Min Operation is Commutative

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Theorem

The Min operation is commutative:

$\map \min {x, y} = \map \min {y, x}$


Proof

To simplify our notation:

Let $\map \min {x, y}$ be (temporarily) denoted $x \underline \vee y$.


There are three cases to consider:

$(1): \quad x \le y$
$(2): \quad y \le x$
$(3): \quad x = y$


$(1): \quad$ Let $x \le y$.

Then:

\(\ds x \underline \vee y\) \(=\) \(\ds x\) \(\ds =y \underline \vee x\)


$(2): \quad$ Let $y \le x$.

Then:

\(\ds x \underline \vee y\) \(=\) \(\ds y\) \(\ds =y \underline \vee x\)


$(3): \quad$ Let $x = y$.

Then:

\(\ds x \underline \vee y\) \(=\) \(\ds x = y\) \(\ds =y \underline \vee x\)


Thus $\underline \vee$, i.e. $\min$, has been shown to be commutative in all cases.

$\blacksquare$


See also