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 $Q)$
