Index Laws/Sum of Indices/Semigroup

Theorem
Let $\left({T, \oplus}\right)$ be a semigroup.

For $a \in T$, let $\oplus^n a = a^n$ be defined as the $n$th power of $a$:


 * $a^n = \begin{cases}

a & : n = 1 \\ a^x \oplus a & : n = x + 1 \end{cases}$

That is:
 * $a^n = \underbrace{a \oplus a \oplus \cdots \oplus a}_{n \text{ copies of } a} = \oplus^n \left({a}\right)$

Then:


 * $\forall m, n \in \N_{>0}: a^{n + m} = a^n \oplus a^m$

Proof
Let $a \in T$.

Because $\left({T, \oplus}\right)$ is a semigroup, $\oplus$ is associative on $T$.

The proof proceeds by the finite induction.

Let $S$ be the set of all $m \in \N_{>0}$ such that:
 * $\forall m, n \in \N_{>0}: a^{n + m} = a^n \oplus a^m$

that is:
 * $\forall m, n \in \N_{>0}: \oplus^{n + m} a = \left({\oplus^n a}\right) \oplus \left({\oplus^m a}\right)$

Basis for the Induction
By the definition of the power of an element, the mapping $\oplus^n: \N_{>0} \to T$ is defined as:


 * $\forall n \in \N_{>0}: \oplus^n a = f_a \left({n}\right)$

where $f_a: \N_{>0} \to T$ is the recursively defined mapping:


 * $\forall n \in \N_{>0}: f_a \left({n}\right) = \begin{cases}

a & : n = 1 \\ f_a \left({r}\right) \oplus a & : n = r + 1 \end{cases}$

Thus:

So $1 \in S$.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $k \in S$ is true, where $k \ge 1$, then it logically follows that $k + 1 \in S$ is true.

So this is our induction hypothesis:
 * $\oplus^{n + k} a = \left({\oplus^n a}\right) \oplus \left({\oplus^k a}\right)$

It is then to be shown that:
 * $\oplus^{n + \left({k + 1}\right)} a = \left({\oplus^n a}\right) \oplus \left({\oplus^{k + 1} a}\right)$

Induction Step
This is our induction step:

So $k + 1 \in S$.

So by the Principle of Finite Induction, $S = \N_{>0}$.

Thus this result is true for all $m, n \in \N_{>0}$:
 * $\forall m, n \in \N_{>0}: \oplus^{n + m} a = \left({\oplus^n a}\right) \oplus \left({\oplus^m a}\right)$

or:
 * $\forall m, n \in \N_{>0}: a^{n + m} = a^n \oplus a^m$