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.

Multiplication is commutative

Statement

For all $m, n \in N$ : $m \cdot n = n \cdot m$ .

Facts & Assumptions

Given: multiplication (Multiplication of natural numbers).

[L1]

$0 \cdot k = 0$ for all $k \in N$ (Zero and one under multiplication).

[L2]

The successor-left law $σ (n) \cdot m = n \cdot m + m$ (Distributivity and the successor law for multiplication).

[L3]

The induction principle (The principle of mathematical induction).

Proof

technique · induction on $n$, with $m$ fixed

1.1baseL1

Base $n = 0$ : $m \cdot 0 = 0$ and $0 \cdot m = 0$ by [L1], so $m \cdot 0 = 0 \cdot m$ .

1.2ih

Inductive hypothesis: $m \cdot n = n \cdot m$ .

2.1step 1.2L2

Step: $m \cdot σ (n) = m \cdot n + m = n \cdot m + m = σ (n) \cdot m$ , using the multiplication recursion, the hypothesis, then the successor-left law [L2].

3.1step 1.1step 2.1discharge-induction∎

By induction [L3], $m \cdot n = n \cdot m$ for all $n$ , hence for all $m, n \in N$ .

Multiplication is commutative

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources