Q. The compound statement $(-(P \wedge Q)) \vee((-P) \wedge Q) \Rightarrow((-P) \wedge(-Q))$ is equivalent to
Solution:
Let $\quad r =(\sim( P \wedge Q )) \vee((\sim P ) \wedge Q ) ; \quad s =$
$((\sim P ) \wedge(\sim Q ))$
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 (A) : $((\sim P ) \vee Q ) \wedge((\sim Q ) \vee P )$
is equivalent to (not of only $P) \wedge($ not of only $Q$ )
$=($ Both $P , Q )$ and (neither P nor $Q )$
| 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 |