Min is Half of Sum Less Absolute Difference

Theorem
For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:
 * $\min \set {a, b} = \dfrac 1 2 \paren {a + b - \size {a - b} }$

Proof
From the definition of min:
 * $\map \min {a, b} = \begin{cases}

a: & a \le b \\ b: & b \le a \end{cases}$

Let $a < b$.

Then:

Let $a \ge b$.

Then: