Second Principle of Finite Induction

Theorem
Let $S \subseteq \N$ be a subset of the natural numbers.

Let $n_0 \in \N$ be given. ($n_0$ is often, but not always, zero or one.)

Suppose that:


 * $(1): \quad n_0 \in S$
 * $(2): \quad \forall n \ge n_0: \left({\forall k: n_0 \le k \le n \implies k \in S}\right) \implies n + 1 \in S$

Then:


 * $\forall n \ge n_0: n \in S$

In particular, if $n_0 = 0$, then $S = \N$.

Proof
Define $T$ as:


 * $T = \left\{{n \in \N : \forall k: n_0 \le k \le n: k \in S}\right\}$

Since $n \le n$, it follows that $T \subseteq S$.

Therefore, it will suffice to show that:


 * $\forall n \ge n_0: n \in T$

Firstly, we have that $n_0 \in T$ the following condition holds:


 * $\forall k: n_0 \le k \le n_0 \implies k \in S$

Since $n_0 \in S$, it thus follows that $n_0 \in T$.

Now suppose that $n \in T$; that is:


 * $\forall k: n_0 \le k \le n \implies k \in S$

By $(2)$, this implies:


 * $n + 1 \in S$

Thus, we have:


 * $\forall k: n_0 \le k \le n + 1 \implies k \in S$

Therefore, $n + 1 \in T$.

Hence, by the Principle of Mathematical Induction:


 * $\forall n \ge n_0: n \in T$

as desired.