102.40  $1576.80 =
$525.60.
96. Writing exercise; answers will vary.
97.
The logic circuit shown is equivalent to the
statement ~(A  A)  B, which is in turn
equivalent to ~A  B, which is equivalent to
A  B.
3.4 Exercises
For each given conditional statement (symbolically
as p  q), write (a) the converse (q  p), (b) the
inverse (~p  ~q), and (c) the contrapositive
(~q  ~p) in if...then forms. Wording may vary in
the answers to Exercises 110.
1. The conditional statement: If beauty were a
minute, then you would be an hour.
(a) Converse: If you were an hour, then
beauty would be a minute.
4. The conditional statement: If I had a nickel
for each time that happened, I would be
rich.
(a) Converse: If I were rich, then I would
have a nickel for each time that
happened.
(b) Inverse: If I did not have a nickel for
each time that happened, then I would
not be rich.
(c) Contrapositive: If I were not rich, then I
would not have a nickel for each time
that happened.
It is helpful to restate the conditional statement in
an “if...then” form for the exercises 58 and 10.
5. The conditional statement: If you walk in
front of a moving car, then it is dangerous to
your health.
(a) Converse: If it is dangerous to your
health, then you walk in front of a
moving car.
(b) Inverse: If beauty were not a minute,
then you would not be an hour.
(b) Inverse: If you do not walk in front of a
moving car, then it is not dangerous to
your health.
(c) Contrapositive: If you were not an
hour, then beauty would not be a
minute.
(c) Contrapositive: If it is not dangerous to
your health, then you do not walk in
front of a moving car.
2. The conditional statement: If you lead, then
I will follow.
(a) Converse: If I follow, then you lead.
(b) Inverse: If you do not lead, then I will
not follow.
(c) Contrapositive: If I do not follow, then
you do not lead.
3. The conditional statement: If it ain’t broke,
don’t fix it.
(a) Converse: If you don’t fix it, then it
ain’t broke.
(b) Inverse: If it’s broke, then fix it.
(c) Contrapositive: If you fix it, then it’s
broke.
6. The conditional statement: If it’s milk, then
it contains calcium.
(a) Converse: If it contains calcium, then
it’s milk.
(b) Inverse: If it’s not milk, then it does not
contain calcium.
(c) Contrapositive: If it does not contain
calcium, then it’s not milk.
7. The conditional statement: If they are birds
of a feather, then they flock together.
(a) Converse: If they flock together, then
they are birds of a feather.
(b) Inverse: If they are not birds of a
feather, then they do not flock together.
Copyright © 2016 Pearson Education, Inc.
3.4: THE CONDITIONAL AND RELATED STATEMENTS 69
(c) Contrapositive: If they do not flock
together, then they are not birds of a
feather.
8. The conditional statement: If it is a rolling
stone, then it gathers no moss.
(a) Converse: If it gathers no moss, then it
is a rolling stone.
(b) Inverse: If it is not a rolling stone, then
it gathers moss.
(c) Contrapositive: If it gathers moss, then
it is not a rolling stone.
9. The conditional statement: If you build it,
then he will come.
(a) Converse: If he comes, then you built it.
(b) Inverse: If you don’t build it, then he
won’t come.
(c) Contrapositive: If he doesn’t come,
then you didn’t build it.
10. The conditional statement: If there’s smoke,
then there’s fire.
(a) Converse: If there’s fire, then there’s
smoke.
(b) Inverse: If there’s no smoke, then
there’s no fire.
(c) Contrapositive: If there’s no fire, then
there’s no smoke.
11. The conditional statement: p  ~q.
(b) Inverse: p  q.
(c) Contrapositive: q  p.
14. The conditional statement: ~q  ~p.
(a) Converse: ~p  ~q.
(b) Inverse: q  p.
(c) Contrapositive: p  q.
15. The conditional statement: p  (q  r).
(a) Converse: (q  r)  p.
(b) Inverse: ~p  ~(q  r) or
~p  (~q  ~r).
(c) Contrapositive: (~q  ~r)  ~p.
16. The conditional statement: (r  ~q)  p.
(a) Converse: p  (r  ~q).
(b) Inverse: ~(r  ~q)  ~p or
(~r  q)  ~p.
(c) Contrapositive: ~p  ~(r  ~q) or
~p  (~r  q).
17. Writing exercise; answers will vary.
18. Writing exercise; answers will vary.
Writing the statements, Exercises 1940, in the
form “if p, then q” we arrive at the following
results.
19. The statement “If the Kings go to the
playoffs, pigs will fly” becomes “If the
Kings go to the playoffs, then pigs will fly.”
(a) Converse: ~q  p
(b) Inverse: ~p  q.
(c) Contrapositive: q  ~p.
12. The conditional statement: ~p  q.
(a) Converse: q  ~p.
20. The statement “If I score 90% or higher on
my test, I’ll go to a movie” becomes “If I
score 90% or higher on my test, then I’ll go
to a movie”.
21. The statement “Legs of 3 and 4 imply a
hypotenuse of 5” becomes “If it has legs of
3 and 4, then it has a hypotenuse of 5.”
(b) Inverse: p  ~q.
(c) Contrapositive: ~q  p.
13. The conditional statement: ~p  ~q.
22. The statement “’This is a leap year’ implies
that next year is not” becomes “If this is a
leap year, then next year is not.”
(a) Converse: ~q  ~p.
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CHAPTER 3 INTRODUCTION TO LOGIC
23. The statement “All whole numbers are
rational numbers” becomes “If a number is a
whole number, then it is a rational number.”
24. The statement “No irrational numbers are
rational” becomes “If a number is a rational
number, then it is not an irrational number.”
25. The statement “Doing logic puzzles is
sufficient for driving me crazy” becomes “If
I do logic puzzles, then I am driven crazy.”
26. The statement “Being in Kalamazoo is
sufficient for being in Michigan” becomes
“If you are in Kalamazoo, then you are in
Michigan.”
27. The statement “Two coats of paint are
necessary to cover the graffiti” becomes “If
the graffiti are to be covered, then two coats
of paint must be used.”
28. The statement “Being an environmentalist is
necessary for being elected” becomes “If
one is elected, then one is an
environmentalist.”
29. The statement “Employment will improve
only if the economy recovers” becomes “If
employment improves, then the economy
recovers.”
30. The statement “The economy will recover
only if employment improves” becomes “IF
the economy recovers, then employment
improves.”
35. The statement “A rectangle is a
parallelogram with perpendicular adjacent
sides” becomes “If the figure is a rectangle,
then it is a parallelogram with perpendicular
adjacent sides.”
36. The statement “A square is a rectangle with
two adjacent sides equal” becomes “If the
figure is a square, then it is a rectangle with
two adjacent sides equal.”
37. The statement “A triangle with two
perpendicular sides is a right triangle”
becomes “If a triangle has two perpendicular
sides, then it is a right triangle.”
38. The statement “A parallelogram is a foursided figure with opposite sides parallel”
becomes “If the figure is a parallelogram,
then it is a four-sided figure with opposite
sides parallel.”
39. The statement “The square of a three-digit
number whose units digit is 5 will end in
25” becomes “If a three-digit number whose
units digit is 5 is squared, then the square
will end in 25.”
40. The statement “An integer whose units digit
is 0 or 5 is divisible by 5” becomes “If an
integer has a units digit of 0 or 5, then it is
divisible by 5.”
41. Option D is the answer since “r is necessary
for s” represents the converse, s  r, of all
of the other statements.
31. The statement “No whole numbers are not
integers” becomes “If a number is a whole
number, then it is an integer.”
42. Writing exercise; answers will vary.
32. The statement “No integers are irrational
numbers” becomes “If a number is an
integer, then it is rational.”
44. Writing exercise; answers will vary.
33. The statement “The Phillies will win the
pennant when their pitching improves”
becomes “If their pitching improves, then
the Phillies will win the pennant.”
43. Writing exercise; answers will vary.
45. The statement “6 = 9  3 if and only if
8 + 2 = 10” is true, since this is a
biconditional composed of two true
statements.
34. The statement “The grass will be greener
when we’re on the other side” becomes “If
we’re on the other side, then the grass will
be greener.”
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3.4: THE CONDITIONAL AND RELATED STATEMENTS 71
46. The statement “3 + 1  7 if and only if
8  8” is false since this is a biconditional consisting of a true and a false statement.
47. The statement “8 + 7  15 if and only if 3  5  8” is false, since this is a biconditional consisting of a
false and a true statement.
48. The statement “6  2 = 18 if and only if 9 + 7  16” is true, since this is a biconditional consisting of two
false statements.
49. The statement “George H.W. Bush was president if and only if George W. Bush was not president” is
false, since this is a biconditional consisting of a true and a false statement.
50. McDonald’s sells Whoppers if and only if Apple manufactures iPods” is false, since this is a
biconditional consisting of a false and a true statement.
51. The statements “Michael Jackson is alive” and “Michael Jackson is dead” are contrary, since both cannot
be true at the same time.
52. The statements “That book is nonfiction” and “That book costs more than $150” are consistent, since
both statements can be true.
53. The statements “This number is a whole number” and “This same number is irrational” are contrary,
since both cannot be true at the same time.
54. The statements “This number is positive” and “This same number is a natural number” are consistent,
since both statements can be true.
55. The statements “This number is an integer” and “This same number is a rational number” are consistent,
since both statements can be true.
56.
The logic circuit shown is expressed symbolically as
~  ~  A  ~  A  B    ~  B  ~  A  B   
  A ~  A  B     B ~  A  B    by DeMorgan's laws 
We then build the truth table for this statement.

