Addition Law of Probability/Proof 2

Theorem
Let $\Pr$ be a probability measure on an event space $\Sigma$.

Let $A, B \in \Sigma$.

Then:
 * $\Pr \left({A \cup B}\right) = \Pr \left({A}\right) + \Pr \left({B}\right) - \Pr \left({A \cap B}\right)$

That is, the probability of either event occurring equals the sum of their individual probabilities less the probability of them both occurring.

Proof
From Set Difference and Intersection form Partition:


 * $A$ is the union of the two disjoint sets $A \setminus B$ and $A \cap B$;
 * $B$ is the union of the two disjoint sets $B \setminus A$ and $A \cap B$.

So, by the definition of probability measure:
 * $\Pr \left({A}\right) = \Pr \left({A \setminus B}\right) + \Pr \left({A \cap B}\right)$
 * $\Pr \left({B}\right) = \Pr \left({B \setminus A}\right) + \Pr \left({A \cap B}\right)$

From Set Difference Disjoint with Reverse:
 * $\left({A \setminus B}\right) \cap \left({B \setminus A}\right) = \varnothing$

Hence:

Hence the result.