Homomorphism of Generated Group

Theorem
Let $$\left({G, \circ}\right)$$ and $$\left({H, \circ}\right)$$ be groups.

Let $$\phi: G \to H$$ and $$\psi: G \to H$$ be homomorphisms.

Let $$\left \langle {S} \right \rangle = G$$ be the group generated by $S$.

Let $$\forall x \in S: \phi \left({x}\right) = \psi \left({x}\right)$$

Then $$\phi = \psi$$.

Proof
Let $$H = \left\{{x \in G: \phi \left({x}\right) = \psi \left({x}\right)}\right\}$$.

From Homomorphisms on the Same Groups, $$H$$ is a subgroup of $$G$$.

But from the definition of the group generated by $S$, the smallest subgroup that contains $$S$$ is $$G$$ itself.

Thus $$G = \left\{{x \in G: \phi \left({x}\right) = \psi \left({x}\right)}\right\}$$ and so $$\phi = \psi$$.