CS 281 Test II

July 23, 1999

1. Find the powerset of each of the following sets: [9]

(a) {1}
(b) {$\emptyset$ , {$\emptyset$ }}
(c) $\emptyset$

2. Find the following Cartesian products: [9]

(a) { $\emptyset , \{ \emptyset \} \} \times \{ \emptyset \}$
(b) { $\emptyset \} \times \{ 1,2 \} \times \{ 3,4 \}$
(c) $\emptyset \times \{ \{ \emptyset \}, \emptyset \}$

3. Indicate which of the following are true and which are false. [15]

(a) $\emptyset \in \{ x \}$
(b) $\{x \} \subseteq \{\{x \}\}$
(c) $\{ x \} \in \{\{x \}\}$
(d) { $x\} \in \{ x\}$
(e) $\emptyset \subseteq \{ x \}$

4. Find the smallest four elements of the set S defined recursively as follows: [8]

Basis Clause: $1 \in S$
Inductive Clause: If $x \in S$, then $3x + 1 \in S$.
Extremal clause: Nothing is in S unless it is obtained from the above two clauses.

5. Recursively define each of the following sets:

(a) The set of non-negative integers that produce remainder 1 when divided by 5. [10]
(b) { $3^{n} - 1 \mid n$ is a natural number.} [12]

6. Prove the following statements on sets A and B.

(a) $A \cap B \subseteq A \cup B$ [10]
(b) If $(A - B) = \emptyset$, then $A \subseteq B$. [13]

7. Which rules of inference are used to establish the conclusion of the following argument ? [14]

Premises:
"All lions are fierce."
"Some lions do climb trees."

Conclusion:
"There are fierce creatures that climb trees."

Give your answer using L(x), C(x) and F(x) to denote "x is a lion", "x climbs trees" and "x is fierce", respectively, and assuming the universe is the set of all creatures. Using these symbols, the statements given above can be expressed as follows:

$\forall x [L(x) \rightarrow F(x) ]$
$\exists x [L(x) \wedge \neg C(x) ]$
---------------------
$\exists x [F(x) \wedge \neg C(x) ]$

You may use the following table.

Logical Equivalences and Implications

1. $P \Leftrightarrow (P \vee P)$	2. $P \Leftrightarrow (P \wedge P)$
3. $(P \vee Q) \Leftrightarrow (Q \vee P)$	4. $(P \wedge Q) \Leftrightarrow (Q \wedge P)$
5. $[(P \vee Q) \vee R] \Leftrightarrow [P \vee (Q \vee R)]$	6. $[(P \wedge Q) \wedge R] \Leftrightarrow [P \wedge (Q \wedge R)]$
7. $\neg (P \vee Q) \Leftrightarrow (\neg P \wedge \neg Q)$	8. $\neg (P \wedge Q) \Leftrightarrow (\neg P \vee \neg Q)$
9. $[P \wedge (Q \vee R)] \Leftrightarrow [(P \wedge Q \vee (P \wedge R)]$	10. $[P \vee (Q \wedge R)] \Leftrightarrow [(P \vee Q) \wedge (P \vee R)]$
11. $(P \vee T) \Leftrightarrow T$	12. $(P \wedge T) \Leftrightarrow P$
13. $(P \vee F) \Leftrightarrow P$	14. $(P \wedge F) \Leftrightarrow F$
15. $(P \vee \neg P) \Leftrightarrow T$	16. $(P \wedge \neg P) \Leftrightarrow F$
17. $P \Leftrightarrow \neg (\neg P )$	18. $(P \rightarrow Q) \Leftrightarrow (\neg P \vee Q)$
19. $(P \leftrightarrow Q) \Leftrightarrow [(P \rightarrow Q) \wedge (Q \rightarrow P)]$	20. $[(P \wedge Q) \rightarrow R] \Leftrightarrow [P \rightarrow (Q \rightarrow R)]$
21. $[(P \rightarrow Q) \wedge (P \rightarrow \neg Q)] \Leftrightarrow \neg P$	22. $(P \rightarrow Q) \Leftrightarrow (\neg Q \rightarrow \neg P)$
1. $P \Rightarrow (P \vee Q)$	2. $(P \wedge Q) \Rightarrow P$
3. $[P \wedge (P \rightarrow Q)] \Rightarrow Q$	4. $[(P \rightarrow Q) \wedge \neg Q] \Rightarrow \neg P$
5. $[\neg P \wedge (P \vee Q)] \Rightarrow Q$	6. $[(P \rightarrow Q) \wedge (Q \rightarrow R)] \Rightarrow (P \rightarrow R)$
7. $(P \rightarrow Q) \Rightarrow [(Q \rightarrow R) \rightarrow (P \rightarrow R)]$	8. $[(P \rightarrow Q) \wedge (R \rightarrow S)] \Rightarrow [(P \wedge R) \rightarrow (Q \wedge S)]$
9. $[(P \leftrightarrow Q) \wedge (Q \leftrightarrow R)] \Rightarrow (P \leftrightarrow R)$	10. $[(P \rightarrow Q) \wedge (R \rightarrow S)] \Rightarrow [(P \vee R) \rightarrow (Q \vee S)]$)

Additional Inference Rule: Conjunction

P
Q
------------------
P $\wedge$ Q