Proof of the addition rule for two events

Proof

1. Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $A, B \in \mathcal{F}$ be events.
2. The sets $A \cup B$ and $A \cap B = (A^c \cup B^c)^c$ are in $\mathcal{F}$, because $\mathcal{F}$ is a σ-algebra.
3. We have $A \subseteq A \cup B$. By monotonicity of probability, $P((A \cup B) \setminus A) = P(A \cup B) - P(A)$.
4. A point of $A \cup B$ that is not in $A$ lies in $B$. So $(A \cup B) \setminus A = B \setminus A$.
5. We have $A \cap B \subseteq B$. By monotonicity of probability again, $P(B \setminus (A \cap B)) = P(B) - P(A \cap B)$.
6. A point of $B$ lies outside $A \cap B$ exactly when it lies outside $A$. So $B \setminus (A \cap B) = B \setminus A$.
7. By steps 3 to 6, both differences equal $P(B \setminus A)$. So $P(A \cup B) - P(A) = P(B) - P(A \cap B)$.
8. Adding $P(A) + P(A \cap B)$ to both sides gives $P(A \cup B) + P(A \cap B) = P(A) + P(B)$. Subtracting $P(A \cap B)$ from both sides gives the equivalent form.