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 $\struct {R, +, \circ}$ be a ring with zero $0_R$.

Let $\RR$ be a relation compatible with $\struct {R, +, \circ}$.

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

Then:
 * $\paren {x \circ z} \mathrel \RR \paren {y \circ z}$
 * $\paren {z \circ x} \mathrel \RR \paren {z \circ y}$

Corollary
Let $\struct {R, +, \circ, <}$ 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 $x \mathrel \RR y$, by the definition of compatibility:
 * $0_R \mathrel \RR \paren {y + \paren {-x} }$

By the definition of compatibility:
 * $0_R \mathrel \RR \paren {\paren {y + \paren {-x} } \circ z}$
 * $0_R \mathrel \RR \paren {z \circ \paren {y + \paren {-x} } }$

Since $\circ$ distributes over $+$:
 * $0_R \mathrel \RR \paren {y \circ z + \paren {-x} \circ z}$
 * $0_R \mathrel \RR \paren {z \circ y + z \circ \paren {-x} }$

By Product with Ring Negative:
 * $-\paren {\paren {-x} \circ z} = x \circ z$
 * $-\paren {z \circ \paren {-x} } = z \circ x$

Thus by compatibility:
 * $\paren {x \circ z} \mathrel \RR \paren {y \circ z}$
 * $\paren {z \circ x} \mathrel \RR \paren {z \circ y}$