Composite of Isomorphisms is Isomorphism/Algebraic Structure

Theorem
Let: be algebraic structures.
 * $\left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right)$
 * $\left({S_2, *_1, *_2, \ldots, *_n}\right)$
 * $\left({S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}\right)$

Let: be isomorphisms.
 * $\phi: \left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({S_2, *_1, *_2, \ldots, *_n}\right)$
 * $\psi: \left({S_2, *_1, *_2, \ldots, *_n}\right) \to \left({S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}\right)$

Then the composite of $\phi$ and $\psi$ is also an isomorphism.

Proof
If $\phi$ and $\psi$ are both isomorphisms, then they are by definition:
 * homomorphisms
 * bijections.

So:


 * From Composite of Homomorphisms in Algebraic Structure we have that $\phi \circ \psi$ and $\psi \circ \phi$ are both homomorphisms
 * From Composite of Bijections we have that $\phi \circ \psi$ and $\psi \circ \phi$ are both bijections

... and hence by definition also isomorphisms.