Extension Theorem for Distributive Operations/Commutativity

Theorem
Then:
 * If $\circ$ is commutative, then so is $\circ'$

Proof
By hypothesis, all the elements of $\struct {R, *}$ are cancellable.

Thus Inverse Completion of Commutative Semigroup is Abelian Group can be applied.

So $\struct {T, *}$ is an abelian group.

Suppose $\circ$ is commutative.

As $\circ'$ distributes over $*$, for all $n \in R$, the mappings:

are endomorphisms of $\struct {T, *}$ that coincide on $R$ by the commutativity of $\circ$ and hence are the same mapping.

Therefore $\forall x \in T, n \in R: x \circ' n = n \circ' x$.

Finally, for all $y \in T$, the mappings:

are endomorphisms of $\struct {T, *}$ that coincide on $R$ by what we have proved and hence are the same mapping.

Therefore $\circ'$ is commutative.