Principle of Mathematical Induction/One-Based

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


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

Proof 1

Let $S$ be the set defined as:

$S := \set {n \in \N_{>0}: \map P n \text { is false} }$

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

From the Well-Ordering Principle it follows that $S$ has a minimal element $m$.

From $(1)$ we have that $\map P 1$ holds.

Hence $1 \notin S$.

Therefore $m \ne 1$.

Therefore $m - 1 \in \N_{>0}$.

But $m$ is the minimal element of $S$.

So $m - 1 \notin S$.

Therefore $\map P {m - 1}$ is true.

Hence by $(2)$ it follows that $\map P m$.

But then $m \notin S$.

This contradicts our supposition that $m \in S$.

Hence there can be no such $m \in S$.

So $S = \O$ and the result follows.


Proof 2

Let $M$ be the set of all $n \in \N_{>0}$ for which $\map P n$ holds.

By $(1)$ we have that $1 \in M$.

By $(2)$ we have that if $k \in M$ then $k + 1 \in M$.

From the Axiomatization of $1$-Based Natural Numbers, Axiom $(\text F)$, it follows that $M = \N_{>0}$.


Proof 3

We have that Natural Numbers are Non-Negative Integers.

Then we have that Integers form Well-Ordered Integral Domain.

The result follows from Induction on Well-Ordered Integral Domain.


Also see

  • Results about Proofs by Induction can be found here.