Sum Less Minimum is Maximum

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

Proof
From Sum of Maximum and Minimum we have that $a + b = \max \set {a, b} + \min \set {a, b}$.

Thus $a + b - \min \set {a, b} = \max \set {a, b}$ follows by subtracting $\min \set {a, b}$ from both sides.

It is clear that this result applies when $a, b$ in $\Z, \Q$ or $\R$, as subtraction is well-defined throughout those number sets.

However, it can be seen to apply in $\N$ as well, despite the fact that $n - m$ is defined on $\N$ only when $n \ge m$.

This is because the fact that $a + b \ge \min \set {a, b}$ follows immediately again from $a + b = \max \set {a, b} + \min \set {a, b}$.