Inverse of Product of Subsets of Group/Proof 2

Superset
We will show that $\forall z \in Y^{-1} \circ X^{-1}: z \in \left({X \circ Y}\right)^{-1}$, from which:
 * $Y^{-1} \circ X^{-1} \subseteq \left({X \circ Y}\right)^{-1}$

Let $z \in Y^{-1} \circ X^{-1}$.

By the definition of subset product:
 * $\exists x' \in X^{-1}, y' \in Y^{-1}: z = y' \circ x'$

Then by Inverse of Group Product:
 * $(2)\quad z^{-1} = x'^{-1} \circ y'^{-1}$

By the definition of inverse of subset:
 * $x'^{-1} \in X$ and $y'^{-1} \in Y$

By the definition of subset product:
 * $x'^{-1} \circ y'^{-1} \in X \circ Y$

Thus by $(2)$:
 * $z^{-1} \in X \circ Y$

By the definition of inverse of subset:
 * $z \in \left({X \circ Y}\right)^{-1}$

Subset
We will show that $\forall z \in \left({X \circ Y}\right)^{-1}: z \in Y^{-1} \circ X^{-1}$, from which:
 * $\left({X \circ Y}\right)^{-1} \subseteq Y^{-1} \circ X^{-1}$

Let $z \in \left({X \circ Y}\right)^{-1}$.

By the definition of inverse of subset:
 * $z^{-1} \in X \circ Y$

From Inverse of Inverse of Subset of Group:
 * $z^{-1} \in \left({ X^{-1} }\right)^{-1} \circ \left({ Y^{-1} }\right)^{-1}$

Thus by the superset proof above:
 * $z^{-1} \in \left({ Y^{-1} \circ X^{-1} }\right)^{-1}$

From the definition of inverse of subset:
 * $z \in Y^{-1} \circ X^{-1}$