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 associative

Statement

For all $a, b, c \in N$ : $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ .

Facts & Assumptions

Given: multiplication and addition (Multiplication of natural numbers, Addition of natural numbers).

[L1]

Left distributivity $a \cdot (b + c) = a \cdot b + a \cdot c$ (Distributivity and the successor law for multiplication).

[L2]

The induction principle (The principle of mathematical induction).

Proof

technique · induction on $c$, with $a, b$ fixed

1.1base

Base $c = 0$ : $(a \cdot b) \cdot 0 = 0$ and $a \cdot (b \cdot 0) = a \cdot 0 = 0$ , so the two sides are equal.

1.2ih

Inductive hypothesis: $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ .

2.1step 1.2L1

Step: $(a \cdot b) \cdot σ (c) = (a \cdot b) \cdot c + a \cdot b = a \cdot (b \cdot c) + a \cdot b = a \cdot (b \cdot c + b) = a \cdot (b \cdot σ (c))$ , using the multiplication recursion, the hypothesis, left distributivity [L1], and $b \cdot σ (c) = b \cdot c + b$ .

3.1step 1.1step 2.1discharge-induction∎

By induction [L2], $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ for all $c$ , hence for all $a, b, c \in N$ .

Multiplication is associative

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources