Constant Operation is Commutative

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Theorem

Let $S$ be a set.

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


Then $\sqbrk c$ is a commutative operation:

$\forall x, y \in S: x \sqbrk c y = y \sqbrk c x$


Proof

\(\ds x \sqbrk c y\) \(=\) \(\ds c\) Definition of Constant Operation
\(\ds y \sqbrk c x\) \(=\) \(\ds c\) Definition of Constant Operation
\(\ds \leadsto \ \ \) \(\ds x \sqbrk c y\) \(=\) \(\ds y \sqbrk c x\)


Hence the result by definition of commutativity.

$\blacksquare$