If $A$ and $B$ are two independent events such that $\Pr(A)\gt 0$ , $\Pr(B)\gt 0$ and none of the two are certain events. Then what does $\Pr(A^\complement\mid B^\complement)$ equals?
The answer is $1 - \Pr(A\mid B)$ isn't it? $1- \Pr(A\mid B^\complement)$ is also possible?
$\begin{align}\mathsf P(A^\complement\mid B^\complement) &= \dfrac{\mathsf P(A^\complement\cap B^\complement)}{\mathsf P(B^\complement)}&&\text{by definition}\\[1ex]&=\dfrac{1-\mathsf P(A\cup B)}{1-\mathsf P(B)}&&\text{probability of complements}\\[1ex]&=\dfrac{1-\mathsf P(A)-\mathsf P(B)+\mathsf P(A\cap B)}{1-\mathsf P(B)}&&\text{Venn's Law}\\[1ex]&=\dfrac{(1-\mathsf P(A))-\mathsf P(B)+\mathsf P(A)\,\mathsf P(B)}{1-\mathsf P(B)}&&\text{via independence}\\[1ex]&=\dfrac{(1-\mathsf P(A))\,(1-\mathsf P(B))}{1-\mathsf P(B)}&&\text{distribution}\\[1ex]&=1-\mathsf P(A)&&\text{cancelation of common factors}\\[1ex]&=\mathsf P(A^\complement)&&\text{probability of complements} \end{align}$
Therefore, if $A$ and $B$ are independent events, then $A^\complement$ and $B^\complement$ are independent events, too.