Math 55 Worksheet
Adapted from worksheets by Rob Bayer, Summer 2009.
Instructions
• Introduce yourselves! Despite popular belief, math is in fact a team sport!
• Decide who will be your first scribe.
• Do the problems below, having a different person be the scribe for each one.
• Try to work out the problems as a group, but feel free to flag me down if you run into a wall.
Predicates and Quantifiers
1. If P (x) is the statement “x > 0” and G(x, y) is the statement x2 ≥ y, determine the truth value of each of the
following
(a) P (1) True
(b) G(2, −3) True
(c) P (−1) → G(−2, 1) True
2. Using the same predicates as above, determine the truth values of each of the following statements if the domain
is the set of all real numbers.
(a) ∀xP (x) False
(b) (∃xP (x)) ∧ (∀xG(x, 0)) True
(c) ∃yG(2, y) True
3. Suppose the domain of P (x) consists of the integers 0,1,2,3. Re-write each of the following statements without
using quantifiers:
(a) ∀xP (x) Solution: P (0) ∧ P (1) ∧ P (2) ∧ P (3)
(b) ∀x (x 6= 3 → ¬P (x)) Solution:
¬P (0) ∧ ¬P (1) ∧ ¬P (2)
4. Let S(x) be the statement “x is a student,” L(x) be “x lives in Japan” J(x) be “x speaks Japanese.” Translate
each of the following into English or into logic symbols as appropriate. The domain is the set of all people.
(a) ∃x (L(x) ∧ S(x)). Solution: Some person is a student and lives in Japan
(b) ∀x((L(x)∧¬S(x)) → J(x)). Solution: Every person who lives in Japan and is not a student speaks Japanese
(c) ∃x(S(x) ∧ L(x) ∧ ¬J(x)).Solution: There is a student who lives in Japan and who does not speak Japanese
(d) There is a Japanese speaking student. Solution: ∃x(J(x) ∧ S(x)
(e) Not all speakers of Japanese live in Japan. Solution: ∃x(J(x) ∧ ¬L(x)
(f) Some students live in Japan, but some don’t. Solution: ∃x(S(x) ∧ J(x)) ∧ ∃x(S(x) ∧ ¬L(x))
(g) The only Japanese residents who don’t speak Japanese are students. Solution: ∀x(L(x) ∧ ¬J(x) → S(x))
5. Determine whether each of the following pairs of sentences are equivalent. If so, explain why. If not, give an
example of predicates and domains where they differ.
(a) ∃x(P (x) ∧ Q(x)); ∃xP (x) ∧ ∃xQ(x). Solution: False. For example, take P (x) to be the statement x was
born in Indiana, and Q(x) to be the statement x was born in California. Suppose our domain consists of
Math 55 students. Then the second statement is true because there is someone who was born in California
and someone who was born in Indiana (me!)
However, the first statement is not true, since there does not exists any person “x” for who was born in
both states!
(b) ***IMPORTANT: On homework and exams, you NEED to give a justification for this type of problem.
And it should be about as long as the one I just wrote. There’s not really a shorter way to do this. So
make sure you’re justifying things clearly and cleanly.
(c) ∀x(P (x) ∧ Q(x)); ∀xP (x) ∧ ∀xQ(x). Solution: True. A proof of this fact (or a similar one) is in the book.
But essentially, both sides say ”P and Q are true for all x”
(d) ∃x(P (x) → Q(x)); ∃xP (x) → ∃xQ(x). Solution: This one is crazy! So let’s talk about it... Call the two
propositions in question 1 and 2. Suppose that P (x) is x > 0. And Q(x) is x = 10. And suppose our
domain is the set {0, 1}. Then let’s see what happens:
Statement 1 : Let’s try 0 and 1.
P (0) = F alse, Q(0) = F alse
P (1) = T rue, Q(1) = F alse In particular, P (0) =⇒ Q(0) is TRUE (remember that a false statement
always implies anything. So Statement 1 is true.
However, Let’s look at statement 2: The first bit is ∃xP (x). This is TRUE, (for example x = 1.
The second bit is ∃xQ(x). This is FALSE (Q(x) is never true :( )
So Statement two is True → F. Which is FALSE.
Sound the fanfare! We’ve found a situation where Statement 1 is True, but Statement 2 is false. Phew.
Story Time: Ok, this one was quite brutal. I actually spent a good amount of time thinking about it,
(and my friends are still working through it) And I’d be lying if I said this counterexample came naturally.
However, what’s important is how I eventually found it. So that’s what today’s Story Time is about.
