CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-02 (claude-opus-5)

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The induction axiom is independent of the other Peano axioms

Statement refuted

Refuted claim: any triple $(N, 0, \sigma)$ satisfying (P1) $0$ is not a successor and (P2) $\sigma$ is injective automatically satisfies (P3) induction. The witness is $N = \mathbb{N} \sqcup \mathbb{Z}$ (a disjoint union of a copy of the naturals and a copy of the integers), with base point $0$ the zero of the $\mathbb{N}$ -copy and $\sigma$ the usual successor $n \mapsto n + 1$ on each copy separately.

Facts & Assumptions

Given: $N = \mathbb{N} \sqcup \mathbb{Z}$ with $\sigma$ acting as successor within each copy; $0 =$ the $\mathbb{N}$ -copy's zero. Write $N_{\mathbb{N}}, N_{\mathbb{Z}}$ for the two copies.

[L1]

The Peano axioms P1, P2, P3 (Peano system).

[L2]

$\mathbb{N}$ itself is the standard model (The natural numbers $\mathbb{N}$ (von Neumann)).

Counterexample

technique · direct

1.1

P1 holds: $\sigma$ is the successor within each copy; $0$ is the $\mathbb{N}$ -copy zero, which is not the successor of any element (nothing in $N_{\mathbb{N}}$ maps to it, and $\sigma$ maps $N_{\mathbb{Z}}$ into $N_{\mathbb{Z}}$ ), so $\sigma(x) \neq 0$ for all $x$ .

givenL1

1.2

P2 holds: $\sigma$ is injective on $N_{\mathbb{N}}$ and on $N_{\mathbb{Z}}$ separately, and $\sigma$ maps each copy into itself, so $\sigma$ is injective on $N$ .

givenL1

1.3

P3 fails: let $S = N_{\mathbb{N}}$ , the $\mathbb{N}$ -copy; then $0 \in S$ and $\sigma$ maps $S$ into $S$ ( $\sigma(N_{\mathbb{N}}) \subseteq N_{\mathbb{N}}$ ), so $S$ contains $0$ and is closed under $\sigma$ ; the $\mathbb{N}$ -copy is the standard model [L2].

givenL1L2

2.1

But $S = N_{\mathbb{N}} \neq N$ , since the $\mathbb{Z}$ -copy $N_{\mathbb{Z}}$ is disjoint from $S$ and nonempty.

step 1.3

3.1

Thus $(N, 0, \sigma)$ satisfies P1 and P2 but not P3, refuting the claim: induction is independent of P1 and P2 and cannot be dropped.

step 1.1step 1.2step 1.3step 2.1∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 17 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Peano axioms (Wikipedia) (standard reference, not scraped)
Mathematical induction (Wikipedia) (standard reference, not scraped)
Peano axioms (Encyclopedia of Mathematics) (standard reference, not scraped)
K. Sutner, Dedekind-Peano Axioms (Carnegie Mellon) (standard reference, not scraped)