Solution to Exam 1, Fall '13 [PDF]

CMPSC 360
Discrete Mathematics for Computer Science
Fall 2013
Exam I: Logic
September 16, 2013
Name: Annotated Solution Last 4 digits of ID: ____ ____ ____ ____
In addition to correct solutions to each problem, this document will also give some commentary and strategies for
understanding how to solve some problems, particularly where that seems like it would be most helpful.
Honor Code Statement
I certify that I have not discussed the contents of the exam with any other student and I will not discuss the contents
of this exam with any student in this class before it is returned to me with a grade. I have not cheated on this exam in
any way, nor do I know of anyone else who has cheated.
Signature: _________________________________________________________
Directions/Notes:
• Write your name just on this cover page. Write your ID on every page of this exam.
(This is very important this time. We will almost certainly disassemble your exam to grade problems in parallel. If we waste time
because you didn't follow this direction, you will be penalized. If we cannot match a sheet of your exam back to the rest of your exam
because you did not follow this direction, you will not earn credit on that sheet.)
• No notes, books, or calculators are permitted.
• Be sure to sign the honor code statement when you are finished.
• All questions on this exam are implicitly prefaced with “As taught in CMPSC 360 lectures this term.”
• Always justify your work and present solutions using the conventions taught in class.
• Use pencil to complete this exam. Use of pen will result in an automatic 10-point deduction and we
reserve the right not to read any problem with cross-outs.
• It is acceptable to use the shorthand “LHS” to refer to the left-hand side of an equivalence in writing a
proof.
• In algebraic substitutions, you may only use laws from our list of equivalences, not anything derived as
a result of some problem or example.
Score Breakdown:
# 1 2 3 4 5 6 7 8 9 10 11 AP Total
Score
Value 8 8 6 12 6 13 10 8 14 10 7 -- 100

CMPSC 360 Exam 1 – Fall 2013 – Page 2
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1. Build a truth table for the following expression: ~(p ∨ q) ↔ (q ∧ ~p) [8 pts.]
p q p ∨ q ~(p ∨ q) ~p (q ∧ ~p) ~(p ∨ q) → (q ∧ ~p) (q ∧ ~p) → ~(p ∨ q) ~(p ∨ q) ↔ (q ∧ ~p)
T T T F F F T T T
T F T F F F T T T
F T T F T T T F F
F F F T T F F T F
2. Prove the following logical equivalence algebraically: (p ∨ q) ∧ ~(p ∧ q) ≡ (p ∨ q) ∧ (p → ~q) ∧ (q → ~p) [8 pts.]
(p ∨ q) ∧ ~(p ∧ q) ≡ (p ∨ q) ∧ (~p ∨ ~q) by De Morgan's law
≡ (p ∨ q) ∧ (~p ∨ ~q) ∧ (~p ∨ ~q)
by idempotent law
≡ (p ∨ q) ∧ (~p ∨ ~q) ∧ (~q ∨ ~p)
by commutative law
≡ (p ∨ q) ∧ (p → ~q) ∧ (q → ~p) by "or" def. of → ❑
Comment: This was, of course, the problem that was given in recitation in Week 2 and you were strongly encouraged to do,
so it should have been very familiar.
3. In each part, you are given a claim that is either true or false. In a complete sentence, either assert it is true with a
concrete proof or assert it is false with concrete disproof.
[6 pts.]
a. Given the sets A = {x ∈ R | 0 < x < 3} and B = {e, π}, B ⊆ A.
B ⊈ A, as π ∈ B but π ∉ A.
b. (6, Z) = ( 3 2 − 3 , {…, -2, -1, 0, 1, 2, …})
(6, Z) = (3 2 - 3, {…, -2, -1, 0, 1, 2, …}), as 6 = 3 2 - 3 and Z = {…, -2, -1, 0, 1, 2, …}
Comments:
1. These questions were to follow up on the quiz from Chapter 1, but also to reinforce the idea of working from
definitions.
2. Your sentences needed to be specific. "B ⊈ A" is much more specific than "the claim is false" (and a lot less writing
on your part too).
3. Remember that the definition for subsethood is a universal conditional. For a generic universal conditional statement
∀x ∈ D [P(x) → Q(x)], the negation is ∃x ∈ D [P(x) ∧ ~Q(x)], and so the strategy for claiming something doesn't
meet a universal conditional is always the same: exhibit an x in the domain for which P(x) is true but for which Q(x)
is false, exactly what the negation says. So, here, you just needed to give a specific element of B that was not in A.
(This question was to be a nice connection from Chapter 1 to Chapter 3. Did you make the connection? Hopefully
you do now!)

