<s>[INST] Let A and B be independent. Show that $A^c$ and $B^c$ c are independent - better solution? My solution seems too long winded. This should take less writing, no? Here it is.
$B^c$ means complement of $B$ (i.e. $P(B) = 1 - P(B^c)$
If $A$ and $B$ are independent events:
$$
P(A \cap B) = P(A)P(B)
$$
So I do:
$$
P(A^c \cap B^c) = P(A^c | B^c)P(B^c) = (1 - P(A|B^c))P(B^c) = (1 - \frac{P(A\cap B^c)}{P(B^c)})P(B^c) = (1 - \frac{P(B^c|A)P(A)}{P(B^c)})P(B^c)
$$
Now, I can use the result that since $A$ and $B$ are independent $P(B^c|A) = 1 - P(B|A) = 1 - P(B) = P(B^c)$. Then:
$$
(1 - \frac{P(B^c|A)P(A)}{P(B^c)})P(B^c) = (1 - \frac{P(B^c)P(A)}{P(B^c)})P(B^c) = (1 - P(A))P(B^c) = P(A^c)P(B^c)
$$
Therefore, since $P((A^c \cap B^c) = P(A^c)P(B^c)$ then $A^c$ and $B^c$ are independent.
 [/INST] Here is another way:
$$\begin{align} P(A^c\cap B^c) &=1-P(A\cup B)=1-P(A)-P(B)+P(A\cap B)\\  &= 1-P(A)-P(B)+P(A)P(B) \\ &=1-P(A)-P(B)(1-P(A))=(1-P(A))(1-P(B)) \\ &= P(A^c)P(B^c)\end{align}$$
</s>