General Distributivity Theorem

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

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


 * $\left({a_1 \circ \cdots \circ a_n}\right) * b = \left({a_1 * b}\right) \circ \cdots \circ \left({a_n * b}\right)$
 * $b * \left({a_1 \circ \cdots \circ a_n}\right) = \left({b * a_1}\right) \circ \cdots \circ \left({b * a_n}\right)$

Consequently, in the context of a ring, this can be translated into:

Let $x, y \in \left({R, +, \circ}\right)$. Then:


 * $\forall n \in \Z^*: \left({n \cdot x} \right) \circ y = n \cdot \left({x \circ y}\right) = x \circ \left({n \cdot y}\right)$

Proof
We will prove that:
 * $\forall n \in \N^*: \left({a_1 \circ \cdots \circ a_n}\right) * b = \left({a_1 * b}\right) \circ \cdots \circ \left({a_n * b}\right)$

Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $\left({a_1 \circ \cdots \circ a_n}\right) * b = \left({a_1 * b}\right) \circ \cdots \circ \left({a_n * b}\right)$

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

Basis for the Induction

 * $P(2)$ is the case:
 * $\left({a_1 \circ a_2}\right) * b = \left({a_1 * b}\right) \circ \left({a_2 * b}\right)$

which is true by dint of $\left({R, \circ, *}\right)$ being a ringoid.

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:


 * $\left({a_1 \circ \cdots \circ a_k}\right) * b = \left({a_1 * b}\right) \circ \cdots \circ \left({a_k * b}\right)$

Then we need to show:


 * $\left({a_1 \circ \cdots \circ a_k \circ a_{k+1}}\right) * b = \left({a_1 * b}\right) \circ \cdots \circ \left({a_k * b}\right) \circ \left({a_{k+1} * b}\right)$

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:
 * $\forall n \in \N^*: \left({a_1 \circ \cdots \circ a_n}\right) * b = \left({a_1 * b}\right) \circ \cdots \circ \left({a_n * b}\right)$

The result:
 * $b * \left({a_1 \circ \cdots \circ a_n}\right) = \left({b * a_1}\right) \circ \cdots \circ \left({b * a_n}\right)$

is proved in the same way.