Principle of Mathematical Induction/Peano Structure

Theorem
Let $\struct {P, s, 0}$ be a Peano structure.

Let $\map Q n$ be a propositional function depending on $n \in P$.

Suppose that:
 * $(1): \quad \map Q 0$ is true


 * $(2): \quad \forall n \in P: \map Q n \implies \map Q {\map s n}$

Then:
 * $\forall n \in P: \map Q n$

Proof
Let $A \subseteq P$ be defined by:
 * $A := \set {n \in P: \map Q n}$

From $(1)$, $0 \in A$.

From $(2)$:
 * $\forall n \in P: n \in A \implies \map s n \in A$

As this holds for all $n \in P$, it holds a fortiori for all $n \in A$.

Thus the condition:
 * $n \in A \implies \map s n \in A$

is satisfied.

So by Axiom $(\text P 5)$ of the Peano Axioms:
 * $A = P$

That is:
 * $\forall n \in P: \map Q n$

Also see

 * Principle of Finite Induction
 * Principle of Mathematical Induction


 * Second Principle of Mathematical Induction for Peano Structure