LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck 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 $m, n \in N$ : if $m \cdot n = 0$ then $m = 0$ or $n = 0$ . Equivalently, if $m \neq 0$ and $n \neq 0$ then $m \cdot n \neq 0$ .

Facts & Assumptions

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

[L1]

Every nonzero natural is a successor: if $x \neq 0$ then $x = σ (a)$ for some $a \in N$ (Every nonzero natural number is a successor).

[L2]

$σ (j) \neq 0$ for all $j \in N$ (Peano axiom P1) (The von Neumann naturals form a Peano system).

Proof

technique · contrapositive

1.1

It suffices to prove the contrapositive: if $m \neq 0$ and $n \neq 0$ then $m \cdot n \neq 0$ .

contrapositive-reduce

1.2

Assume $m \neq 0$ and $n \neq 0$ .

assume-hyp

2.1

By [L1] write $m = σ (a)$ and $n = σ (b)$ for some $a, b \in N$ .

L1step 1.2

3.1

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

step 2.1given

4.1

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

step 3.1L2discharge-contrapositive∎

Depends on

Used by

Order is compatible with multiplication Lemma
ℕ × ℕ ≈ ℕ Theorem

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

T. Tao, Analysis I, 3rd ed., §2.1-2.3 (Peano axioms, recursion, arithmetic) (standard reference, not scraped)
Peano axioms (Wikipedia) (standard reference, not scraped)
W. Aitken, MATH 378 Ch. 1: The Peano Axioms (CSU San Marcos) (standard reference, not scraped)