Sum of Summations equals Summation of Sum

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Theorem

Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.

Let $\ds \sum_{\map R i} x_i$ denote a summation over $R$.


Let the fiber of truth of $R$ be finite.

Then:

$\ds \sum_{\map R i} \paren {b_i + c_i} = \sum_{\map R i} b_i + \sum_{\map R i} c_i$


Infinite Sequence

Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers $\Z$.

Let $\ds \sum_{\map R i} x_i$ denote a summation over $R$.


Let the fiber of truth of $R$ be infinite.


Let $\ds \sum_{\map R i} b_i$ and $\ds \sum_{\map R i} c_i$ be convergent.


Then:

$\ds \sum_{\map R i} \paren {b_i + c_i} = \sum_{\map R i} b_i + \sum_{\map R i} c_i$


Proof

Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$.

Then:

\(\ds \sum_{\map R i} \paren {b_i + c_i}\) \(=\) \(\ds \sum_{\map R i} \paren {a_{i 1} + a_{i 2} }\) Definition of $b_i$ and $c_i$
\(\ds \) \(=\) \(\ds \sum_{\map R i} \paren {\sum_{j \mathop = 1}^2 a_{i j} }\) Definition of Summation
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^2 \paren {\sum_{\map R i} a_{i j} }\) Exchange of Order of Summation
\(\ds \) \(=\) \(\ds \sum_{\map R i} a_{i 1} + \sum_{\map R i} a_{i 2}\) Definition of Summation
\(\ds \) \(=\) \(\ds \sum_{\map R i} b_i + \sum_{\map R i} c_i\) Definition of $b_i$ and $c_i$

$\blacksquare$


Sources

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