Min Operation is Commutative

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.

$\blacksquare$

Also see

• Max Operation is Commutative