CMPSC 360 Exam 1 – Fall 2013 – Page 3
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4. Suppose you're walking on campus observing and you pause near a tree by a sidewalk to observe the set of animals in
your view. Let's call that set A. We have lots of squirrels and rabbits here, so let S be the set of squirrels in your view and
let R be the set of rabbits in your view. (By definition, S ⊆ A and R ⊆ A.)
[12 pts.]
Define the following predicates on any input x from A:
UnderTree(x) ≡ x is under the tree
OnSidewalk(x) ≡ x is on the sidewalk
Note: Only the final number to the top right of this problem matters in grading. We left some other numbers on papers
to help sort papers into piles when grading – don't worry about those. The final score was determined by your work on
all parts of the problem as a whole.
a. Using the sets and predicates defined above, translate each statement to symbolic logic notation.
i. All squirrels are on the sidewalk.
∀x ∈ A, x ∈ S → OnSidewalk(x)
Comment: You could also write this as "∀x ∈ S, OnSidewalk(x)" and many did. However, for it to make
any sense for you to answer part (b), your response here would have to be a universal conditional
statement.
ii. There are at most two squirrels under the tree.
~∃x ∈ S ∃y ∈ S ∃z ∈ S [UnderTree(x) ∧ UnderTree(y) ∧ UnderTree(z) ∧ x ≠ y ∧ y ≠ z ∧ x ≠ z]
Comment: Remember that "at most two" is equivalent to "not at least three," and that's exactly the mindset
of the translation here. This can be done with universal quantifiers too – and you can feed the negation
through and apply the "or" definition of the conditional to get that form – but it's trickier. Be very careful:
"at most two" only states an upper bound on the number of squirrels; it does not say anything about a
lower bound. A common error here was to use existential quantifiers without the negation out front, but that
nearly always says there's at least one squirrel under the tree. Indeed, there could be none.
iii. If there are at least two squirrels on the sidewalk, the rabbit is under the tree.
∃x ∈ S ∃y ∈ S [OnSidewalk(x) ∧ OnSidewalk(y) ∧ x ≠ y] → ∃z ∈ R [(∀x ∈ R → x = z) ∧ UnderTree(z)]
Comment: This one was really two independent logical statements, connected by a conditional. The
variables for the squirrels and the rabbit are unrelated. Placement of parentheses or brackets can affect
meaning significantly. Also, note that we don't say our "at least two" squirrels exist; their existence is
conditional. The rabbit definitely exists, and you could speak of the rabbit first if you want, but vacuous
truth makes the rabbit exist in the form here.
b. Express the contrapositive of Statement (i) in Part (a) in English.
All animals that are not on the sidewalk are not squirrels.
5. What is the difference between an inclusive "or" and an exclusive "or"? Give a real-life example where exclusive "or"
makes sense but inclusive does not. Express the notion of an exclusive "or" of two atomic statements symbolically in
four distinctly different ways.
[6 pts.]
Sentences formed using both operators are true when exactly one of the operands is true. An inclusive "or" is true when
both operands are true, whereas an exclusive "or" is not. Real-life examples vary (and were interesting to read).
Ways to express exclusive "or":
(p ∨ q) ∧ ~(p ∧ q)
(p ∧ ~q) ∨ (q ∧ ~p) [This is disjunctive normal form]
(p ∨ q) ∧ (p → ~q) ∧ (q → ~p) [As per problem left for you in recitation / #2]
~(p ↔ q)
[As per problem left for you in recitation]
Comment: The exclusive "or" forms needed to be distinctly different, and this problem partly served to see if you got
the takeaway lessons from the problems that were recommended to you at the end of Week 2's recitation.

