Construction of Inverse Completion/Image of Quotient Mapping is Subsemigroup

Theorem
Let the mapping $\psi: S \to T'$ be defined as:
 * $\forall x \in S: \psi \left({x}\right) = \left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_\boxtimes$

Let $S'$ be the image $\psi \left({S}\right)$ of $S$.

Then $\left({S', \oplus'}\right)$ is a subsemigroup of $\left({T', \oplus'}\right)$.

Proof
We have that $S'$ is the image $\psi \left({S}\right)$ of $S$.

For $\left({S', \oplus'}\right)$ to be a subsemigroup of $\left({T', \oplus'}\right)$, by Subsemigroup Closure Test we need to show that $\left({S', \oplus'}\right)$ is closed.

Let $x, y \in S'$.

Then $x = \phi \left({x'}\right), y = \phi \left({y'}\right)$ for some $x', y' \in S$.

But as $\phi$ is an isomorphism, it obeys the morphism property.

So $x \oplus' y = \phi \left({x'}\right) \oplus' \phi \left({y'}\right) = \phi \left({x' \circ y'}\right)$.

Hence $x \oplus' y$ is the image of $x' \circ y' \in S$ and hence $x \oplus' y \in S'$.

Thus by the Subsemigroup Closure Test, $\left({S', \oplus'}\right)$ is a subsemigroup of $\left({T', \oplus'}\right)$