Intersection of Relations Compatible with Operation is Compatible

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

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

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

Let $\mathcal Q = \bigcap \mathscr F$.

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

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

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

Then for each $\mathcal R \in \mathscr F$,
 * $x \mathop{\mathcal R} y$.

Then since $\mathcal R$ is a relation compatible with $\circ$,
 * $\left({x \circ z}\right) \mathop {\mathcal R} \left({y \circ z}\right)$.

Since this holds for each $\mathcal R \in \mathscr F$,
 * $\left({x \circ z}\right) \mathop {\mathcal Q} \left({y \circ z}\right)$.

We have shown that
 * $\forall x,y,z \in S: x \mathop {\mathcal Q} y \implies \left({x \circ z}\right) \mathop {\mathcal Q} \left({y \circ z}\right)$.

A precisely similar argument shows that
 * $\forall x, y, z \in S: x \mathop {\mathcal Q} y \implies \left({z \circ x}\right) \mathop {\mathcal Q} \left({z \circ y}\right)$,

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