Independence of $A$ and $B$ implies the independence of $\neg A$ and $B$ Does the following apply? $$P(A\mid B)=P(A)\implies P(\neg A\mid B)=P(\neg A)$$
My rough answer is that suppose $A$ is the probability of rainy and $B$ is the probability of toothache. Then both $P(A\mid B)=P(A)$ and $P(\neg A\mid B)=P(\neg A)$ apply. But can we prove this mathematically?

Firstly, $$p(A|B)=p(A)\implies \frac{p(A\cap B)}{p(B)}=p(A)\implies p(A\cap B)=p(A)p(B)$$
Then, $$p(A'|B)=\frac{p(A'\cap B)}{p(B)}$$
$$=\frac{p(B)-p(A\cap B)}{p(B)}$$
$$=1-\frac{p(A)p(B)}{p(B)}$$
$$=1-p(A)=p(A')$$
