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.

Multiplication is associative

Statement

For all a,b,c∈Na, b, c \in \mathbb{N}: (a⋅b)⋅c=a⋅(b⋅c)(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⋅(b+c)=a⋅b+a⋅ca \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
1.1

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

base
1.2

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

ih
2.1

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

step 1.2L1
3.1

By induction [L2], (a⋅b)⋅c=a⋅(b⋅c)(a \cdot b) \cdot c = a \cdot (b \cdot c) for all cc, hence for all a,b,c∈Na, b, c \in \mathbb{N}.

step 1.1step 2.1discharge-induction∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 22 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