$P(A \mid B) + P(A \mid \bar{B}) = P(A)$

Solution 1:

To show that $P(A\cap B)+P(A\cap B^C)=P(A)$ you can argue that since $B\cap B^C=\emptyset$ then also $(A\cap B)\cap (A\cap B^C)=\emptyset$ and so by the probability additivity axiom for disjoint sets $$P(A\cap B)+P(A\cap B^C) = P((A\cap B)\cup(A\cap B^C))$$ By the distributive law for sets this equals $$P(A\cap(B\cup B^C))=P(A\cap\Omega)=P(A).$$

On the other hand the statement $P(A|B)+P(A|B^C)=P(A)$ is not correct in general because it neglects to take into account the probability of the events $B,B^C$. When you used the definition of conditional probability you should have gotten $\frac{P(A\cap B)}{P(B)} + \frac{P(A\cap B^C)}{P(B^C)}$ rather than $\frac{P(A\cap B)}{P(B)}P(B) + \frac{P(A\cap B^C)}{P(B^C)}P(B^C)$.