Second Principle of Mathematical Induction

Theorem
Let $P \left({n}\right)$ be a propositional function depending on $n \in \N$.

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

Suppose that:


 * $(1): \quad P \left({n_0}\right)$ is true
 * $(2): \quad \forall k \in \N: k \ge n_0: P \left({n_0}\right) \land P \left({n_0 + 1}\right) \land \ldots \land P \left({k-1}\right) \land P \left({k}\right) \implies P \left({k+1}\right)$

Then:


 * $P \left({n}\right)$ is true for all $n \ge n_0$.

This process is called proof by (mathematical) induction.

Proof
Suppose that the two given conditions hold:


 * $(1): \quad P \left({n_0}\right)$ is true
 * $(2): \quad \forall k \in \N: k \ge n_0: P \left({n_0}\right) \land P \left({n_0 + 1}\right) \land \ldots \land P \left({k-1}\right) \land P \left({k}\right) \implies P \left({k+1}\right)$

Let $S = \left\{{x \in \N: P \left({x}\right)}\right\}$.

That is, the set of all $x \in \N$ for which $P \left({x}\right)$ holds.

From Subset of Set with Propositional Function, $S \subseteq \N$.

We have that $n_0 \in S$ from $(1)$.

Let $k \in \N$.

Now suppose that $\forall j \in \N: n_0 \le j \le k: j \in S$.

That is, $P \left({n_0}\right), P \left({n_0 + 1}\right), P \left({n_0 + 2}\right), \ldots, P \left({k}\right)$ all hold.

Then that means $\left({\N_k \setminus \N_{n_0}}\right) \subseteq S$

From $(2)$ it follows that $P \left({k + 1}\right)$ holds, and so $\left({\N_{k + 1} \setminus \N_{n_0}}\right) \subseteq S$.

Thus we have established:
 * $S \subseteq \N$
 * $n_0 \in S$
 * $\left({\N_k \setminus \N_{n_0}}\right) \subseteq S \implies \left({\N_{k+1} \setminus \N_{n_0}}\right) \subseteq S$

From the Second Principle of Finite Induction it follows that $\N \setminus \N_{n_0} \subseteq S$.

That is, for every element $k$ of $\N \setminus \N_{n_0}$, it follows that $P \left({k}\right)$ holds.

But $\N \setminus \N_{n_0}$ is precisely the set of all $n \in \N$ such that $n \ge n_0$.

Hence the result.