I saw this exercise in Herstein's *Abstract Algebra*. I can do two proofs:

**Non-rigorous "proof" with words:**

To find $m(A \cup B)$, we can try adding $m(A)$ and $m(B)$. But any element which is in $A \cap B$ is counted twice, so we must subtract $m(A \cap B)$.

**Super boring and annoying proof**

Let the elements of $A$ be $a\_1, a\_2, ..., a\_n$ and let the elements of $B$ be $b\_1, b\_2, ..., b\_k$.

Case $1$: $A \cap B = \phi$.

$\vdots$

Case 2: $A \cap B \not = \phi$. Then suppose $a\_{i\_1} = b\_{i\_1, a\_{i\_2}}=b\_{i\_2}$, ...

My question is, how can I construct a proof "in between"? The first one doesn't seem like a very good proof.

If there is no proof in between, how could I improve the first one? (The second is just too painful to do any version of it.