LemmaStatement: AI-generatedProof: AI-generated

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.

An elementary sufficient budget for the synchronization criterion

Statement

Let $n > 6$ , $0 < a \leq 1 / 24$ and $0 < b < 1 / 4$ . Set $q = 1 / (2 b) - 1 > 1$ . If $π / (96 a) > \log n / \log q + 1$ , then $a < 1 / 12$ and $\frac{π / 4}{\arcsin (12 a)} > \frac{\log (n / 6)}{\log (1 / (2 b) - 1)} + 1$ . The inverse sine is its principal branch in radians.

Facts & Assumptions

Given: The data and hypotheses in the Statement.

Goal: Establish the conclusions in the Statement.

[F1]

The principal inverse sine is continuous on $[- 1, 1]$ and is the inverse of sine restricted to $[- π / 2, π / 2]$ ; logarithms have domain $(0, \infty)$ : Principal inverse sine and inverse cosine and The natural logarithm as the inverse of the exponential function.

[F2]

For $- 1 < t < 1$ , $(\arcsin t)' = 1 / \sqrt{1 - t^{2}}$ : For -1<y<1, (arcsin y)ᵖʳⁱᵐᵉ=1/√1-y² and (arccos y)ᵖʳⁱᵐᵉ=-1/√1-y².

[F3]

Sine is odd, so $\sin 0 = - \sin 0$ and hence $\sin 0 = 0$ : Parity and the Pythagorean identity for sine and cosine.

[F4]

If $f$ is continuous on $[u, v]$ with $u < v$ and differentiable on $(u, v)$ , then $f (v) - f (u) = f' (c) (v - u)$ for some $c \in (u, v)$ : The mean value theorem, as the case g(x) = x of Cauchy's: for f continuous on [a,b] with a < b and differentiable on (a,b) there is c ∈ (a,b) with f(b) - f(a) = f'(c)(b-a).

[F5]

The natural logarithm is strictly increasing and $\log 1 = 0$ : Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm.

Proof

1.1givenF1F2F3F4algebra

By the principal inverse convention and $\sin 0 = 0$ , $\arcsin 0 = 0$ . For $0 < x \leq 1 / 2$ , the principal inverse sine is continuous on $[0, x]$ and differentiable on $(0, x)$ , since $[0, x] \subset (- 1, 1)$ . The mean value theorem therefore gives $\arcsin x = x / \sqrt{1 - c^{2}}$ for some $0 < c < x$ . Since $1 - c^{2} \geq 3 / 4 > 1 / 4$ , this yields $0 < \arcsin x \leq 2 x$ . In particular, $0 < 12 a \leq 1 / 2$ gives $0 < \arcsin (12 a) \leq 24 a$ .

2.1givenF1F5step 1.1algebra∎

The positivity in step 1.1 permits taking reciprocals and gives $(π / 4) / \arcsin (12 a) \geq π / (96 a)$ . The hypothesis $0 < b < 1 / 4$ gives $q = 1 / (2 b) - 1 > 1$ , so $\log q > \log 1 = 0$ . Also $0 < n / 6 < n$ , hence $\log (n / 6) < \log n$ . Thus $(π / 4) / \arcsin (12 a) \geq π / (96 a) > \log n / \log q + 1 > \log (n / 6) / \log q + 1$ , as required; and $a \leq 1 / 24 < 1 / 12$ .

Depends on

Used by · 2

Dependency tree · 2 levels

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