A probability problem involving independence. Problem: Given are events $A$,$B$ and $C$. Knowing that: $A$ and $B$ are independent,$(A\cap B)\subset C$ and $(A'\cap B')\subset C'$ prove, that $P(A\cap C)\ge P(A)P(C)$. I can't seem to find a way to get the $P(A)P(C)$ when i start from the left side of the inequation. I've tried to write $A\cap C$ in some weird ways but that didn't help that much, and also I noticed that $(A'\cap B')\subset C'$ implies $C\subset(A\cup B)$, but couldn't find a way to use it, other than getting something like $P(A\cup B)\ge P(C)\ge P(A\cap B)=P(A)P(B)$.

By assumption, $A \cap C$ is the disjoint union of $A \cap B$ and $A \cap B' \cap C$, so
$$P(A \cap C) = P(A \cap B) + P(A \cap B' \cap C).$$
In the same way, $C$ is the disjoint union of $A \cap B$ and $A \cap B' \cap C$ and $A' \cap B \cap C$, so
$$P(A)P(C) = P(A)P(A \cap B) + P(A)P(A \cap B' \cap C) + P(A)P(A' \cap B \cap C).$$
By difference,
$$P(A \cap C)-P(A)P(C) = P(A')P(A \cap B) + P(A')P(A \cap B' \cap C) - P(A)P(A' \cap B \cap C).$$
Therefore,
$$P(A \cap C)-P(A)P(C) \ge P(A')P(A \cap B) + 0 - P(A)P(A' \cap B).$$
Using the independence of $A$ and $B$, we see that the right-hand side of the last inequality is $0$, which yields $P(A \cap C) \ge P(A)P(C)$.
