Inverse of Algebraic Structure Isomorphism is Isomorphism

Theorem
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.

Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a mapping.

Then $\phi$ is an isomorphism $\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ is also an isomorphism.

Proof
Let $\phi$ be an isomorphism.

Then by definition $\phi$ is a bijection.

Thus $\exists \phi^{-1}$ such that $\phi^{-1}$ is also a bijection from Bijection iff Inverse is Bijection.

That is:
 * $\exists \phi^{-1}: \struct {T, *} \to \struct {S, \circ}$

It follows that:

So $\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ is a homomorphism.

$\phi^{-1}$ is also (from above) a bijection.

Thus, by definition, $\phi^{-1}$ is an isomorphism.

Let $\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ be an isomorphism.

Applying the same result as above in reverse, we have that $\paren {\phi^{-1} }^{-1}: \struct {S, \circ} \to \struct {T, *}$ is also an isomorphism.

But by Inverse of Inverse of Bijection:
 * $\paren {\phi^{-1} }^{-1} = \phi$

and hence the result.