Summation of i from 1 to n of Summation of j from 1 to i

Theorem

 * $\displaystyle \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i a_{i j} = \sum_{j \mathop = 1}^n \sum_{i \mathop = j}^n a_{i j}$

Proof

 * $\displaystyle \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i$

can be expressed as:
 * $\displaystyle \sum_{R \left({i}\right)} \sum_{S \left({i, j}\right)} a_{i j}$

where:
 * $R \left({i}\right)$ is the propositional function $1 \le i \le n$
 * $S \left({i, j}\right)$ is the propositional function $1 \le j \le i$

We wish to find a propositional function $S' \left({j}\right)$ which is to be:
 * there exists an $i$ such that both $1 \le i \le n$ and $1 \le j \le i$

This is satisfied by the propositional function:
 * $S' \left({j}\right) := 1 \le j \le n$

Next we wish to find a propositional function $R' \left({i, j}\right)$ which is to be:
 * both $1 \le i \le n$ and $1 \le j \le i$

This is satisfied by the propositional function:
 * $R' \left({i, j}\right) := j \le i \le n$

Hence the result, from Exchange of Order of Summation with Dependency on Both Indices.