Quotient Mapping of Inverse Completion

Theorem
Let $$\left({T, \circ'}\right)$$ be an inverse completion of a commutative semigroup $$\left({S, \circ}\right)$$, where $$C$$ is the set of cancellable elements of $$S$$.

Let $$f: S \times C: T$$ be the mapping defined as:

$$\forall x \in S, y \in C: f \left({x, y}\right) = x \circ' y^{-1}$$

Let $$\mathcal{R}_f$$ be the equivalence relation induced by $f$.

Then:

$$\left({x_1, y_1}\right) \mathcal{R}_f \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$$

Proof
By the definition of $$\mathcal{R}_f$$:

$$\left({x_1, y_1}\right) \mathcal{R}_f \left({x_2, y_2}\right) \iff x_1 \circ' y_1^{-1} = x_2 \circ' y_2^{-1}$$

Now:

... which leads us to:

$$\left({x_1, y_1}\right) \mathcal{R}_f \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$$