Question

The number of choices for $\Delta \in \{\wedge ,\vee ,\Rightarrow ,\iff \}$, such that $\left(p\Delta q\right)\Rightarrow \left(\right(p\Delta \sim q)\vee ((\sim p)\Delta q\left)\right)$ is a tautology, is

• A. $2$
• B. $3$
• C. $1$
• D. $4$

Answer 1

The correct option is A $2$

Let $x:\left(p\Delta q\right)\Rightarrow (p\Delta \sim q)\vee (\sim p\Delta q)$

Case $-$ I
When $\Delta$ is same as $\vee ,$ then $(p\Delta \sim q)\vee (\sim p\Delta q)$ becomes
$(p\vee \sim q)\vee (\sim p\vee q)$ which is always true. So $x$ becomes a tautology.

Case $-$ II
When $\Delta$ is same as $\wedge ,$ then
$(p\wedge q)\Rightarrow (p\wedge \sim q)\vee (\sim p\wedge q)$

If $p\wedge q$ is $T$, then $(p\wedge \sim q)\vee (\sim p\wedge q)$ is $F.$ So $x$ cannot be a tautology.

Case $-$ III
When $\Delta$ is same as $\Rightarrow ,$ then
$(p\Rightarrow \sim q)\vee (\sim p\Rightarrow q)$ is same as $(\sim p\vee \sim q)\vee$ $(p\vee q)$, which is always true. So $x$ becomes a tautology.

Case $-$ IV
When $\Delta$ is same as $\iff ,$ then
$(p\iff q)\Rightarrow (p\iff \sim q)\vee (\sim p\iff q)$

$p\iff q$ is true when $p$ and $q$ have same truth values, then $p\iff \sim q$ and $\sim p\iff q$ both are false. Hence $x$ cannot be a tautology.

So, finally $\Delta$ can be $\vee$ or $\Rightarrow$.