Principle of Least Counterexample

Theorem
Suppose $$P \left({n}\right)$$ is a condition on $$n \in \left\{{x \in \Z: x \ge m \in \Z}\right\}$$.

Suppose next that: $$\neg \left({\forall n \ge m: P \left({n}\right)}\right)$$.

(That is, not all $$n \ge m$$ satisfy $$P \left({n}\right)$$.)

Then there is a least counterexample, that is a lowest integral value of $$n$$ for which $$\neg P \left({n}\right)$$.

Proof
Let $$S = \left\{{n \in \Z: n \ge m \in \Z: \neg P \left({n}\right)}\right\}$$.

That is, $$S$$ is the set of all elements in $$\Z$$ not less than $$m$$ for which the condition is false.

Since:
 * $$\neg \left({\forall n \ge m: P \left({n}\right)}\right)$$

it follows that:
 * $$S \ne \varnothing$$

Also, $$S \subseteq \Z$$ and is bounded below (by $$m$$).

Therefore $S$ has a least member, which proves the result.