Legs?

Instead of finding the complement of $A$ and $B$ and then taking their union, we can use De Morgan's Laws to just find $A \cap B$ and take the complement.

Because the number of 4-legged animals is 3, we can be sure that there are 9 objects that aren't 4-legged animals!

$\begin{array}{l l}|A^C \cup B^C| &= |(A \cap B)^C| \\ &=|S| - |(A \cap B)| \\ &= 12 - 3 \\ &= 9\end{array}$

$|A^{C}\cup B^{C}|$ = $|A^{C}|$ + $|B^{C}|$ - $|A^{C} intersection B^{C}|$ = 7 + 6 - 4 = 9, althought the best solution( for me) is from Andrew Ellinor.

Since $A^C$ means "elements that don't have 4 legs" and $B^C$ means "elements that aren't animals,"
$A^C \cup B^C$ means "things that don't have 4 legs OR that aren't animals" which is most things, all but the 3 that are 4-legged animals. $12-3 = \fbox{9}$ .