This question is for an intro class in probability theory :)
Let $$A, B, C$$ be three events that are pairwise & jointly independent. Show that $$A^c$$ and $$B \cup C^c$$ are also independent.
I started with some definitions. Pairwise & joint independence give us the following: $$P(A \cap B) = P(A)P(B)$$ $$P(B \cap C) = P(B)P(C)$$ $$P(A \cap C) = P(A)P(C)$$ $$P(A \cap B \cap C) = P(A)P(B)P(C)$$
I could also show that pairwise independence of the complements of the events holds, ie. $$P(A^c \cap B^c) = P(A^c)P(B^c)$$ $$P(B^c \cap C^c) = P(B^c)P(C^c)$$ $$P(A^c \cap C^c) = P(A^c)P(C^c)$$
The way I'm thinking about this problem is to simply follow the definition of independence. Show that: $$P(A^c \cap (B \cup C^c)) = P(A^c) \cdot P(B \cup C^c) = P(A^c) \cdot (P(B) + P(C^c) - P(B \cap C^c))$$ And the work I've done so far: $$P(A^c \cap (B \cup C^c)) = P((A^c \cap B) \cup (A^c \cap C^c)) = P(A^c \cap B) + P(A^c \cap C^c) - P((A^c \cap B) \cap (A^c \cap C^c) = P(A^c \cap B) + P(A^c \cap C^c) - P((A^c \cap (B \cap C^c)) = P(A^c)P(B) + P(A^c)P(C^c) - P(A^c \cap (B \cap C^c))$$ This is where I get stuck. I think this method reduces the problem to proving $$A^c, (B \cap C^c)$$ are independent, which I can't seem to do. Any hints?