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

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:
 * The Principle of Finite Induction as PFI
 * The Principle of Complete Finite Induction as PCI.

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 $S = \N$.

Therefore PFI implies PCI.