Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-02 (claude-opus-5)
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

No natural number equals its own successor

Statement

For every n∈Nn \in \mathbb{N}, n≠σ(n)n \neq \sigma(n).

Facts & Assumptions

Given: N\mathbb{N} with distinguished element 00 and successor σ\sigma.

[L1]

σ(n)≠0\sigma(n) \neq 0 for all nn (P1) and σ\sigma is injective (P2) (The von Neumann naturals form a Peano system).

[L2]

The induction principle (The principle of mathematical induction).

Proof

technique · induction
1.1

Base case n=0n = 0: σ(0)≠0\sigma(0) \neq 0 by P1, so 0≠σ(0)0 \neq \sigma(0).

baseL1
1.2

Inductive hypothesis: n≠σ(n)n \neq \sigma(n).

ih
2.1

If σ(n)=σ(σ(n))\sigma(n) = \sigma(\sigma(n)) then injectivity of σ\sigma gives n=σ(n)n = \sigma(n), contradicting the hypothesis; hence σ(n)≠σ(σ(n))\sigma(n) \neq \sigma(\sigma(n)).

step 1.2L1
3.1

By induction [L2], n≠σ(n)n \neq \sigma(n) for all n∈Nn \in \mathbb{N}.

step 1.1step 2.1discharge-induction∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 16 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