Relation Induced by Strict Positivity Property is Compatible with Addition

Theorem
Let $\left({D, +, \times}\right)$ be an ordered integral domain where $P$ is the positivity property.

Let the relation $<$ be defined on $D$ as:


 * $\forall a, b \in D: a < b \iff P \left({-a + b}\right)$

Then $<$ is compatible with $+$, that is:


 * $\forall x, y, z \in D: x < y \implies \left({x + z}\right) < \left({y + z}\right)$
 * $\forall x, y, z \in D: x < y \implies \left({z + x}\right) < \left({z + y}\right)$

Corollary
Let $\le$ be the relation defined on $D$ as:
 * $\le \ := \ < \cup \Delta_D$

where $\Delta_D$ is the diagonal relation.

Then $\le$ is compatible with $+$.

Proof
Let $a < b$:

And so $<$ is seen to be compatible with $+$.

Proof of Corollary
Let $a \le b$.

If $a \ne b$ then $a < b$ and the main result applies.

Otherwise $a = b$.

But $\left({D, +}\right)$ is the additive group of $\left({D, +, \times}\right)$ and the Cancellation Laws apply:
 * $a + c = b + c \iff a = b \iff c + a = c + b$

So $\le$ is seen to be compatible with $+$.