Question text is in black, solutions in blue.
Q1: 20 points Q2: 20 points Q3: 10 points Q4: 15 points Q5: 15 points Q6: 20+10 points Total: 100+10 points
Here are definitions of sets, predicates, and statements used in Questions 1-5 on this exam.
Remember that the score of any quantifier is always to the end of the statement it is in.
Let D be a finite set of dogs including the named dogs Cardie (c), Duncan (d), Mia (m), and Toby (t), along possibly with others.
Let DW by the set of the seven days of the week: {Sun, Mon, Tue, Wed, Thu, Fri, Sat}.
Let N be the set of natural numbers {0, 1, 2, 3,...}.
Let L(x, y, r) be the predicate defined by "dog x likes dog y on day of week r".
Let w be the function from D to N defined so that w(x) is the weight of dog x in kilograms, rounded to the nearest integer.
Let SW(x, y) ("same weight") be the binary relation on D defined by "w(x) = w(y)".
Let LEW(x, y) ("less than or equal weight") be the binary relation on D defined by "w(x) ≤ w(y)".
L(d, t, Sun) → ¬(L(d, m, Wed) ∧ L(d, t, Tue))
If it is not the case that Duncan likes Toby both on Sundays and on Tuesdays,
then Duncan likes both Mia on Wednesdays and Toby on Sundays.
(The most common error here was to misplace the ¬ and say "It is not
the case that if Duncan likes...")
(∀r:L(d, t, r)) ∨ (∀s:¬L(d, t, s))
Probably a majority of you used only one quantifier and wrote something
like "∀r:L(d, t, r) ∨ ¬L(d, t, r)" which is the tautology
"On any given day, Duncan either likes Toby or he doesn't".
Cardie likes every dog on every day of the week.
∃x:∃y:(w(x) = w(y)) and (x ≠ y)
I took off a point for forgetting the "x ≠ y". You could also
just put a ¬ in front of the definition of "one-to-one", as long
as you adapted it correctly to talk about the function w.
It is not the case that every dog has a day on which Duncan likes it. Or, more usefully for other questions, "There exists a dog that Duncan dislikes every day". Lots of people negated the quantifiers incorrectly to get statements like "There is a day on which Duncan dislikes every dog" or "There is a day on which Duncan likes some dog".
∀x: LEW(m, x) → ¬L(d, x, Fri)
On Wednesdays, Duncan likes every dog that weighs less than or as much as Toby.
I tried to do you a favor here by making VII and VIII so similar.
TRUE. It is easily shown to be reflexive, symmetric, and transitive.
TRUE: LEW(x, x) means "w(x) = w(x)", and the transitivity follows from the transitivity of ≤.
TRUE. It is not antisymmetric because V says that there are two distinct dogs x and y with the same weight. For those dogs, LEW(x, y) and LEW(y, x) are both true but x = y is false.
TRUE. The set of dogs is finite and thus may collectively have only a finite set of weights, whereas the naturals are infinite.
FALSE. If it were total, Duncan would like each dog on at least one day. But VI says that this is not the case. It is true that R is not well-defined, since Duncan likes Toby on more than one day.
FALSE. It is total because Cardie likes at least one dog on every day. It is not well-defined because she likes more than one dog on at least one day.
FALSE. For any day r, X(r, n) means that n is the minimum weight of any dog that Cardie likes on day r. Since Cardie likes all dogs on all days, for every r this is true if and only if n is the minimum weight of any dog in D. So X is a function, but since it has the same value for each of the seven days, it is not one-to-one.
TRUE. This was perhaps the trickiest problem on the test. As we find in Question 3, Duncan likes Toby every day. So each day, the set of dogs he likes includes Toby and so is non-empty, and there is a minimum weight of the dogs in that set. So Y is a function. If, for example, Toby were the lightest dog in D, then Y(r, w(t)) would be true for every r and Y would not be one-to-one. But alternatively, there could be seven different dogs with different weights, such that each was the minumum-weight dog Duncan likes on a separate day. In this case Y would be a one-to-one function. So we cannot tell from the statements whether Y is one-to-one.
TRUE. The combined relation is reflexive because for every element x of A ∪ B, (x, x) is in either U or V. It is symmetric because if (x, y) is in U ∪ V, then x and y are in the same set A or B, and thus (y, x) must be in the same relation as (x, y) because that relation (U or V) is itself symmetric. Similarly if (x, y) and (y, z) are both in U ∪ V, then x, y, and z are either all in A or all in B, and (x, z) must be in the same relation (U or V) as (x, y) and (y, z) because that relation is itself transitive.
FALSE. The relation U ∩ V contains no pairs at all since A and B are disjoint. So it is not reflexive, since it does not have (x, x) for any element x of either A or B. (And both A and B have at least one element each.)
(Note: The point values of Questions 3 and 5 were reversed on the test paper, though they were correct on page 2.)
From Statements I, II, and III, there is only one possible setting of the truth values of the three propositions L(d, t, Sun), L(d, t, Tue), and L(d, m, Wed). Determine this setting and justify it, either with a truth table or a deductive argument.
Let p = L(d, t, Sun), q = L(d, t, Tue), and r = L(d, m, Wed). Then
Statement I is "p → ¬(r ∧ q)", II is ¬(p ∧ q) →
(r ∧ p)", and III implies "p ↔ q".