B))

(B

~
(A

B) )
T
~ (A
F T
T
T
T
T
T
F
T
T
T
T
F
T
T
F
F
F
F
F
T
T
F
F
F
F
F
T
T
F
T
T
F
F
T
T
F
T
T
F
F
F
T
F
T
T
F
F
F
4
3
1
2
1
5
1
4
3
1
2
1
A
B
(A

T
T
T
T
F
T
F
T
F
F
1
The final column (step 5) shows truth values that match those for the biconditional (true when A and B
have the same truth value, false when A and B have different truth values). The truth tables for the
biconditional p  q and the statement ~  p  q  are shown side-by-side on the next page.
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CHAPTER 3 INTRODUCTION TO LOGIC
p
q
p

q
~
(p

q)
T
T
T
T
T
T
T
F
T
T
F
T
F
F
F
T
T
F
F
T
F
F
T
F
F
T
T
F
F
F
T
F
T
F
F
F
1
2
1
3
1
2
1
From their truth tables, we see that p  q and ~  p  q  are equivalent.
57. Let p represent “the troll is Truthful Troll” and let q represent “the door to the east leads to freedom.”
Then (1) the statement “if the troll is Truthful Troll, then the conditional statement he uttered is true” is
symbolized by p   p  q  , and (2) the statement “if the conditional statement he uttered is true, then
he is Truthful Troll” is symbolized  p  q   p .
58. The biconditional formed from the statements in Exercise 58 is p   p  q  .
(a) The truth table for this statement is shown below.
p

