Intersection of Relations Compatible with Operation is Compatible

Theorem
Let $\left({S, \circ}\right)$ be a closed algebraic structure.

Let $\mathscr F$ be a indexed family of relations on $S$.

Suppose that each element of $\mathscr F$ is compatible with $\circ$.

Let $\mathcal Q = \bigcap \mathscr F$ be the intersection of $\mathscr F$.

Then $\mathcal Q$ is a relation compatible with $\circ$.

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

Suppose that $x \mathrel{\mathcal Q} y$.

Then for each $\mathcal R \in \mathscr F$:


 * $x \mathrel{\mathcal R} y$

Then since $\mathcal R$ is a relation compatible with $\circ$:


 * $\left({x \circ z}\right) \mathrel{\mathcal R} \left({y \circ z}\right)$

Since this holds for each $\mathcal R \in \mathscr F$:


 * $\left({x \circ z}\right) \mathrel{\mathcal Q} \left({y \circ z}\right)$

We have shown that:


 * $\forall x, y, z \in S: x \mathrel{\mathcal Q} y \implies \left({x \circ z}\right) \mathrel{\mathcal Q} \left({y \circ z}\right)$

A precisely similar argument shows that:


 * $\forall x, y, z \in S: x \mathrel{\mathcal Q} y \implies \left({z \circ x}\right) \mathrel{\mathcal Q} \left({z \circ y}\right)$,

so $Q$ is a relation compatible with $\circ$.