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.
$\blacksquare$