[proofplan] Unpack the definition of conditional probability and compare it with the multiplicative definition of independence. [/proofplan] [step:Condition on B] Since $\mathbb P(B)>0$, \begin{align*} \mathbb P(A\mid B)=\frac{\mathbb P(A\cap B)}{\mathbb P(B)}. \end{align*} Thus $\mathbb P(A\mid B)=\mathbb P(A)$ if and only if \begin{align*} \mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B), \end{align*} which is exactly independence of $A$ and $B$. [/step] [step:Condition on A] If also $\mathbb P(A)>0$, the same computation gives \begin{align*} \mathbb P(B\mid A)=\frac{\mathbb P(A\cap B)}{\mathbb P(A)}. \end{align*} Therefore $\mathbb P(B\mid A)=\mathbb P(B)$ if and only if \begin{align*} \mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B). \end{align*} This is again the independence condition. [/step]