Relation Induced by Strict Positivity Property is Compatible with Addition

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 compatible with $+$, that is:

$\forall x, y, z \in D: x < y \implies \paren {x + z} < \paren {y + z}$
$\forall x, y, z \in D: x < y \implies \paren {z + x} < \paren {z + y}$

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 $a < b$:

 $\ds$  $\ds a < b$ $\ds$ $\leadsto$ $\ds \map P {-a + b}$ by definition of $<$ $\ds$ $\leadsto$ $\ds \map P {-a + b + \paren {-c} + c}$ for any $c \in D$ $\ds$ $\leadsto$ $\ds \map P {\paren {-a + \paren {-c} } + \paren {b + c} }$ properties of $+$ in $D$ $\ds$ $\leadsto$ $\ds \map P {-\paren {a + c} + \paren {b + c} }$ properties of $+$ in $D$ $\ds$ $\leadsto$ $\ds a + c < b + c$ by definition of $<$ $\ds$ $\leadsto$ $\ds c + a < c + b$ commutativity of $+$

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

$\blacksquare$