Inductive Semigroup whose Inductive Elements Commute is Commutative Semigroup

Theorem
Let $\struct {S, \circ}$ be a semigroup.

Let there exist $\alpha, \beta \in S$ which fulfil the condition for $\struct {S, \circ}$ to be an inductive semigroup:
 * the only subset of $S$ containing both $\alpha$ and $x \circ \beta$ whenever it contains $x$ is $S$ itself.

Let $\alpha$ and $\beta$ commute.

Then $\struct {S, \circ}$ is a commutative semigroup.

Proof
Suppose $\struct {S, \circ}$ is a semigroup.

Suppose there exist $\alpha, \beta \in S$ such that $\struct {S, \circ}$ is an inductive semigroup.

That is, suppose there exist $\alpha, \beta \in S$ such that the only subset of $S$ containing both $\alpha$ and $x \circ \beta$ whenever it contains $x$ is $S$ itself.

That is:
 * $\exists \alpha, \beta \in S: \forall A \subseteq S: \paren {\alpha \in A \land \paren {\forall x \in A: x \circ \beta \in A} } \implies A = S$

By Form of Elements of Inductive Semigroup, the elements of $S$ are of the form:
 * $\alpha \circ \beta \circ \beta \circ \cdots \circ \beta$

Then:
 * $\forall x, y \in S: x \circ y = y \circ x$

Hence, $\struct {S, \circ}$ is a commutative semigroup, as required.