ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-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.

Numerical cancellation can zero an entry while the filled graph retains its edge

Example

Work over $Q$ , which is a field, and take

$A = (\begin{array}{ccc} 3 & 1 & 1 / 2 \\ 1 & 2 & 1 \\ 1 / 2 & 1 & 3 \end{array})$

with vertices ${0, 1, 2}$ and elimination order $(1, 0, 2)$ . The initial structural graph is a triangle. After eliminating vertex $1$ , the numerical entry between vertices $0$ and $2$ cancels to zero, although the symbolic graph retains that edge. Every pivot in this ordering is nonzero.

Facts & Assumptions

Given: The displayed symmetric matrix over $Q$ , the vertex labels, and the elimination order $(1, 0, 2)$ .

[L1]

Structural graphs and ordered symmetric elimination: the graph update retains existing edges between remaining vertices and completes the later-neighbor set to a clique.

[L2]

The rationals form a field: the rational numbers with their standard operations form a field.

[L3]

The filled graph contains every off-diagonal Schur-complement nonzero: assume each pivot $p_{k}$ is nonzero and define $S_{i j}^{k + 1} = S_{i j}^{k} - S_{i v_{k}}^{k} S_{v_{k} j}^{k} / p_{k}$ .

[L4]

The filled graph contains every off-diagonal Schur-complement nonzero: every nonzero off-diagonal entry of each Schur complement is an edge of the corresponding filled graph.

Verification

1.1givenL1algebra

All three off-diagonal positions of $A$ are nonzero, so the initial graph has edges ${0, 1}$ , ${0, 2}$ , and ${1, 2}$ . Eliminating vertex $1$ leaves ${0, 2}$ as a later-neighbor pair, and [L1] retains its edge in the filled graph.

2.1givenL2L3step 1.1algebra

The first pivot is $2$ , and the update in [L3] gives $S_{02}^{1} = 1 / 2 - (1) (1) / 2 = 0$ . The two remaining diagonal entries are $3 - 1 / 2 = 5 / 2$ , so the last two pivots are both $5 / 2$ and are nonzero. Thus all pivots are nonzero while the numerical support loses ${0, 2}$ .

3.1L1L4step 1.1step 2.1∎

By [L4], the filled graph contains the numerical support; step 1.1 shows its edge ${0, 2}$ remains, while step 2.1 shows the corresponding numerical entry is zero. Thus the inclusion is strict in this example.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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

Yousef Saad, Iterative Methods for Sparse Linear Systems, 2nd ed., Chapter 3 §§3.2.1, 3.3.2 and 3.6, printed pp. 77–83 and 96–98 (standard reference, not scraped)