Cardinality of intersection of sets Consider the following problem: find $n(A \cap B)$ if $n(A)=10$, $n(B)=13$ and $n(A \cup B) = 15$.
I know if I want to find the union I use the Cardinal Number formula:
$$n(A\cup B) = n(A) + n(B) - n(A\cap B)$$ 
But how do I do it the other way: to find $n(A\cap B)$? 
Would it be $n(A\cap B) = n(A) + n(B) - n(A\cup B)$ ?
 A: Yes... $n(A\cap B) = n(A) + n(B) - n(A\cup B)$ is the way to go.
$n(A\cap B) = 8$ according to this.
A little Venn to visualize the formula:

A: We know that: $n(A)=10$, $n(B)=13$ and $n(A \cup B) = 15$.
Using the Cardinal Number formula:
$$\underbrace{n(A\cup B)}_{15} = \underbrace{n(A)}_{10} + \underbrace{n(B)}_{13} - n(A\cap B)\tag{1}$$ 
So indeed, we can write the equivalent to $(1)$: $$n(A\cap B) = \underbrace{n(A)}_{10} + \underbrace{n(B)}_{13} - \underbrace{n(A\cup B)}_{15}\tag{2}$$
$$\text{So, given what we know}:\;\; n(A\cap B) = 10 + 13 - 15 = 8$$
A: Yes.  The reason why this works lies in that "n" consists of a function which maps sets to cardinal numbers (which are sets too in set theory, but that doesn't matter here).  So, for n(A), n(B), and so on, we can treat n(A) just like any other sort of number, and thus use variables, and write things n(A)=x, n(B)=y, n(A∪B)=z and so on.  Thus, we can prove that n(A∩B)=n(A)+n(B)−n(A∪B) follows from n(A∪B)=n(A)+n(B)−n(A∩B) using the basic properties of ordinary algebra.
