I am trying to see if the following holds: if $A,B,C$ are independent so is $P(A^c\cap (B\cup C^c)) $.
I wanted to do a proof by contradiction so I started by assuming that this holds and tried to find if it does not necessarily hold.
\begin{equation} \begin{aligned} P(A^c\cap (B\cup C^c)) & = P(A^c)P(B\cup C^c)\\ & = [1- P(A)] [P(B) + 1-P(C) -P(B)(1-P(C)]\\ & = [1- P(A)] [ 1-P(C) -P(B)P(C)]\\ & = [1-P(A)][1-P(C)(1-P(B))] \end{aligned} \end{equation}
Now I tried to expand this expression hoping that something will cancel out but it resulted just in the expression:
\begin{equation} 1-P(A)-P(C)-P(A)P(C)-P(B)P(C)-P(A)P(B)P(C) \end{equation} Which does not look like any definition of independence I ever saw. But then I know that all individual terms of that expression are independent and if I subtract just a bunch of independent events from the event space (1) it feels like the results should be independent too.
I wouldn't know by contradiction (note that if you want to prove it by contradiction, then you should instead start by assuming that they are NOT independent), but note that: \begin{align} P(A^c \cap (B\cup 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)P( B)P( C^c) \\ & = P(A^c) ( P(B ) + P(C^c) - P( B)P( C^c)) \\ & = P(A^c) P( B\cup C^c ) \end{align} where in the second equality I used the independence of $A,B$, and $C$, otherwise the well known formula for the probability of the union of two events.