Proof
Notice that we can write \(A \cup B\) as the disjoint union
\begin{equation*}
A \cup B = (A-B) \cup (A \cap B) \cup (B-A).
\end{equation*}
We can also write disjointly
\begin{gather*}
A = (A-B) \cup (A \cap B)\\
B = (A \cap B) \cup (B-A)
\end{gather*}
Hence,
\begin{align*}
P(A) & + P(B) - P(A \cap B) \\
& = [P(A-B) + P(A \cap B)] \\
& + [P(A \cap B) + P(B-A)] - P(A \cap B)\\
& = P(A-B) + P(A \cap B) + P(B-A)\\
& = P(A \cup B)
\end{align*}