Alphabeta Math
LemmaStatement: Literature-sourcedProof: 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.

Cancellation for multiplication by a nonzero factor

Statement

For all m,n,k∈N with k≠0: if m⋅k=n⋅k then m=n.

Facts & Assumptions

Given: multiplication defined by m⋅0=0 and m⋅σ(j)=m⋅j+m; the order m≤n  ⟺  ∃j (m+j=n), with m<n meaning m≤n and m≠n.

[L1]

Trichotomy: for all m,n∈N exactly one of m<n, m=n, m>n holds (Trichotomy of the order on N).

[L2]

Strict monotonicity: if k≠0 and m<n then m⋅k<n⋅k (Order is compatible with multiplication).

Proof

technique · contradiction
1.1

Suppose m⋅k=n⋅k with k≠0 but m≠n.

assume-contra
2.1

By trichotomy [L1], m≠n forces m<n or n<m.

step 1.1L1
3.1

If m<n then m⋅k<n⋅k by [L2] (since k≠0); if n<m then n⋅k<m⋅k by [L2]; either way m⋅k≠n⋅k.

step 2.1L2
4.1

This contradicts m⋅k=n⋅k, so the assumption m≠n is impossible and m=n.

step 1.1step 3.1discharge-contradiction∎

Depends on

Used by

Dependency tree · two levels

19 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