Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
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.

Sine chord and linear bounds

Statement

For 0≤u≤1/2 and v≥0, sin⁡(πu)≥2u,sin⁡(v)≤v.

Facts & Assumptions

Given: u∈[0,1/2], v≥0, and π as defined by Pi as twice the smallest positive zero of cosine.

[F2]

sin⁡′=cos⁡, and cos⁡ is nonincreasing on [0,π/2] (The derivatives of sine and cosine are cosine and minus sine, Signs, monotonicity intervals, and ranges of sine and cosine).

[F6]

Sine is 1-Lipschitz on R: ∣sin⁡a−sin⁡b∣≤∣a−b∣ (Sine and cosine are 1-Lipschitz on R).

Proof

1.1givenF1F3F4construct

Define g(t)=sin⁡t−2t/π on [0,π/2]. By [F1], [F3], and [F4], g(0)=g(π/2)=0 and g′(t)=cos⁡t−2/π on the interior.

2.1F2step 1.1

By [F2], g′ is nonincreasing: if s<t, then cos⁡s≥cos⁡t, so g′(s)≥g′(t).

3.1F1F5step 2.1choosealgebra

Fix t∈(0,π/2). Apply [F5] on [0,t] and [t,π/2] to obtain a∈(0,t) and b∈(t,π/2) with g(t)=tg′(a) and −g(t)=(π/2−t)g′(b). Since a<b, step 2.1 gives g′(a)≥g′(b). Dividing by the positive numbers t and π/2−t, then multiplying by their product, gives g(t)π/2≥0; as π>0, g(t)≥0.

4.1F1step 3.1algebra

If 0<u<1/2, take t=πu∈(0,π/2) in step 3.1 to get sin⁡(πu)≥2u. At u=0 and u=1/2, the same inequality is equality by [F1].

5.1F1F6step 4.1algebra∎

For v≥0, [F6] and sin⁡0=0 give sin⁡v=sin⁡v−sin⁡0≤∣sin⁡v−sin⁡0∣≤v. Thus both stated inequalities hold.

Depends on

Used by

Dependency tree · two levels

32 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