A truth table easily shows that there is exactly one solution to all three
of these compound statements, namely "p ∧ q ∧ ¬r".
If you attack this problem by deductive proof, you need to either show that
at most one solution is possible (in which case I have asserted that one exists)
or find and verify a solution (in which case I have asserted that there
aren't any others. Many people got 8/10 on this problem because they started
a deductive proof, found an apparent solution partway through, and failed to
verify it.
We can begin a deductive proof by assuming p, whereupon we get
¬(q ∧ r) by Modus Ponens on I. Using III, we get q and thus the
only possible solution with p true is "p ∧ q ∧ ¬r".
However, this doesn't rule out the situation that this possible solution is
wrong, and that there is another solution with p false. In order to get full
credit you had to complete your deductive argument in one of two ways: (a)
explicitly show that "p ∧ q ∧ ¬r" satisfies I, II, and II, or (b)
show that is no other solution by showing that ¬p leads to a contradiction.
(The latter is easy -- if p is false, so is q by III, and then II evaluates to
"¬(0 ∧ 0) → (r ∧ 0)", which is false whatever r is.)
Prove, using any or all of Statements I through VIII, that Toby weighs less than Mia. (Hint: One way to proceed is Proof by Contradiction.)
Taking the hint, we begin by assuming the negation of "Toby weighs
less than Mia", which is LEW(m, t). (In general I took three points
off from people who messed up this negation or otherwise did not deal
with the case that Toby and Mia have equal weight.)
From LEW(m, t) we can derive L(d, m, Wed) from Statement VIII, or
¬L(m, t, Fri) from Statement VII. The first directly contradicts a
conclusion from Question 3, and since the premises of Question 3 still hold
here, this is a contradiction. The second statement along with Statement
III implies that ∀r:¬L(d, t, r), and this also contradicts the
answer to Question 3.
Even without using Question 3, we can take the obvious fact LEW(t, t)
and use it to derive L(d, t, Wed). This and ¬L(d, t, Fri) together
contradict Statement III.
Any of these Contradiction proofs can also be rephrased as a Contrapositive
proof.
Prove, using any or all of Statements I through VIII, the following statement:
∃a:∀r: (L(d, a, r) ⊕ L(c, a, r)) &and' ¬LEW(a, t)
Make your use of quantifier proof rules clear.
Far too many people began by applying quantifier proof rules to the goal statement, which is not valid because we don't know the goal statement to be true. The outer quantifer of the goal is ∃, meaning that our last rule will be Existence. So our first task is to determine for which dog we will try to prove the rest of the statement. We need to determine, "off-line", which dog is liked by Cardie or Duncan but not both on each day, and weighs more than Toby. We know from IV that any dog is liked by Cardie on any day, so we will need our dog to be disliked by Duncan on any day. It is not possible from the axioms to prove this for any of the named dogs, because we have information about Duncan's preferences only for Toby and Mia, and neither of those work. (Duncan likes Toby on all days, as we saw in Q3, and we have no information about whether he likes Mia on days other than Wednesday or Friday.) But Statement VI, after one use of the negation of quantifiers rule, tells us that there exists a dog that Duncan dislikes on all days. By Instantiation, let this dog be named z. Now we can begin a Generalization proof that dog z satisfies the required statement. We let r be an arbitrary day and use Specification to prove that L(c, z, r) and ¬L(d, z, r) are both true, making "L(d, z, r) ⊕ L(c, z, r)" true by the definition of XOR. We only have left to prove that LEW(z, t) is false. This is easy by Contradiction -- if LEW(z, t) were true, Specification from Statement VIII and Modus Ponens tell us that L(d, z, Wed) is true, but we know the latter to be false by Specification from ∀r:¬L(d, z, r). Since r was arbitrary, we have proved ∀r: (L(d, z, r) ⊕ L(c, z, r)) ∧ ¬LEW(z, t). The desired conclusion now follows by Existence.
49 is 7 times 7, and 100 is 2 times 2 times 5 times 5. We can tell that the two numbers are relatively prime because no prime occurs in both factorizations.
100 = 2(49) + 2, 49 = 24(2) + 1, 2 = 2(1) + 0. Since 1 was that last nonzero number produced, it is the gcd of 49 and 100, and the definition of "relatively prime" is that the gcd is 1.
So x = 49, y = -24, and x is the inverse of 49, modulo 100.
My solution was to first find a way for the machine to give me 13 cents by
giving me stamps in exchange for bills. 24 bills to it and 49 stamps from it
combine to give me 1 cent, so it can give me 13 cents if I give it 13(24) =
312 bills and get 13(49) = 637 stamps. If I increase the bills by 13 to
325, the machine gets a net amount of 1287 cents.
There is a solution involving fewer bills and stamps, which many people
found by an enlightened version of trial and error. This was 31 bills in
exchange for 37 stamps. This can be derived from my solution, as the machine
could not give me 600 of the 637 stamps and I could not give it 6(49) = 294
of the bills.
Several people multiplied the equation 49x + 100y = 1 by 1287, and proposed
that I give the machine 1287(24) bills and receive 1287(49) stamps in exchange.
This is a good idea, but unfortunately results in transferring the 1287 cents
from the machine to me rather than from me to the machine. I gave 7 points out
of 10 for this answer.
Last modified 23 October 2015