Inclusion-Exclusion Proof without inverses If $f$ is additive then it satisfies the inclusion-exclusion
principle
$$
   f(\bigcup A) = \sum_{B \subset A} f(\bigcap B) \cdot (-1)^{|B|+1}
$$
where $B$ is (here and henceforth) assumed to be nonempty.
See \url{http://www.proofwiki.org/wiki/Inclusion-Exclusion_Principle#Comment}
All proofs I've seen rely on using subtraction and I am unable to avoid it in my attempted proofs.
Could you please proof the equivalent restatement (?),
$$
   f(\bigcup A) +  \sum_{\text{ even }B \subset A} f(\bigcap B)
   = \sum_{\text{ odd } B \subset A} f(\bigcap B)
$$
without using the notion of subtraction.
Motavation: we are in a setting where + has no inverse -.
Thank you :)
 A: You could do an induction proof just like in the link; however I think this should follow from the "usual case" by an "abstract-nonsense" sort of argument. Here is an attempt. (I will use the notation from the linked proofwiki page, since the one in the question is not consistent.)
We have $\cup_{i} A_i = \sqcup_{\alpha \neq \emptyset} A_\alpha$ where $\alpha \subset \{1, \ldots, n\}$ is an indexing set and $A_\alpha = \left(\cap_{i \in \alpha} A_i \right) \cap \left(\cap_{j \notin \alpha} A_j^\complement\right)$  - this just says that each element of the union of $A_i$'s is in some non-empty collection $\alpha$ of individual $A_i$'s and this collection is uniquely determined by the element. We have similar formulas for every subset $s$ of $\{1, \ldots, n\}$ 
$\cap_{i \in s} A_i = \sqcup_{s\subset \alpha} A_\alpha$ 
This  says that an element which is in all $A_i$'s (for $i$ in $s$) must simply be contained in the set of $A_i$'s that includes $s$.
Thus on both sides of the equation 
$f(\cup_{i} A_i)+\sum_{|s| even} f( \cap_{i \in s} A_i) = \sum_{|s| odd} f( \cap_{i \in s} A_i)$ 
$f$ is applied to some disjoint unions of $A_\alpha$'s. By additivity of $f$, this means that both sides are (positive integer) linear combinations of $f(A_\alpha)$'s. The only thing we need to know is that the coefficients of each $A_\alpha$ on both sides are the same. This is a purely combinatorial statement which is equivalent to the "usual" inclusion-excusion. 
(For example, consider {subsets of $1,\ldots, n\}$, and define A_i={those subsets  that don't contain i}. Then $A_\alpha$ has one element for each $\alpha$ ("the $\alpha$ itself), and applying usual inclusion-exclusion to additive $f$ with value $1$ on singleton set containing that element and $0$ on other singleton sets, proves that the coefficient of $A_\alpha$ on both sides of the equation is the same. Since this is true for all $\alpha$ the result follows.)
A: Let $\Omega$ be a set and let $\mathcal A$ be a finite collection of subsets of $\Omega$.
Then we have the equality:$$\mathbf{1}_{\bigcup\mathcal{A}}+\sum_{\varnothing\neq\mathcal{B\subseteq\mathcal{A}}\text{ and }\left|\mathcal{B}\right|\text{ even}}\mathbf{1}_{\bigcap\mathcal{B}}=\sum_{\mathcal{B\subseteq\mathcal{A}}\text{ and }\left|\mathcal{B}\right|\text{ odd}}\mathbf{1}_{\bigcap\mathcal{B}}\tag1$$
To prove this it is enough to take an arbitrary element $\omega\in\Omega$ and to show that both sides of $(1)$ give the same outcome if $\omega$ is substituted.
Let $\mathcal A=\{A_1,\dots, A_n$} where the $A_i$ are distinct subsets of $\Omega$, choose some $\omega\in\Omega$ and let: $$I=\{i\in\{1,\dots,n\}\mid \omega\in A_i\}$$
Observe that $\mathbf1_{\bigcap\mathcal B}(\omega)=1$ iff $\mathcal B\subseteq\{A_i\mid i\in I\}$.
If $I=\varnothing$ then substituting $\omega$ gives $0$ on both sides
If $I\neq\varnothing$ then the collection $I$ has exactly $2^{|I|-1}-1$ subcollections that have a positive even number of elements and also has exactly $2^{|I|-1}$ subcollections that have an odd number of elements.
So on LHS we find $1+(2^{|I|-1}-1)=2^{|I|-1}$ and on RHS we find $2^{|I|-1}$
Proved is now that $(1)$ is correct.
If e.g. $\mu$ is a measure and the elements of $\mathcal A$ are measurable then taking the expectation on both sides we find:$$\mu\left(\bigcup\mathcal{A}\right)+\sum_{\mathcal{\varnothing\neq B\subseteq\mathcal{A}}\text{ and }\left|\mathcal{B}\right|\text{ even}}\mu\left(\bigcap \mathcal{B}\right)=\sum_{\mathcal{B\subseteq\mathcal{A}}\text{ and }\left|\mathcal{B}\right|\text{ odd}}\mu\left(\bigcap \mathcal{B}\right)$$
