Two events are independent ...
But what about these two, my textbook isn't clear on whether it's [if] or [if and only if]:
On
The events are independent if and only if $P(A\cap B)=P(A)P(B)$. This is the definition of independence.
The other conditions, $P(A\mid B)=P(A)$ etc., are not quite equivalent.
With the additional assumption that both events have positive probability, then they would be equivalent since $P(A\mid B)=P(A\cap B)/P(B)$.
However, it is possible that $P(B)=0$, in which case the events are independent since $P(A\cap B)=P(A)P(B)$, but $P(A)\neq P(A\mid B)$ since the latter is undefined. So they are really just "if" statements, and to get an "if and only if" you would need to say
$A$ and $B$ are independent if and only if either $P(B)=0$ or $P(A\mid B)=P(A)$.
On
Independence definition by $P(A\mid B)=P(A)$ holds asymmetrical information about dependency. Then, as is written, only in case $P(B)>0$ and $P(A\mid B)=\frac{P(A \cap B)}{P(B)}$ we obtain $P(A \cap B)=P(A)P(B)$. And because $P(A\mid B)=\frac{P(A \cap B)}{P(B)}$ is not defined for $P(B)=0$ we "forget" about initial definition of independence and take as basis $P(A \cap B)=P(A)P(B)$. But this last does not hold the initial asymmetry and becomes symmetrical and from independence for $(A,B)$ we obtain all pairs $(A,B^c),(A^c,B),(A^c,B^c)$ independence.
Let me repeat, that happened loss leaves room for desire improve definition by $P(A\mid B)=P(A)$. Desire define $P(A\mid B)$ as object for which holds $P(A\mid B)P(B)=P(A \cap B)$ gives uncertainty in $P(B)=0$ case, but I not lost hope one day meet acceptable decision in this case.
$A,B$ (with non-zero probability) are independent iff $\Pr(A\cap B)=\Pr(A)\Pr(B)$ iff $\Pr(A\vert B)=\Pr(A)$ iff $\Pr(B\vert A)=\Pr(B)$.
The proofs in all directions are simple and us the formula $\Pr(A\vert B)\Pr(B)=\Pr(A\cap B)$ in all possible directions.