Size of a union of two sets We were ask to prove that $|A \cup B| = |A| + |B| - |A \cap B|‫‪$.
It was easy to prove it using a Venn diagram, but I think we might
be expected to do if more formally. Is there a formal way?
 A: $A\cup B = (A\setminus B) \cup (B\setminus A) \cup (A\cap B)$. These three sets are disjoint, so
$$
|A\cup B| = |A\setminus B| + |B\setminus A| + |(A\cap B)| 
$$
But $A\setminus B = A\setminus(A\cap B)$, so $|A\setminus B|=|A|-|A\cap B|$. A similar equality holds for $|B \setminus A|$. Substitution of these into the displayed equation above yields your result. 
Of course, one might need to formally show that  $|A\setminus (A\cap B)| = |A|-|A\cap B|$. I can't decide if this is any less obvious than the original proposition...
A: Just elaborating on the last post,
$\begin{array}{l}
A \cup B\\
 = (A \cup {A^c}) \cap (A \cup B)\\
 = A \cup ({A^c} \cap B)\\
 = (A \cap (B \cup {B^c})) \cup ({A^c} \cap B)\\
 = (A \cap B) \cup (A \cap {B^c}) \cup ({A^c} \cap B)
\end{array}$ 
Note that these 3 sets are mutually disjoint, since
$\begin{array}{l}
(A \cap B) \cap (A \cap {B^c}) = A \cap (B \cap {B^c}) = A \cap \Phi  = \Phi \\
(A \cap B) \cap ({A^c} \cap B) = (A \cap {A^c}) \cap B = \Phi  \cap B = \Phi 
\end{array}$
Hence, by basic addition rule for counting,
$\begin{array}{l}
|A \cup B|\\
 = |A \cap B| + |A \cap {B^c}| + |{A^c} \cap B|\;\;\;\;\;\;\ldots (1)
\end{array}$
Again, we have
$\begin{array}{l}
A = A \cap (B \cup {B^c}) = (A \cap B) \cup (A \cap {B^c})\\
B = (A \cup {A^c}) \cap B = (A \cap B) \cup ({A^c} \cap B)
\end{array}$ 
Again, these 3 sets are mutually disjoint, as shown earlier.
Hence, by basic addition rule for counting,
$\begin{array}{l}
|A| = |A \cap B| + |A \cap {B^c}| \Rightarrow |A \cap {B^c}| = |A| - |A \cap B|\\
|B| = |A \cap B| + |{A^c} \cap B| \Rightarrow |{A^c} \cap B| = |B| - |A \cap B|
\end{array}$
Hence, from (1), we get
$\begin{array}{l}
|A \cup B|\\
 = |A \cap B| + |A \cap {B^c}| + |{A^c} \cap B|\\
 = |A \cap B| + |A| - |A \cap B| + |B| - |A \cap B|\\
 = |A| + |B| - |A \cap B|
\end{array}$  (Proved)
