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
Suppose the existence of:
 * $S \subseteq \N: n_0 \in S \land \left({\left({\N_k \setminus \N_{n_0}}\right) \subseteq S \implies \left({\N_{k+1} \setminus \N_{n_0}}\right) \subseteq S}\right)$

such that $S \supsetneq \N \setminus \N_{n_0}$.

Now, consider the set $T = \left({\N \setminus \N_{n_0}}\right) \setminus S$.

We know $T \ne \varnothing$ from our assumption $S \supsetneq \N \setminus \N_{n_0}$ and Set Difference with Superset is Empty Set.

By the Well-Ordering Principle, $T$ must contain a smallest element: call it $a$.

We have that:
 * $n_0 \in S \implies n_0 \notin T \implies a \ne n_0 \implies a > n_0$

From the definition of $T$, we know that $a \notin S$, and as $a$ is the least element of $T$, it follows that:
 * $\forall k \in \N: n_0 \le k < a: \left({\N_k \setminus \N_{n_0}}\right) \subseteq S$

Now consider the set $\N_b: 0 < b = a-1 < a$.

As $b < a$ and $a$ is the least element of $T$, $b \notin T \implies b \in S \implies \left({\N_b \setminus \N_{n_0}}\right) \subseteq S$.

But from the property of $S$ that $\left({\N_k \setminus \N_{n_0}}\right) \subseteq S \implies \left({\N_{k+1} \setminus \N_{n_0}}\right) \subseteq S$, we have:
 * $\left({\N_b \setminus \N_{n_0}}\right) \subseteq S \implies \left({\N_{b+1} \setminus \N_{n_0}}\right) \subseteq S \implies \left({\N_a \setminus \N_{n_0}}\right) \subseteq S \implies a \in S$

So we have obtained a contradiction, and the assumption we made that $T \ne \varnothing$ is false, and therefore:
 * $\left({\N \setminus \N_{n_0}}\right) \setminus S = \varnothing \implies \left({\N \setminus \N_{n_0}}\right) \subseteq S$

from Set Difference with Superset is Empty Set.

Hence the result.