Alphabeta Math
LemmaStatement: Literature-sourcedProof: 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.

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

The natural numbers have no zero divisors

Statement

For all m,n∈Nm,n\in\mathbb{N}: if m⋅n=0m\cdot n=0 then m=0m=0 or n=0n=0. Equivalently, if m≠0m\ne 0 and n≠0n\ne 0 then m⋅n≠0m\cdot n\ne 0.

Facts & Assumptions

Given: multiplication defined by m⋅0=0m\cdot 0=0 and m⋅σ(k)=m⋅k+mm\cdot\sigma(k)=m\cdot k+m; addition defined by m+0=mm+0=m and m+σ(k)=σ(m+k)m+\sigma(k)=\sigma(m+k).

[L1]

Every nonzero natural is a successor: if x≠0x\ne 0 then x=σ(a)x=\sigma(a) for some a∈Na\in\mathbb{N} (Every nonzero natural number is a successor).

[L2]

σ(j)≠0\sigma(j)\ne 0 for all j∈Nj\in\mathbb{N} (Peano axiom P1) (The von Neumann naturals form a Peano system).

Proof

technique · contrapositive
1.1

It suffices to prove the contrapositive: if m≠0m\ne 0 and n≠0n\ne 0 then m⋅n≠0m\cdot n\ne 0.

contrapositive-reduce
1.2

Assume m≠0m\ne 0 and n≠0n\ne 0.

assume-hyp
2.1

By [L1] write m=σ(a)m=\sigma(a) and n=σ(b)n=\sigma(b) for some a,b∈Na,b\in\mathbb{N}.

L1step 1.2
3.1

Then m⋅n=m⋅σ(b)=m⋅b+m=m⋅b+σ(a)=σ(m⋅b+a)m\cdot n=m\cdot\sigma(b)=m\cdot b+m=m\cdot b+\sigma(a)=\sigma(m\cdot b+a), using the multiplication recursion, m=σ(a)m=\sigma(a), and c+σ(a)=σ(c+a)c+\sigma(a)=\sigma(c+a).

step 2.1given
4.1

By [L2], σ(m⋅b+a)≠0\sigma(m\cdot b+a)\ne 0, so m⋅n≠0m\cdot n\ne 0; this proves the contrapositive, hence if m⋅n=0m\cdot n=0 then m=0m=0 or n=0n=0.

step 3.1L2discharge-contrapositive∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 17 results over 12 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