Mathematical Reasoning Question 1

Question 1 - 24 January - Shift 1

The compound statement $(\sim(P \wedge Q)) \vee((\sim P) \wedge Q) \Rightarrow((\sim P) \wedge(\sim Q))$ is equivalent to

(1) $((\sim P) \vee Q) \wedge((\sim Q) \vee P)$

(2) $(\sim Q) \vee P$

(3) $((\sim P) \vee Q) \wedge(\sim Q)$

(4) $(\sim P) \vee Q$

Show Answer

Answer: (1)

Solution:

Let $\mathbf{r}=(\sim(P \wedge Q)) \vee((\sim P) \wedge Q) ; s = ((\sim P) \wedge(\sim Q))$

$P$ $Q$ $\sim(P \wedge Q)$ $(-P) \wedge Q$ $r$ $S$ $r \to s$
$T$ $T$ $F$ $F$ $F$ $F$ $T$
$T$ $F$ $T$ $F$ $T$ $F$ $F$
$F$ $T$ $T$ $T$ $T$ $F$ $F$
$F$ $F$ $T$ $F$ $T$ $T$ $T$

Option (1) : $((\sim P) \vee Q) \wedge((\sim Q) \vee P)$

is equivalent to (not of only $P) \wedge($ not of only $Q$ ) $=($ Both $P, Q)$ and (neither P nor $Q)$