Principle of Finite Induction/Proof 1

Proof
Let $\Z_{\ge n_0} := \set {n \in \Z: n \ge n_0}$.

$S \ne \Z_{\ge n_0}$.

Let $S' = \Z_{\ge n_0} \setminus S$.

Because $S \ne \Z_{\ge n_0}$ and $S \subseteq \Z_{\ge n_0}$, we have that $S' \ne \O$.

By definition, $\Z_{\ge n_0}$ is bounded below by $n_0$.

From Set of Integers Bounded Below by Integer has Smallest Element, $S'$ has a minimal element.

Let $k$ be this minimal element of $S'$.

By $(1)$ we have that:
 * $n_0 \in S$

and so:
 * $n_0 \notin S'$

Hence:
 * $k \ne n_0$

and so:
 * $k > n_0$

It follows that:
 * $k - 1 \le n_0$

Because $k$ is the minimal element of $S'$:
 * $k - 1 \notin S'$

and so:
 * $k - 1 \in S$

But by $(2)$:
 * $\paren {k - 1} + 1 = k \in S$

So we have:
 * $k \in S$

and:
 * $k \notin S$

Hence by Proof by Contradiction $S = \Z_{\ge n_0}$.