Sum over Complement of Finite Set
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Theorem
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f: S \to \mathbb A$ be a mapping.
Let $T \subseteq S$ be a subset.
Let $S \setminus T$ be its relative complement.
Then we have the equality of summations over finite sets:
- $\ds \sum_{s \mathop \in S \setminus T} \map f s = \sum_{s \mathop \in S} \map f s - \sum_{t \mathop \in T} \map f t$
Proof
Note that by Subset of Finite Set is Finite, $T$ is indeed finite.
By Set is Disjoint Union of Subset and Relative Complement, $S$ is the disjoint union of $S \setminus T$ and $T$.
The result now follows from Sum over Disjoint Union of Finite Sets.
$\blacksquare$