Principle of Least Counterexample

Theorem
Let $\map P n$ be a condition on $n \in \set {x \in \Z: x \ge m \in \Z}$.

Suppose that:
 * $\neg \paren {\forall n \ge m: \map P n}$

(That is, not all $n \ge m$ satisfy $\map P n$.)

Then there exists a least counterexample, that is a smallest integral value of $n$ for which $\neg \map P n$.

Proof
Let $S = \set {n \in \Z: n \ge m \in \Z: \neg \map P n}$.

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

Since:
 * $\neg \paren {\forall n \ge m: \map P n}$

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

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

Therefore $S$ has a smallest element, which proves the result.

Also known as
Some sources refer to the least counterexample as the least rascal.