CMPSC 360 Exam 1 – Fall 2013 – Page 4
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6. Define AL to be the set of Major League Baseball teams in the American League. The American League is partitioned
into three subsets: E, the AL East Division; C, the AL Central Division; and W, the AL West Division. We use the
notation w-l to give a team's record, where w is the number of wins and l is the number of losses. As of September 4 1 ,
here are the standings for the three divisions of the American League:
East (E) Central (C) West (W)
Boston (84-57)
Tampa Bay (76-61)
New York (75-64)
Baltimore (73-65)
Toronto (64-76)
Detroit (81-59)
Cleveland (74-65)
Kansas City (72-67)
Minnesota (61-77)
Chicago (56-82)
Oakland (80-59)
Texas (80-59)
LA (64-73)
Seattle (63-76)
Houston (46-93)
Determine whether each statement is true or false (worth no credit but important to help us know your intentions) and
prove or disprove each statement (worth all the credit).
[13 pts.]
a. ∃t ∈ AL s.t. t has at least 80 wins.
True.
Boston ∈ AL and Boston has at least 80 wins.
Comment: You need to prove the statement by providing one example. Part of that is asserting the relevance of the
example: the first part of the sentence should be noting that whatever team you want to discuss indeed is in the
domain, i.e. it's an element of AL.
b. ∀t ∈ AL, t has at least 80 losses → t ∈ C or t ∈ W
True.
Chicago has at least 80 losses and Chicago ∈ C or Chicago ∈ W.
Houston has at least 80 losses and Houston ∈ C or Houston ∈ W.
c. ∀t ∈ AL, t has at least twice as many wins as losses → t ∈ E
True.
There is no team t ∈ AL s.t. t has twice as many wins as losses, so this statement is vacuously true.
Comment: Don't misinterpret vacuous truth as false; if no element of the domain satisfies the hypothesis of a
universal conditional statement, that statement is indeed true – vacuously true.
d. ∀t ∈ E, ∃s ∈ C s.t. s has fewer wins than t
True.
Boston ∈ E and Detroit ∈ C and Detroit has fewer wins than Boston.
Tampa Bay ∈ E and Chicago ∈ C and Chicago has fewer wins than Tampa Bay.
New York ∈ E and Chicago ∈ C and Chicago has fewer wins than New York.
Baltimore ∈ E and Chicago ∈ C and Chicago has fewer wins than Baltimore.
Toronto ∈ E and Chicago ∈ C and Chicago has fewer wins than Toronto.
Comment: For this statement and the next, you need to prove the universal claim true via exhaustion. In each case,
you must spell out all five teams in E. Note that the order of quantifiers is different, and indeed, the interpretations
of the sentences differ. We wanted you to exhaust the details completely to show your understanding of quantifiers.
e. ∃s ∈ C s.t. ∀t ∈ E, s has fewer wins than t
True.
Chicago ∈ C and Boston ∈ E and Chicago has fewer wins than Boston
and Tampa Bay ∈ E and Chicago has fewer wins than Tampa Bay
and New York ∈ E and Chicago has fewer wins than New York
and Baltimore ∈ E and Chicago has fewer wins than Baltimore
and Toronto ∈ E and Chicago has fewer wins than Toronto.
1 when most of the exam was written, to give plenty of time to proofread and copy

CMPSC 360 Exam 1 – Fall 2013 – Page 5
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7. Determine, via formal proof or disproof, whether the following argument form is valid: [10 pts.]
p ∨ ~q
q → r
p ∨ r → q ∧ p
∴ r ∨ p
We construct a truth table to check the validity of this argument form:
premises
conclusion
p q r ~q p ∨ r q ∧ p p ∨ ~q q → r p ∨ r → q ∧ p r ∨ p
T T T F T T T T T T
T T F F T T T F T
T F T T T F T T F
T F F T T F T T F
F T T F T F F T F
F T F F F F F F T
F F T T T F T T F
F F F T F F T T T F
The first and eighth rows are critical rows. As the conclusion is false in one of the critical rows (the bottom row), the
argument form is not valid.
Comment: As happens pretty much every semester, results on this question were at the extremes. The vast majority of
you did rather well. Some did not try a technique that was right, but remember that there's only one technique we used
for proving the validity of an argument form and it is this one. See the beginning of the class notes from Section 2.3 to
review if you've forgotten the technique.
8. We say a statement is in conjunctive normal form (CNF) when [8 pts.]
a. All logical connectives that are not in parentheses are "and."
b. All logical connectives that in parentheses are "or."
c. The "not" operation may be used inside parentheses, but not outside.
In other words, the statement is a conjunction ("and" statement) of disjunctions ("or statements) of atomic sentences.
Here are some examples:
p ∧ q (p ∨ q) ∧ (~q ∨ r) (p ∨ q ∨ ~r) ∧ (~p ∨ ~q ∨ r)
Convert the following statement to conjunctive normal form, justifying all work and presenting work like in a proof:
p ↔ q ∧ r
p ↔ q ∧ r ≡ (p → q ∧ r) ∧ (q ∧ r → p) by "or" def. of →
≡ (~p ∨ (q ∧ r)) ∧ (~(q ∧ r) ∨ p) by "or" def. of →
≡ (~p ∨ (q ∧ r)) ∧ ((~q ∨ ~r) ∨ p) by De Morgan's law
≡ ((~p ∨ q) ∧ (~p ∨ r)) ∧ ((~q ∨ ~r) ∨ p) by distributive law
≡ (~p ∨ q) ∧ (~p ∨ r) ∧ (~q ∨ ~r ∨ p) as dropping parentheses doesn't introduce ambiguity
Comment: Most responses fell into one of three categories: completely right or nearly right, suffering from problems
with parentheses, and problems with the order of operations. As far as parenthesis go, pay very close attention to add
them in applying the definition of the biconditional in the first line. To say "p → q ∧ r ∧ q ∧ r → p" is unclear;
parentheses are needed. Some forgot the order of operations; remember that conditonals and biconditionals are evaluated
last in a line. Unfortunately, while getting the order of operations wrong might be a single error, doing so makes it that
you miss the ability to do most of the derivation.

