CS 381 Solutions to Homework 3

pp. 19 - 21

4 b)

$p$	$q$	$r$	$(p \wedge q) \wedge r$	$p \wedge (q \wedge r)$
F	F	F	F	F
F	F	T	F	F
F	T	F	F	F
F	T	T	F	F
T	F	F	F	F
T	F	T	F	F
T	T	F	F	F
T	T	T	T	T

6

$p$	$q$	$p \wedge q$	$\neg (p \wedge q)$	$\neg p \vee \neg q$
F	F	F	T	T
F	T	F	T	T
T	F	F	T	T
T	T	T	F	F

10
a) $(\neg p \wedge (p \vee q)) \Leftrightarrow ((\neg p \wedge p) \vee (\neg p \wedge q))$
$\Leftrightarrow (F \vee (\neg p \wedge q))$
$\Leftrightarrow (\neg p \wedge q)$
$\Leftrightarrow (q \wedge \neg p)$
By "simplification" this implies $q$.

b) If we can show (( $p \rightarrow q) \wedge (q \rightarrow r) \wedge p) \rightarrow r$ then by "Exportation" we get the desired relation.
$( (p \rightarrow q) \wedge (q \rightarrow r) \wedge p)$
$\Leftrightarrow p \wedge (p \rightarrow q) \wedge (q \rightarrow r)$
$\Leftrightarrow p \wedge (\neg p \vee q) \wedge (q \rightarrow r)$
$\Leftrightarrow ((p \wedge \neg p) \vee (p \wedge q)) \wedge (q \rightarrow r)$
$\Leftrightarrow ((False) \vee (p \wedge q)) \wedge (q \rightarrow r)$
$\Leftrightarrow (p \wedge q) \wedge (q \rightarrow r)$
$\Leftrightarrow (p \wedge q) \wedge (\neg q \vee r)$
$\Leftrightarrow (p \wedge q \wedge \neg q) \vee (p \wedge q \wedge r)$
$\Leftrightarrow False \vee (p \wedge q \wedge r)$
$\Leftrightarrow p \wedge q \wedge r$
$\Rightarrow r$

20
b) $(p \vee q \vee r) \wedge s$
c) $(p \wedge T) \vee (q \wedge F)$

pp. 183

2 a) $[P \wedge Q ] \Rightarrow Q$ -- Simplification
b) $[[ P \vee Q ] \wedge \neg P ] \Rightarrow Q$ -- Disjunctive syllogism
c) $[ [P \rightarrow Q ] \wedge P ] \Rightarrow Q$ -- Modus ponens
d) $[ P \Rightarrow [ P \vee Q ] ]$ -- Addition
e) $[[ P \rightarrow Q ] \wedge [Q \rightarrow R ]] \Rightarrow [P \rightarrow R]$ -- Hypothetical syllogism

4. Let $R, F, S, D$ and $T$ represent "It rains", "It is foggy", "The sailing race will be held", "The life saving demonstration will go on", and "The trophy will be awarded", respectively. Then the argument as stated can be represented as follows:

$[ \neg R \vee \neg F ] \rightarrow [ S \wedge D]$
$S \rightarrow T$
$\neg T$
-----------
$R$

The inferencing goes as follows:

$S \rightarrow T$
$\neg T$
-----------
$\neg S$ by disjunctive syllogism.

Since $[ S \wedge D ] \rightarrow S$,
$[ \neg R \vee \neg F ] \rightarrow [ S \wedge D]$
$[ S \wedge D ] \rightarrow S$
--------------
$[ \neg R \vee \neg F ] \rightarrow S$

Hence
$[ \neg R \vee \neg F ] \rightarrow S$
$\neg S$
----------
$R \wedge F$ , since $\neg [ \neg R \vee \neg F ] \Leftrightarrow R \wedge F$.

Hence by simplification $R$ is concluded from $R \wedge F$, that is, it rained.