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:

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

Proof

\(\ds \paren {x \sqbrk c y} \sqbrk c z\)

\(=\)

\(\ds c \sqbrk c z\)

Definition of Constant Operation

\(\ds \)

\(=\)

\(\ds c\)

Definition of Constant Operation

\(\ds x \sqbrk c \paren {y \sqbrk c z}\)

\(=\)

\(\ds x \sqbrk c c\)

Definition of Constant Operation

\(\ds \)

\(=\)

\(\ds c\)

Definition of Constant Operation

$\blacksquare$