Alphabeta Math
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.

Order is compatible with addition

Statement

For all m,n,k∈N: m≤n  ⟺  m+k≤n+k; and m<n  ⟺  m+k<n+k (Order on the natural numbers).

Facts & Assumptions

Given: The order m≤n  ⟺  ∃j (m+j=n) (Order on the natural numbers) and addition with m+0=m, m+σ(n)=σ(m+n) (Addition of natural numbers).

[L1]

Addition is associative and commutative (Addition is associative, Addition is commutative).

[L2]

Cancellation: m+k=n+k⇒m=n (Addition is cancellative).

Proof

technique · direct
1.1

Forward: if m≤n, say m+j=n, then (m+k)+j=(m+j)+k=n+k, so m+k≤n+k.

givenL1
1.2

Backward: if m+k≤n+k, say (m+k)+j=n+k, then (m+j)+k=n+k, so m+j=n by cancellation, giving m≤n.

givenL1L2
2.1

The strict forms follow: m<n means m≤n and m≠n, while m+k=n+k  ⟺  m=n by cancellation, so m<n  ⟺  m+k<n+k.

step 1.1step 1.2L2∎

Depends on

Used by

Dependency tree · two levels

12 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