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MTH 231, Posted Saturday of Week 1 at 12:00am
Week 1, Written Homework Key
Complete the following problems from the book by putting your answers in the space provided. Submit the
entire document in gradescope. Problems in Bold will be graded for correctness. The rest of the problems
will be graded for completion.
Section
Section
Section
Section
1.1:
1.2:
1.3:
1.4:
17 a,b, c, d, e, f, 30
4, 8
8, 10, 22
8 a,b, c, d, 32
17. Let p, q and r be the propositions
p: Grizzly bears have been seen in the area.
q: Hiking is safe on the trail.
r: Berries are ripe along the trail.
Write these propositions using p, q, and r and logical connectives (including negations).
(a) Berries are ripe along the trail, but grizzly bears have not been seen in the area.
r ∧ ¬p
(b) Grizzly bears have not been seen in the area and hiking on the trail is safe, but berries are ripe along
the trail.
¬p ∧ q ∧ r
(c) If berries are ripe along the trail, hiking is safe if and only if grizzly bears have not been seen in the
area.
r → (q ←→ ¬p)
(d) It is not safe to hike on the trail, but grizzly bears have not been seen in the area and the berries
along the trail are ripe.
¬q ∧ ¬p ∧ r
(e) For hiking on the trail to be safe, it is necessary but not sufficient that berries not be ripe along the
trail and for grizzly bears not to have been seen in the area.
(q → (¬r ∧ ¬p)) ∧ ¬((¬r ∧ ¬p) → q)
(f) Hiking is not safe on the trail whenever grizzly bears have been seen in the area and berries are ripe
along the trail.
(p ∧ r) → ¬q
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30. State the converse, contrapositive, and inverse of each of these conditional statements.
(a) If it snows tonight, then I will stay home. (p → q)
Converse: (q → p) If I stay home, then it snows tonight.
Contrapositive: (¬q → ¬p) If I do not stay home, then it does not snow tonight.
Inverse:(¬p → ¬q) If it does not snow tonight, then I do not stay home.
(b) I go to the beach whenever it is a sunny summer day. (That is, if it is a sunny summer day, then I
go to the beach.)
Converse: If I go to the beach, then it is a sunny summer day.
Contrapositive: If I do not go to the beach, then it is not a sunny summer day.
Inverse: If it is not a sunny summer day, then I do not go to the beach.
(c) When I stay up late, it is necessary that I sleep until noon. (That is, if I stay up late, then I sleep
until noon.)
Converse: If I sleep until noon, then I stay up late.
Contrapositive: If I do not sleep until noon, then I do not stay up late.
Inverse: If I do not stay up late, then I do not sleep until noon.
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4. Translate the given statement into propositional logic using the propositions provided.
“To use the wireless network in the airport you must pay the daily fee unless you are a subscriber to the
service.” (That is, if you use the wireless network in the airport, then you must pay the daily fee; unless
you are a subscriber to the service.)
Express your answer in terms of
w: You can use the wireless network in the airport
d: You pay the daily fee
s: You are a subscriber to the service
¬s → (w → d)
8. Express these system specifications using the given propositions and logical connectives (including negations).
p: The user enters a valid password
q: Access is granted
r: The user has paid the subscription fee
(a) The user has paid the subscription fee, but does not enter a valid password.
r ∧ ¬p
(b) Access is granted whenever the user has paid the subscription fee and enters a valid password.
(r ∧ p) → q
(c) Access is denied if the user has not paid the subscription fee.
¬r → ¬q
(d) If the user has not entered a valid password but has paid the subscription fee, then access is granted.
(¬p ∧ r) → q
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8. Use DeMorgan’s Laws to find the negation of each of the following statements.
(a) Kwame will take a job in industry or go to graduate school.
Solution: Kwame will not take a job in industry and will not go to graduate school.
(b) Yoshiko knows Java and calculus.
Solution: Yoshiko does not know Java or does not know calculus.
(c) James is young and strong.
Solution: James is not young or is not strong.
(d) Rita will move to Oregon or Washington.
Solution: Rita will not move to Oregon and will not move to Washington.
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10. For each of these compound propositions, use the conditional-disjunction equivalence (Example 3) to
find an equivalent compound proposition that does not involve conditionals.
