Extension Theorem for Distributive Operations/Associativity

Theorem
Then:
 * If $\circ$ is associative, 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 associative.

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

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

Therefore:
 * $\forall x \in T, n, p \in R: \paren {x \circ' n} \circ' p = x \circ' \paren {n \circ' p}$

Similarly, for all $x \in T, p \in R$, the mappings:

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

Therefore:
 * $\forall x, y \in T, p \in R: \paren {x \circ' y} \circ' p = x \circ' \paren {y \circ' p}$

Finally, for all $x, y \in T$, 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 associative.