Which of the following Boolean expressions is not a tautology?

\left(1\right) ( p ⇒ q ) ∨ ( ~ q ⇒ p )$</p><p>By applying$a ⇒ b ≡ ~ a ∨ b \text{ and } ~ \left(~ a\right) = a,$we get</p><p>$= ( ~ p ∨ q ) ∨ ( q ∨ p )$</p><p>Since, the operation disjunction is commutative, hence, we have</p><p>$= ( ~ p ∨ q ) ∨ ( p ∨ q )$</p><p>Now, applying the distributive law i.e.$a ∨ \left(b ∨ c\right) ≡ $\left(a ∨ b\right) ∨ \left(a ∨ c\right),$ we get

$= ( ~ p ∨ p ) ∨ q$

Also, we know that $~ a ∨ a ≡ t \text{ and } t ∨ a ≡ t,$ where $t$ represents tautology.

Thus, we have

$= t ∨ q = t .$

Hence, the option \left(A\right)$is a tautology.</p><p>\left(2\right) ( q ⇒ p ) ∨ ( ~ q ⇒ p )$

Again, using the same definitions, we get

$= ( ~ q ∨ p ) ∨ ( q ∨ p )$

$= ( ~ q ∨ q ) ∨ p$

$= t ∨ p = t .$

Hence, the option \left(B\right)$is also a tautology.</p><p>\left(3\right) ( p ⇒ ~ q ) ∨ ( ~ q ⇒ p )$

$= ( ~ p ∨ ~ q ) ∨ ( q ∨ p )$

$= ( ~ p ∨ p ) ∨ ( ~ q ∨ q )$

$= t ∨ t = t .$

Hence, the option \left(C\right)$is also a tautology.</p><p>\left(4\right) ( ~ p ⇒ q ) ∨ ( ~ q ⇒ p )$

$= ( p ∨ q ) ∨ ( q ∨ p )$

$= ( p ∨ q ) ∨ ( p ∨ q )$

$= p ∨ q .$

Which is not a tautology.