Unravelling $P(A^c\cap(B\cup C))$ So I'm being asked to show that $$P(A^c\cap(B\cup C)) = P(B) + P(C) - P(B\cap C) - P(C \cap A) - P(A \cap B) + P(A \cap B \cap C)$$
Here's what I've tried so far:
From the left side: 
$P(A^c\cap(B\cup C))$
= $P(A^c) + P(B\cup C) - P(A^c\cup B\cup C)$
= $P(A^c) + P(B\cup C) - [P(A^c\cup B) + P(C) - P((A^c\cup B) \cap C))]$
= $P(A^c) + P(B\cup C) - P(A^c\cup B) - P(C) + P((A^c\cup B) \cap C))$
I know its not much but I've been going around in circles for a while. I'm guessing in order to get rid of the $P(A^c)$ I need to turn them into $1 - P(A)$, but I'm not sure how to isolate the $A^c$s when they're in a union or intersection with another event :/
 A: Alternatively, I would prefer to remove $A^c$ from the first move,
\begin{align}
&P(A^c \cap (B \cup C)) \\&= P(B \cup C) - P(A \cap (B \cup C))\\
&= P(B)+P(C)-P(B\cap C) - P((A \cap B) \cup (A \cap C))\\
&= P(B)+P(C)-P(B\cap C) - P(A \cap B)-P(A\cap C)+P(A \cap B\cap C)\\
\end{align}
A: If you are allowed to consider the following formula for the union of three events as already known, the calculation becomes quite short:


*

*$P(A\cup B\cup C) = P(A) + P(B) + P(C) - P(A\cap B)  - P(B\cap C)  - P(A\cap C) + P(A\cap B \cap C)$
Looking on the RHS, you only have to show
$$P(A^c \cap (B \cup C)) = P(A\cup B\cup C) - P(A)$$
But this is easy using


*

*$(1)$: $P(A\cup (B\cup C)) = P(A) + P(B\cup C) - P(A \cap (B\cup C))$ and

*$(2)$: $P(B\cup C) = P(A \cap (B\cup C)) + P(A^c \cap (B\cup C))$
\begin{eqnarray*} P(A\cup B\cup C) - P(A) 
& \stackrel{(1)}{=} & \color{blue}{P(A)} + P(B\cup C) - P(A \cap (B\cup C))  \color{blue}{-P(A)}\\
& \stackrel{(2)}{=} & \color{blue}{P(A \cap (B\cup C))} + P(A^c \cap (B\cup C))\color{blue}{-P(A \cap (B\cup C))}\\
& = &  \boxed{P(A^c \cap (B\cup C))}
\end{eqnarray*}
