Question

Which of the following is not true for any two statements p and q?
(a) $\sim \left[ p\vee \left( \sim q \right) \right]=\left( \sim p \right)\wedge q$
(b) $\sim \left( p\vee q \right)=\left( \sim p \right)\vee \left( \sim q \right)$
(c) $q\wedge \sim q$ is a contradiction
(d) $\sim \left( p\wedge \left( \sim p \right) \right)$ is a tautology.

Answer 1

Draw a truth table for each option using the properties of Boolean algebra. For option (c), check if all the values of the expression are false or not. If all are false, then only it will be a contradiction. For option (d), check if all the values of the expression are true or not. If all are true then only it will be a tautology. Check all the truth tables and select the option in which the incorrect statement is given.

Complete step-by-step solution:
We have been provided with two statements and we have to check the option in which the wrong condition is given. Let us draw the truth table for each one of them one by one.
(a) $\sim \left[ p\vee \left( \sim q \right) \right]=\left( \sim p \right)\wedge q$
Drawing the truth table, we get,

$p$	$q$	$\sim p$	$\sim q$	$L.H.S=\left( \sim \left[ p\vee \left( \sim q \right) \right] \right)$	$R.H.S=\left( \sim p \right)\wedge q$
T	T	F	F	F	F
T	F	F	T	F	F
F	T	T	F	T	T
F	F	T	T	F	F

Clearly, we can see that $L.H.S=R.H.S$. Hence option (a) is true.
(b) $\sim \left( p\vee q \right)=\left( \sim p \right)\vee \left( \sim q \right)$
Drawing the truth table, we get,

$p$	$q$	$\sim p$	$\sim q$	$L.H.S=\sim \left( p\vee q \right)$	$R.H.S=\left( \sim p \right)\vee \left( \sim q \right)$
T	T	F	F	F	F
T	F	F	T	F	T
F	T	T	F	F	T
F	F	T	T	T	T

Clearing we can see that $L.H.S\ne R.H.S$. Hence option (b) is not true.
(c) Here, we have to check if $q\wedge \sim q$ is a contradiction.
Drawing the truth table, we get,

$q$	$\sim q$	$q\wedge \sim q$
T	F	F
F	T	F

Clearly, we can see that both the conditions are false and we know that if all the conditions of a Boolean expression are false, then it is called a contradiction. Hence, $q\wedge \sim q$ is a contradiction and option (c) is true.
(d) Here, we have to check if $\sim \left( p\wedge \left( \sim p \right) \right)$ is a tautology.
Drawing the truth table, we get,

$p$	$\sim p$	$p\wedge \left( \sim p \right)$	$\sim \left( p\wedge \left( \sim p \right) \right)$
T	F	F	T
F	T	F	T

Clearly, we can see that both the conditions are true and we know that if all the conditions of a Boolean expression is true, then it is called tautology. Hence, $\sim \left( p\wedge \left( \sim p \right) \right)$ is a tautology and option (d) is true.
Now, on checking all the options, we can conclude that the Boolean expression in option (b) is not true.
Hence, option (b) is our answer.

Note: One may note that we can also solve this question without using a truth table, we can use several properties of Boolean algebra to check the options. We must understand the meaning of terms like ‘contradiction’ and ‘tautology’ to check options (c) and (d) respectively. The main purpose of drawing a truth table is to make us understand the expressions more clearly.