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