Q1: 15 points Q2: 30 points Q3: 30 points Q4: 25+10 points Total: 100+10 points
Question text is in black, solutions in blue.
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.
Let S be a finite set of animals consisting of exactly the five distinct animals Cardie (c), Duncan (d), Floyd (f), Scout (s), Whistle (w).
Let D be the unary relation on A defined so that D(x) means "x is a dog".
Let F be the unary relation on A defined so that F(x) means "x lives on the farm".
Let R be the unary relation on A defined so that R(x) means "x is a retriever".
Let M be the binary relation on A defined so that M(x, y) means "animal x met animal y during the morning walk". Note that two animals could be together on the walk without meeting during it.
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".
Let Σ be the alphabet {0, 1}, so that Σ* is the set of binary strings.
Let f be the function from Σ* to Σ* defined so that f(w) = ww (the concatenation of the string w with itself). For example, f(011) = 011011.
Let g be the function from Σ* to Σ* defined so that g(λ) = λ and for any letter a in Σ and any string w, g(aw) = w. (So g deletes the first letter of its input if there is one.) For example, g(011) = 11.
¬D(f) ∧ ∀x: ¬F(x) → M(f, x)
Every animal either is not a retriever or is a dog.
¬(F(f) → M(d, c)) (The parentheses are necessary. Many of you translated this to "F(f) ∧ ¬M(d, c)", which was useful later.)
No animal met itself, and if any animal met another, the other also met it.
∀x: (M(c, x) ↔ M(d, x)) ∧ (F(x) → (M(c, x) ∧ M(d, x))
Some retriever who lives on the farm met Whistle, who does not live on the farm.
∀x: F(x) → M(w, x) (Note that I took off points if you had M(x, w) instead of M(w, x) -- M is only symmetric if we are assuming that Statement IV is true.)
From Statement I we get ¬D(f) by Left Separation. From Statement II we
get ¬R(f) ∨ D(f) by Specification, and we can rewrite this
if we like as R(f) → D(f) by Definition of Implication. By
Definition of Implication and DeMorgan, we can rewrite Stateement
III as F(f) ∧ ¬M(d, c), which implies F(f) by Left
Separation.
So we have ¬D(f), R(f) → D(f), and F(f). This gives
two of the three desired truth values directly. To get the third,
we can rewrite the implication as its contrapositive ¬D(f)
→ ¬R(f), and then derive ¬R(f) by Modus Ponens. So
F(f) is true and the other two are false.
As above, we can rewrite Statement III (now our goal) as F(f) ∧
¬M(d, c). We will prove the two statements F(f) and ¬M(d,
c) and then use Conjunction. We first prove F(f) by
contradiction. Assume ¬F(f). If we specify Statenent I to f we
get ¬F(f) → M(f,
f). Modus Ponens gives us M(f, f). But by specifying the first
part of Statement IV to f, we get ¬M(f, f), and this is a
contradiction. There are other valid ways to prove this, but some
of you tried to prove it by quoting part (a) of this problem. Of
course that is no good because in (a) you are assuming Statement
III, the goal of this part, to be true.
It remains to prove ¬M(d, c), and again we can use
contradiction.
Assume M(d, c) and specify the first part of Statement V to c, getting
M(d, c) ↔ M(c, c). But specifying the first part of Statement
IV to c gives us ¬M(c, c), a contradiction.
Since our goal is a universal statement, we use Generalization. Let x
be an arbitrary animal. We want to prove F(x) → M(w, x), and we
break this proof into five cases depending on the identity of animal
x.
If x = c or x = d, we specify the second part of Statement V to x and get
F(x) → (M(c, x) ∧ M(d, x). If F(x) were true, by Modus
Ponens and Separation we could get M(c, c) (if x = c) or M(d, d)
(if x = d). In either case we can get a contradiction by
specifying the first part of Statement IV to c or d. Since F(x)
must be false in these cases, the desired implication follows by
Vacuous Proof.
If x = w, we have ¬F(x) by Separation from Statement VI, and
this gives us the desired implication by Vacuous Proof.
If x = f, we can specify the second part of Statement I to w and
get ¬F(w) → M(f, w). Since we have already seen that ¬F(w)
follows from Statement VI, by Modus Ponens we know that M(f, w) is
true. Specifying the second part of Statement IV to f and w gives
us that M(w, f) is true, so the desired implication follows by
Trivial Proof.
If x = s, we have more work to do because none of the statements
explicitly mention Scout. By Instantiation on the second part of
Statement VI, we know that some animal y satisfies F(y) ∧ R(y)
∧ M(y, w). This animal cannot be c, d, or w because we know
that F(c), F(d), and F(w) are all false. It cannot be f because we
know ¬D(f) from Statement I, and ¬R(f) follows using
Statement II as in part (a) of this problem. (Quoting part (a) is
legitimate because its assumptions, Statements I-III, are still
assumed to be true here.) This means y = s, and thus that F(s),
R(s), and M(s, w) are all true. M(w, s) follows by specifying the
second part of Statement IV to s and w, then using Modus Ponens.
And the desired implication follows by Trivial Proof.
I've been careful about quoting rules and using symbols above, but
a convincing proof can use just English, for example: "Let x be an
arbitrary dog. If x is Cardie or Duncan, x does not live on the
farm because if they did V would obligate them
to meet themselves, which they cannot by IV. If x were Floyd, he
met Whistle by I, since Whistle does not live on the farm by VI.
