ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-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 LP primal-dual residual and exact gap certificate

Example

For $\min x_{1} + 2 x_{2}$ subject to $x_{1} + x_{2} = 1$ and $x \geq 0$ , take candidate $x = (3 / 5, 2 / 5)$ , dual scalar $y = 1$ , and dual slack $s = (0, 1)$ .

Facts & Assumptions

Given: $A = (1, 1)$ , $b = 1$ , $c = (1, 2)$ , and the candidate variables above.

[F1]

The algorithm's residuals are $r_{p} = A x - b$ and $r_{d} = c - A^{T} y - s$ for this LP (Finite-budget primal-dual Newton methods for LP, convex QP, and symmetric cones).

[F2]

Exact primal feasibility, dual feasibility, stationarity, and gap $x^{T} s$ imply $0 \leq f_{0} (x) - p^{*} \leq x^{T} s$ (Fraction-to-boundary steps preserve positivity and exact primal-dual feasibility certifies a gap).

Verification

1.1givenF1algebra

The primal residual is $(1, 1) \cdot (3 / 5, 2 / 5) - 1 = 0$ . The dual stationarity residual is $c - A^{T} y - s = (1, 2) - (1, 1) - (0, 1) = (0, 0)$ , and $x \geq 0$ , $s \geq 0$ , so the candidate pair is exactly feasible.

2.1F2step 1.1algebra∎

The primal objective is $3 / 5 + 2 (2 / 5) = 7 / 5$ , the dual value is $b y = 1$ , and the gap is $x^{T} s = 2 / 5$ . Every primal feasible vector has $x_{1} + x_{2} = 1$ , $x_{i} \geq 0$ , and objective $x_{1} + 2 x_{2} = 1 + x_{2} \geq 1$ ; the feasible point $(1, 0)$ attains $1$ , so $p^{*} = 1$ . Thus the objective error is exactly $7 / 5 - 1 = 2 / 5 = x^{T} s$ , and [F2] gives the same certified upper bound.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 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

Stephen Boyd and Lieven Vandenberghe, Convex Optimization (standard reference, not scraped)
Hamza Fawzi, Topics in Convex Optimisation, Lecture 16 (standard reference, not scraped)