Question

The statement $(p\wedge \sim q)\vee q\vee (\sim p\wedge q)$ is equivalent to

• A. $p\wedge q$
• B. $p\vee q$
• C. $p\vee \sim q$
• D. $\sim p\wedge q$

Answer 1

The correct option is B $p\vee q$

$(p\wedge \sim q)\vee q\vee (\sim p\wedge q)=\left\{\right(p\vee q)\wedge (\sim q\vee q\left)\right\}\vee (\sim p\wedge q)$
$=\left\{\right(p\vee q)\wedge T\}\vee (\sim p\wedge q)$
$=(p\vee q)\vee (\sim p\wedge q)$
$=\left(\right(p\vee q)\vee \sim p)\wedge (p\vee q\vee q))\cdots (1)$

From associative law , we know

$(p\vee q)\vee q\equiv p\vee (q\vee q)$
$\equiv p\vee q$

Similarly, $(p\vee q)\vee \sim p\equiv \sim p\vee (p\vee q)$

Again from associative law , the above expression can be written as $(\sim p\vee p)\vee q$
$\equiv T\vee q\equiv T$

$\therefore$ Equation $\left(1\right)$ can written as

$\left(\right(p\vee q)\vee \sim p)\wedge (p\vee q\vee q))\equiv T\wedge (p\vee q)$
$\equiv p\vee q$

$\text{Alternate Solution}$
Let $p,q$ substatements represent $A$ and $B$ respectively.
Using sets analogy,
$(p\wedge \sim q)\vee (\sim p\wedge q)$ is same as $(A\cap B^{\prime })\cup \left(B\right)\cup (A^{\prime }\cap B)$
Now from venn diagrams,
$\left(1\right)A\cap B^{\prime }$


$\left(2\right)B$


$\left(3\right)A^{\prime }\cap B$


$\therefore \left(1\right)\cup \left(2\right)\cup \left(3\right)$


$=A\cup B\equiv p\vee q$