# 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?

$$𝑃(¬𝐴|𝐵)$$
= $$1 - 𝑃(𝐴|𝐵)$$
= $$1 - 𝑃(𝐴)$$, from given $$P(A|B) = P(A)$$
= $$𝑃(¬𝐴)$$, Q.E.D.
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, $$p(A'|B)=\frac{p(A'\cap B)}{p(B)}$$ $$=\frac{p(B)-p(A\cap B)}{p(B)}$$ $$=1-\frac{p(A)p(B)}{p(B)}$$ $$=1-p(A)=p(A')$$