# Equivalence of Well-Ordering Principle and Induction/Proof/PFI implies PCI

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## Theorem

The Principle of Finite Induction implies the Principle of Complete Finite Induction.

That is:

- Principle of Finite Induction: Given a subset $S \subseteq \N$ of the natural numbers which has these properties:
- $0 \in S$
- $n \in S \implies n + 1 \in S$

- then $S = \N$.

implies:

- Principle of Complete Finite Induction: Given a subset $S \subseteq \N$ of the natural numbers which has these properties:
- $0 \in S$
- $\set {0, 1, \ldots, n} \subseteq S \implies n + 1 \in S$

- then $S = \N$.

## Proof

To save space, we will refer to:

Let us assume that the **PFI** is true.

Let $S \subseteq \N$ which satisfy:

- $(A): \quad 0 \in S$
- $(B): \quad \set {0, 1, \ldots, n} \subseteq S \implies n + 1 \in S$.

We want to show that $S = \N$, that is, the **PCI** is true.

Let $P \paren n$ be the propositional function:

- $P \paren n \iff \set {0, 1, \ldots, n} \subseteq S$

We define the set $S'$ as:

- $S' = \set {n \in \N: P \paren n \text { is true} }$

$P \paren 0$ is true by $(A)$, so $0 \in S'$.

Assume $P \paren k$ is true where $k \ge 0$.

So $k \in S'$, and by hypothesis:

- $\set {0, 1, \ldots, k} \subseteq S$

So by $(B)$:

- $k + 1 \in S$

Thus:

- $\set {0, 1, \ldots, k, k + 1} \subseteq S$.

That last statement means $P \paren {k + 1}$ is true.

This means $k + 1 \in S'$.

Thus we have satisfied the conditions:

- $0 \in S'$
- $n \in S' \implies n + 1 \in S'$

That is, $S' = \N$, and $P \paren n$ holds for all $n \in \N$.

Hence, by definition:

- $S = \N$

So **PFI** gives that $S = \N$.

$\blacksquare$

## Sources

- 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers