Sum of Summations over Overlapping Domains/Proof

Theorem

$\ds \sum_{\map R j} a_j + \sum_{\map S j} a_j = \sum_{\map R j \mathop \lor \map S j} a_j + \sum_{\map R j \mathop \land \map S j} a_j$

where $\lor$ and $\land$ signify logical disjunction and logical conjunction respectively.


Proof

\(\ds \sum_{\map R j} a_j + \sum_{\map S j} a_j\) \(=\) \(\ds \sum_{j \mathop \in \Z} a_j \sqbrk {\map R j} + \sum_{j \mathop \in \Z} a_j \sqbrk {\map R j}\) Definition of Summation by Iverson's Convention
\(\ds \) \(=\) \(\ds \sum_{j \mathop \in \Z} a_j \paren {\sqbrk {\map R j} + \sqbrk {\map R j} }\)

Let:

$A := \set {j \in \Z: \map R j}$
$B := \set {j \in \Z: \map R j}$

The result then follows from Cardinality of Set Union.

$\blacksquare$


Sources

  • 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: Exercise $21$