Principle of Finite Induction/One-Based/Proof 2

Proof
Let $T$ be the set of all $1$-based natural numbers not in $S$:
 * $T = \N_{>0} \setminus S$

$T$ is non-empty.

From the Well-Ordering Principle, $T$ has a smallest element.

Let this smallest element be denoted $a$.

We have been given that $1 \in S$.

So:
 * $a > 1$

and so:
 * $0 < a - 1 < a$

As $a$ is the smallest element of $T$, it follows that:
 * $a - 1 \notin T$

That means $a - 1 \in S$.

But then by hypothesis:
 * $\paren {a - 1} + 1 \in S$

But:
 * $\paren {a - 1} + 1 = a$

and so $a \notin T$.

This contradicts our assumption that $a \in T$.

It follows by Proof by Contradiction that $T$ has no such smallest element.

Hence it follows that $T$ can have no elements at all.

That is:
 * $\N_{>0} \setminus S = \O$

That is:
 * $S = \N_{>0}$