Principle of Finite Induction/One-Based/Proof 2

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

Suppose that:

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


$S = \N_{>0}$


Let $T$ be the set of all $1$-based natural numbers not in $S$:

$T = \N_{>0} \setminus S$

Aiming for a contradiction, suppose $T$ 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 $1 \in S$.


$a > 1$

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:

$\paren {a - 1} + 1 \in S$


$\paren {a - 1} + 1 = a$

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:

$S = \N_{>0}$