Second Principle of Finite Induction

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

Let $n_0 \in \N$ be given.

Suppose that:


 * $(1): \quad n_0 \in S$


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

Then:


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

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

The second principle of mathematical induction is usually stated and demonstrated for $n_0$ being either $0$ or $1$.

This is often dependent upon whether the analysis of the fundamentals of mathematical logic are zero-based or one-based.

Proof
Define $T$ as:


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

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.