General Distributivity Theorem/Lemma 2

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

Then for every sequence $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ of elements of $R$, and for every $b \in R$:


 * $\displaystyle b * \paren {\sum_{j \mathop = 1}^n a_j} = \sum_{j \mathop = 1}^n \paren {b * a_j}$

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
The proof proceeds by the Principle of Mathematical Induction.

Recall that as $\struct {R, \circ, *}$ is a ringoid, $*$ is distributive over $\circ$:
 * $\forall a, b, c \in R: a * \paren {b \circ c} = \paren {a * b} \circ \paren {a * c}$

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

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

Basis for the Induction
$P(1)$ is true, as this just says:
 * $b * a_1 = b * a_1$

$P(2)$ is the case:

This is our basis for the induction.

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

So this is our induction hypothesis:


 * $\displaystyle b * \paren {\sum_{j \mathop = 1}^k a_j} = \sum_{j \mathop = 1}^k \paren {b * a_j}$

Then we need to show:


 * $\displaystyle b * \paren {\sum_{j \mathop = 1}^{k + 1} a_j} = \sum_{j \mathop = 1}^{k + 1} \paren {b * a_j}$

Induction Step
This is our induction step:

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

Therefore:
 * $\displaystyle \forall n \in \Z_{> 0}: b * \paren {\sum_{j \mathop = 1}^k a_j} = \sum_{j \mathop = 1}^k \paren {b * a_j}$