Alternative implies powers up to the fifth are well-defined

Statement
Suppose $$(S,*)$$ is an alternative magma, i.e., it satisfies the following two identities:

$$x * (x * y) = (x * x) * y \ \forall \ x,y \in S$$

and

$$x * (y * y) = (x * y) * y \ \forall \ x,y \in S$$

These are respectively termed the left-alternative law and right-alternative law.

Then cubes, fourth powers, and fifth powers are well-defined in $$S$$. In other words, if $$x^2$$ denotes $$x*x$$, we have the following for all $$a \in S$$:


 * 1) Cubes are well-defined: $$a^2 * a = a * a^2$$, and this is denoted as $$a^3$$.
 * 2) Fourth powers are well-defined: $$a^3 * a = a^2 * a^2 = a * a^3$$, and this is denoted as $$a^4$$.
 * 3) Fifth powers are well-defined: $$a^4 * a = a^3 * a^2 = a^2 * a^3 = a * a^4$$ and this is denoted as $$a^5$$.

Related facts

 * Jordan implies powers up to the fifth are well-defined
 * Left alternative and flexible implies powers up to the fifth are well-defined
 * Right alternative and flexible implies powers up to the fifth are well-defined

Proof
As above, we assume $$(S,*)$$ is an alternative magma and $$a$$ is an arbitrary element of $$S$$.

Cubes are well-defined
To prove: $$a * a^2 = a^2 * a$$.

Proof: This follows from either of the alternative laws, setting $$x = y = a$$.

Fourth powers are well-defined


To prove: $$a^3 * a = a^2 * a^2 = a * a^3$$.

Proof: We first show that $$a^3 * a = a^2 * a^2$$. Consider:

$$\! a^3 * a = (a^2 * a) * a$$

By the right-alternative law, setting $$x = a^2, y = a$$, we obtain that:

$$\! a^3 * a = a^2 * (a * a) = a^2 * a^2$$

We now show that $$a^2 * a^2 = a * a^3$$. We have:

$$\! a^2 * a^2 = (a * a) * a^2$$

By the left-alternative law, setting $$x = a, y = a^2$$, we obtain:

$$\! a^2 * a^2 = a * (a * a^2) = a * a^3$$.

Thus, we have shown that all three expressions are equal.

Fifth powers are well-defined
To prove: $$a^4 * a = a^3 * a^2 = a^2 * a^3 = a * a^4$$.

Proof:

We first show that $$a^4 * a = a^3 * a^2$$. Note that:

$$\! a^4 * a = (a^3 * a) * a = a^3 * (a * a) = a^3 * a^2$$

where, in an intermediate step, we used the right-alternative law with $$x = a^3, y = a$$.

We next show that $$a^3 * a^2 = a * a^4$$. Note that:

$$\! a^3 * a^2 = (a * a^2) * a^2 = a * (a^2 * a^2) = a * a^4$$

where, in an intermediate step, we used the right-alternative law with $$x = a, y = a^2$$.

We finally show that $$a * a^4 = a^2 * a^3$$. Note that:

$$\! a * a^4 = a * (a * a^3) = (a * a) * a^3 = a^2 * a^3$$

where, in an intermediate step, we used the left-alternative law with $$x = a, y = a^3$$.

This completes the proof.