Second Principle of Finite Induction/One-Based

Theorem
Let $S \subseteq \N_{>0}$ be a subset of the $1$-based natural numbers.

Suppose that:


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


 * $(2): \quad \forall n \in \N_{>0}: \paren {\forall k: 1 \le k \le n \implies k \in S} \implies n + 1 \in S$

Then:


 * $S = \N_{>0}$

Proof
Define $T$ as:


 * $T = \set {n \in \N_{>0}: \forall k: 1 \le k \le n: k \in S}$

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

Therefore, it will suffice to show that:


 * $\forall n \ge 1: n \in T$

Firstly, we have that $1 \in T$ the following condition holds:


 * $\forall k: 1 \le k \le 1 \implies k \in S$

Since $1 \in S$, it thus follows that $1 \in T$.

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


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

By $(2)$, this implies:


 * $n + 1 \in S$

Thus, we have:


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

Therefore, $n + 1 \in T$.

Hence, by the Principle of Finite Induction:


 * $\forall n \ge 1: n \in T$

That is:


 * $T = \N_{>0}$

and as $S \subseteq \N_{>0}$ it follows that:


 * $S = N_{>0}$