ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck pass

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.

A manufactured shear field on a clamped rectangle

Statement

Assume Countable Choice. On $D = (0, 1)^{2}$ , clamp the left edge $Γ_{D} = {0} \times (0, 1)$ and prescribe traction on the other three open edges $Γ_{N}$ . Let $μ > 0$ and $λ \geq 0$ be constants and set $u (x, y) = (0, x)$ . The stress $σ_{λ} (u) = 2 μ ε (u) + λ (div u) I_{2}$ is constant and satisfies $- div σ_{λ} (u) = 0$ , $u ∣_{Γ_{D}} = 0$ , and $σ_{λ} (u) n = (0, μ)$ on the right edge, $(μ, 0)$ on the top edge, and $(- μ, 0)$ on the bottom edge. This is a manufactured linear-elasticity solution; no uniqueness result for partial Dirichlet data is asserted.

Facts & Assumptions

Given: Countable Choice; the linear-elasticity strain and stress conventions of Linear isotropic elasticity: displacement form and data; the square $D = (0, 1)^{2}$ ; constant $μ > 0$ ; and $λ \geq 0$ .

[A1]

Countable Choice is inherited through the imported model definition; the direct matrix and traction computations use no choice. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

The symmetric gradient is $ε (u) = (\nabla u + \nabla u^{T}) / 2$ and $σ_{λ} (u) = 2 μ ε (u) + λ (div u) I_{2}$ . (Linear isotropic elasticity: displacement form and data)

Verification

1.1A1givenF1algebra

The gradient of $u = (0, x)$ is $(\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array})$ , so $ε (u) = (\begin{array}{cc} 0 & 1 / 2 \\ 1 / 2 & 0 \end{array})$ and $div u = 0$ . Hence [F1] gives $σ_{λ} (u) = (\begin{array}{cc} 0 & μ \\ μ & 0 \end{array})$ , independent of $λ$ .

2.1givenstep 1.1algebra∎

The stress is constant, so its row-wise divergence is zero. The displacement vanishes at $x = 0$ . Multiplication by the outward unit normals $(1, 0)$ , $(0, 1)$ , and $(0, - 1)$ on the right, top, and bottom edges gives the three displayed tractions.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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

Long Chen, Introduction to Linear Elasticity and Its Finite Element Approximation (complete 13-page notes) (standard reference, not scraped)
Daniele Boffi, Franco Brezzi, Michel Fortin, Mixed Finite Element Methods and Applications (complete 691-page text) (standard reference, not scraped)