General Distributivity Theorem/Lemma 2

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

Then for every sequence $\left \langle {a_k} \right \rangle_{1 \le k \le n}$ of elements of $R$, and for every $b \in R$:


 * $\displaystyle b * \left({\sum_{j \mathop = 1}^n a_j}\right) = \sum_{j \mathop = 1}^n \left({b * a_j}\right)$

where:
 * $\displaystyle \sum_{j \mathop = 1}^n a_j$ is the summation $a_1 \circ a_2 \circ \cdots \circ a_n$
 * $n$ is a strictly positive integer: $n \in \Z_{> 0}$.

Proof
Proof by induction:

For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle b * \left({\sum_{j \mathop = 1}^n a_j}\right) = \sum_{j \mathop = 1}^n \left({b * a_j}\right)$

$P(1)$ is true, as this just says $b * a_1 = b * a_1$.

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

Basis for the Induction
$P(2)$ is the case:

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:


 * $\displaystyle b * \left({\sum_{j \mathop = 1}^k a_j}\right) = \sum_{j \mathop = 1}^k \left({b * a_j}\right)$

Then we need to show:


 * $\displaystyle b * \left({\sum_{j \mathop = 1}^{k + 1} a_j}\right) = \sum_{j \mathop = 1}^{k + 1} \left({b * a_j}\right)$

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:
 * $\displaystyle \forall n \in \Z_{> 0}: b * \left({\sum_{j \mathop = 1}^k a_j}\right) = \sum_{j \mathop = 1}^k \left({b * a_j}\right)$