Opposite Ring is Ring

Theorem
Let $R = \struct {S, +, \circ}$ be a ring.

Let $R^O$ be the opposite ring.

Then $R^O$ is a ring.

Proof
By definition of the opposite ring:
 * $R^O = \struct {S, +, *}$

where $* : S \times S \to S$ is defined by:
 * $\forall x, y \in S: x * y = y \circ x$.

By definition of the ring $R$, $\struct {S, +}$ is an abelian group.

To complete the proof, it remains to be shown that $\struct{S, *}$ is a semigroup.

That is, it remains to show that $\struct{S, *}$ is associative.

Let $a, b, c \in S$.

The result follows.