CMPSC 360 Exam 1 – Fall 2013 – Page 6
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9. Consider the following premises: [14 pts.]
(a) x
(b) r ∨ s → y
(c) ~w ∨ ~x
(d) y → (z → ~p)
(e) r ∧ x → (q ∨ z)
(f) ~r → w
(g) q → ~p
a. Use the premises given above to deduce the conclusion ~p.
If you are unable to solve the problem as written, instead aim for "(q ∨ z) ∧ (z → p)" and circle this sentence (for up to 10/12 points on this part).
(1) ~w by ∨ Elim: c, a
(2) ~~r by modus tollens: f, 1
(3) r by double negation: 2
(4) r ∨ s → (z → ~p) by transitivity: b, d
(5) r ∨ s by ∨ Intro: 3
(6) z → ~p by modus ponens: 4, 5
(7) r ∧ x by definition of ∧: 3, a
(8) q ∨ z by modus ponens: e, 7
(9) ~p by cases: 8, g, 6
Comment: Here we had several perfect or nearly perfect solutions, which is good news. But, like the problem set,
some responses did not use the rules of inference correctly or used rules that don't exist, which does not work. For
example, consider statement (4): r ∨ s → (z → ~p). To be able to use modus ponens, you must first have the exact
form of the hypothesis. This is why the ∨ Introduction step (Step 5) is necessary. If you struggled here, work more
with the rules of inference. This problem was designed to require you to use most of the rules (with transitivity be
optional).
b. Suppose premise (f) were replaced with the statement "~w → r." Would you be able to deduce the same conclusion?
Explain.
This statement is simply the contrapositive of premise (f). As a statement is logically equivalent to its contrapositive,
making this substitution would allow us to deduce the same conclusion.

CMPSC 360 Exam 1 – Fall 2013 – Page 7
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10. Prove the following logical equivalence algebraically: [10 pts.]
∀ε > 0 (∃N (∀n (n > N → |x n – L| < ε))) ≡ ~∃ε > 0 (∀N (∃n (n > N ∧ |x n – L| ≥ ε)))
[For one extra-credit point, what is the meaning of this? It came up in an indirect prerequisite of this course.]
∀ε > 0 (∃N (∀n (n > N → |x n – L| < ε))) ≡ ~~∀ε > 0 (∃N (∀n (n > N → |x n – L| < ε))) by double negation
≡ ~∃ε > 0 ~(∃N (∀n (n > N → |x n – L| < ε))) by De Morgan's for quant.
≡ ~∃ε > 0 (∀N ~(∀n (n > N → |x n – L| < ε))) by De Morgan's for quant.
≡ ~∃ε > 0 (∀N (∃n ~(n > N → |x n – L| < ε))) by De Morgan's for quant.
≡ ~∃ε > 0 (∀N (∃n ~(~(n > N) ∨ |x n – L| < ε))) by "or" def. of →
≡ ~∃ε > 0 (∀N (∃n (~~(n > N) ∨ ~(|x n – L| < ε)))) by De Morgan's law
≡ ~∃ε > 0 (∀N (∃n (n > N ∨ ~(|x n – L| < ε)))) by double negation
≡ ~∃ε > 0 (∀N (∃n (n > N ∨ |x n – L| ≥ ε))) by negating "