User:Dfeuer/Multiplying Compatible Relationship by Zero-Related Element

Note: this theorem generalizes the sixth property of ordered rings and a strict total version for ordered integral domains to compatible relations.

Theorem
Let $(R,+,\circ)$ be a ring with zero $0_R$.

Let $\mathcal R$ be a relation compatible with $(R,+,\circ)$.

Let $x,y,z\in R$ such that:
 * $x \mathrel{\mathcal R} y$
 * $0_R \mathrel{\mathcal R} z$

Then:
 * $(x \circ z) \mathrel{\mathcal R} (y \circ z)$
 * $(z \circ x) \mathrel{\mathcal R} (z \circ y)$

Corollary
Let $(R,+,\circ,<)$ be a strictly ordered ring with zero $0_R$.

Let $x,y,z \in R$ such that:
 * $x < y$
 * $0 < z$

Then
 * $x \circ z < y \circ z$
 * $z \circ x < z \circ y$

Proof
Since $x \mathrel{\mathcal R} y$, by the definition of compatibility:
 * $0_R \mathrel{\mathcal R} (y + (-x))$

By the definition of compatibility:
 * $0_R \mathrel{\mathcal R} ((y + (-x)) \circ z)$
 * $0_R \mathrel{\mathcal R} (z \circ (y + (-x)))$

Since $\circ$ distributes over $+$:
 * $0_R \mathrel{\mathcal R} (y \circ z + (-x) \circ z)$
 * $0_R \mathrel{\mathcal R} (z \circ y + z \circ (-x))$

By Product with Ring Negative:
 * $-((-x) \circ z) = x \circ z$
 * $-(z \circ (-x)) = z \circ x$

Thus by compatibility:
 * $(x \circ z) \mathrel{\mathcal R} (y \circ z)$
 * $(z \circ x) \mathrel{\mathcal R} (z \circ y)$