Inclusion-Exclusion Principle

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

Let $A_1, A_2, \ldots, A_n$ be finite sets.

Let $f: \mathcal S \to \R$ be an additive function.

Then:

Proof
Proof by induction:

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:

$P(1)$ is true, as this just says $f \left({A_1}\right) = f \left({A_1}\right)$.

Basis for the Induction
$P(2)$ is the case:
 * $f \left({A_1 \cup A_2}\right) = f \left({A_1}\right) + f \left({A_2}\right) - f \left({A_1 \cap A_2}\right)$

which is the result Additive Function is Strongly Additive.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({r}\right)$ is true, where $r \ge 2$, then it logically follows that $P \left({r+1}\right)$ is true.

So this is our induction hypothesis:

Then we need to show:

Induction Step
This is our induction step:

Consider $\displaystyle f \left({\bigcup_{i \mathop = 1}^r A_i \cap A_{r+1}}\right)$.

By the fact that Intersection Distributes over Union, this can be written:


 * $\displaystyle f \left({\bigcup_{i \mathop = 1}^r \left({A_i \cap A_{r+1}}\right)}\right)$

To this, we can apply the induction hypothesis:

At the same time, we have the expansion of the term $\displaystyle f \left({\bigcup_{i \mathop = 1}^r A_i}\right)$ to take into account.

So we can consider the general term of $s$ intersections in the expansion of $\displaystyle f \left({\bigcup_{i \mathop = 1}^{r+1} A_i}\right)$:


 * $\displaystyle \left({-1}\right)^{s-1} \sum_{{I} \subseteq \left[{1 \,.\,.\,r}\right] \atop {\left|{I}\right| = s}} f \left({\bigcap_{i \mathop \in I} A_i}\right) - \left({-1}\right)^{s-2} \sum_{{J} \subseteq \left[{1 \,.\,.\,r}\right] \atop {\left|{J}\right| = s-1}} f \left({\bigcap_{i \mathop \in J} A_i \cap A_{r+1}}\right)$

where:
 * $I$ ranges over all sets of $s$ elements out of $\left[{1 \,.\,.\, r}\right]$
 * $J$ ranges over all sets of $s-1$ elements out of $\left[{1 \,.\,.\, r}\right]$
 * $1 \le s \le r$

Messy though it is, it can be seen that this expression is merely:


 * $\displaystyle \left({-1}\right)^{s-1} \sum_{{I'} \subseteq \left[{1 \,.\,.\,r+1}\right] \atop {\left|{I'}\right| = s}} f \left({\bigcap_{i \mathop \in I'} A_i}\right)$

where this time, $I'$ ranges over all sets of $s$ elements out of $\left[{1 \,.\,.\, r+1}\right]$ and $1 \le s \le r+1$.

This is the required term in $s$ intersections in the expansion of $\displaystyle f \left({\bigcup_{i \mathop = 1}^{r+1} A_i}\right)$.

Just to check, we can see the first term is:
 * $\displaystyle \sum_{i \mathop = 1}^r f \left({A_i}\right) + f \left({A_{r+1}}\right) = \sum_{i \mathop = 1}^{r+1} f \left({A_i}\right)$

As the expression $\displaystyle f \left({\bigcup_{i \mathop = 1}^r A_i \cap A_{r+1}}\right)$ consists only of intersections of two or more elements of $\mathcal S$, we see it does not contribute to this first term.

Finally, let us make sure of the last term - this is:
 * $\displaystyle - \left({-1}\right)^{r-1} f \left({\bigcap_{i \mathop = 1}^r A_i \cap A_{r+1}}\right)$

which works out as:
 * $\displaystyle \left({-1}\right)^r f \left({\bigcap_{i \mathop = 1}^{r+1} A_i}\right)$

We've done enough.

So $P \left({r}\right) \implies P \left({r+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

Comment
This result is usually quoted in the context of combinatorics, where $f$ is the cardinality function.

It is also seen in the context of probability theory, in which $f$ is taken to be a probability measure.

Historical Notes
This formula, in various forms, has been attributed to: