# Second Principle of Finite Induction

## Theorem

Let $S \subseteq \Z$ be a subset of the integers.

Let $n_0 \in \Z$ 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:

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

The **second principle of finite 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.

### Zero-Based

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

Suppose that:

- $(1): \quad 0 \in S$

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

Then:

- $S = \N$

### One-Based

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:

- $T = \set {n \in \Z : \forall k: n_0 \le k \le n: k \in S}$

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 $n_0 \in T$ if and only if 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:

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

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Therefore, $n + 1 \in T$.

Hence, by the Principle of Finite Induction:

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

as desired.

$\blacksquare$

## Terminology

### Basis for the Induction

The step that shows that the integer $n_0$ is an element of $S$ is called the **basis for the induction**.

### Induction Hypothesis

The assumption that $\forall k: n_0 \le k \le n: k \in S$ for some $n \in \Z$ is the **induction hypothesis**.

### Induction Step

The step which shows that $n + 1 \in S$ follows from the assumption that $k \in S$ for all values of $k$ between $n_0$ and $n$ is called the **induction step**.

## Also known as

Some sources refer to the **Second Principle of Finite Induction** as the **Second Principle of Mathematical Induction**, and gloss over the differences between the two proof techniques if they discuss them both at all.

Hence the word **finite** may well not appear in the various published expositions of this technique.

The **Second Principle of Finite Induction** is also known as the **Principle of Complete (Finite) Induction**.

Both terms are used on $\mathsf{Pr} \infty \mathsf{fWiki}$.

The abbreviations **PCI** or **PCFI** can sometimes be seen.

Some sources call it the **Principle of Strong (Finite) Induction**.

Such sources may similarly refer to the **(First) Principle of Finite Induction** as the **Principle of Weak (Finite) Induction**.

These names are misleading, as both principles are equivalent, and so neither is weaker or stronger than the other.

The process of demonstrating a proof by means of the **Second Principle of Finite Induction** is often referred to as **Proof by Complete (Finite) Induction**.

## Also see

- Results about
**Proofs by Induction**can be found**here**.

## Algorithmic Proof Procedure

The **Second Principle of Mathematical Induction** can be implemented as an algorithm as follows.

Let $\map P n$ be a propositional function depending on $n \in \N$.

Let it be established that:

- $\text{(a)}: \quad \map P 1$ is true
- $\text{(b)}: \quad$ If all of $\map P 1, \map P 2, \ldots, \map P k$ are true, then $\map P {k + 1}$ is true.

Then the following algorithm provides a proof of $\map P k$ for all $n \in \N$:

- $\mathbf 1:$ Prove $\map P 1$.
- Set $k \gets 1$ and prove $\map P 1$ according to $\text{(a)}$.

- $\mathbf 3:$ Prove $\map P {k + 1}$.

- $\mathbf 4:$ Increase $k$.
- $k \gets k + 1$ and to to step $\mathbf 2$.