(p

q)
T
T
T
T
T
T
F
T
F
T
F
F
F
T
F
F
F
T
T
F
F
F
F
F
T
F
1
3
1
2
1
p
q
T
T
(b) Since this biconditional must be true, the first row of the truth table corresponds to the situation.
Thus, p and q are both true. That is, the troll is Truthful Troll, and the eastern door leads to freedom.
3.5 Exercises
1. Draw an Euler diagram where the region representing “amusement parks” must be inside the region
representing “locations that have thrill rides” so that the first premise is true.
Locations that have thrill rides
Amusement parks
x
x represents Universal Orlando
Let x represent the amusement park Universal Orlando. By the second premise, x must lie in the
“amusement parks” region. Since this forces the conclusion to be true, the argument is valid.
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3.5: ANALYZING ARGUMENTS WITH EULER DIAGRAMS 73
2. Draw an Euler diagram where the region representing “disc jockeys” must be inside the region
representing “those who play music” so that the first premise is true.
Those who play music
Disc jockeys
x
x represents Calvin
Let x represent Calvin. By premise 2, x must lie in the “disc jockeys” region. Since this forces the
conclusion to be true, the argument is valid.
3. Draw an Euler diagram where the region representing “celebrities” must be inside the region
representing “those who have problems” so that the first premise is true.
Those who have problems
x
Celebrities
x
x represents that man
Let x represent “that man.” By the second premise, x must lie in the “those who have problems” region.
Thus, he could be inside or outside the inner region. Since this allows for a false conclusion (he doesn’t
have to be in the “celebrities” region for both premises to be true), the argument is invalid.
4. Draw an Euler diagram where the region representing “Southerners” lies inside the region representing
“those who speak with an accent.” Let x represent Nick. By the second premise, Nick must lie inside the
region of those who speak with an accent and hence may lie inside the inner or the outer region. But, for
the conclusion to be true, he must lie only inside the inner region. Thus, the argument is invalid.
Those who speak with an accent
x
Southerners
x
x represents Nick
5. Draw an Euler diagram where the region representing “dogs” must be inside the region representing
“creatures that love to bury bones” so that the first premise is true.
Creatures that love to bury bones
Dogs
x
x represents Puddles
Let x represent “Puddles.” By the second premise, x must lie outside the region representing “creatures
that love to bury bones.” Since this forces the conclusion to be true, the argument is valid.
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CHAPTER 3 INTRODUCTION TO LOGIC
6. Draw an Euler diagram where the region representing “vice-presidents” must be inside the region
representing “those who use cell phones” so that the first premise is true.
Let x represent “Bob.” By the second premise, x must lie outside the region representing “those who use
cell phones.” Since this forces the conclusion to be true, the argument is valid.
7. Draw an Euler diagram where the region representing “residents of Colorado” must be inside the region
representing “those who know how to breathe thin air” so that the first premise is true.
Let x represent “Julie.” By the second premise, x must lie in the “those who know how to breathe thin
air” region. Thus, she could be inside or outside the inner region. Since this allows for a false conclusion
(she doesn’t have to be in the “residents of Colorado” region for both premises to be true), the argument
is invalid.
8. Draw an Euler diagram where the region representing “people who drive” must be inside the region
representing “people who have a photo I.D.” so that the first premise is true.
Let x represent “Kay.” By the second premise, x must lie in the “people who have a photo I.D.” region.
Thus, she could be inside or outside the inner region. Since this allows for a false conclusion (she
doesn’t have to be in the “people who drive” region for both premises to be true), the argument is
invalid.
9. Draw an Euler diagram where the region representing “dinosaurs” intersects the region representing
“plant-eaters.” This keeps the first premise true.
Let x represent “Danny.” By the second premise, x must lie in the region representing “plant-eaters.”
Thus, he could be inside or outside the region “dinosaurs.” Since this allows for a false conclusion, the
argument is invalid.
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3.5: ANALYZING ARGUMENTS WITH EULER DIAGRAMS 75
10. Draw an Euler diagram where the region representing “philosophers” intersects the region representing
“those who are absent minded.” This keeps the first premise true.
Let x represent “Nicole.” By the second premise, x must lie in the region representing “philosophers.”
Thus, she could be inside or outside the region representing “people who are absent minded.” Since this
allows for a false conclusion, the argument is invalid.
11. Draw an Euler diagram where the region representing “nurses” intersects the region representing “those
who belong to a union.” This keeps the first premise true.
Let x represent “Heather.” By the second premise, x must lie in the region representing “nurses.” Thus,
she could be inside or outside the region “those who belong to a union.” Since this allows for a false
conclusion, the argument is invalid.
12. The following represents one way to diagram the premises so that they are true; however, the argument
is invalid since, according to the diagram, no trucks have both sound systems and gun racks making the
conclusion false.
13. Interchanging the second premise and the conclusion of Example 3 (in the text) yields the following
argument.
All magnolia trees have green leaves.
That plant is a magnolia tree.
That plant has green leaves.
Draw an Euler diagram where the region representing “Magnolia trees” must be inside the region
representing “Things that have green leaves” so that the first premise is true.
Let x represent “That plant.” By the second premise, x must lie inside the region representing “Magnolia
trees.” Since this forces the conclusion to be true, the argument is valid, which makes the answer to the
question yes.
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CHAPTER 3 INTRODUCTION TO LOGIC
14. The valid argument of Example 4 (in the text) is,
All expensive things are desirable.
All desirable things make you feel good.
All things that make you feel good make
you live longer.
All expensive things make you live longer.
Another possible conclusion, which will keep the argument valid is “All expensive things make you feel
good.” The argument remains valid since the premises diagrammed (Figure 13, in the text) force this
conclusion to be true also.
15. The following is a valid argument which can be constructed from the given Euler diagram.
All people with blue eyes have blond hair.
Erin does not have blond hair.
Erin does not have blue eyes.
16. The following is a valid argument which can be constructed from the given Euler diagram.
RNs can perform all procedures that LVNs can perform.
LVNs can give vaccinations.
RNs can give vaccinations.
17. The following represents one way to diagram the premises so that they are true; however, the argument
is invalid since, according to the diagram, all birds are planes, which is false even though the stated
conclusion is true.
18. The following Euler diagram represents true premises.
Since the diagram forces the conclusion to be true also, the argument is valid.
19. The following Euler diagram yields true premises. It also forces the conclusion to be true.
Thus, the argument is valid. Observe that the diagram is the only way to show true premises.
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3.5: ANALYZING ARGUMENTS WITH EULER DIAGRAMS 77
20. The following Euler diagram represents true premises.
The argument is invalid even though the conclusion is true since the diagram implies that no chickens are
birdsa false statement.
21. The following Euler diagram represents true premises.
But x can reside inside or outside of the “Cities that are northeast of Deming” diagram. In the one case (x
inside) the conclusion is true. In the other case (x outside) the conclusion is false. Since true premises
must always give a true conclusion, the argument is invalid.
22. The following Euler diagram yields true premises. It also forces the conclusion to be true.
Thus, the argument is valid. Observe that the diagram is the only way to show true premises.
23. The following Euler diagram represents the two premises as being true and we are forced into a true
conclusion.
Thus, the argument is valid.
24. The following Euler diagram represents true premises.
No information, however, is given regarding the relationship between the largest angle and the longest
side. The argument is invalid even though the conclusion is true.
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CHAPTER 3 INTRODUCTION TO LOGIC
The premises marked A, B, and C are followed by
several possible conclusions (Exercises 2530).
Take each conclusion in turn, and check whether
the resulting argument is valid or invalid.
A. All people who drive contribute to air
pollution.
B. All people who contribute to air
pollution make life a little worse.
C. Some people who live in a suburb make
life a little worse.
Diagram the three premises to be true.
25. We are not forced into the conclusion,
“Some people who live in a suburb
contribute to air pollution” since option (A)
represents true premises and a false
conclusion. Thus, the argument is invalid.
26. We are not forced into the conclusion,
“Some people who live in a suburb drive”
since diagrams (A) and (B) represent true
premises where this conclusion is false.
Thus, the argument is invalid.
27. We are not forced into the conclusion,
“Suburban residents never drive” since
diagram (C) represents true premises where
this conclusion is false. Thus, the argument
is invalid.
28. We are not forced into the conclusion,
“Some people who contribute to air
pollution live in a suburb” since option (A)
represents true premises and a false
conclusion. Thus, the argument is invalid.
29. The conclusion, “Some people who make
life a little worse live in a suburb” yields a
valid argument since all three options (AC)
represent true premises and force this
conclusion to be true.
30. The conclusion, “All people who drive
make life a little worse” yields a valid
argument since all three options (AC)
represent true premises and force this
conclusion to be true.
3.6 Exercises
1. Let p represent “Rascal Flatts comes to
town,” q represent “I will go to the concert,”
and r represent “I’ll call in sick for work.”
The argument is then represented
symbolically by:
pq
qr
p  r.
This is the valid argument form “reasoning
by transitivity.”
2. Let p represent “you use binoculars,” q
represent “you get a glimpse of the bald
eagle,” and r represent “you will be
amazed.”
The argument is the represented
symbolically by:
pq
qr
p  r.
This is the valid argument form “reasoning
by transitivity.”
3. Let p represent “Marina works hard
enough” and q represent “she will get a
promotion.” The argument is then
represented symbolically by:
pq
p
q.
This is the valid argument form “modus
ponens.”
4. Let p represent “Isaiah’s ankle heals on
time” and q represent “he will play this
season.” The argument is then represented
symbolically by:
pq
p
q.
This is the valid argument form “modus
ponens.”
5. Let p represent “he doesn’t have to get up at
3:00 A.M.” and q represent “he is ecstatic.”
The argument is then represented
symbolically by:
pq
q
p.
Since this is the form “fallacy of the
converse,” it is invalid and considered a
fallacy.
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3.6: ANALYZING ARGUMENTS WITH TRUTH TABLES 79
6. Let p represent “you are a mathematician”
and q represent “you turn coffee into
theorems.” The argument is then represented
symbolically by:
pq
q
p.
Since this is the form “fallacy of the
converse,” it is invalid and considered a
fallacy.
7. Let p represent “Clayton pitches” and q
represent “the Dodgers win.” The argument
is then represented symbolically by:
pq
~q
~p.
This is the valid argument form “modus
tollens.”
11. Let p represent “she uses e-commerce” and
q represent “she pays by credit card.” The
argument is then represented symbolically
by:
p  q (or q  p)
~q
p.
Since this is the form “disjunctive
syllogism,” it is a valid argument.
12. Let p represent “Mia kicks” and q represent
“Drew passes.” The argument is then
represented symbolically by:
p  q (or q  p)
~p
p.
Since this is the form “disjunctive
syllogism,” it is a valid argument.
8. Let p represent “Josh plays” and q represent
“the opponent gets shut out.” The argument
is then represented symbolically by:
pq
~q
~p.
This is the valid argument form “modus
tollens.”
9. Let p represent “you’re going through hell”
and q represent “keep going.” The argument
is then represented symbolically by:
pq
~p
~q.
Since this is the form “fallacy of the
inverse,” it is invalid and considered a
fallacy.
10. Let p represent “you can’t get rid of the
skeleton in your closet” and q represent
“you’d best teach it to dance.” The argument
is then represented symbolically by:
pq
~p
~q.
Since this is the form “fallacy of the
inverse,” it is invalid and considered a
fallacy.
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CHAPTER 3 INTRODUCTION TO LOGIC
To show validity for the arguments in the following exercises, we must show that the conjunction of the
premises implies the conclusion. That is, the conditional statement  P1  P2  ...  Pn   C must be a tautology.
For exercises 13 and 14 we will use the standard (long format) to develop the corresponding truth tables. For
the remainder of the exercises we will use the alternate (short format) to create the truth tables.
13. Form the conditional statement
[(p  q)  p]  ~q
from the argument. Complete a truth table.
p
q
p  q
(p  q)  p
~q
[(p  q)  p]  ~q
T
T
T
T
F
F
T
F
T
T
T
T
F
T
T
F
F
T
F
F
F
F
T
T
Since the conditional, formed by the conjunction of premises implying the conclusion, is not a tautology,
the argument is invalid.
14. Form the conditional statement
[(p  ~q)  p]  ~q
from the argument. Complete a truth table.
p
q
~q
p  ~q
(p  ~q)  p
[(p  ~q)  p]  ~q
T
T
F
F
F
T
T
F
T
T
T
T
F
T
F
F
F
T
F
F
T
F
F
T
Since the conditional, formed by the conjunction of premises implying the conclusion, is a tautology, the
argument is valid.
15. Form the conditional statement
[(~p  ~q)  q]  p
from the argument. Complete a truth table.
p
q
[(~p

