Product of Positive Element and Element Greater than One

Theorem
Let $\left({R, +, \circ, \le}\right)$ be an ordered ring with unity $1_R$.

Let $x,y \in R$.

Suppose that $x > 0$ and $y > 1_R$.

Then $x \circ y > x$.

Proof
By the definition of an ordering compatible with a ring, $\ge$ is a relation compatible with $+$.

Hence the premise that $y > 1_R$ implies that $y + (-1_R) > 1_R+(-1_R)$.

Thus $y - 1_R > 0$.

Since $x>0$, Properties of Ordered Ring (property 6) implies that $x \circ (y - 1_R) > x \circ 0$.

By Ring Product with Zero, $x \circ 0 = 0$.

Thus $x \circ (y-1_R) > 0$.

By the definition of a ring, $\circ$ is distributive over $+$, so $x \circ y - x \circ 1_R > 0$.

By the definition of the unity of a ring, $x \circ 1_R = x$, so