General Distributivity Theorem

Theorem
Let $\left({R, \circ, *}\right)$ be a ringoid.

Then for every pair of sequences $\left \langle {a_i} \right \rangle_{1 \le i \le m}, \left \langle {b_j} \right \rangle_{1 \le j \le n}$ of elements of $R$:


 * $\displaystyle \left({\sum_{i \mathop = 1}^m a_i}\right) * \left({\sum_{j \mathop = 1}^n b_j} \right) = \sum_{{1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le n}} \left({a_i * b_j}\right)$

where:
 * $\displaystyle \sum_{i \mathop = 1}^m a_i$ 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 $P \left({n}\right)$ be the proposition:
 * $\displaystyle \forall m \in \Z_{> 0}: \left({\sum_{i \mathop = 1}^m a_i}\right) * \left({\sum_{j \mathop = 1}^n b_j} \right) = \sum_{{1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le n}} \left({a_i * b_j}\right)$

We have that $\left({R, \circ, *}\right)$ is a ringoid, and so:
 * $\forall a, b, c \in S: \left({a \circ b}\right) * c = \left({a * c}\right) \circ \left({b * c}\right)$
 * $\forall a, b, c \in R: a * \left({b \circ c}\right) = \left({a * b}\right) \circ \left({a * c}\right)$

Basis for the Induction
$P(1)$ is the case:


 * $\displaystyle \forall m \in \Z_{> 0}: \left({\sum_{i \mathop = 1}^m a_i}\right) * b_1 = \sum_{1 \mathop \le i \mathop \le m} \left({a_i * b_1}\right)$

This is demonstrated in Lemma 1.

This is our basis for the induction.

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

So this is our induction hypothesis:


 * $\displaystyle \forall m \in \Z_{> 0}: \left({\sum_{i \mathop = 1}^m a_i}\right) * \left({\sum_{j \mathop = 1}^k b_j} \right) = \sum_{{1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le k}} \left({a_i * b_j}\right)$

Then we need to show:


 * $\displaystyle \forall m \in \Z_{> 0}: \left({\sum_{i \mathop = 1}^m a_i}\right) * \left({\sum_{j \mathop = 1}^{k + 1} b_j} \right) = \sum_{{1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le {k + 1}}} \left({a_i * b_j}\right)$

Induction Step
This is our induction step:

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

Therefore:
 * $\displaystyle \forall m, n \in \Z_{> 0}: \left({\sum_{i \mathop = 1}^m a_i}\right) * \left({\sum_{j \mathop = 1}^n b_j} \right) = \sum_{{1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le n}} \left({a_i * b_j}\right)$

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