Dr. Timo de Wolff
Institute of Mathematics
www.math.tamu.edu/~dewolff/Fall14/math302.html
MATH 302 – Discrete Mathematics – Section 501
Homework 3
Fall 2014
Due: Friday, September 26th, 2014, 9:10 a.m.
When you hand in your homework, do not forget to add your name and your UIN.
Exercise 1. Each of the following arguments is wrong. Figure out the mistake and explain,
which rule has been violated or applied wrongly.
1.
2.
3.
1
2
3
4
5
6
7
8
9
∃x : ¬P (x) ∨ Q(x)
∀x : P (x) ∧ R(x)
P (a) ∧ R(a)
P (a)
R(a)
¬P (a) ∨ Q(a)
Q(a)
Q(a) ∧ R(a)
∀x : Q(x) ∧ R(x)
Premise
Premise
2
3
3
1
4, 6
5, 7
8 Conclusion
1
2
3
4
5
6
P (a)
∀x : P (x) → Q(x)
P (a) → Q(a)
Q(a)
Q(a) ∨ S(a)
∀x : Q(x) ∨ S(x)
Premise
Premise
2
1, 3
4
5 Conclusion
1
2
3
4
5
∀x∃y : P (x, y)
∃y : P (a, y)
P (a, b)
∀x : P (x, b)
∃y∀x : P (x, y)
1
Premise
1
2
3
4 Conclusion
Hint: Recall the rules of inference regarding quantifiers, particularly those, which we
remarked as “critical” during the lecture.
Exercise 2. Consider the following predicates: P (x) : “being a pirate”, T (x): “like treasures”, S(x): “like to sail”, R(x, y): “drink more rum than”. Our domain is the set of
persons.
1. Translate the following sentences into a proposition using quantifiers.
(a) “Every pirate drinks more rum than everyone, who is not a pirate.”
(b) “Billy Bones drinks more rum than every pirate, who does not like treasures.”
(c) “There exists no pirate, who does not like treasures and to sail.”
2. Translate the following propositions into regular sentences.
(a) ∃x : P (x) ∧ ∀y : P (y) → R(x, y).
(b) ¬∀x : R(x, Jack Sparrow) ∧ S(x) → P (x).
Exercise 3. Prove the following statement using rules of inference and logical equivalences.
∃x : Q(x) ∧ ¬R(x)
Premise
∀x∀y : ¬P (x, y) → R(y) Premise
∃y∀x : P (x, y)
Conclusion
Exercise 4. If two symbols a, b denote the same object in the universe, then we say “a and
b are identical”, denoted as a = b (and a 6= b if a and b are not identical).
Let P (x) be an arbitrary predicate. We define a new, third quantifier ∃! (for the predicate
P (x)) in the following way:
∃!x : P (x) :⇔ ∃x : P (x) ∧ ∀y : P (y) → x = y.
1. Explain in own words what the expression “∃!x : P (x)” means. Give a reason using
the (meaning of the) upper definition.
2. Assume we have the following premises:
1 ∃!x : P (x)
2 ∀x : P (x) → Q(x)
Can we conclude that ∃!x : Q(x)? Give either a proof or give a counterexample and
explain in own words why one cannot conclude such a proposition in general.
2