~q)

q]

p
T
T
F
T
F
T
T
T
T
T
F
F
T
T
F
F
T
T
F
T
T
F
F
F
T
T
F
F
F
T
T
T
F
F
T
F
1
2
1
3
2
4
3
Since the conditional, formed by the conjunction of premises implying the conclusion, is a tautology, the
argument is valid.
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3.6: ANALYZING ARGUMENTS WITH TRUTH TABLES 81
16. Form the conditional statement
[(p  ~q)  p]  ~q
from the argument. Complete a truth table.
p
q
[(p

~q)

p]

~q
T
T
T
T
F
T
T
F
F
T
F
T
T
T
T
T
T
T
F
T
F
F
F
F
F
T
F
F
F
F
T
T
F
F
T
T
1
2
1
3
2
4
3
Since the conditional, formed by the conjunction of premises implying the conclusion, is not a tautology,
the argument is invalid.
17. Form the conditional statement
[(p  q)  (q  p)]  (p  q)
from the argument. Complete a truth table.
(q  p)]

(p  q)
T
T
T
T
F
T
T
F
T
F
F
T
F
T
T
T
F
F
1
3
2
4
3
p
q
[(p  q)

T
T
T
T
F
F
F
T
F
F
Since the conditional, formed by the conjunction of premises implying the conclusion, is not a tautology,
the argument is invalid.
18. Form the conditional statement
[(~p  q)  p]  ~q
from the argument. Complete a truth table.
p
q
[(~p

q)

p]

~q
T
T
F
T
T
T
T
F
F
T
F
F
T
F
T
T
T
T
F
T
T
T
T
F
F
T
F
F
F
T
F
F
F
F
T
T
1
2
1
3
2
4
3
Since the conditional, formed by the conjunction of premises implying the conclusion, is not a tautology,
the argument is invalid. Note: If you are completing the truth table along rows (rather than down
columns), you could stop after completing the first row, knowing that with a false conditional, the
statement will not be a tautology.
Copyright © 2016 Pearson Education, Inc.
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CHAPTER 3 INTRODUCTION TO LOGIC
19. Form the conditional statement
[(p  ~q)  q]  ~p
from the argument. Complete a truth table.
p
q
[(p

~q)

q]

~p
T
T
T
F
F
F
T
T
F
T
F
T
T
T
F
F
T
F
F
T
F
T
F
T
T
T
T
F
F
F
T
T
F
F
T
T
1
2
1
3
2
4
3
Since the conditional, formed by the conjunction of premises implying the conclusion, is a tautology, the
argument is valid.
20. Form the conditional statement
[(p  ~q)  ~p]  ~q
from the argument. Complete a truth table.
~q)

~p]

F
F
F
F
T
F
T
T
F
F
T
T
T
F
T
T
F
F
F
T
T
T
T
T
T
1
2
1
3
2
4
3
p
q
[(p
T
T
T
T
F
T
F
T
F
F
F

~q
Since the conditional, formed by the conjunction of premises implying the conclusion, is not a tautology,
the argument is invalid.
21. Form the conditional statement
([(p  q)  (q  p)]  p)  (p  q)
from the argument. Complete a truth table.
p
q
([(p

q)

(q

p)]

p)

(p

q)
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
F
T
F
F
F
F
T
T
F
T
T
T
T
F
F
T
F
T
T
F
T
F
F
F
F
T
F
T
T
F
F
F
T
F
T
F
T
F
F
F
T
F
F
F
1
2
1
3
1
2
1
4
3
5
3
4
3
Since the conditional, formed by the conjunction of premises implying the conclusion, is a tautology, the
argument is valid.
Copyright © 2016 Pearson Education, Inc.
3.6: ANALYZING ARGUMENTS WITH TRUTH TABLES 83
22. Form the conditional statement ([(p  q)  (p  q)]  q)  p from the argument. Complete a truth
table.
p
q
([(p

q)

(p

q)]

q)

p
T
T
T
T
T
T
T
T
T
T
T
T
T
T
F
T
F
F
T
T
T
F
F
F
T
T
F
T
F
F
T
T
F
T
T
T
T
F
F
F
F
F
F
F
F
F
F
F
F
F
T
F
1
2
1
3
1
2
1
4
1
5
1
Since the conditional formed by the conjunction of premises implying the conclusion, is not a tautology,
the argument is invalid.
23. Form the conditional statement
[(~p  q)  (~p  q)  p]  ~q
from the argument. Complete a truth table.
p
q
[(~p

q)

(~p

q)]

p]

~q
T
T
F
T
T
T
F
T
T
T
T
F
F
T
F
F
F
F
F
F
T
F
F
T
T
T
F
T
T
T
T
T
T
T
T
F
F
T
F
F
F
T
T
F
F
T
F
F
F
F
T
T
1
2
1
3
4
3
5
4
2
Since the conditional, formed by the conjunction of premises implying the conclusion, is not a tautology,
the argument is invalid.
24. Form the conditional statement
([(r  p)  (r  q)]  (q  p))  (r  p)
from the argument.
p
q
r
([(r

p)

(r

q)]

(q

p))

