Question

The Boolean expression $ \sim \left( {p \vee q} \right) \vee \left( { \sim p \wedge q} \right)$ is equivalent to:
A) $q$
B) $ \sim q$
C) $ \sim p$
D) $p$

Answer 1

Hint: Here the Boolean expression is given that is $ \sim \left( {p \vee q} \right) \vee \left( { \sim p \wedge q} \right)$. Boolean expression is a logical statement that returns either true or false. We need to draw the Boolean table.

Complete step-by-step answer:
Boolean expression is the expression of logic. It deals with variables that can have two discrete values, $0$ means False $\left( {\text{F}} \right)$ and $1$ means True $\left( {\text{T}} \right)$. So here, Boolean expression is given
$ \sim \left( {p \vee q} \right) \vee \left( { \sim p \wedge q} \right)$.
So first find out $ \sim \left( {p \vee q} \right)$.
Now, let us draw the table.
Here the symbol $ \vee $ is of OR and we follow additional property. If $p$ is true, then it is treated as $1$ and if $q$ is false, it is treated as $0$. Then, $p \vee q = p + q = 1 + 0 = 1$.

$p$	$q$	$p \vee q$	$ \sim \left( {p \vee q} \right)$
${\text{T}}$	${\text{T}}$	${\text{T}}$	${\text{F}}$
${\text{T}}$	${\text{F}}$	${\text{T}}$	${\text{F}}$
${\text{F}}$	${\text{T}}$	${\text{T}}$	${\text{F}}$
${\text{F}}$	${\text{F}}$	${\text{F}}$	${\text{T}}$

$ \sim $ is the negation mark. For example, if we get $p \vee q$ as true , then $ \sim \left( {p \vee q} \right)$ must be false.
Now let us find $\left( { \sim p \wedge q} \right)$.
Here $ \wedge $ is the symbol of AND and we follow multiplication property.

$p$	$ \sim p$	$q$	$p \wedge q$	$ \sim p \wedge q$
${\text{T}}$	${\text{F}}$	${\text{T}}$	${\text{T}}$	${\text{F}}$
${\text{T}}$	${\text{F}}$	${\text{F}}$	${\text{F}}$	${\text{F}}$
${\text{F}}$	${\text{T}}$	${\text{T}}$	${\text{F}}$	${\text{T}}$
${\text{F}}$	${\text{T}}$	${\text{F}}$	${\text{F}}$	${\text{F}}$

Now we have to find $ \sim \left( {p \vee q} \right) \vee \left( { \sim p \wedge q} \right)$
Now let us arrange the table

$p$	$q$	$ \sim \left( {p \vee q} \right)$	$\left( { \sim p \wedge q} \right)$	$ \sim \left( {p \vee q} \right) \vee \left( { \sim p \wedge q} \right)$
${\text{T}}$	${\text{T}}$	${\text{F}}$	${\text{F}}$	${\text{F}}$
${\text{T}}$	${\text{F}}$	${\text{F}}$	${\text{F}}$	${\text{F}}$
${\text{F}}$	${\text{T}}$	${\text{F}}$	${\text{T}}$	${\text{T}}$
${\text{F}}$	${\text{F}}$	${\text{T}}$	${\text{F}}$	${\text{T}}$

Here we got just opposite of $p$. So the answer is $ \sim p$.

So option C is the correct answer.

Note: If we write $p \wedge q$, then it is an AND operator but if $ \sim \left( {p \wedge q} \right)$is given, then it becomes a NAND operator. Now, if we write $p \vee q$, then it is an OR operator and similarly, if $ \sim \left( {p \vee q} \right)$is given, then it becomes a NOR operator.