General Distributivity Theorem

Theorem
Let $\struct {R, \circ, *}$ be a ringoid.

Then for every pair of sequences $\sequence {a_i}_{1 \mathop \le i \mathop \le m}, \sequence {b_j}_{1 \mathop \le j \mathop \le n}$ of elements of $R$:
 * $\ds \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$

where:
 * $\ds \sum_{i \mathop = 1}^m a_i$ is the summation $a_1 \circ a_2 \circ \cdots \circ a_m$
 * $m$ and $n$ are strictly positive integers: $m, n \in \Z_{> 0}$

Lemmata
The proof requires the following lemmata:

Proof
Proof by induction:

For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
 * $\ds \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$

We have that $\struct {R, \circ, *}$ is a ringoid, and so:
 * $\forall a, b, c \in S: \paren {a \circ b} * c = \paren {a * c} \circ \paren {b * c}$
 * $\forall a, b, c \in R: a * \paren {b \circ c} = \paren {a * b} \circ \paren {a * c}$

Basis for the Induction
$\map P 1$ is the case:


 * $\ds \forall m \in \Z_{>0}: \paren {\sum_{i \mathop = 1}^m a_i} * b_1 = \sum_{1 \mathop \le i \mathop \le m} \paren {a_i * b_1}$

This is demonstrated in Lemma 1.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:


 * $\ds \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^k b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le k} } \paren {a_i * b_j}$

Then we need to show:


 * $\ds \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^{k + 1} b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le {k + 1} } } \paren {a_i * b_j}$

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\ds \forall m, n \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$

The same result can be obtained by fixing $n$ and using induction on $m$, which requires Lemma 2 to be used for its base case.

Also see

 * Multiple of Ring Product


 * Product Rule for Sequences for the infinite case in the context of the standard number fields