(r

p)
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
F
F
F
T
T
F
T
T
T
T
T
T
T
F
T
T
T
F
T
T
T
T
T
T
T
T
F
F
F
T
T
T
T
T
T
F
F
F
F
T
T
F
T
T
F
F
F
T
T
F
T
T
F
T
T
T
F
F
T
T
T
F
F
T
F
F
T
T
T
F
F
T
F
F
F
F
T
F
F
F
F
T
F
F
T
F
F
F
F
F
T
T
F
F
T
T
T
F
F
F
F
F
T
T
T
F
F
F
F
F
F
F
T
F
F
F
F
F
F
F
T
F
F
F
1
2
1
3
1
2
1
4
2
3
2
5
3
4
3
Since the conditional, formed by the conjunction of premises implying the conclusion, is a tautology, the
argument is valid.
Copyright © 2016 Pearson Education, Inc.
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CHAPTER 3 INTRODUCTION TO LOGIC
25. Form the conditional statement
([(~p  r)  (p  q)]  (~r  p))  (q  r)
from the argument.
p
q
r
([(~p

r)

(p  q)]

(~r

p))

(q  r)
T
T
T
F
F
T
T
T
T
F
T
T
T
T
T
T
F
F
F
F
T
T
T
T
T
T
F
F
T
F
T
F
F
T
T
T
T
F
T
T
T
T
T
F
F
F
F
F
T
T
T
T
T
T
T
T
F
T
T
T
T
T
T
T
T
F
T
F
T
T
F
T
F
T
F
F
T
T
F
T
F
F
T
F
F
F
T
T
T
T
F
F
F
F
T
F
T
T
F
F
F
T
F
F
T
F
F
T
F
F
T
T
1
2
1
3
2
4
2
3
2
5
4
The F in the final column 5 shows us that the statement is not a tautology and hence, the argument is
invalid.
26. Form the conditional statement ([(p  ~q)  (q  ~r)]  (p  ~r))  (r  p) from the argument.
Complete a truth table.
p
q
r
([(p

~q)

(q

~r)]

(p

~r))

(r

p)
T
T
T
T
F
F
F
T
F
F
F
T
T
F
T
T
T
T
T
T
F
T
F
F
T
T
T
T
T
T
T
T
T
F
T
T
T
F
T
T
T
T
T
F
T
F
T
T
T
F
T
T
T
T
T
F
F
T
T
T
T
F
T
T
T
T
T
T
T
F
T
T
F
T
T
F
T
F
T
T
F
F
F
F
F
F
T
T
F
F
F
T
F
F
T
F
T
T
T
T
T
F
T
T
T
F
T
F
F
F
T
F
T
T
T
F
T
F
F
F
F
F
T
T
F
F
F
F
F
F
T
T
T
F
T
T
T
F
T
T
T
F
T
F
1
2
1
3
1
2
1
6
4
5
4
9
7
8
7
Since the conditional, formed by the conjunction of premises implying the conclusion, is a tautology, the
argument is valid.
27. Every time something squeaks, I use WD-40.
Every time I use WD-40, I must go to the hardware store.
Every time something squeaks, I go to the hardware store.
28. Writing exercise; answers will vary.
Copyright © 2016 Pearson Education, Inc.
3.6: ANALYZING ARGUMENTS WITH TRUTH TABLES 85
29. Let p represent “Joey loves to watch movies,” q represent “Terry likes to jog,” and r represent “Carrie
drives a school bus.” The argument is then represented symbolically by:
p
q  ~p
~q  r
r.
Construct the truth table for [p  (q  ~p)  (~q  r)]  r.
p
q
r
[p

(q

~p)

(~q

r)]

r
T
T
T
T
F
T
F
F
F
F
T
T
T
T
T
T
F
T
F
T
F
F
F
F
T
F
T
F
T
F
T
T
T
F
T
F
T
T
T
T
T
T
T
F
F
T
T
F
T
F
F
T
F
F
T
F
F
T
T
F
F
T
T
T
F
F
T
T
T
T
F
T
F
F
F
T
T
T
F
F
T
F
T
F
F
F
T
F
F
F
T
T
F
T
T
T
T
T
F
F
F
F
F
F
T
T
F
T
F
F
T
F
2
3
1
2
1
4
2
3
2
5
4
Since the conditional, formed by the conjunction of premises implying the conclusion, is a tautology, the
argument is valid.
30. Let p represent “Hurricane Gustave hit that grove of trees,” q represent “trees are devastated,” and r
represent “people plant trees when disasters strike.” The argument is then represented symbolically by:
pq
r ~ q
r  ~p.
Construct the truth table for [(p  q)  (r  ~q)]  (r  ~p).

(r
T
F
T
T
F
F
F
F
T
F
F
F
T
T
F
F
T
T
F
F
F
T
F
T
T
F
T
F
F
F
1
2
1
3
1
[(p

p
q
r
T
T
T
T
T
T
T
F
T
T
T
F
T
T
F
T
F
F
T
F
T
T
F
F
T
F
F
F
T
F
F
F
q)

~q)]

(r

F
F
T
T
F
F
F
F
T
F
T
F
T
T
T
T
F
F
F
F
T
T
F
T
F
T
F
F
T
T
T
T
F
F
T
F
T
T
T
T
T
T
T
T
F
T
T
F
T
T
2
1
4
2
3
2
~p)
Since the conditional, formed by the conjunction of premises implying the conclusion, is a tautology, the
argument is valid.
Copyright © 2016 Pearson Education, Inc.
86
CHAPTER 3 INTRODUCTION TO LOGIC
31. Let p represent “Yoda is my favorite Star Wars character,” q represent “I hate Darth Vader,” and r
represent “I hate Luke Skywalker.” The argument is then represented symbolically by:
pq
r  q
~r
~p.
Construct the truth table for [(p  q)  (r  q)  (~r)]  ~p. (Note: we do not have to complete a
column under each simple statement p, q, and r, as we did in exercises above, since it is easy to compare
the appropriate index columns to create the truth value for each connective.)
p
q
r
[(p  q)

(r  q)

(~r)]

~p
T
T
T
T
T
T
F
F
T
F
T
T
F
T
T
T
T
T
F
F
T
F
T
F
F
T
F
F
T
F
T
F
F
F
F
F
F
T
T
F
F
T
T
T
T
T
F
F
T
T
F
T
F
T
T
T
T
T
T
T
F
F
T
T
T
T
F
F
T
T
F
F
F
T
F
F
F
T
T
T
1
2
1
3
2
4
3
Since the conditional, formed by the conjunction of premises implying the conclusion, is not a tautology,
the argument is invalid. Note: If you are completing the truth table along rows (rather than down
columns), you could stop after completing the second row, knowing that with a false conditional, the
statement will not be a tautology.
32. Let p represent “Carrie Underwood sings,” q represent “Joe Jonas is a teen idol,” and r represent
“Jennifer Hudson wins a Grammy.” The argument is then represented symbolically by:
p  ~q
~q  ~r
r
~p.
Construct the truth table for [(p  ~q)  (~q  ~r)  r]  ~p.
p
q
r
[(p

~q)

(~q

~r)

r]

