CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-02 (claude-opus-5)

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 a well-defined total operation

Statement

For each $m \in N$ there is a unique function $p_{m} : N \to N$ with $p_{m} (0) = 0$ and $p_{m} (σ (n)) = p_{m} (n) + m$ . Writing $m \cdot n : = p_{m} (n)$ , multiplication $\cdot : N \times N \to N$ is a well-defined total binary operation satisfying $m \cdot 0 = 0$ and $m \cdot σ (n) = m \cdot n + m$ .

Facts & Assumptions

Given: $N$ with successor $σ$ and addition $+$ (Addition of natural numbers).

[L1]

The recursion theorem (The recursion theorem).

[L2]

For fixed $m$ , the map ${a d d}_{m} : x \mapsto x + m$ is a function $N \to N$ (Addition of natural numbers).

[L3]

$(N, 0, σ)$ is a Peano system (The von Neumann naturals form a Peano system).

Proof

technique · direct

1.1L2

Fix $m \in N$ ; the map ${a d d}_{m} : N \to N$ , ${a d d}_{m} (x) = x + m$ , is a well-defined function.

2.1L1L3step 1.1

Since $(N, 0, σ)$ is a Peano system [L3], apply the recursion theorem with $A = N$ , $a = 0$ , $f = {a d d}_{m}$ : there is a unique $p_{m} : N \to N$ with $p_{m} (0) = 0$ and $p_{m} (σ (n)) = {a d d}_{m} (p_{m} (n)) = p_{m} (n) + m$ .

3.1step 2.1∎

Define $m \cdot n : = p_{m} (n)$ ; as $p_{m}$ is total for each $m \in N$ , the assignment $(m, n) \mapsto m \cdot n$ is a well-defined function $N \times N \to N$ with $m \cdot 0 = 0$ and $m \cdot σ (n) = m \cdot n + m$ .

Multiplication is a well-defined total operation

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources