CmpE 220 Fall 2012
PS #1
Fundamentals of Logic
October 10, 2012
1. (Rosen: Section 1.1, pp. 17 Question 10) Let p, q and r be the propositions
p : You get an A on the final exam.
q : You do every exercise in this book.
r : You get an A in this class.
Write these propositions using p, q and r and logical connectives.
(a) You get an A in this class, but you do not do every exercise in this book.
(b) You get an A on the final, you do every exercise in this book, and you get an A in
this class.
(c)To get an A in this class, it is necessary for you to get an A on the final.
(d) You get an A on the final, but you do not do every exercise in this book; nevertheless you get an A in this class.
(e) Getting an A on the final and doing every exercise in this book is sufficient for
getting an A in this class.
(f) You will get an A in this class if and only if you either do every exercise in this
book or you get an A on the final.
Solution
(a) r ∧ ¬q
(b) p ∧ q ∧ r
(c) r → p
(d) p ∧ ¬q ∧ r
(e) (p ∧ q) → r
(f) r ↔ (p ∨ q)
2. (Grimaldi: Section 2.1, pp. 54 Question 5) Let p,q,r denote the following statements about a particular triangle ABC.
p : Triangle ABC is isosceles.
q : Triangle ABC is equilateral.
r : Triangle ABC is equiangular.
Translate each of the following into an English sentence.
(a) q → p
(b) ¬p → ¬q
(c) q ↔ r
(d) p ∧ ¬q
(e) r → p
Solution
(a) If triangle ABC is equilateral, then it is isosceles.
(b) If triangle ABC is not isosceles, then it is not equilateral.
(c) Triangle ABC is equilateral if and only if it is equiangular.
(d) Triangle ABC is isosceles but it is not equilateral.
(e) If triangle ABC is equiangular, then it is isosceles.
3. (Grimaldi: Section 2.1, pp. 54 Question 6) Determine the truth value of each of
the following implications.
(a) If 3 + 4 = 12 then 3 + 2 = 6
(b) If 3 + 3 = 6 then 3 + 4 = 9
Solution
(a) F → F ≡ T
(b) T → F ≡ F
4. (Rosen: Section 1.1, pp. 18 Question 18) Write each of these statements in the
form “if p, then q”in English.
(a) It is necessary to wash the boss’s car to get promoted.
(d) Willy gets caught whenever he cheats.
(e) You can access the website only if you pay a subscription fee.
Solution
(a) p: You wash the boss’s car. q: You get promoted. (q → p)
If you get promoted then you must have washed the boss’s car.
(d) p: Willy gets caught. q: Willy cheats. (q → p)
If Willy cheats, he gets caught.
(e) p: You can access the website. q: You pay a subscription fee. (p → q)
If you can access the website then you must have paid a subscription fee.
5. (Grimaldi: Section 2., pp. 67 Question 19c) Simplify the following compound
statement by giving reasons at each step.
[(¬p ∨ ¬q) → (p ∧ q ∧ r)] ⇐⇒ p ∧ q Solution
[(¬p ∨ ¬q) → (p ∧ q ∧ r)]
⇐⇒
⇐⇒
⇐⇒
⇐⇒
¬(¬p ∨ ¬q) ∨ (p ∧ q ∧ r) s → t ⇐⇒ ¬s ∨ t
(¬¬p ∧ ¬¬q) ∨ (p ∧ q ∧ r) DeMorgan’s Laws
(p ∧ q) ∨ (p ∧ q ∧ r) Law of Double Negation
(p ∧ q) Absorption Law
6. (Grimaldi: Section 2., pp. 67 Question 18b) Give the reasons for each step in
the following simplifications of compound statement.
(p → q) ∧ [¬q ∧ (r ∨ ¬q)]
⇐⇒
⇐⇒
⇐⇒
⇐⇒
⇐⇒
⇐⇒
⇐⇒
(p → q) ∧ ¬q
(¬p ∨ q) ∧ ¬q
¬q ∧ (¬p ∨ q)
(¬q ∧ ¬p) ∨ (¬q ∧ q)
(¬q ∧ ¬p) ∨ F0
(¬q ∧ ¬p)
¬(q ∨ p)
Solution
(p → q) ∧ [¬q ∧ (r ∨ ¬q)]
⇐⇒
⇐⇒
⇐⇒
⇐⇒
⇐⇒
⇐⇒
⇐⇒
(p → q) ∧ ¬q Absorption & Commutative Law
(¬p ∨ q) ∧ ¬q s → t ⇐⇒ ¬s ∨ t
¬q ∧ (¬p ∨ q) Commutative Law over ∧
(¬q ∧ ¬p) ∨ (¬q ∧ q) Distributive Law of ∧ over ∨.
