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.
$\blacksquare$