Sum of Summations equals Summation of Sum/Infinite Sequence/Proof 1

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Theorem

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:

\(\ds \sum_{\map R i} \paren {b_i + c_i}\) \(=\) \(\ds \sum_{\map R i} \paren {a_{i 1} + a_{i 2} }\) by definition
\(\ds \) \(=\) \(\ds \sum_{\map R i} \paren {\sum_{1 \mathop \le j \mathop \le 2} a_{i j} }\) Definition of Summation
\(\ds \) \(=\) \(\ds \sum_{1 \mathop \le j \mathop \le 2} \paren {\sum_{\map R i} a_{i j} }\) Exchange of Order of Summation: Finite and Infinite Series
\(\ds \) \(=\) \(\ds \sum_{\map R i} a_{i 1} + \sum_{\map R i} a_{i 2}\) Definition of Summation
\(\ds \) \(=\) \(\ds \sum_{\map R i} b_i + \sum_{\map R i} c_i\) by definition

$\blacksquare$