Complement of Relation Compatible with Group is Compatible

Theorem
Let $\struct {G, \circ}$ be a group.

Let $\RR$ be a relation on $G$.

Let $\RR$ be compatible with $\circ$.

Let $\QQ = \complement_{G \times G} \RR$, so that:
 * $\forall a, b \in G: a \mathrel \QQ b \iff \neg \paren {a \mathrel \RR b}$

Then $\QQ$ is a relation compatible with $\circ$.

Proof
Let $x, y, z \in G$.

Suppose that $\neg \paren {\paren {x \circ z} \mathrel \QQ \paren {y \circ z} }$.

Then by the definition of $\QQ$:
 * $\paren {x \circ z} \mathrel \RR \paren {y \circ z}$

Because $\RR$ is compatible with $\circ$:


 * $\paren {x \circ z} \circ z^{-1} \mathrel \RR \paren {y \circ z} \circ z^{-1}$

By and the :


 * $x \mathrel \RR y$

so by the definition of $\QQ$:


 * $\neg \paren {x \mathrel \QQ y}$

By the Rule of Transposition:
 * $\forall x, y, z \in G: x \mathrel \QQ y \implies \paren {x \circ z} \mathrel \QQ \paren {y \circ z}$

A similar argument shows that:
 * $\forall x, y, z \in G: x \mathrel \QQ y \implies \paren {z \circ x} \mathrel \QQ \paren {z \circ y}$

Thus, by definition, $\QQ$ is a relation compatible with $\circ$.