Question:
Let $p, q, r$ be three statements such that the truth value of $(p \wedge q) \rightarrow(\sim q \vee r)$ is $\mathrm{F}$. Then the truth values of $p, q, r$ are respectively:
Correct Option: , 4
Solution:
$(p \wedge q) \rightarrow(\sim q \vee r)$
$=\sim(p \wedge q) \vee(\sim q \vee r)$
$=(\sim p \vee \sim q) \vee(\sim q \vee r)$
$=(\sim p \vee \sim q \vee r)$
$\because(\sim p \vee \sim q \vee r)$ is false, then $\sim p, \sim q$ and $r$ all these must
be false.
$\Rightarrow p$ is true, $q$ is true and $r$ is false.