I figured ”well if these two statements are the same, then their negations should be the same”. So I looked
at their negations. Remember that negation switches the quantifiers ∃ and ∀, and that the negation:
¬(P → Q) = P ∧ ¬Q. Work out on your own what the negations of statements 1 and 2 should be... (this
is active story time now, because you’ll learn more by trying this yourself.)
...
... No seriously, you should work it out. Because it took me three tries to get the right answer, and this is
great practice! Here’s what I got:
¬ (Statement 1) = ¬(∃x(P (x) → Q(x))) = ∀x(P (x) ∧ ¬Q(x))
¬ (Statement 2) = ¬(∃xP (x) → ∃xQ(x)) = ∃xP (x) ∧ ∀x¬Q(x)
Now then I started thinking... and realized that the for statement one, we needed to have P (x) be true for
EVERY x. Whereas the second statement only required it to be true for some x. And we needed Q(x) to
be always false in both cases.
So I made up a truth table like in the example above.
P Q
F F
T F
And then cooked up some actual propositions P and Q and a domain which had this truth table. I crossed
my fingers and checked it.... and it worked!
This is the type of thinking you should expect to do in this class. Getting stuck the first second, and third
times doesn’t mean you’re lost, it just means you’re feeling the problem out! end of StoryTime.
Nested Quantifiers and Restricted Domains
1. Let T (x, y) be “x is taking y”, L(x, y) be “x likes y”, R(x, y) be “x is required to take y.” If the domain for x is
all students and they domain for y is all classes, translate each of the following between English and Logic:
(a) ∀x∃yL(x, y). Solution: For every student, there is some class that they like
(b) ∃y∀x¬L(x, y). Solution: There is a class which no student likes
(c) ∀y∃xR(x, y). Solution: For every class there is a student who is required to take it
(d) ∃x∃y(T (x, y) ∧ ¬L(x, y)). Solution: There is a student taking some class that (s)he doesn’t like
(e) Every student is required to take at least one class. Solution: ∀x∃yR(x, y)
(f) Some student likes all of his/her current classes. Solution: ∃x∀y(T (x, y) → L(x, y))
(g) Some student only likes courses which he/she isn’t required to take. Solution: ∃x∀y(L(x, y) → ¬R(x, y))
2. Determine the truth value of each of the following statements. The domain is the set of all real numbers
(a) ∀x∃y(x > y). True
(c) ∀x∃y(x = y 2 ).False, let x = −1
(b) ∃x∃y(x ≥ y ∧ y ≥ x). True
(d) ∀x∀y∃z(x > y → x > z > y). True
(e) ∃x∃y(x + y = 1 ∧ x − y = 3). True
(f) ∀ > 0 ∃δ > 0 ∀α (|x − 3| < δ → |x2 − 9| < ).True
3. Re-write each of the following so that negation symbols appear only after predicates.
(a) ¬∃x∀yP (x, y). ∀x∃y¬P (x, y)
(b) ¬∃x¬P (x). ∀xP (x)
(c) ¬∀x∀y(P (x, y) → ¬Q(x, y)). ∃x∃y(¬P (x, y) ∧ ¬Q(x, y))
(d) ¬((∀xP (x)) ∨ ∃y∀z(R(y, z) ∧ S(y, z))). ∃x¬P (x) ∧ ∀y∃z(¬R(y, z) ∨ ¬S(y, z))
4. F The symbol ∃!xP (x) stands for “there exists one and only one x such that P(x) is true” and is often pronounced
“there is a unique x...” Show that ∃!xP (x) can be rewritten using just regular quantifiers.
Puzzles
1. Consider the following set of four statements:
(a) One of these statements is false
(b) Two of these are false
(c) Three of these are false
(d) Four of these are false
Which of the above, if any, are true?
2. Two bicyclists enter opposite ends of a 100-foot long tunnel that is only wide enough for one bike. One is
travelling 10 ft/s and the other is travelling 5 ft/s in the opposite direction. A bird flying 20 ft/s enters the
tunnel just in front of the 10 ft/s cyclist. When the bird gets to the other cyclist, it immediately turns around
and flies back towards the 10 ft/s cyclist. When the bird gets back to him, it turns around again, etc, etc. How
many feet does the bird fly before the bicyclists collide?
3. F Four people travelling at night come to a small footbridge and need to cross to the other side. Unfortunately,
they only have 1 flashlight and the bridge can only hold the weight of two people at once. One person takes 10
minutes to cross, another 5, another 2, and the last takes 1 minute. Anyone crossing must have the flashlight
and when travelling together, they must go the pace of the slower person. What is the quickest time in which
that can all get across?