Q. The negation of $p \to (\sim p \vee q) $ is
Solution:
Truth table
p
q
$\sim p$
$\sim q$
$(\sim p \vee q)$
$p \rightarrow(\sim p \vee q )$
$( - p \vee- q )$
$p \vee(\sim p \vee q )$
$\sim( p \vee q )$
$p \rightarrow\sim( p \vee q )$
$p \rightarrow q$
$p \wedge \sim q$
$\sim[p \rightarrow(\sim p \vee q)]$
F
F
T
T
T
T
T
T
T
T
T
F
F
F
T
T
F
T
T
F
F
F
T
T
F
F
T
F
F
T
F
F
T
T
F
F
F
T
T
T
T
F
F
T
T
T
T
F
F
T
F
F
Here, we observe that the truth value of column $\sim[p \rightarrow(\sim p \vee q)]$ and $(p \wedge \sim q)$ are same.
Hence,$\sim[p \rightarrow(\sim p \vee q)] \equiv(p \wedge \sim q)$
| p | q | $\sim p$ | $\sim q$ | $(\sim p \vee q)$ | $p \rightarrow(\sim p \vee q )$ | $( - p \vee- q )$ | $p \vee(\sim p \vee q )$ | $\sim( p \vee q )$ | $p \rightarrow\sim( p \vee q )$ | $p \rightarrow q$ | $p \wedge \sim q$ | $\sim[p \rightarrow(\sim p \vee q)]$ |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| F | F | T | T | T | T | T | T | T | T | T | F | F |
| F | T | T | F | T | T | F | F | F | T | T | F | F |
| T | F | F | T | F | F | T | T | F | F | F | T | T |
| T | T | F | F | T | T | T | T | F | F | T | F | F |