~p
T
T
T
T
T
F
T
F
T
F
T
T
F
F
T
T
F
T
T
F
T
F
T
T
F
F
T
F
T
F
T
T
T
T
F
T
F
F
F
T
T
F
T
F
F
T
T
T
T
T
T
T
F
F
T
F
F
T
T
F
F
F
F
F
T
F
F
T
T
T
F
T
F
F
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
T
F
T
F
F
F
T
T
T
F
F
F
F
T
T
T
T
T
T
F
F
T
T
1
2
1
3
1
2
1
4
3
5
4
Since the conditional, formed by the conjunction of premises implying the conclusion, is not a tautology,
the argument is invalid. Note: If you are completing the truth table along rows (rather than down
columns), you could stop after completing the first row, knowing that with a false conditional, the
statement will not be a tautology.
Copyright © 2016 Pearson Education, Inc.
3.6: ANALYZING ARGUMENTS WITH TRUTH TABLES 87
33. Let p represent “The Cowboys will make the playoffs,” q represent “Troy comes back to play,” and r
represent “Jerry coaches the Cowboys.” The argument is then represented symbolically by
pq
~r  q
r
~p.
Construct the truth table for [(p  q)  (r  q)  (~r)]  ~p.
p
q
r
[(p  q)

(~r  q)

r]

~p
T
T
T
T
T
T
T
T
F
F
T
T
F
T
T
T
F
F
T
F
T
F
T
F
F
F
F
T
T
F
T
F
F
F
F
T
F
F
T
F
F
T
T
F
F
T
F
T
T
T
F
T
F
F
F
T
F
F
T
T
F
F
T
T
T
F
T
T
T
T
F
F
F
T
T
T
F
F
T
T
1
2
1
3
2
4
3
Since the conditional, formed by the conjunction of premises implying the conclusion, is not a tautology,
the argument is invalid. (Note: If you are completing the truth table along rows, rather than down
columns, you could stop after completing the first row, knowing that with a false conditional, the
statement will not be a tautology.
34. Let p represent “I’ve got you under my skin,” q represent “you are deep in the heart of me,” and r
represent “you are really a part of me.” The argument is then represented symbolically by:
pq
q  ~r
qr
p  r.
Construct the truth table for [(p  q)  (q  ~r)  (q  r)]  (p  r).
p
[(p  q)

(q
(q  r)]

(p  r)
F
T
T
T
T
T
F
F
F
T
T
T
T
F
F
T
F
F
F
F
T
T
T
T
T
T
T
T
T
T
F
T
T
T
T
T
T
F
F
T
T
1
4
3
5
4

~r)
q
r
T
T
T
T
F
T
F
F
T
T
F
T
T
T
T
T
T
F
T
F
F
F
T
F
T
F
F
F
F
F
T
F
T
T
T
F
T
F
T
F
T
T
T
F
F
T
T
T
F
F
F
F
T
T
F
2
3
1
2

Since the conditional, formed by the conjunction of premises implying the conclusion, is not a tautology,
the argument is invalid. (Note: If you are completing the truth table along rows, rather than down
columns, you could stop after completing the second row, knowing that with a false conditional, the
statement will not be a tautology.)
Copyright © 2016 Pearson Education, Inc.
88
CHAPTER 3 INTRODUCTION TO LOGIC
35. Let p represent “Dr. Hardy is a department chairman,” q represent “he lives in Atlanta,” and r represent
“his first name is Larry.” The argument is then represented symbolically by
pq
qr
~r  ~p.
Construct the truth table for [(p  q)  (q  r)]  (~r  ~p).
p
q
r
[(p  q)

(q  r)]

(~r

~p)
T
T
T
T
T
T
T
F
T
F
T
T
F
T
F
F
T
T
F
F
T
F
T
F
F
F
T
F
T
F
T
F
F
F
F
F
T
T
F
F
F
T
T
T
T
T
T
F
T
T
F
T
F
T
F
F
T
T
T
T
F
F
T
T
F
F
T
F
T
T
F
F
F
T
F
F
T
T
T
T
1
2
1
3
1
2
1
Since the conditional, formed by the conjunction of premises implying the conclusion, is a tautology, the
argument is valid.
36. Let p represent “I am your women,” q represent “you are my man,” and r represent “I stop loving you.”
The argument is then represented symbolically by:
(p  q)  ~r
r
~p  ~q.
Construct the truth table for {[(p  q)  ~r]  r}  (~p  ~q).
p
q
r
([(p  q)

~r}

r)

