Q. Identify the false statement
A) $\sim\left\lbrack p \vee (\sim q) \right\rbrack \equiv (\sim p) \land q$
B) $\lbrack p \vee q\rbrack \vee (\sim p)$ is a tautology
C) $\lbrack p \land q\rbrack \land (\sim p)$ is a contradiction
D) $\sim(p \vee q) \equiv (\sim p) \vee (\sim q)$
Explanation
(d)
[table] $p$ | $q$ | $\sim p$ | $\sim q$ | $p \vee$
$( \sim q)$ | $( \sim p)$
$\land q$ | $p \vee q$ | $\sim (p \vee q)$ | $( \sim p)$
$\vee$
$( \sim q)$ | $(p \vee q)$
$\vee$
$( \sim p)$ ; T
F
T
F | T
T
F
F | F
T
F
T | F
F
T
T | T
F
T
T | F
T
F
F | T
T
T
F | F
F
F
T | F
T
T
T | T
T
T
T [/table]
It is clear from the table that columns 8 and 9 are not equal, ie, $\sim (p \vee q)$ is not equivalent to$\ ( \sim p) \vee ( \sim q)$.
Hence, option (e) is false statement.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.