CS 381 Test

February, 2001

1. Fill in the blanks with the SHORTEST string of characters so that the resultant proposition is valid. [20]

$\neg (P \wedge ( \neg P \vee Q))$ $\Leftrightarrow$ ( $\framebox [1.0in]{$\neg P$}\vee \neg($ $\framebox [1.0in]{$\neg P$}\vee Q ))$

$\Leftrightarrow ($ $\framebox [1.0in]{$\neg P$}\vee ($ $\framebox [1.0in]{P}\wedge \neg Q ))$

$\Leftrightarrow (($ $\framebox [1.0in]{$\neg P$}\vee P) \wedge ($ $\framebox [1.0in]{$\neg P$}\vee \neg Q ))$

$\Leftrightarrow ($ $\framebox [1.0in]{T}\wedge ($ $\framebox [1.0in]{$\neg P$}\vee \neg Q ))$

$\Leftrightarrow ( \neg P \vee \framebox [1.0in]{$\neg Q$}$ $)$

$\Leftrightarrow \neg ( P$ $\framebox [1.0in]{$\wedge$}$ $Q )$

2. State each of the following formulas in English, if it is a wff. If it is not a wff, then give a reason why it is not a wff. Here $L(x,y)$ means $x$ likes $y$ and $N(x,y)$ means $x \neq y$ and the universe is the set of people: [15]

(a) $\forall x \exists y L(x,y)$
Everyone likes someone.
(b) $\exists x \forall y [N(x,y) \rightarrow L(x,y)]$
Someone likes everyone else.
(c) $\exists x \exists y L(x, N(x, y))$
Not a wff. $N(x,y)$ can not be an argument of $L$.
(d) $\exists x \forall y L(x,y) \rightarrow \exists x \exists y L(x,y)$
If someone likes everyone, then someone likes someone.
(e) $\exists x L(x, \forall y)$
Not a wff. $\forall y$ cannot be an argument of $L$.

3 (a) Express the argument given below using the symbol suggested for each proposition. [8]
(b) Check whether or not the reasoning is correct using inference rules on the wffs (symbolic form) of (a). [15]

Argument: If A took the laptop (A) or B lied (B), then a crime was committed (C). If a crime was committed, then D must have been in town (D). But D was not in town. Therefore, A did not take the laptop.

(a)
$A \vee B \rightarrow C$
$C \rightarrow D$
$\neg D$
----------------
$\neg A$

(b)
Reasoning is correct because:

$C \rightarrow D$
$\neg D$
----------------
$\neg C$ (Modus Tollens)

$A \vee B \rightarrow C$
$\neg C$
----------------
$\neg (A \vee B)$ (Modus Tollens)

$(\neg A \wedge \neg B)$ from $\neg (A \vee B)$ by De Morgan.

Hence $\neg A$ from $(\neg A \wedge \neg B)$ by simplification.

4. Express the assertions given below as a proposition of a predicate logic using the following predicates. The universe is the set of objects.[20]

$K(x, y)$: $x$ likes $y$.
$L(x)$: $x$ is a lion.
$P(x)$: $x$ is a person.
$S(x)$: $x$ is strong.

(a) Everyone likes a (any) lion if it is strong.
$\forall x [[ L(x) \wedge S(x)] \rightarrow \forall y [ P(y) \rightarrow K(y,x)]]$, which is equivalent to
$\forall x \forall y [[ L(x) \wedge S(x) \wedge P(y)] \rightarrow K(y,x)]$
(b) Some strong lion likes only people.
$\exists x [ L(x) \wedge S(x) \wedge \forall y [K(x,y) \rightarrow P(y)]]$
(c) Sam likes a (some) lion.
$\exists x [L(x) \wedge K(Sam, x)]$
(d) Some person likes a (any) lion only if it is strong.
$\exists x [P(x) \wedge \forall y [[L(y) \wedge K(x,y)] \rightarrow S(y)]]$
(e) Not everyone likes a (any) lion.
$\neg \forall x \forall y [[P(x) \wedge L(y) ] \rightarrow K(x,y)]$

5. Find the power set of each of the following sets: [7]

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

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

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