- Answer the problems on the exam pages.
- There are four problems for 100 total points. Scale will be determined after the exam but a good guess is A = 90, C = 60.
- If you need extra space use the back of a page.
- No books, notes, calculators, or collaboration.

Exam text is in black, solutions in blue.

Corrections in green made 25 September 2015.

Q1: 15 points Q2: 30 points Q3: 25 points Q4: 30 points Total: 100 points

**Question 1 (20):**Briefly identify the following terms or concepts (3 points each):- (a) the converse of an implication
The converse of an implication "p → q" is the implication "q → p". It is not equivalent to the original.

- (b) the contrapositive of an implication
The contrapositive of an implication "p → q" is the implication "¬q → ¬p". It is equivalent to the original.

- (c) an existential quantifier
The existential quantifier ∃ is put before a predicate along with a variable that is free in the predicate. If P(x) is a statement with free variable x, "∃x: P(x)" means "P(x) is true for some x" or "there exists an x such that P(x) is true".

- (d) the DeMorgan rules for ∧ and ∨
"¬(p ∧ q) ↔ (¬p ∨ ¬q)" and "¬(p ∨ q) ↔ (¬p ∧ ¬q)"

- (e) a contradiction (as a kind of compound proposition)
A contradiction is a compound proposition that is false for all possible values of its atomic propositions. Its negation is a tautology.

- (a) the converse of an implication
**Question 2 (30):**For each of these three number sequences, give the next three numbers in the sequence, find a recursive formula, and find a closed formula. (10 points each)- a
_{1}= 3, a_{2}= -2, a_{3}= -7, a_{4}= -12, a_{5}= -17a

_{6}= -22, a_{7}= -27, a_{8}= -32. Recursive: a_{1}= 3, a_{n+1}= a_{n}= 5. Closed form: a_{n}= -5n + 8. - b
_{1}= 3, b_{2}= 0, b_{3}= -1, b_{4}= 0, b_{5}= 3b

_{6}= 8, b_{7}= 15, b_{8}= 24. Recursive: b_{1}= 3, b_{n+1}= b_{n}+ 2n - 7. Closed: b_{n}= n^{2}- 6n + 8. (How to get this? The solution is a quadratic, of the form xn^{2}+ yn + z. You can plug in n = 1 to get x + y + z = 3, n=2 to get 4x + 2y + z = 0, and n = 3 to get 9x + 3y + z = -1. Subtracting the first equation from the others gives 3x + y = -3 and 8x + 2y = -4, from which you can get 6x + 2y = -6 ,2x = 2, x = 1, and y = -2. Then you solve for z in any of the three equations.) - c
_{1}= 3, c_{2}= 4, c_{3}= 6, c_{4}= 10, c_{5}= 18c

_{6}= 34, c_{7}= 66, c_{8}= 130. Recursive: c_{1}= 3, c_{n+1}= 2c_{n}- 2. Closed: c_{n}= 2^{n-1}+ 2. (How to get this? The solution is going to involve 2^{n}because of the doubling. So look for a similarity with 2, 4, 8, 16, 32, 64, 128, and see that c_{n}is the n-1'st entry of this sequence, plus 2.

- a
**Question 3 (25):**Translate the following statements as indicated. The dogs Cardie and Duncan are denoted symbolically by c and d respectively. The proposition p means "Cardie is soft", and q means "Duncan is fierce". If x and y are dogs, the predicate S(x, y) means "dog x is smaller than dog y". (5 points each)- (a) (to symbols) Cardie is soft, and if Duncan is fierce
then Cardie is not soft.
p ∧ (q → ¬p)

- (b) (to English) p → ∃x: S(d, x)
If Cardie is soft, then there exists a dog larger than Duncan.

- (c) (to symbols) Every dog is smaller than some dog
that is larger than Duncan.
∀x: &exists;y: S(x, y) ∧ S(y, d)

- (d) (to English) (p ⊕ (q ∧ S(c, d))) →
∀x: S(x, d)
Either Cardie is soft, or if Duncan is fierce and Cardie is smaller than Duncan, than all dogs are smaller than Duncan, but not both.

- (e) (to symbols) Cardie is either soft or smaller than
every dog except Duncan, or both.
p ∨ ∀x: (x ≠ d) → S(x, d)

- (a) (to symbols) Cardie is soft, and if Duncan is fierce
then Cardie is not soft.
**Question 4 (30):**Establish the following facts with truth tables.- (a, 15) The compound proposition "p → (q → p)"
is a tautology.
`p --> (q --> p) 0 1 0 1 0 0 1 1 0 0 1 1 0 1 1 1 1 1 1 1`

Since the second column is all ones, the statement is a tautology.

- (b, 15) The compound propositions
"p ∧ (q ∨ ¬r)" and "(p ∨ q) ∧ (¬q ∨ p)
∧ (p ∨ ¬r)" are logically equivalent.
`(p and (q or not r) <--> (p or q) and (not q or p) and (q or not r) 0 0 0 1 1 0 1 0 0 0 0 1 0 1 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 1 1 1 0 1 0 1 1 0 0 1 0 0 0 1 1 1 0 0 0 1 1 0 1 1 0 1 1 0 0 1 0 0 0 1 1 0 1 1 1 0 1 1 0 1 1 1 0 1 1 0 1 1 1 0 1 1 0 1 0 0 0 0 1 1 1 1 0 1 1 0 1 1 0 0 0 0 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 0 1`

`The ↔ operation is applied to the second column and the fifth from last column.`

- (a, 15) The compound proposition "p → (q → p)"
is a tautology.

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```Last modified 25 September 2015