Principle of Least Counterexample

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

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

(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 $$\lnot P \left({n}\right)$$.

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

Since $$\lnot \left({\forall n \ge m: P \left({n}\right)}\right)$$, $$S \ne \varnothing$$.

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

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