(~p

~q)
T
T
T
T
F
F
F
T
T
F
F
F
T
T
F
T
T
T
F
F
T
F
F
F
T
F
T
F
T
F
T
T
T
F
T
T
T
F
F
F
T
T
F
F
T
F
T
T
F
T
T
F
T
F
T
T
T
T
T
F
F
T
F
F
T
T
F
F
T
T
T
F
F
F
T
F
T
F
T
T
T
T
T
T
F
F
F
F
T
T
F
F
T
T
T
T
1
2
1
3
2
4
2
3
2
Since the conditional, formed by the conjunction of premises implying the conclusion, is a tautology, the
argument is valid.
Copyright © 2016 Pearson Education, Inc.
3.6: ANALYZING ARGUMENTS WITH TRUTH TABLES 89
The following exercises involve Quantified
arguments and can be analyzed, as such, by Euler
diagrams. However, the quantified statements can
be represented as conditional statements as well.
This allows us to use a truth tableor recognize a
valid argument formto analyze the validity of the
argument.
42. The statement “none of your sons can do
logic” becomes “if he is your son, then he
can’t do logic.”
37. Let p represent “you are a man,” q represent
“you are created equal,” and r represent
“you are a woman.” The argument is then
represented symbolically by:
pq
qr
p  r.
This is a “reasoning by Transitivity”
argument form and, hence, is valid.
44. The statement “no teetotalers are
pawnbrokers” becomes “if the person is a
teetotaler, then the person is not a
pawnbroker.”
38. Let p represent “you are a man,” q represent
“you are mortal,” and r represent “you are
Socrates.” The argument is then represented
symbolically by:
pq
rp
r  q.
By interchanging the first premise with the
second premise the argument becomes a
“reasoning by transitivity” form and, hence,
is valid.
39. (a) We apply reasoning by repeated
transitivity to the seven premises. A
conclusion from this reasoning, which
makes the argument valid, is reached by
linking the first antecedent to the last
consequent. This conclusion is “When
the cable company keeps you on hold,
your dad gets punched over a can of
soup.”
(b) If the premise “Your dad does not get
punched over a can of soup” is added,
then we can apply reasoning by
transitivity to the contrapositives of the
original premises to draw the conclusion
that “The cable company does not keep
you on hold.”
40. Writing exercise; answers will vary.
Answers in Exercises 4148 may be replaced by
their contrapositives.
41. The statement “all my poultry are ducks”
becomes “if it is my poultry, then it is a
duck.”
43. The statement “guinea pigs are hopelessly
ignorant of music” becomes “if it is a guinea
pig, then it is hopelessly ignorant of music.”
45. The statement “no teachable kitten has green
eyes” becomes “if it is a teachable kitten,
then it does not have green eyes.”
46. The statement “opium-eaters have no selfcommand” becomes “if it is an opium-eater,
then it has no self-command.”
47. The statement “I have not filed any of them
that I can read” becomes “if I can read it,
then I have not filed it.”
48. The statement “all of them written on blue
paper are filed” becomes “if it is written on
blue paper, then it is filed.”
49. (a) “No ducks are willing to waltz”
becomes “if it is a duck, then it is not
willing to waltz” or p  ~s.
(b) “No officers ever decline to waltz”
becomes “if one is an officer, then one
is willing to waltz” or r  s.
(c) “All my poultry are ducks” becomes “if
it is my poultry, then it is a duck” or
q  p.
(d) The three symbolic premises are
p  ~s
rs
q  p.
Begin with q, which only appears once.
Replacing r  s with its contrapositive,
~s  ~r, rearrange the three premises.
qp
p  ~s
~s  ~r
By repeated use of reasoning by
transitivity, the conclusion which
provides a valid argument is q  ~r.
In words, “if it is my poultry, then it is
not an officer,” or “none of my poultry
are officers.”
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CHAPTER 3 INTRODUCTION TO LOGIC
50. (a) “Everyone who is sane can do logic”
becomes “if one is sane, then one is
able to do logic” or r  p.
(b) “No lunatics are fit to serve on a jury”
becomes “if one is a lunatic (or not
same), then one is not fit to serve on a
jury” or ~r  ~q.
(c) “None of your sons can do logic”
becomes “if he is your son, then he
cannot do logic” or s  ~p.
(d) The three symbolic premises are
rp
~r  ~q
s  ~p.
Replacing r  p with its contrapositive,
~p  ~r, rearrange the three premises.
s  ~p
~p  ~r
~r  ~q
By repeated use of reasoning by
transitivity, the conclusion which
provides a valid argument is s  ~q.
In words, “if he is your son, then he is
not fit to serve on a jury,” or “your sons
are not fit to serve on a jury.”
51. (a) “Promise-breakers are untrustworthy”
becomes “if one is a promise-breaker,
then one is not trustworthy” or r  ~s.
(b) “Wine-drinkers are very
communicative” becomes “if one is a
wine-drinker, then one is very
communicative” or u  t.
(c) “A person who keeps a promise is
honest” becomes “if one is not a
promise-breaker, then one is honest” or
~r  p.
(d) “No teetotalers are pawnbrokers”
becomes “if one is not a wine-drinker,
then one is not a pawnbroker” or
~u  ~q.
(e) “One can always trust a very
communicative person” becomes “if
one is very communicative, then one is
trustworthy” or t  s.
(f) The symbolic premise statements are
r  ~s
ut
~r  p
~u  ~q
t  s.
Begin with q, which only appears once.
Using the contrapositive of ~u  ~q,
(q  u), and r  ~s, (s  ~r),
rearrange the five premises as follows:
qu
ut
ts
s  ~r
~r  p.
By repeated use of reasoning by
transitivity, the conclusion which
provides a valid argument is q  p.
In words, this conclusion can be stated
as “if one is a pawnbroker, then one is
honest,” or “all pawnbrokers are
honest.”
52. Let p be “it is a guinea pig,” q be “it is
hopelessly ignorant of music,” r be “it keeps
silent while the Moonlight Sonata is being
played,” and s be “it appreciates
Beethoven.”
(a) “Nobody who really appreciates
Beethoven fails to keep silent while the
Moonlight Sonata is being played”
becomes “if one appreciates Beethoven,
then one keeps silent while the
Moonlight Sonata is being played” or
s  r.
(b) “Guinea pigs are hopelessly ignorant of
music” becomes “if you are a guinea
pig, then you are hopelessly ignorant of
music” or p  q.
(c) “No one who is hopelessly ignorant of
music ever keeps silent while the
Moonlight Sonata is being played,”
becomes “if one is hopelessly ignorant
of music, then one fails to keep silent
while the Moonlight Sonata is being
played” or q  ~r.
(d) In symbols, the statements are
(a) s  r
(b) p  q
(c) q  ~r.
Using the contrapositive of the premise
s  r, (~r  ~s), rearrange the
premises as follows:
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3.6: ANALYZING ARGUMENTS WITH TRUTH TABLES 91
pq
q  ~r
~r  ~s.
By repeated use of reasoning by
transitivity, the conclusion which
provides a valid argument is p  ~s.
In words, this conclusion can be stated
as “if you are a guinea pig, then you do
not appreciate Beethoven,” or
equivalently, “guinea pigs don’t
appreciate Beethoven.”
53. Begin by changing each quantified premise
to a conditional statement.
(a) The statement “all the dated letters in
this room are written on blue paper”
becomes “if it is dated, then it is on blue
paper” or r  w.
(b) The statement “none of them are in
black ink, except those that are written
in the third person” becomes “if it is not
in the third person, then it is not in
black ink” or ~u  ~t.
(c) The statement “I have not filed any of
them that I can read” becomes “if I can
read it, then it is not filed” or v  ~s.
(d) The statement “none of them that are
written on one sheet are undated”
becomes “if it is on one sheet, then it is
dated” or x  r.
(e) The statement “all of them that are not
crossed are in black ink” becomes “if it
is not crossed, then it is in black ink” or
~q  t.
(f) The statement “all of them written by
Brown begin with ‘Dear Sir’” becomes
“if it is written by Brown, then it begins
with ‘Dear Sir’” or y  p.
(g) The statement “all of them written on
blue paper are filed” becomes “if it is
on blue paper, then it is filed” or w  s.
(h) The statement “none of them written on
more than one sheet are crossed”
becomes “if it is not on more than one
sheet, then it is not crossed” or
~x  ~q.
(i) The statement “none of them that begin
with ‘Dear Sir’ are written in the third
person” becomes “if it begins with
‘Dear Sir,’ then it is not written in the
third person” or p  ~u.
(j) The symbolic premise statements are
(a) r  w
(b) ~u  ~t
(c) v  ~s
(d) x  r
(e) ~q  t
(f) y  p
(g) w  s
(h) ~x  ~q
(i) p  ~u.
Begin with y, which appears only once.
Using contrapositives of v  ~s
(s  ~v), ~q  t (~t  q), and
~x  ~q (q  x), rearrange the nine
statements.
yp
p  ~u
~u  ~t
~t  q
qx
xr
rw
ws
s  ~v.
By repeated use of reasoning by
transitivity, the conclusion that makes
the argument valid is y  ~v.
In words, the conclusion can be stated
as “if it is written by Brown, then I
can’t read it,” or equivalently “I can’t
read any of Brown’s letters.”
54. (a) “No one who is going to a party ever
fails to brush his hair” becomes “if one
is going to a party, then he brushes his
hair” or p  q.
(b) “No one looks fascinating if he is
untidy” becomes “if one is untidy, then
he does not look fascinating” or
~u  ~s.
(c) “Opium-eaters have no self-command”
becomes “if one is an opium-eater; then
he has no self command” or t  ~r.
(d) “Everyone who has brushed his hair
looks fascinating” becomes “if one has
brushed his hair, then he looks
fascinating” or q  s.
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CHAPTER 3 INTRODUCTION TO LOGIC
(e) “No one wears white kid gloves unless
he is going to a party” becomes “if he
wears white kid gloves, then he is going
to the party” or v  p.
(f) “A man is always untidy if he has no
self-command” becomes “if a man has
no self-command, then a man is not
tidy” or ~r  ~u.
(g) The symbolic premise statements are
(a) p  q
(b) ~u  ~s
(c) t  ~r
(d) q  s
(e) v  p
(f) ~r  ~u
Begin with t, which only occurs once.
Using the contrapositives of q  s,
(~s  ~q), p  q, (~q  ~p), and
v  p, (~p  ~v), rearrange the five
premises as follows:
t  ~r
~r  ~u
~u  ~s
~s  ~q
~q  ~p
~p  ~v.
By repeated use of reasoning by
transitivity, the conclusion which
provides a valid argument is t  ~v.
In words, this conclusion can be stated
as “if he is an opium-eater then he
doesn’t wear white kid gloves” or
equivalently, “opium-eaters do not wear
white kid gloves.”
Chapter 3 Test
1. The negation of “6  3 = 3” is “6  3  3.”
2. The negation of “all men are created equal”
is “some men are not created equal.”
3. The negation of “some members of the class
went on the field trip” is “no members of the
class went on the field trip.” An equivalent
answer would be “all members of the class
did not go on the field trip.”
4. The negation of “If I fall in love, it will be
forever” is “I fall in love and it will not be
forever.” Remember that
~(p  q)  (p  ~q).
5. The negation of “she applied and did not get
a student loan” is “she did not apply or she
got a student loan.” Remember that
~(p  q)  (~p  ~q).
Let p represent “you will love me” and let q
represent “I will love you.”
6. The symbolic form of “If you won’t love
me, then I will love you” is “~p  q.”
7. The symbolic form of “I will love you if you
will love me.” is “p  q.”
8. The symbolic form of “I won’t love you if
and only if you won’t love me” is
“~q  ~p.”
9. Writing the symbolic form “~p  q” in
words, we get “you won’t love me and I will
love you.”
10. Writing the symbolic form “~(p  ~q)” in
words, we get “it is not the case that you
will love me or I won’t love you” (or
equivalently, by De Morgan’s, “you won’t
love me and I will love you”).
Assume that p is true and that q and r are false for
Exercises 1114.
11. Replacing q and r with the given truth
values, we have
~F  ~F
T  T
T.
The compound statement ~q  ~r is true.
12. Replacing p, q and r with the given truth
values, we have
F  (T  ~F)
F  (T  T)
F T
T.
The compound statement r  (p  ~q) is
true.
13. Replacing r with the given truth value (s not
known), we have
F  (s  F)
F  not known
T.
The compound statement r  (s  r) is
true.
Copyright © 2016 Pearson Education, Inc.
CHAPTER 3 TEST 93
The wording may vary in the answer in
Exercises 2124.
14. Replacing p and q with the given truth
values, we have
T  (T  F)
T  (F)
F.
The compound statement p  (p  q) is
false.
21. “All integers are rational numbers” can be
stated as “if the number is an integer, then it
is a rational number.”
22. “Being a rhombus is sufficient for a polygon
to be a quadrilateral” can be stated as “if a
polygon is a rhombus, then it is a
quadrilateral.”
15. Writing exercise; answers will vary.
16. The necessary condition for
23. “Being divisible by 2 is necessary for a
number to be divisible by 4” can be stated as
“if a number is divisible by 4, then it is
divisible by 2.” Remember that the
“necessary” part of the statement becomes
the consequent.
(a) a conditional statement to be false is
that the antecedent must be true and the
consequent must be false.
(b) a conjunction to be true is that both
component statements true.
24. “She digs dinosaur bones only if she is a
paleontologist” can be stated as “if she digs
dinosaur bones, then she is a
paleontologist.” Remember that the “only
if” part of the statement becomes the
consequent.
(c) a disjunction to be false is that both
component statements must be false.
(d) a biconditional to be true is that both
component statements have the same
truth value.
17.
18.
p

