Second Principle of Mathematical Induction/One-Based/Proof 2

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Let $\map P n$ be a propositional function depending on $n \in \N_{>0}$.

Suppose that:

$(1): \quad \map P 1$ is true
$(2): \quad \forall k \in \N_{>0}: \map P 1 \land \map P 2 \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$


$\map P n$ is true for all $n \in \N_{>0}$.


Let $S \subseteq \N_{>0}$ containing those $n \in \N_{>0}$ for which $\map P n$ does not hold.

Aiming for a contradiction, suppose $S \ne \O$.

Then by the Well-Ordering Principle $S$ contains a minimal element $s$.

We have that $s \ne 1$ because $\map P 1$ is true from $(1)$.

Thus there must exist some $k \in \N_{>0}$ such that $s = k + 1$.

As $k + 1$ is the minimal element of $S$ it follows that all $n$ such that $n < k + 1 = s$ are not in $S$

But this means that $\map P n$ is true for all $n < s$

But by $(2)$ it follows that $\map P s$ is true.

That is, $s \notin S$.

This contradicts our assertion that $s \in S$.

Hence our assumption that $S \ne \O$ is false.

It follows by Proof by Contradiction that $S = \O$.

So $\map P n$ holds for all $n \in \N_{>0}$.