Group Homomorphism Preserves Identity/Proof 3

Proof
From Group Homomorphism of Product with Inverse, we have:
 * $\forall x, y \in G: \phi \left({x \circ y^{-1}}\right) = \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1}$

Putting $x = y$ we have: