ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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 two-column matrix where basis pursuit is not unique

Example

Let $A = [1 1] \in M_{1 \times 2} (R)$ and $s = 1$ , using one-based coordinate labels $j = 1, 2$ corresponding to library indices $j - 1$ . The vector $h = (1, - 1)$ lies in $\ker A$ , and for $S = {1}$ the strict nullspace inequality fails because $∥ h_{S} ∥_{1} = ∥ h_{S^{c}} ∥_{1} = 1$ . Under the constraint $z_{1} + z_{2} = 1$ , the function $∣ z_{1} ∣ + ∣ z_{2} ∣$ has minimum $1$ , attained both by $x^{†} = (1, 0)$ and by $z = (0, 1)$ ; therefore the displayed optimization problem has two distinct minimizers.

Example references

Strict nullspace property and coordinate conventions

Finite-dimensional one-norm

Triangle inequality for real absolute values

Facts & Assumptions

Given: $A = [1 1]$ , $s = 1$ , and the vectors $h = (1, - 1)$ , $x^{†} = (1, 0)$ , and $z = (0, 1)$ in $R^{2}$ .

[F1]

The strict nullspace property: strict NSP requires $∥ h_{S} ∥_{1} < ∥ h_{S^{c}} ∥_{1}$ for every nonzero kernel vector and every $S$ of size at most $s$ .

[F2]

The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ : on $R^{2}$ , $∥ u ∥_{1} = ∣ u_{1} ∣ + ∣ u_{2} ∣$ .

[F3]

The triangle inequality: $∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣$ for real $a, b$ .

[F4]

The strict nullspace property: the matrix-vector product is computed by the one-based row formula in the definition.

Verification

1.1givenF1F2F4algebra

By [F4], $A h = 1 - 1 = 0$ , and $h \neq 0$ . For $S = \emptyset$ , $h_{S} = (0, 0)$ and $h_{S^{c}} = h$ , so [F2] gives $∥ h_{S} ∥_{1} = 0 < 2 = ∥ h_{S^{c}} ∥_{1}$ . For $S = {1}$ , $h_{S} = (1, 0)$ and $h_{S^{c}} = (0, - 1)$ , so [F2] gives $∥ h_{S} ∥_{1} = 1 = ∥ h_{S^{c}} ∥_{1}$ . Thus the required strict inequality fails for this admissible support, as claimed by [F1].

2.1givenF2F3F4algebra∎

By [F4], every vector $w = (w_{1}, w_{2})$ feasible for this datum satisfies $w_{1} + w_{2} = 1$ , so [F3] gives $1 = ∣ w_{1} + w_{2} ∣ \leq ∣ w_{1} ∣ + ∣ w_{2} ∣$ . Both $x^{†}$ and $z$ are feasible and have objective value $1$ by [F2]; they are distinct, so both attain the minimum of the displayed objective and it is not unique.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

22 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.