I am used to seeing, for dependent events $A,B$, $$P(A \wedge B) = P(B)P(A\vert B) = P(B\vert A) P(A)$$
What if I want to condition on $B$ not happening. Then is it just $$ P(A\wedge \neg B) = P(\neg B)P(A\vert \neg B)? $$
Is there a relationship between $P(A\wedge B)$ and $P(A\wedge \neg B)$? Such as maybe $P(A\wedge B) + P(A\wedge \neg B) = P(A)$? If so, can a proof be provided?
Hint: $$A = (A \cap B) \cup(A\cap B^c),$$ $$\Phi = (A \cap B) \cap (A\cap B^c),$$ and $$\Pr(X \cup Y) = \Pr(X) + \Pr(Y) - \Pr (X \cap Y). $$