Question text is in black, solutions in font.
Q1: 20 points Q2: 30 points Q3: 30 points Q4: 20+10 points Total: 100+10 points
Here are definitions of sets, predicates, and statements used on this exam.
Remember that the score of any quantifier is always to the end of the statement it is in.
Question 2 deals with the following scenario. One day Cardie and Duncan were joined on their morning walk by several other dogs. The set S of dogs on this group walk included Bingley (b), Cardie (c), Duncan (d), Guinness (g), Whistle (w), and perhaps others.
Let the binary predicate JB on S be defined so that JB(x, y) means "dog x joined the walk before dog y". Assume that the relation corresponding to JB is antireflexive, antisymmetric, and transitive.
Let N be the set of natural numbers {0, 1, 2, 3,...}.
If a, b, and m are naturals, with m > 0, the notation "a ≡ b (mod m)" means "a is congruent to b, modulo m".
The operator "%" on naturals, as in Java, refers to integer division, so that "x % y" is the remainder on dividing x by y.
Given any dog, Cardie joined the walk before it if and only if Duncan
joined the walk before it.
Many answers to this one were wrong because they "split the
quantifier" by saying "Cardie joined the walk before all dogs if
and only if Duncan joined the walk before all dogs", which
translates to the inequivalent "(∀x: JB(c, x)) ↔
(∀x: JB(d, x))".
JB(b, c) → (JB(c, g) ∧ JB(g, b))
Many answers omitted the parentheses, which did not cost
them credit because we did say that the → operator has higher
precedence than ∧.
Guinness joined the walk before Bingley, and either Bingley joined before Whistle, or Cardie joined before Guinness, but not both.
∃x:∃y: JB(w, x) ∧ JB(w, y) ∧ (x ≠ y)
The "x ≠ y" is necessary to make the two dogs "distinct".
It's arguable that we should include "x ≠ w" and "y ≠ w" to
convey that they are "other dogs", but since JB is antireflexive
it's natural to assume that. There's no need to say anything to
cover the "at least", though you could -- saying that there are two
such dogs does not say that there are or are not any others.
∀x: (x ≠ b) → JB(x, b)
We gave full credit for saying "↔" instead of "→", as
it is debatable whether the English statement implies that JB(b, b)
is false. A common wrong answer was "∀x: JB(x, b) and
¬JB(b, b)", which contradicts itself.
It is not the case that for every dog, it joined the walk before
Whistle or Bingley joined the walk before it.
Many answers gave the correct alternate form "There exists a dog
who did not join the walk before Whistle and did not join the walk
after Bingley". Splitting the quantifier was also a problem here,
as when this was translated as "it is not the case that all dogs
joined before Whistle or that all dogs joined after Bingley".
Here is the truth table, letting p = JB(c, g), q = JB(b, w), and r =
JB(g, b). The only setting making both II and III true has p and r
true and q false.
To prove this by deductive sequence, we begin by assuming II and
III. By Left Separation on III, we get r. We can then prove ¬q
by Contradiction as follows: if q is true, (p and r) is true by Modus
Ponens on II, which implies p by Left Separation. But then (p xor q)
is false, and this contradicts (p xor q), which follows from III by
Right Separation. Since q is false and (p xor q) is true, q must be true.
(q --> p and r) and (r and (p xor q)
0 1 0 0 0 0 0 0 0 0 0
0 1 0 0 1 0 1 0 0 0 0
1 0 0 0 0 0 0 0 0 1 1
1 0 0 0 1 0 1 1 0 0 0
0 1 1 0 0 0 0 0 1 1 0
0 1 1 1 1 1 1 1 1 1 0
1 0 1 0 0 0 0 0 1 0 1
1 1 1 1 1 0 1 0 1 0 1
In order to use our quantifier proof rules, we need to transform our
goal (Statement VI) so it begins with a quantifier instead of a
negation. The equivalent form is "∃y:¬JB(y, w) ∧
¬JB(b, y)". We can prove this by Existence, if we can find a
concrete example of a dog that did not walk either before Whistle or
after Bingley. (Note that we need a single dog that meets both
these conditions. Many of you found separate counterexamples for each
condition, which does not imply that there is a common counterexample
for both. This is the same error of "splitting the quantifier"
mentioned in Question 1 parts (a) and (f).)
Fortunately there are several such dogs available.
We can take y to be Whistle, since JB(w, w) is false by
antireflexivity and JB(b, w) was found to be false in part (a). Or
we could take y to be Bingley, since JB(b, w) is false and JB(b, b)
is also false by antireflexivity. Statement IV tells us that there
are two dogs that joined after Whistle. Either of these (if we
Instantiate it and call it a) can also
serve as a possible y, because JB(w, a) implies ¬JB(a, w) by
antisymmetry, and JB(b, a) is false either by Specialization from
Statement V (if a ≠b) or antireflexivity (if a = b). (One can
also derive from IV that there exists a dog other than
Bingley that Whistle joined before.)