(a) ¬p → ¬q
Solution:
¬p → ¬q ≡ ¬(¬p) ∨ ¬q
by conditional-disjunction
≡ p ∨ ¬q
by double negation
(b) (p ∨ q) → ¬p
Solution:
(p ∨ q) → ¬p ≡ ¬(p ∨ q) ∨ ¬p
by conditional-disjunction
≡ (¬p ∧ ¬q) ∨ ¬p
by DeMorgan
≡ ¬p
by absorption
(c) (p → ¬q) → (¬p → q)
Solution:
(p → ¬q) → (¬p → q) ≡ ¬(p → ¬q) ∨ (¬p → q)
≡ ¬(¬p ∨ ¬q) ∨ (¬(¬p) ∨ q)
≡ (¬(¬p) ∧ ¬(¬q)) ∨ (¬(¬p) ∨ q)
by conditional-disjunction
by conditional-disjunction
by DeMorgan
≡ (p ∧ q) ∨ (p ∨ q)
by double negation
≡ ((p ∧ q) ∨ p) ∨ q
by associativity
≡p∨q
by absorption
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22. Show that two compound propositions
p→q
and
¬q → ¬p
are logically equivalent. To do this, either show that both sides are true, or that both sides are false, for
exactly the same combinations of truth values of the propositional variables in these expressions (whichever
is easier).
Solution: p → q is false only when p is true and q is false. ¬q → ¬p is false only when ¬q is true and ¬p
is false, equivalently, only when q is false and p is true. Thus, p → q and ¬q → ¬p are logically equivalent
because they are both false for exactly the same combinations of truth values of the propositional variables.
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8. Translate these statements into English, where R(x) is “x is a rabbit” and H(x) is “x hops” and the
domain consists of all animals.
(a) ∀x(R(x) → H(x))
Solution: Every animal that is a rabbit hops. Alternatively, for every animal, if it’s a rabbit, then it
hops.
(b) ∀x(R(x) ∧ H(x))
Solution: Every animal is a rabbit and hops.
(c) ∃x(R(x) → H(x))
Solution: There is an animal which, if it’s a rabbit, then it hops.
(d) ∃x(R(x) ∧ H(x))
Solution: There is an animal that is a rabbit and hops.
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32. Express each of these statements using quantifiers. Then form the negation of the statement so that no
negation is to the left of a quantifier. Next, express the negation in simple English. (Do not simply use the
phrase “It is not the case that.”)
Solution: Let
P (x) : x has fleas;
Q(x) : x can add;
R(x) : x can climb;
S(x) : x can speak French;
U (x) : x can swim;
V (x) : x can catch fish;
Domain of x will vary for different parts.
(a) All dogs have fleas. ∀x in the domain of all dogs, P (x)
Negation: ¬(∀x in the domain of all dogs, P (x)) ≡ ∃x in the domain of all dogs, ¬P (x).
Therefore, the negation of the statement is “There is at least one dog that does not have fleas.”
(b) There is a horse that can add. ∃x in the domain of all horses, Q(x)
Negation: ¬(∃x in the domain of all horses, Q(x)) ≡ ∀x in the domain of all horses, ¬Q(x).
Therefore, the negation of the statement is “All horses cannot add,” equivalently, “No horse can add.”
(c) Every koala can climb. ∀x in the domain of all koalas, R(x)
Negation: ¬(∀x in the domain of all koalas, R(x)) ≡ ∃x in the domain of all koalas, ¬R(x).
Therefore, the negation of the statement is “There is at least one koala that cannot climb.”
(d) No monkey can speak French. ¬(∃x in the domain of all monkeys, S(x))
Negation: ¬ (¬(∃x in the domain of all monkeys, S(x))) ≡ ∃x in the domain of all monkeys, S(x)
Therefore, the negation of the statement is “There is at least one monkey that can speak French.”
(e) There exists a pig that can swim and catch fish. ∃x in the domain of all pigs, (U (x) ∧ V (x))
Negation: ¬(∃x in the domain of all pigs, (U (x) ∧ V (x))) ≡ ∀x in the domain of all pigs,
¬(U (x) ∧ V (x)) ≡ ¬U (x) ∨ ¬V (x).
Therefore, the negation of the statement is “Every pig cannot either swim or catch fish,” equivalently,
“No pig can both swim and catch fish.”
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