# First Midterm Exam Spring 2016

### Directions:

• Answer the problems on the exam pages.
• There are six problems, some with multiple parts, for 100 total points plus 10 extra credit. Actual scale A = 88, C = 58.
• Some useful definitions precede the questions below.
• No books, notes, calculators, or collaboration.
• In case of a numerical answer, an arithmetic expression like "217 - 4" need not be reduced to a single integer.

```  Q1: 20 points
Q2: 20 points
Q3: 10 points
Q4: 20 points
Q5: +10 points
Q6: 30 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.

Let D be a finite set of dogs consisting of exactly the four distinct dogs Cardie (c), Duncan (d), Mia (m), and Scout (s).

Let N be the set of natural numbers {0, 1, 2, 3,...}.

There are three owners O1, O2, and O3.

Let BT(u, n) be the relation from D to N defined so that BT(n, u) means "dog u belongs to owner On".

Let E be the binary relation on D defined so that E(u, v) means "dog u goes out earlier than dog v".

Let Y be the binary relation on D defined so that Y(u, v) means "dog u is younger than dog v". You may assume that no two dogs in D are exactly the same age.

Some of the true/false questions also refer to an arbitrary set D of dogs with predicates T(x) meaning "dog x is a terrier" and S(x, y) meaning "dog x is sillier than dog y".

• Question 1 (20): Translate each statement as indicated, using the set of dogs D = {c, d, m, s}, the set of naturals N, the "belongs to" predicate BY(u, n). the "earlier" predicate E(u, v), and the "younger" predicate Y(u, v). All these are defined above.

• (a, 2) (to English) (Statement I) (E(c, d) ∨ E(d, c)) → Y(m, d)

• (b, 3) (to symbols) (Statement II) Mia is younger than Cardie, and if either Cardie or Duncan goes out earlier than the other, then Mia is not younger than Duncan.

• (c, 2) (to English) (Statement III) ¬(Y(m, c) ∧ Y(m, d)) → (E(c, d) ⊕ E(d, c))

• (d, 3) (to symbols) (Statement IV) The relation Y is antireflexive and antisymmetric.

• (e, 2) (to English) (Statement V) ∀u:∀v: E(u, v) ↔ [∃i:∃j: BT(u, i) ∧ BT(v, j) ∧ (i < j)]

• (f, 3) (to symbols) (Statement VI) Every dog has exactly one owner, and O3 owns exactly one dog.

• (g, 2) (to English) (Statement VII) ∀x: (x ≠ s) ↔ (E(s, x) ∧ Y(s, x))

• (h, 3) (to symbols) (Statement VIII) There is a dog u such that for any dog v, v is younger than u if and only if v is Scout. (Equivalently, Scout and only Scout is younger than u.)

• Question 2 (20): The following are ten true/false questions, with no explanation needed or wanted, no partial credit for wrong answers, and no penalty for guessing. They use the sets and relations defined above. Many depend as well on the statements I through VIII given in Question 1.

• (a) If we know that p → q is true, and r is any proposition at all, then (p ∨ r) → q must be true.

• (b) If we know that p is true, and r is any proposition at all, then p ∨ r must also be true.

• (c) The relation BT is a function from D to N, assuming that Statements I-VIII are all true,

• (d) The relation BT from D to N is neither onto nor one-to-one, assuming that Statements I-VIII are all true.

• (e) Define the relation Y' by the rule Y'(u, v) ↔ ((u = v) ∨ Y(u, v)). Then Y' is a total order on D, assuming the English meaning of "younger" and that no two dogs are exactly the same age.

• (f) Define the relation E' by the rule E' (u, v) ↔ "dog u goes out either earlier than dog v or at the same time". Then E' is a partial order on D, assuming the English meaning of "earlier" and the truth of Statements I-VIII.

• (g) If u and v are any two binary strings, and u is a prefix of v, then uR is a suffix of vR. (Here uR denotes the reversal of u.)

• (h) Consider the statement ∀x: ¬T(x) → S(x, d), meaning "every non-terrier is sillier than Duncan". If there exists any terrier in the set of dogs, then this implication is true because its premise "∀x: ¬T(x)" is false.

• (i) If the statements "all terriers like snacks" and "all dogs who are not terriers like snacks" are both true, then the statement "all dogs like snacks" must be true.

• (j) If x and y are naturals, and 143 divides the product xy, then 143 must also divide either x or y (or both).

• Question 3 (10): This question uses the sets, definitions, and predicates above, and the statements from Question 1.

(Note: The point value of Questions 3 was given as 15 on the test paper, though it was correct on page 2.)

From Statements I, II, and III, there is only one possible setting of the truth values of the four propositions Y(m, c), Y(m, d), E(c, d), and E(d, c). Determine this setting and show that it is the only setting that works. You may use either a truth table or a deductive argument.

• Question 4 (20): This question also uses the sets, definitions, and predicates from above and the statements from Question 1.

Prove, using any or all of Statements I through VII, that Statement VIII is true. Do not assume anything about the English meaning of the predicates, except what you are given in the statements. Make your use of quantifier proof rules clear.

• Question 5 (10XC): This question also uses the sets, definitions, and predicates from above and the statements from Question 1.

Assuming the truth of Statements I-VIII, determine which owner owns which dog, and determine the value of the predicate E(u, v) for all possible pairs of dogs u and v. Make your use of quantifier proof rules clear.

• Question 6 (20+10): Here are some straightforward number theory questions.

• (a, 5) Give prime factorizations of the naturals 14, 16, 21, and 24.

• (b, 5) Based on your prime factorizations, explain which of the four numbers in part (a), if any, have inverses modulo which of the others.

• (c, 10) Find an inversse of 28. modulo 39, and an inverse of 39, modulo 28. Be sure to make clear which is which.

• (d, 10XC) Find which naturals x satisfy both the congruences x ≡ 25 (mod 28) and x ≡ 20 (mod 39). You may give an arithmetic expression in your answer without resolving it to an explicit integer.