By IV, then, Whistle met him. If x were Whistle, he would not live
on the farm by VI. Finally, we can show that Scout must be the
animal from VI because Floyd is not a retriever and the other three
do not live on the farm, Whistle by VI and Cardie and Duncan as we
showed from V above. So Scout met Whistle, and thus Whistle met
Scout by IV. So every animal either does not live on the farm or
was met by Whistle."
The statements do not settle all the atomic predicates. We know
that Floyd and Scout live on the farm and the others do not. We
know that Scout is both a dog and a retriever, and that Floyd is
neither, but we know nothing about the caninity or retrieverness of
the other three. We know most of the values of the M predicate,
but not whether Cardie and Duncan met Whistle (and vice versa), and
not whether Floyd met Scout (and vice versa).
FALSE. It says that M is antireflexive and symmetric.
FALSE. If A is empty, the only possible relation has no elements. But the defining properties of equivalence relations and partial orders are all universal statements, so they are all true if the base set is empty. We've also observed that the equality relation on any set is both an equivalence relation and a partial order, and the unique relation on an empty set is the equality relation.
FALSE. There is no string w such that f(w) = 0, for example.
FALSE. If two strings u and v are different, f(u) = uu and f(v) = vv are also different.
TRUE. Any string w is equal to both g(w0) and g(w1).
TRUE. Every string is the image under g of two strings. For example, λ is both g(0) and g(1).
FALSE. There is exactly one counterexample, w = λ. The length of f(w) is even for any string w, and g(w) reduces the length by one unless w = λ.
TRUE. Certainly λ is such a string. The only other two are 0 and 1., since if w is non-empty the length of g(f(w)) is 2|w| - 1.
TRUE. Let x be an element common to all the sets, and let each of the sets have k-1 other elements of its own, that are not in any of the other sets. This can be done as long as there are at least 1 + n(k-1) elements available.
FALSE. The empty relation on any non-empty set is symmetric (vacously) but not reflexive.
FALSE. A finite total order must have two such elements. But a non-total order can have them as well. Consider the set {a, b, c, d} where P(a, b), P(a, c), P(a, d), and P(c, d) are all true but neither P(b, c) nor P(c, b) is true. Here a is the minimum element and d is the maximum, but the order is not total.
FALSE. It says that it is a subset, but it does not say or imply that there exists at least one dog that is not a retriever.
TRUE. The second part says exactly that. Those of you who took the exam early on Tuesday or through Disabilities Services had a slightly different statement that finished with "the set of animals who met Cardie and Duncan". Since Statement IV is not assumed to be true here, the correct answer for those people was FALSE, as I discovered only when someone in the early session Wednesday pointed this out. I graded those exams with TRUE as the right answer. If you lost points because you caught this mistake, show me your exam and we'll talk.
FALSE. This looks like the unique factorization theorem but it isn't. A counter example is p = s = 2 and q = r = 3.
FALSE. The smallest such natural is 3×5×7 = 105. The smallest odd primes are 3, 5, and 7, because 1 is not prime.
32 = 2×2×2×2×2, and 55 = 5×11. The two numbers are relatively prime to one another because they share no prime factors. Of the ten given numbers, all but 5 and 10 are relatively prime to 55 and the five odd ones are relatively prime to 32. That leaves 1, 3, 7, and 9 as relatively prime to both. Note that 1 is relatively prime to any positive integer.
The Euclidean Algorithm produces remainders 55, 32, 23, 9, 5, 4, and
1. The successive linear combinations are:
The inverse of 55 modulo 32 is 7, and the inverse of 32 modulo
55 is -12, or 43.
The Chinese Remainder Theorem will allow us to determine the
equivalence class of the number x of coconuts, modulo 32×55 =
1760. This might or might not be enough to determine the number
of coconuts when we combine it with the extra information 0 ≤ x
< 2000. So we really need to find the exact number in order to
answer the question. Many of you found via trial and error that
250 satisfies both congruences. Some of you got full credit
without using the CRT at all, since you showed that there were no
other solutions in the given range. (This was aided by arguing
that any number that met both congruences had to be divisible by
10, which I suppose is actually an implicit use of the CRT in
itself.)
But the conventional way to determine the equivalence
class of x works just fine. Using the inverses found in part (b),
we compute 55×7×26 + 32×(-12)×30 = 10010 -
11520 = -1510. The number of coconuts must then be 250, which we
find by adding 1760 to -1510. However you found 250, if you used
the CRT you needed
to note that the next largest number in the
equivalence class is 250 + 1760 = 2010, which is too large to be a
solution. The most common error in this method was switching the 26 and
the 30.
Many people gave unconvincing answers that amounted to "Relatively
prime means that the numbers don't divide one another, so it's true by
definition."
It isn't. The definition of R(x, y) is that no third number z
with z > 1 divides both x and y. Note that A(4, 6), for example,
is true while R(4, 6) is not.
A correct proof that uses the definitions is as follows.
Assume (for a proof by contradiction) that R(x, y), x > 1, and y > 1
are all true but A(x, y) is false. That means that either D(x, y)
or D(y, x) is true. But if x divides y, it is a common factor of x
and y because we know it divides x, and thus RP(x, y) is false.
Similarly, if y divides x, y is the forbidden common factor.
Either way, there is a contradiction, so the given implication is
true.
My proof does not use the FTA, but you were certainly allowed to do so. I
got some convincing proofs along the lines of "RP(x, y) means that
no prime occurs in both factorizations, but if one of x or y
divided the other there would be a prime in both factorizations."
Last modified 20 October 2017