# Principle of Finite Induction

## Theorem

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

Let $n_0 \in \Z$ be given.

Suppose that:

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

- $(2): \quad \forall n \ge n_0: n \in S \implies n + 1 \in S$

Then:

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

That is:

- $S = \set {n \in \Z: n \ge n_0}$

The **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:

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

- $(2): \quad \forall n \in \N : n \in S \implies n + 1 \in S$

Then:

- $S = \N$

### One-Based

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

Suppose that:

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

- $(2): \quad \forall n \in \N_{>0} : n \in S \implies n + 1 \in S$

Then:

- $S = \N_{>0}$

## Proof 1

Let $\Z_{\ge n_0} := \set {n \in \Z: n \ge n_0}$.

Aiming for a contradiction, suppose $S \ne \Z_{\ge n_0}$.

Let $S' = \Z_{\ge n_0} \setminus S$.

Because $S \ne \Z_{\ge n_0}$ and $S \subseteq \Z_{\ge n_0}$, we have that $S' \ne \O$.

By definition, $\Z_{\ge n_0}$ is bounded below by $n_0$.

From Set of Integers Bounded Below by Integer has Smallest Element, $S'$ has a minimal element.

Let $k$ be this minimal element of $S'$.

By $(1)$ we have that:

- $n_0 \in S$

and so:

- $n_0 \notin S'$

Hence:

- $k \ne n_0$

and so:

- $k > n_0$

It follows that:

- $k - 1 \le n_0$

Because $k$ is the minimal element of $S'$:

- $k - 1 \notin S'$

and so:

- $k - 1 \in S$

But by $(2)$:

- $\paren {k - 1} + 1 = k \in S$

So we have:

- $k \in S$

and:

- $k \notin S$

Hence by Proof by Contradiction $S = \Z_{\ge n_0}$.

$\blacksquare$

## Proof 2

The validity of the material on this page is questionable.This only takes on board a subset of $\N$, where we need a subset of $\Z$You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Questionable}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

Consider $\N$ defined as a naturally ordered semigroup.

The result follows directly from Principle of Mathematical Induction for Naturally Ordered Semigroup: General Result.

$\blacksquare$

## Contexts

The Principle of Finite Induction can be introduced in a formal development of abstract algebra or mathematical logic in various contexts, and proved from first principles in each.

### Peano Structure

Let $\struct {P, s, 0}$ be a Peano structure.

Let $S \subseteq P$.

Suppose that:

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

- $(2): \quad \forall n: n \in S \implies \map s n \in S$

Then:

- $S = P$

## 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 made that $n \in S$ for some $n \in \Z$ is the **induction hypothesis**.

### Induction Step

The step which shows that $n \in S \implies n + 1 \in S$ is called the **induction step**.

## Also known as

Some sources refer to the **Principle of Finite Induction** as the **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.

Some sources refer to the **Principle of Finite Induction** as the ** First Principle of (Finite) Induction**, to distinguish it from the

**Second Principle of Finite Induction**.

Some call it the **induction principle**.

The abbreviation **PFI** is often seen.

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

Such sources may similarly refer to the **Second Principle of Finite Induction** as the **Principle of Strong (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 **Principle of Finite Induction** is often referred to as **Proof by (Finite) Induction**.

## Also see

## Sources

- 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 5$ Applications to natural numbers: Exercise $5.1$

This page may be the result of a refactoring operation.As such, the following source works, along with any process flow, will need to be reviewed. When this has been completed, the citation of that source work (if it is appropriate that it stay on this page) is to be placed above this message, into the usual chronological ordering.If you have access to any of these works, then you are invited to review this list, and make any necessary corrections.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{SourceReview}}` from the code. |

- 1972: A.G. Howson:
*A Handbook of Terms used in Algebra and Analysis*... (previous) ... (next): $\S 4$: Number systems $\text{I}$: Peano's Axioms - 2008: Paul Halmos and Steven Givant:
*Introduction to Boolean Algebras*... (previous) ... (next): Appendix $\text{A}$: Set Theory: Induction