(~p

q)
T
T
T
F
T
T
F
T
F
F
F
F
F
T
F
F
T
T
T
F
F
F
F
T
T
F
2
3
1
2
1
p
q
T
T
25. The conditional statement: If a picture paints
a thousand words, then the graph will help
me understand it.
(a) Converse: If the graph will help me
understand it, then a picture paints a
thousand words.
p
q
~
(p  q)

(~p

~q)
T
T
F
T
T
F
F
F
T
F
T
F
T
F
T
T
F
T
T
F
T
T
T
F
F
F
T
F
T
T
T
T
2
1
3
1
2
1
Since the last completed column (3) is all
true, the conditional is a tautology.
19. The statement “some negative integers are
whole numbers” is false, since all whole
numbers are non-negative.
(b) Inverse: If a picture doesn’t paint a
thousand words, then the graph won’t
help me understand it.
(c) Contrapositive: If the graph doesn’t
help me understand it, then a picture
doesn’t paint a thousand words.
26. The conditional statement: ~p  (q  r).
(a) Converse: (q  r)  ~p.
(b) Inverse: p  ~(q  r), or
p  (~q  ~r).
(c) Contrapositive: ~(q  r)  p, or
(~q  ~r)  p.
20. The statement “all irrational numbers are
real numbers” is true, because the real
numbers are made up of both the rational
and irrational numbers.
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CHAPTER 3 INTRODUCTION TO LOGIC
27. Complete an Euler diagram as:
Since, when the premises are diagrammed as being true, and we are forced into a true conclusion, the
argument is valid.
28. (a) Let p represent “he eats liver” and q represent “he will eat anything.” The argument is then
represented symbolically by:
pq
p
q.
This is the valid argument form “modus ponens,” hence the answer is A.
(b) Let p represent “you use your seat belt” and q represent “you will be safer.” The argument is then
represented symbolically by:
pq
~p
~q.
The answer is F, a fallacy of the inverse.
(c) Let p represent “I hear Mr. Bojangles,” q represent “I think of her,” and r represent “I smile.” The
argument is then represented symbolically by:
pq
qr
p  r.
This is the valid argument form “reasoning by transitivity,” hence the answer is C.
(d) Let p represent “she sings” and q represent “she dances.”
The argument is then represented symbolically by:
p  q
~p
q.
This is the valid argument form “disjunctive syllogism,” hence the answer is D.
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CHAPTER 3 TEST 95
29. Let p represent “I write a check,” q represent “it will bounce,” and r represent “the bank guarantees it.”
The argument is then represented symbolically by:
pq
r  ~q
r
~p.
Construct the truth table for ([(p  q)  (r  ~q)]  r)  (~p).

(r

~q)]

r)

(~p)
T
F
T
T
T
F
F
F
T
T
F
F
T
F
F
F
T
F
F
F
T
T
T
F
T
T
F
F
F
F
F
T
T
F
F
T
F
T
T
T
F
T
F
F
F
T
T
T
F
T
F
T
T
F
T
F
F
F
T
T
F
F
T
T
T
T
T
T
T
T
T
T
F
F
F
T
T
F
T
T
F
F
T
T
2
3
1
2
1
4
3
5
4
p
q
r
T
T
T
T
T
F
T
F
T
T
F
F
([(p  q)
Since the conditional, formed by the conjunction of premises implying the conclusion, is a tautology, the
argument is valid.
30. Construct the truth table for [(~p  ~q)  (q  p)]  (p  q).
p
q
[(~p

~q)

(q

p)]

(p

q)
T
T
F
T
F
T
T
T
T
T
T
T
T
T
F
F
T
T
T
F
T
T
T
T
T
F
F
T
T
F
F
F
T
F
F
T
F
T
T
F
F
T
T
T
T
F
T
F
F
F
F
F
1
2
1
3
1
2
1
4
2
3
2
Since the conditional, formed by the conjunction of premises implying the conclusion, is not a tautology,
the argument is invalid.
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