MATH3030 Homework
Hong Yue,
Georgia College, Department of Mathematics,
Milledgeville, Georgia, 31061, USA, [email protected]
September 14, 2013
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Week 2 Homework
HW: P32 1.4.1. (3) (5) (6)
Exercise 1.4.1. For each of the following arguments, if it is valid, give a
derivation, and if it is not valid, show why.
3. E → F
¬G → ¬F
H→I
E∨H
G∨I
Solution. This argument is valid. See the derivation below:
(1) E → F
(2) ¬G → ¬F
(3) H → I
(4) E ∨ H
(5) F → G,
(6) E → G,
(7) G ∨ I,
(2), Contrapositive.
(1),(5) Hypothetical Syllogism.
(6), (3), (4) Constructive Dilemma.
5. P → Q
¬R → (S → T )
R ∨ (P ∨ T )
¬R
G∨S
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Solution. This argument is not valid. See the following counterexample. For
P, Q, R, S, T take the given values, respectively, all the premises are true, but
the conclusion is false.
P
F
Q
F
R
F
S
F
T
T
P →Q
T
S→T
T
¬R → (S → T )
T
R ∨ (P ∨ T )
T
¬R
T
6. ¬A → (B → ¬C)
C → ¬A
(¬D ∨ A) → ¬¬C
¬D
¬B
Solution. This argument is valid by the following derivation:
(1) ¬A → (B → ¬C)
(2) C → ¬A
(3) (¬D ∨ A) → ¬¬C
(4) ¬D
(5)
(6)
(7)
(8)
(¬D ∨ A) → C,
(3), Double Negation
¬A → (C → ¬B),
(1) Contrapositive, Double Negation
(¬D ∨ A),
(4), Addition
¬B,
(7), (5), (2), (6), Hypothetical Syllogism.
Exercise 1.4.2. For each of the following arguments, if it is valid, give a
derivation, and if it is not valid, show why.
(1) If Fishville is boring, then it is hard to find. If Fishville is not small,
then it is not hard to find. Fishville is boring. Therefore Fishville is small.
Solution:
Let P = Fishville is boring, Q = Fishville is hard to find, R = Fishville
is small. Translate the argument into symbols:
P →Q
¬R → ¬Q
P
R
It is valid by the following derivation:
(1) P → Q
(2) ¬R → ¬Q
2
Q∨S
F
(3) P
(4) Q → R,
(2) Contrapositive
(5) R,
(3), (1), (4) Hypothetical Syllogism.
(2) If the new CD by The Geeks is loud or tedious, then it is not long
and not cacophonous. The new CD by The Geeks is tedious. Therefore the
CD is not long.
Let P = the new CD by The Geeks is loud, Q = the new CD by The
Geeks is tedious, R = the new CD by The Geeks is long. S = the new CD
by The Geeks is cacophonous Translate the argument into symbols:
(P ∨ Q) → (¬R ∧ ¬S)
Q
¬R
It is valid by the following derivation:
(1) (P ∨ Q) → (¬R ∧ ¬S)
(2) Q
(3) (P ∨ Q),
(2), Addition
(4) (¬R ∧ ¬S)
(3), (1)
(5) ¬R,
(4), Simplification.
(3) If the food is green, then it is undercooked. If the food is smelly, then
it is stale. The food is green or it is stale. Therefore the food is undercooked
or it is smelly.
Let P = the food is green, Q = the food is undercooked, R = the food
is smelly, S = the food it is stale Translate the argument into symbols:
P →Q
R→S
P ∨S
Q∨R
Solution. This argument is not valid. See the following counterexample. For
P, Q, R, S take the given values, respectively, all the premises are true, but
the conclusion is false.
P
F
Q
F
R
F
S
T
P →Q
T
R→S
T
P ∨S
T
3
Q∨R
F
(5) It is not the case that Fred plays both guitar and flute. If Fred does
not play guitar and he does not play flute, then he plays both organ and
harp. If he plays harp, then he plays organ. Therefore Fred plays organ.
Let P = Fred plays guitar, Q = Fred plays flute, R = Fred plays organ,
S = Fred plays harp. Translate the argument into symbols:
¬(P ∧ Q)
¬P ∧ ¬Q → (R ∧ S)
S→R
R
Solution. This argument is not valid. See the following counterexample. For
P, Q, R, S take the given values, respectively, all the premises are true, but
the conclusion is false.
P
F
Q
T
R
F
S
T
R∧S
F
¬P ∧ ¬Q
F
¬(P ∧ Q)
T
(¬P ∧ ¬Q) → (R ∧ S)
T
S→R
F
Exercise 1.5.1. Suppose that the possible values of x are all people. Let
Y (x) = x has green hair, let Z(x) = xlikes pickles and let W (x) = x has a
pet frog. Translate the following statements into words.
(1) (∀x)Y (x): All people have green hair. or Everyone has green hair,
(2) (∃x)Z(x). Someone likes pickles. or There is at least a person likes
pickles.
(3) (∀x)[W (x) ∧ Z(x)]. Everyone has a pet frog and likes pickles
(4) (∃x)[Y (x) → W (x)]. There exist a person if he/she has green hair,
then he/she has a pet frog.
(5) (∀x)[W (x) ↔ ¬Z(x)]. Everyone has a pet frog if only if he/she
doesn’t like pickles.
Exercise 1.5.4. Suppose that the possible values of p and q are all fruit. Let
A( p, q) = p tastes better than q, let B(p,q) = p is riper than q and let
C(p,q) = p is the same species as q. Translate the following statements into
symbols.
(3) There is a fruit such that all fruit taste better than it and is not riper
than it: (∃q)(∀p)[A(p, q) ∧ ¬B(p.q)].
(4) For every fruit, there is a fruit of the same species that does not taste
better than it: (∀q)(∃p)[C(p, q) ∧ ¬A(p.q)].
Exercise 1.5.5. Convert the following statements, which do not have their
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quantifiers explicitly given, into statements with explicit quantifiers, both
in symbols and in English.
Suppose that the possible values of x are all people.
(1) People are nice:
All people are nice, Let P (x) = x is nice. (∀x)P (x).
(2) Someone gave me a present.
There is a person who gave me a present. Let Q(x) = x gave a me
present. (∃x)Q(x).
Exercise 1.5.6. Write a negation of each statement. Do not write the word
not applied to any of the objects being quantified (for example, do not write
Not all boys are good for Part (1) of this exercise).
3. The equation x2 − 2x > 0 holds for all real numbers x:
There is a real number x satisfying x2 − 2x ≤ 0.
4. Every parent has to change diapers:
There is a parent who does not have to change diapers.
5. Every flying saucer is aiming to conquer some galaxy:
Some flying saucer is not aiming to conquer some galaxy.
6. There is an integer n such that n2 is a perfect number:
For any integer n, n2 is not a perfect number.
7. There is a house in Kansas such that everyone who enters the house
goes blind:
For each house in Kansas, there is a person who enters the houses does
not go blind.
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