General Distributivity Theorem

Theorem
Let $$\left({R, \circ, \ast}\right)$$ 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 $$b \in R$$:


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

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

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

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

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

This can be proved by induction.

Proof by induction:

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


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

Basis for the Induction

 * $$P(2)$$ is the case $$\left({a_1 \circ a_2}\right) \ast b = \left({a_1 \ast b}\right) \circ \left({a_2 \ast b}\right)$$, which is true by dint of $$\left({R, \circ, \ast}\right)$$ 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:

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

Then we need to show:

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

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

$$b \ast \left({a_1 \circ \cdots \circ a_n}\right) = \left({b \ast a_1}\right) \circ \cdots \circ \left({b \ast a_n}\right)$$ is proved in the same way.