Min is Half of Sum Less Absolute Difference

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

Proof
From the definition of min:
 * $\min \left({a, b}\right) = \begin{cases}

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


 * Let $a < b$.

Then:


 * Let $a \ge b$.

Then: