Answer is C.
Set of variables for $A$={$q,r,s$}
Set of variables for $B$={$q,s,t$}
$\therefore$Variables in $A \cap B$ ={$q,s$}
$φ =q \vee (r \wedge s)$
$Ψ= \neg q (s \vee t)$
$Ψ=\neg \neg q \vee (s \vee t)$
$Ψ=q \vee s \vee t$
According to problem we have two conditions:
- $φ \implies μ$
So, when $φ$ is true $μ$ must be true.
Therefore, $μ=q \vee s$ (Since $r$ is not in the domain of $μ$)
ii. $μ \implies Ψ$
Taking $μ$ as $q \vee s$ also satisfy the above constraint as
$q \vee s \implies q \vee s \vee t$