Once we have such a dog y, we can use Existence to prove the
alternate form of VI. Equivalently, we can begin the proof by
assuming ∀y: JB(y, w) ∨ JB(b, y), and get a contradiction
by Specializing this statement to the dog y for which JB(y, w) and
JB(b, y) are both false.
FALSE. It cannot be determined from Statements I-VI whether Whistle joined before Guinness. But if he did, and there are no other dogs, the order of Cardie/Duncan, Whistle, Guiness, Bingley satisfies all the statements.
TRUE. Statement I means that JB(c, d) and JB(d, c) are both false since they have the same truth values as JB(d, d) and JB(c, c) respectively, and JB is antisymmetric. This violates antisymmetry for the negation of JB.
FALSE, BUT QUESTION WITHDRAWN. It is easy to see that R is reflexive (by antireflexivity of JB) and symmetric (by its definition). If we interpret JB as in the scenario, where it represents events in a linear sequence, R(x, y) means "x and y joined at the same time", which is transitive. But the question only gives Statements I-VI and the three conditions on JB as premises. It is possible to have a scenario where JB is true only when it is required to be by these premises, and in this scenario R is not transitive -- this does not correspond to a possible time sequence of dogs joining the walk, but does show that the transitivity of R cannot be proved from I-VI and the three conditions. We decided that these two possible interpretations render the question unfair, and thus we are withdrawing it.
TRUE, as seen by the left half of the truth table above in the solution to Question 2 (a).
FALSE. If all three atomic variables are false, the left half of the ↔ is true but the right is false.
FALSE. Elements that are in C, but not in either A or B, are in the set on the left but not in the set on the right.
FALSE. Suppose the set in question has just two elements a and b, and only P(a) and Q(b) are true. Then the left-hand side of the ↔ is true and the right-hand side is false. This is, by the way, yet another instance of the "splitting the quantifier" error.
FALSE. If R(x, y) and R(y, x) alway have the same truth value, then their negations also always have the same truth value.
TRUE. We can argue by contradiction. If g were not one-to-one, then g(x) = g(y) for two elements x and y with x ≠ y. Since f is onto, there exist elements u and v for which f(u) = x and f(v) = y, and clearly u ≠ v. But then h = g ∘ f fails to be bijective because it is not one-to-one: h(u) = g(f(u)) = g(x) = g(y) = g(f(v) = h(v).
FALSE. X might be the empty set (such as the set of all unicorns), so that ∀x: P(x) is true, but there is no example v such that P(v) is true.
TRUE. The length of any string in XX is the sum of the lengths of two strings in X, which is the sum of two even numbers and thus is even.
FALSE. We could have u = "ab", v = "b", and w = "bab".
TRUE, BUT QUESTION WITHDRAWN. We meant to write "x ≡ y" instead of "x = y". Had we done that, R would be an equivalence relation, with the class of x consisting of x and its unique inverse (if any). It would be reasonable to interpret the "x = y", with the "(mod n)" following, as "x ≡ y", and answer "false". But as written we could have, for example, n = 5, x = 3, y = 2, and z = 8, so that R(x, y) and R(y, z) are true but R(3, 8) is false. Since there are valid interpretations for both answers, we are withdrawing the question. Since we discovered the issue after returning the exams, we are implementing this by given two points for either answer -- Marius' email gives instructions for claiming your extra points.
FALSE. 115 = 51231 and 119 = 71171 can be so written, but not 97 which is prime. If i or j were allowed to be 0 we could do it, but 0 is not a "positive natural".
TRUE. We could have x = 7 and y = 17. If we replaced 119 by a prime, the statement would be false by the Atomicity Lemma.
The three numbers are pairwise relatively prime.
He gives each of his daughters partial information about x, so that none of them can determine x on her own. He tells Cordelia the remainder x % 97 from dividing x by 97. He tells Goneril the number x % 115, and Regan the number x % 119.
Explain why any two of the daughters, by combining their information, can determine x. You may quote the Chinese Remainder Theorem without proof if you do so accurately.
The Chinese Remainder Theorem in its simple form says that if m and n are
relatively prime, the two congruences x ≡ a (mod m) and x ≡
b (mod n) are equivalent to a single congruence x ≡ c (mod mn).
The proof gives a way to calculate c from a, b, m, and n.
Cordelia and Goneril can together use their two congruences to compute
a number c such that x ≡ c (mod 97 × 115). Since 97
× 119 > 10000, there is at most one number in the range from
0 through 9999 that is congruent to c modulo 97 &time; 115, and
that must be Mr. Lear's PIN number. Cordelia and Goneril can find
the PIN number by adding or subtracting multiples of 97 × 115
to c as necessary to get a number in the proper range. (Since Mr.
Lear's number x was actually in the range, they will wind up with a
number in the range.)
Similarly Cordelia and Regan could together
compute another number c such
that x ≡ c (mod 97 × 119), and since 97 × 119 >
10000, there is at most one x in the range that satisfies this
congruence. Finally, Goneril and Regan together can find the
remainder of x modulo 115 × 119, which is enough to find x
itself because 115 × 119 > 10000.
Last modified 18 October 2018