Constant Operation is Associative

Theorem
Let $S$ be a set.

Let $x \sqbrk c y = c$ be a constant operation on $S$.

Then $\sqbrk c$ is an associative operation:


 * $\forall x, y, z \in S: \paren {x \sqbrk c y} \sqbrk c z = x \sqbrk c \paren {y \sqbrk c z}$