Ring Homomorphism Preserves Subrings/Proof 3

Proof
Let $S$ be a subring of $R_1$.

Since $S \ne \O$ it follows that $\phi \sqbrk S \ne \O$.

Let $x, y \in \phi \sqbrk S$.

Then:
 * $\exists s, t \in S: x = \map \phi s, y = \map \phi t$

So:

As $S$ is a subring of $R_1$, it is closed under $+_1$ and the taking of negatives.

Thus $s +_1 \paren {-t} \in S$ and so $x +_2 \paren {-y} \in \phi \sqbrk S$.

Similarly:

Because $S$ is a subring of $R_1$, it is closed under $\circ_1$.

Thus $s \circ_1 t \in S$ and so $x \circ_2 y \in \phi \sqbrk S$.

The result follows from Subring Test.