Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (openai/gpt-5.4)audited 2026-07-25
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.

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Zero and one under multiplication

Statement

For all n∈Nn \in \mathbb{N}: 0⋅n=00 \cdot n = 0, 1⋅n=n1 \cdot n = n, and n⋅1=nn \cdot 1 = n. Here 1=σ(0)1 = \sigma(0), and the clause n⋅0=0n \cdot 0 = 0 is definitional.

Facts & Assumptions

Given: multiplication m⋅0=0m \cdot 0 = 0 and m⋅σ(n)=m⋅n+mm \cdot \sigma(n) = m \cdot n + m (Multiplication of natural numbers), and addition with m+0=mm + 0 = m and m+σ(k)=σ(m+k)m + \sigma(k) = \sigma(m + k) (Addition of natural numbers).

[L1]

0+k=k0 + k = k for all k∈Nk \in \mathbb{N} (Left identity for addition).

[L2]

The induction principle (The principle of mathematical induction).

Proof

technique · induction on $n$
1.1

Right unit n⋅1=nn \cdot 1 = n: n⋅1=n⋅σ(0)=n⋅0+n=0+n=nn \cdot 1 = n \cdot \sigma(0) = n \cdot 0 + n = 0 + n = n, using the multiplication recursion then [L1].

givenL1
1.2

Base of the left inductions: 0⋅0=00 \cdot 0 = 0 and 1⋅0=01 \cdot 0 = 0 by the clause m⋅0=0m \cdot 0 = 0.

base
1.3

Inductive hypothesis: 0⋅n=00 \cdot n = 0 and 1⋅n=n1 \cdot n = n.

ih
2.1

Step: 0⋅σ(n)=0⋅n+0=0+0=00 \cdot \sigma(n) = 0 \cdot n + 0 = 0 + 0 = 0, using the hypothesis 0⋅n=00 \cdot n = 0 and 0+0=00 + 0 = 0 (the instance k=0k = 0 of [L1]); and 1⋅σ(n)=1⋅n+1=n+1=σ(n)1 \cdot \sigma(n) = 1 \cdot n + 1 = n + 1 = \sigma(n), where n+1=n+σ(0)=σ(n+0)=σ(n)n + 1 = n + \sigma(0) = \sigma(n + 0) = \sigma(n) by the addition recursion m+σ(k)=σ(m+k)m + \sigma(k) = \sigma(m + k) and n+0=nn + 0 = n.

step 1.3givenL1
3.1

By induction [L2], 0⋅n=00 \cdot n = 0 and 1⋅n=n1 \cdot n = n for all nn; together with step 1.1, n⋅1=nn \cdot 1 = n for all nn.

step 1.1step 2.1discharge-induction∎

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Dependency tree · next 3 levels

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