Sum Less Maximum is Minimum

Theorem

For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:

$a + b - \max \left({a, b}\right) = \min \left({a, b}\right)$.

Proof

From Sum of Maximum and Minimum we have that $a + b = \max \left({a, b}\right) + \min \left({a, b}\right)$.

Thus $a + b - \max \left({a, b}\right) = \min \left({a, b}\right)$ follows by subtracting $\max \left({a, b}\right)$ 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 $a + b \ge \max \left({a, b}\right)$ follows immediately again from $a + b = \max \left({a, b}\right) + \min \left({a, b}\right)$.

$\blacksquare$