Prove that if events A and B are independent, then the complement events of A and B are also independent. I know that:
$$\begin{gathered}P\left(A\cap B\right)=P\left(A\right)P\left(B\right)\\
P\left(A^{C}\right)=1-P\left(A\right)\\
P\left(B^{C}\right)=1-P\left(B\right)
\end{gathered}
$$

My proof so far:
$$\begin{gathered}P\left(A^{C}\cap B^{C}\right)=\left(1-P\left(A\right)\right)\left(1-P\left(B\right)\right)=\\
1-P\left(B\right)-P\left(A\right)+P\left(A\right)P\left(B\right)=1-P\left(B\right)-P\left(A\right)+P\left(A\cap B\right)
\end{gathered}
$$

After that, I'm stuck.  Any help would be appreciated.
Assume $A$ and $B$ are independent. Then
\begin{align}
P(A^c \cap B^c) 
&amp;= 1 - P(A \cup B) \\
&amp;= 1 - P(A) - P(B) + P(A \cap B) \\
&amp;= 1 - P(A) - P(B) + P(A)P(B) \\
&amp;= (1-P(A))(1-P(B)) \\
&amp;= P(A^c)P(B^c).
\end{align}
By definition if two events are independent then $P(A | B)=P(A)=P(A/B')$.

So, $P(A/B)=P(A \cap B)/P(B)=P(A)$ and hence by multiplying both sides by $P(B)$ we get $P(A \cap B)= P(A)P(B)$.

Hence proved.