Alphabeta Math
TheoremStatement: 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.

The principle of mathematical induction

Statement

Let SNS \subseteq \mathbb{N}. If 0S0 \in S and σ(n)S\sigma(n) \in S whenever nSn \in S, then S=NS = \mathbb{N}. Equivalently (property form): if a property PP satisfies P(0)P(0) and (P(n)P(σ(n))P(n) \Rightarrow P(\sigma(n))) for all nn, then P(n)P(n) holds for all nNn \in \mathbb{N}. This is the induction principle, the basis of proof by induction.

Facts & Assumptions

Given: N=ω\mathbb{N} = \omega is the smallest inductive set (Inductive set, The natural numbers N\mathbb{N} (von Neumann), The natural numbers exist: a smallest inductive set); 0=0 = \varnothing, σ(n)=n{n}\sigma(n) = n \cup \{n\}.

[L1]

ω\omega is contained in every inductive set (The natural numbers exist: a smallest inductive set).

Proof

technique · direct
1.1

Suppose SNS \subseteq \mathbb{N} with 0=S0 = \varnothing \in S and (xSσ(x)=x{x}Sx \in S \Rightarrow \sigma(x) = x \cup \{x\} \in S).

given
2.1

Then SS contains \varnothing and is closed under xx{x}x \mapsto x \cup \{x\}, so SS is an inductive set.

step 1.1
3.1

N=ωS\mathbb{N} = \omega \subseteq S because ω\omega is a subset of every inductive set.

L1step 2.1
4.1

With SNS \subseteq \mathbb{N} by hypothesis, S=NS = \mathbb{N}.

step 3.1given
5.1

The property form is the set case with S={nN:P(n)}S = \{n \in \mathbb{N} : P(n)\}: P(0)P(0) and P(n)P(σ(n))P(n) \Rightarrow P(\sigma(n)) give 0S0 \in S and σ\sigma-closure, so S=NS = \mathbb{N}, i.e. PP holds for all nn; conversely the set form is the property form applied to P(n):nSP(n) :\equiv n \in S, so the two are equivalent.

step 4.1

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 13 results over 6 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.

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