(¬q ∧ ¬p) ∨ F0 Inverse law
(¬q ∧ ¬p) Identity law
¬(q ∧ p) DeMorgan’s Laws
7. (Rosen: Section 1.3, pp. 47 Question 10) Let C(x) be the statement “x has a
cat”, D(x) be the statement “x has a dog,”and F (x) be the statement “x has a ferret”.
Express each of these statements in terms of C(x), D(x), and F (x) quantifiers, and
logical connectives. Let the universe of discourse consist of all the students in your
course.
(a) A student in your class has a cat, a dog, and a ferret.
(b) All students in your class have a cat, a dog, or a ferret.
(c) Some student in your class has a cat and a ferret, but not a dog.
(d) No student in your class has a cat, a dog, and a ferret.
(e) For each of the three animals, cats, dogs, and ferrets, there is a student in your
class who as one of these animals as a pet.
Solution
(a) ∃x [C(x) ∧ D(x) ∧ F (x)]
(b) ∀x [C(x) ∨ D(x) ∨ F (x)]
(c) ∃x [C(x) ∧ ¬D(x) ∧ F (x)]
(d) ∀x ¬[C(x) ∧ D(x) ∧ F (x)] ≡ ¬∃x [C(x) ∧ D(x) ∧ F (x)]
(e) ∃x C(x) ∧ ∃y D(y) ∧ ∃z F (z)
8. (Rosen: Section 1.4, pp. 59 Question 12) Let I(x) be the statement “x has
an Internet connection”and C(x, y) be the statement “x and y have chatted over the
Internet ”, where the universe of discourse for the variables x and y consists of all the
students in your class. Use quantifiers to express each of these statements.
(a) Jerry does not have an Internet connection.
(b) Rachel has not chatted over the Internet with Chelsea.
(c) Jan and Sharon have never chatted over the Internet.
(d) No one in the class has chatted with Bob.
(e) Sanjay has chatted with everyone except Joseph.
(f ) Someone in your class does not have an Internet connection.
(g) Not everyone in your class has an Internet connection.
(h) Exactly one student in your class has an Internet connection.
(i) Everyone except one student in your class has an Internet connection.
(j) Everyone in your class with an Internet connection has chatted over the Internet
with at least one other student in your class.
(k) Someone in your class has an Internet connection but has not chatted with anyone
else in your class.
(l) There are two students in your class who have not chatted with each other over the
Internet.
(m) There is a student in your class who has chatted with everyone in your class over
the Internet.
(n) There are at least two students in your class who have not chatted with the same
person in your class.
(o) There are two students in your class who between them have chatted with everyone
else in the class.
Solution
(a) ¬I(Jerry)
(b) ¬C(Rachel, Chelsea)
(c) ¬∃x [C(Jan, x) ∨ C(Sharon, x)]
(d) ¬∃x C(Bob, x)
(e) ∀x [¬C(Sanjay, x) ↔ (x = Sanjay ∨ x = Joseph)]
(f ) ∃x ¬I(x)
(g) ¬∀x I(x) ≡ ∃x ¬I(x)
(h) ∃x [I(x) ∧ ∀y(I(y) → x = y)]
(i) ∃x [¬I(x) ∧ ∀y(¬I(y) → x = y)]
(j) ∀x [I(x) → ∃y C(x, y)]
(k) ∃x [I(x) ∧ ∀y (x = y ∨ ¬C(x, y))]
(l) ∃x∃y [x 6= y ∧ ¬C(x, y)]
(m) ∃x∀y [x 6= y → C(x, y)]
(n) ∃x∃y [x 6= y ∧ ∀z ¬(C(x, z) ∧ C(y, z))]
(o) ∃x∃y [x 6= y ∧ ∀z (C(x, z) ∨ C(y, z))]
9. (Rosen: Section 1.5, pp. 73 Question 10) For each of these sets of premises, what
relevant conclusions can be drawn. Explain the rules of inference used to obtain each
conclusion from the premises.
