Alphabeta Math
TheoremStatement: AI-adaptedProof: 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.

Strong (complete) induction

Statement

Let PP be a property of naturals such that for every nNn \in \mathbb{N}, if P(m)P(m) holds for all m<nm < n then P(n)P(n). Then P(n)P(n) holds for all nNn \in \mathbb{N}. (At n=0n = 0 the hypothesis is vacuous, so P(0)P(0) is forced.)

Facts & Assumptions

Given: A property PP satisfying the strong-induction hypothesis HH: for every nn, (P(m) for all m<n)P(n)\big(P(m) \text{ for all } m < n\big) \Rightarrow P(n).

[L1]

Well-ordering: every nonempty subset of N\mathbb{N} has a least element (The well-ordering principle).

[L2]

Exactly one of x<yx < y, x=yx = y, y<xy < x holds (Trichotomy of the order on N\mathbb{N}).

Proof

technique · contradiction
1.1

Suppose PP fails somewhere; let S={nN:P(n) is false}S = \{\, n \in \mathbb{N} : P(n) \text{ is false} \,\}, a nonempty subset of N\mathbb{N}.

assume-contra
2.1

By well-ordering [L1], SS has a least element \ell, so s\ell \le s for every sSs \in S; hence every m<m < \ell lies outside SS, since mSm \in S would give m\ell \le m, which [L2] does not allow alongside m<m < \ell; that is, P(m)P(m) holds for all m<m < \ell.

step 1.1L1L2
3.1

By the hypothesis HH applied at \ell, P()P(\ell) holds, so S\ell \notin S.

step 2.1given
4.1

But S\ell \in S by construction, a contradiction; hence SS is empty and P(n)P(n) holds for all nNn \in \mathbb{N}.

step 2.1step 3.1discharge-contradiction

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 29 results over 15 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