Additive Function is Strongly Additive

Theorem
Let $$\mathcal S$$ be an algebra of sets.

Let $$f: \mathcal S \to \overline {\R}$$ be an additive function on $$\mathcal S$$.

Then:
 * $$\forall A, B \in \mathcal S: f \left({A \cup B}\right) = f \left({A}\right) + f \left({B}\right) - f \left({A \cap B}\right)$$

Proof
From Set Difference Union Intersection we have that:
 * $$A = \left({A \setminus B}\right) \cup \left({A \cap B}\right)$$

and:
 * $$B = \left({B \setminus A}\right) \cup \left({A \cap B}\right)$$

From Set Difference with Intersection we have that:
 * $$\left({A \setminus B}\right) \cap B = \varnothing$$

and:
 * $$\left({B \setminus A}\right) \cap A = \varnothing$$

from which it immediately follows that:
 * $$\left({A \setminus B}\right) \cap \left({A \cap B}\right) = \varnothing$$

and:
 * $$\left({B \setminus A}\right) \cap \left({A \cap B}\right) = \varnothing$$

So:
 * $$A$$ 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 additive function:
 * $$f \left({A}\right) = f \left({A \setminus B}\right) + f \left({A \cap B}\right)$$
 * $$f \left({B}\right) = f \left({B \setminus A}\right) + f \left({A \cap B}\right)$$

We also have from Set Difference Disjoint with Reverse that $$\left({A \setminus B}\right) \cap \left({B \setminus A}\right) = \varnothing$$.

Hence:

$$ $$ $$

Hence the result.