(a) “If I play hockey, then I am sore the next day.” “I use the whirlpool if I am sore.”
“I did not use the whirlpool.”
(c) “All insects have six legs.” “Dragonflies are insects” “Spiders do not have six legs.”
“Spiders eat dragonflies”
(e) “All foods that are healthy to eat do not taste good. ” “Tofu is healthy to eat”
“You only eat what tastes good.” “cheeseburgers are not healthy to eat”
Solution
(a) Let
p: I play hockey. q: I am sore the next day. r: I use the whirlpool
1) p → q
2) q → r
3)
¬r
4)
¬q
5)
¬p
premise
premise
premise
2 and 3 modus tollens
1 and 4 modus tollens
Hence the conclusions are “I am not sore the next day”, and “I did not play hockey”
(c) “All insects have six legs.” “Dragonflies are insects” “Spiders do not have six legs.”
“Spiders eat dragonflies”
Let
I(x) : x are insects. S(x) : x have six legs. E(x, y) : x eat y. The domain of discourse
consists of animal classes. i.e. Domain={Dragonflies, Spiders, Horses, ...}
1)
∀x (I(x) → S(x))
premise
2)
I(Dragonflies)
premise
3)
¬S(Spiders)
premise
4) E(Spiders, Dragonflies)
premise
5) (I(Spiders) → S(Spiders) 1 universal instantiation
6)
¬I(Spiders)
3 and 5 modus tollens
Hence, one possible conclusion is “Spiders are not insects ”.
1)
∀x (I(x) → S(x))
2)
I(Dragonflies)
3)
¬S(Spiders)
4)
E(Spiders, Dragonflies)
5) I(Dragonflies ∧ E(Spiders, Dragonflies)
6)
∃x [I(x) ∧ E(Spiders, x)]
premise
premise
premise
premise
2 and 4 conjunction
5 existential generalization
Another conclusion is “Spiders eat some insects ”.
1)
∀x (I(x) → S(x))
premise
2)
I(Dragonflies)
premise
3)
¬S(Spiders)
premise
4)
E(Spiders, Dragonflies)
premise
5) (I(Dragonflies) → S(Dragonflies) 1 universal instantiation
6)
S(Dragonflies)
2 and 5 modus ponens
Hence, another conclusion is “Dragonflies have six legs”.
(e) “All foods that are healthy to eat do not taste good. ” “Tofu is healthy to eat”
“You only eat what tastes good.” “cheeseburgers are not healthy to eat”
Let
H(x) : x is healthy to eat. T (x) : x tastes good. E(x) : You eat x. The domain of
discourse consists of foods.
1) ∀x [H(x) → ¬T (x)] premise
2)
H(tofu)
premise
3) ∀x [E(x) → T (x)]
premise
4) ¬H(cheeseburgers) premise
5) H(tofu) → ¬T (tofu) 1 universal instantiation
6)
¬T (tofu)
2 and 5 modus ponens
7) E(tofu) → T (tofu)
3 universal instantiation
8)
¬E(tofu)
6 and 7 modus tollens
Tw of the possible conclusions are “Tofu does not taste good”, and “You don’t eat
tofu ”
10. (Rosen: Section 1.5, pp. 73 Question 16) For each of these arguments determine
whether the argument is correct or incorrect and explain why.
(a) Everyone enrolled in the university has lived in a dormitory. Mia has never
lived in a dormitory. Therefore, Mia is not enrolled in the university. Correct,
Universal Instantiation and Modus Tollens
(b) A convertible car is fun to drive. Isaac’s car is not a convertible. Therefore,
Isaac’s car is not fun to drive. Incorrect, Fallacy of denying the hypothesis
(c) Quincy likes all action movies. Quincy likes the movies Eight Men Out. Therefore, Eight Men Out is an action movie. Incorrect, Fallacy of affirming the
conclusion
11. (Grimaldi: Section 2.5, pp. 117 Question 18) Let m, n be two positive integers.
Prove that if m, n are perfect squares, then m · n is a perfect square.
12. (Grimaldi: Section 2.5, pp. 117 Question 20) Prove or disprove: There exists
positive integers m, n where m, n and m + n are all perfect squares.
13. (Grimaldi: Section 2.5, pp. 117 Question 22) Prove that for every integer n,
4n + 7 is odd.