Q1: 10 points Q2: 15 points Q3: 15 points Q4: 15 points Q5: 20+10 points Q6: 25 points Total: 100+10 points
Question text is in black, solutions in blue.
¬(H(c, t) → H(d, t)). This translates to H(c, t) ∧ ¬H(d, t) by Definition of Implication and DeMorgan. It does not translate to ¬H(c, t) → ¬H(d, t) or to any of the many other things people tried using incorrect rules.
If Cardie has the tennis ball, then Ace has it if and only if Biscuit has it.
¬(H(d, t) ∧ ¬H(a, t)), which translates to ¬H(d, t) ∨ H(a, t) by DeMorgan and then, if you like, to H(d, t) → H(a, t) by Definition of Implication.
Let's use A, B, C, and D to refer to the four propositions, so that A means H(a, t), etc. Statement (a) tells us that C is true and that D is false. Since C is true, statement (b) tells us that A and B have the same truth value. Statement (c) is made true by ¬D, so it gives us no further information. The two solutions are thus (1) only C is true, and (2) A, B, and C are true and D is false. No other solution is possible because all other settings violate (a) or (b).
There exist two different dogs such that for any plaything, one of the two dogs has that plaything.
∃g: ∀p: H(g, p)
There exists a plaything that all of the dogs have.
∀p1: ∀p2: (p1 ≠ p2) → (H(a, p1) → H(a, p2)). Handling the "two different playthings" is somewhat tricky. You can't just assert that the two arbitrary playthings are different, because they might not be. You need to say that if they are different, the fact about Ace is true. Also, several people used ↔ rather than → between the two values of H. This leads to an equivalent proposition but I took off a point because the English clearly says "if... then".
Let's work backwards from the conclusion. Only statement (d) talks
about Ace -- it says that Ace has the stuffed chicken if
he has any other plaything. So we need to show that he has
some plaything. Statements (a) and (b) don't help us do
this, at least not directly, because there is no reason that Ace
should be one of the two dogs in (a) or the dog in (b). But (c)
says that one plaything belongs to all the dogs, so it must
belong to Ace.
Now that we have the idea of our proof, we need to use the
rules. To use statement (c), we must work from the outside in. So
we first use Instantiation and say "Let x be the plaything such that
∀g: H(g, x) is true". Then by Specification we can say H(a,
x).
(We can't go directly from ∃p: ∀g: H(g, p) to
∃p: H(a, p), because the quantifier rules can only be used on
the outermost quantifier. But if we wanted this statement we could
get it using three rules.)
Now that we have H(a, p), we want to use Specification on
statement (d), letting p1 be x and p2 be s, to
get (x ≠ s) → (H(a, x) → H(a, s)). But is (x ≠ s)
necessarily true? No, because x was created by Instantiation and we
know only its type and the fact we created it for. (Typo in red fixed
3 April 2013.)
Fortunately we can use Proof by Cases. If x = s, then we can
get our goal of H(a, s) by substitution from H(a, x). If x ≠ s,
we have H(a, x) → H(a, s) by Modus Ponens and then H(a, s) by
Modus Ponens again. Since we can prove H(a, s) in both cases, we
have proved it without an assumption and we are done.
Statement (a) cannot be true because if H is a function, each dog has
exactly one plaything, and thus any two dogs can have at most
two playthings between them. But P has at least the three
elements r, s, and t, and (a) says the two dogs together have
all of them.
Statement (b) cannot be true because it says that some dog
has all the playthings, of which there are at least three. But
if H were a function, each dog would have exactly one.
Statement (c) could be true if H were a function,
because there is a valid function that assigns each dog the same
plaything.
Statement (d) cannot be true because it says that if Ace has
one plaything, he has all the others as well (since he has each
other one by a separate Specifation from (d)). But he must have
at least one if H is total, and he cannot have more than one if
H is well-defined.
Each plaything belongs to a dog, and at least one belongs to more than one dog. So there are fewer playthings than dogs, meaning that there are exactly three elements in P.
Each dog has a different plaything, and at least one plaything is unowned. So there are more playthings than dogs, meaning that there are at least two others in addition to r, s, and t.
Statement (c) is the only one that can be true if H is a function. But if (c) is true, all dogs are assigned the same plaything and H is not one-to-one.
Again statement (c) is the only candidate. If (c) is true, only one plaything is owned by dogs, and thus the function H is not onto, as there is at least one (in fact at least two) playthings that are not owned by any dogs.
25 = 1·18 + 7, 18 = 2·7 + 4, 7 = 1·4 + 3, 4 = 1·3 + 1; 3 = 3·1 + 0. Since 1 was the last nonzero number produced by the Euclidean Algorithm, it is the gcd of 18 and 25 and thus these numbers are relatively prime.
25 = 1·25 + 0·18, 18 = 0·25 + 1·18, 7 = 1·25 - 1·18, 4 = -2·25 + 3·18, 3 = 3·25 - 4·18, 1 = -5·25 + 7·18. So we may take x to be 7 and y to be -5.
The proof I had in mind was that if a prime p divides either 18 or 25,
it must divide 450 and thus 451 % p is necessarily 1, meaning
that p does not divide 451. This is reminiscent of the
construction in the Infinitely Many Primes theorem.
It's also valid to factor 18 and 25, and show directly that
neither 2, 3, nor 5 divide 451.
A student came up with a nice third proof. If a prime p
divides 451, then for some natural r, pr = 451 and thus pr ≡ 1
(mod 18) and pr ≡ 1 (mod 25). Thus p has an inverse mod 18
and mod 25, which by the Inverse Theorem means that p must be
relatively prime to both 18 and 25, and thus can't be a divisor of either.
By the Chinese Remainder Theorem proof, we can use the facts that 7·18 ≡ 1 (mod 25) and -5·25 ≡ 1 (mod 18), and set z to be 13·7·18 + 7·(-5)·25 = 1638 - 875 = 763. Any number that differs from 763 by a multiple of 450 is also a solution, so the smallest natural solution is 313. Without the CRT proof, you could find 313 by dividing each of 13, 38, 63, 88,..., 313 by 18 until you find one with remainder of 7. Or, perhaps more easily, you could start with 7 and add 18's until you get a number congruent to 13 modulo 25, which you could recognize because its last two digits would be "13", "38", "63", or "88".
Last modified 3 April 2013