Relation Induced by Strict Positivity Property is Asymmetric and Antireflexive

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

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


 * $\forall a, b \in D: a < b \iff \map P {-a + b}$

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:


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

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

Then by definition:
 * $\map P {-a + b}$

Suppose also that $b < a$.

Then by definition:
 * $\map P {-b + a}$

and so:
 * $\map P {-\paren {-a + b} }$

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

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

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

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 $\map P {-a + a}$, that is, $\map P {0_D}$.

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.