Proof by induction: For all $n \in \N_{>0}$, let $\map P N$ be the proposition: {{begin-eqn}} {{eqn | l = \map f {\bigcup_{i \mathop = 1}^n A_i}       | r = \sum_{i \mathop = 1}^n \map f {A_i}       | c =  }} {{eqn | o =        | ro= -       | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \map f {A_i \cap A_j}       | c =  }} {{eqn | o =        | ro= +       | r = \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le n} \map f {A_i \cap A_j \cap A_k}       | c =  }} {{eqn | o =        | ro= \cdots       | c =  }} {{eqn | o =        | ro= +       | r = \paren {-1}^{n - 1} \map f {\bigcap_{i \mathop = 1}^n A_i}       | c =  }} {{end-eqn}} $\map P 1$ is true, as this just says $\map f {A_1} = \map f {A_1}$. === Basis for the Induction === $\map P 2$ is the case: :$\map f {A_1 \cup A_2} = \map f {A_1} + \map f {A_2} - \map f {A_1 \cap A_2}$ 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 $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true. So this is our induction hypothesis: {{begin-eqn}} {{eqn | l = \map f {\bigcup_{i \mathop = 1}^r A_i}       | r = \sum_{i \mathop = 1}^r \map f {A_i}       | c =  }} {{eqn | o =        | ro= -       | r = \sum_{1 \mathop \le i \mathop < j \mathop \le r} \map f {A_i \cap A_j}       | c =  }} {{eqn | o =        | ro= +       | r = \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le r} \map f {A_i \cap A_j \cap A_k}       | c =  }} {{eqn | o =        | ro= \cdots       | c =  }} {{eqn | o =        | ro= +       | r = \paren {-1}^{r - 1} \map f {\bigcap_{i \mathop = 1}^r A_i}       | c =  }} {{end-eqn}} Then we need to show: {{begin-eqn}} {{eqn | l = \map f {\bigcup_{i \mathop = 1}^{r + 1} A_i}       | r = \sum_{i \mathop = 1}^{r + 1} \map f {A_i}       | c =  }} {{eqn | o =        | ro= -       | r = \sum_{1 \mathop \le i \mathop < j \mathop \le r + 1} \map f {A_i \cap A_j}       | c =  }} {{eqn | o =        | ro= +       | r = \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le r + 1} \map f {A_i \cap A_j \cap A_k}       | c =  }} {{eqn | o =        | ro= \cdots       | c =  }} {{eqn | o =        | ro= +       | r = \paren {-1}^r \map f {\bigcap_{i \mathop = 1}^{r + 1} A_i}       | c =  }} {{end-eqn}} === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \map f {\bigcup_{i \mathop = 1}^{r + 1} A_i}       | r = \map f {\bigcup_{i \mathop = 1}^r A_i \cup A_{r + 1} }       | c =  }} {{eqn | r = \map f {\bigcup_{i \mathop = 1}^r A_i} + \map f {A_{r + 1} } - \map f {\bigcup_{i \mathop = 1}^r A_i \cap A_{r + 1} }       | c = Basis for the Induction }} {{end-eqn}} Consider $\ds \map f {\bigcup_{i \mathop = 1}^r A_i \cap A_{r + 1} }$. By the fact that Intersection Distributes over Union, this can be written: :$\ds \map f {\bigcup_{i \mathop = 1}^r {A_i \cap A_{r + 1} } }$ To this, we can apply the induction hypothesis: {{begin-eqn}} {{eqn | l = \map f {\bigcup_{i \mathop = 1}^r {A_i \cap A_{r + 1} } }       | r = \sum_{i \mathop = 1}^r \map f {A_i \cap A_{r + 1} }       | c =  }} {{eqn | o =        | ro= -       | r = \sum_{1 \mathop \le i \mathop < j \mathop \le r} \map f {A_i \cap A_j \cap A_{r + 1} }       | c =  }} {{eqn | o =        | ro= +       | r = \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le r} \map f {A_i \cap A_j \cap A_k \cap A_{r + 1} }       | c =  }} {{eqn | o =        | ro= \cdots       | c =  }} {{eqn | o =        | ro= +       | r = \paren {-1}^{r - 1} \map f {\bigcap_{i \mathop = 1}^r A_i \cap A_{r + 1} }       | c =  }} {{end-eqn}} At the same time, we have the expansion of the term $\ds \map f {\bigcup_{i \mathop = 1}^r A_i}$ to take into account. So we can consider the general term of $s$ intersections in the expansion of $\ds \map f {\bigcup_{i \mathop = 1}^{r + 1} A_i}$: :$\ds \paren {-1}^{s - 1} \sum_{\substack {I \subseteq \closedint 1 r \\ \card I = s} } \map f {\bigcap_{i \mathop \in I} A_i} - \paren {-1}^{s - 2} \sum_{\substack {J \subseteq \closedint 1 r \\ \card J = s - 1} } \map f {\bigcap_{i \mathop \in J} A_i \cap A_{r + 1} }$ where: :$I$ ranges over all sets of $s$ elements out of $\closedint 1 r$ :$J$ ranges over all sets of $s - 1$ elements out of $\closedint 1 r$ :$1 \le s \le r$ Messy though it is, it can be seen that this expression is merely: :$\ds \paren {-1}^{s - 1} \sum_{\substack {I' \subseteq \closedint 1 {r + 1} \\ \card {I'} = s} } \map f {\bigcap_{i \mathop \in I'} A_i}$ where this time, $I'$ ranges over all sets of $s$ elements out of $\closedint 1 {r + 1}$ and $1 \le s \le r + 1$. This is the required term in $s$ intersections in the expansion of $\ds \map f {\bigcup_{i \mathop = 1}^{r + 1} A_i}$. Just to check, we can see the first term is: :$\ds \sum_{i \mathop = 1}^r \map f {A_i} + \map f {A_{r + 1} } = \sum_{i \mathop = 1}^{r + 1} \map f {A_i}$ As the expression $\ds \map f {\bigcup_{i \mathop = 1}^r A_i \cap A_{r + 1} }$ consists only of intersections of two or more elements of $\SS$, we see it does not contribute to this first term. Finally, let us make sure of the last term. This is: :$\ds -\paren {-1}^{r - 1} \map f {\bigcap_{i \mathop = 1}^r A_i \cap A_{r + 1} }$ which works out as: :$\ds \paren {-1}^r \map f {\bigcap_{i \mathop = 1}^{r + 1} A_i}$ We've done enough. So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: {{begin-eqn}} {{eqn | l = \map f {\bigcup_{i \mathop = 1}^n A_i}       | r = \sum_{i \mathop = 1}^n \map f {A_i}       | c =  }} {{eqn | o =        | ro= -       | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \map f {A_i \cap A_j}       | c =  }} {{eqn | o =        | ro= +       | r = \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le n} \map f {A_i \cap A_j \cap A_k}       | c =  }} {{eqn | o =        | ro= \cdots       | c =  }} {{eqn | o =        | ro= +       | r = \paren {-1}^{n - 1} \map f {\bigcap_{i \mathop = 1}^n A_i}       | c =  }} {{end-eqn}} {{qed}}