By the law of total probability, I know that $P(A) = P(A|C)P(C) + P(A|C^c)P(C^c)$. Applying the same logic, I would like to say that $$P(A|B) = P(A|B,C)P(C) + P(A|B,C^c)P(C^c)$$ However, I know this conclusion is incorrect because when you expand the probabilities - the LHS does not match the RHS.
How could I properly expand $P(A|B)$ by conditioning on another event, say $C$?
$$ P(A|B) = \frac{P(A,B)}{P(B)} = \frac{P(A,B,C) + P(A,B,C^c)}{P(B)} = \frac{P(A|B,C)P(B,C) + P(A | B, C^c)P(B,C^c)}{P(B)} $$
$$ = P(A|B,C)P(C|B) + P(A | B, C^c)P(C^c|B) $$