ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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.

Gram–Schmidt on an explicit basis of $C^{2}$ with conjugation visible

Example

For the linear-first standard inner product on $C^{2}$ , Gram–Schmidt applied to $v_{0} = (1, i)$ and $v_{1} = (1, 1)$ produces

$e_{0} = \frac{(1, i)}{\sqrt{2}}, e_{1} = (\frac{1 + i}{2}, \frac{1 - i}{2}) .$

Facts & Assumptions

Given: The vectors $v_{0} = (1, i)$ and $v_{1} = (1, 1)$ in $C^{2}$ .

[L1]

The standard complex inner product is $⟨ x, y ⟩ = \sum_{k} x_{k} \overline{y_{k}}$ (The standard formulas $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and $\sum_{k < n} x_{k} \overline{y_{k}}$ on $C^{n}$ are inner products).

[L2]

Gram–Schmidt uses the coefficient $⟨ v_{k}, e_{j} ⟩$ under the linear-first convention (Gram–Schmidt turns every finite independent list into an orthonormal list with the same successive spans).

[L3]

Complex conjugation sends $i$ to $- i$ and satisfies $∣ 1 \pm i ∣^{2} = 2$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Verification

technique · computation

1.1L1L2L3

From [L1], $∥ v_{0} ∥ = \sqrt{2}$ , so $e_{0} = v_{0} / \sqrt{2}$ . The projection coefficient is $⟨ v_{1}, e_{0} ⟩ = (1 - i) / \sqrt{2}$ .

2.1step 1.1L1L3algebra

Subtraction gives $u_{1} = v_{1} - ⟨ v_{1}, e_{0} ⟩ e_{0} = ((1 + i) / 2, (1 - i) / 2)$ . By [L3], $∥ u_{1} ∥^{2} = 1$ , so $e_{1} = u_{1}$ .

3.1step 1.1step 2.1L1L3∎

Direct use of [L1] gives $⟨ e_{1}, e_{0} ⟩ = 0$ and, by conjugate symmetry, $⟨ e_{0}, e_{1} ⟩ = 0$ . Both vectors have norm one, so the displayed list is orthonormal.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Gram–Schmidt on an explicit basis of C2C2 with conjugation visible