Relation Induced by Strict Positivity Property is Asymmetric and Antireflexive

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 asymmetric and antireflexive.

Proof
From the trichotomy law of ordered integral domains, for all $x \in D$ exactly one of the following applies:


 * $(1): \quad P \left({x}\right)$
 * $(2): \quad P \left({- x}\right)$
 * $(3): \quad x = 0$

Let $a, b \in D: a < b$.

Then by definition, $P \left({-a + b}\right)$.

Suppose also that $b < a$.

Then by definition, $P \left({-b + a}\right)$ and so $P \left({-\left({-a + b}\right)}\right)$.

But then that contradicts the trichotomy law of ordered integral domains:

Either:
 * $(1): \quad P \left({-a + b}\right)$

or:
 * $(2): \quad P \left({- \left({-a + b}\right)}\right)$

So $(1)$ and $(2)$ can not both apply, and so if $a < b$ it is not possible that $b < a$.

Thus $<$ is asymmetric.

Now suppose $\exists a \in D: a < a$.

Then that would mean $P \left({-a + a}\right)$, that is, $P \left({0}\right)$.

But this also contradicts the conditions of the trichotomy law of ordered integral domains.

Hence:
 * $\forall a \in D: a \not < a$

and so $<$ is antireflexive.