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

## Theorem

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

Suppose that:

- $(1): \quad \map P 1$ is true

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

Then:

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

## Proof

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 $n \in \N_{>0}$.

$\blacksquare$

## Sources

- 1951: Nathan Jacobson:
*Lectures in Abstract Algebra: I. Basic Concepts*... (previous) ... (next): Introduction $\S 4$: The natural numbers - 1964: W.E. Deskins:
*Abstract Algebra*... (previous) ... (next): $\S 2.3$: Theorem $2.18$ - 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $3$: The Integers: $\S 9$. The Principles of Induction: Theorem $15$ - 2000: James R. Munkres:
*Topology*(2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Theorem $4.2$