ExampleConstruction: AI-generatedVerification: 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 validated mass-action chemical equilibrium root

Example

Consider the reversible reaction $2 A ⇌ B$ with equilibrium constant $1$ and conserved amount $[A] + 2 [B] = 1$ . Writing $a = [A]$ , the mass-action equilibrium relation is $[B] = a^{2}$ , so the reduced residual is $f (a) = 2 a^{2} + a - 1$ . The dyadic box $X = [3 / 8, 5 / 8]$ certifies one positive equilibrium with $a = 1 / 2$ and $[B] = 1 / 4$ .

Facts & Assumptions

Given: The stated mass-action relation, conservation law, residual $f (a) = 2 a^{2} + a - 1$ , and box $X = [3 / 8, 5 / 8]$ .

[F1]

The scalar procedure forms an outward interval Newton image from an enclosure of $f (m)$ and an interval enclosure of $f^{'} (X)$ ; strict inclusion returns a unique-root status (Finite-budget scalar interval Newton root certification).

[F2]

If $0 \notin f^{'} (X)$ and the interval Newton image lies in $int X$ , then $f$ has exactly one root in $X$ (Scalar interval Newton exclusion and strict inclusion certificates).

[F3]

Finite real interval arithmetic encloses exact additions, multiplications, divisions away from zero, and squares (Finite real interval operations enclose their exact values).

[F4]

The derivative on a real interval is the ordinary difference-quotient limit (Continuity and derivative on a real interval).

Verification

1.1givenalgebra

The conservation law and $[B] = a^{2}$ reduce to $f (a) = a + 2 a^{2} - 1 = 0$ . At $a = 1 / 2$ , $[B] = 1 / 4$ , so $[B] = a^{2}$ and $a + 2 [B] = 1 / 2 + 1 / 2 = 1$ ; both concentrations are positive.

1.2givenF3F4algebra

Natural interval evaluation gives $E_{f} (X) = 2 [3 / 8, 5 / 8]^{2} + [3 / 8, 5 / 8] - 1 = [- 11 / 32, 13 / 32]$ , which contains zero. The difference quotient gives $f^{'} (a) = \lim_{h \to 0} (4 a h + 2 h^{2} + h) / h = 4 a + 1$ . The midpoint of $X$ is $m = 1 / 2$ , with $f (m) = 0$ and $f^{'} (X) = 4 [3 / 8, 5 / 8] + 1 = [5 / 2, 7 / 2]$ , which excludes zero.

2.1step 1.1step 1.2F1F2F3∎

The interval Newton image is $N = m - [0, 0] / [5 / 2, 7 / 2] = {1 / 2}$ by exact interval arithmetic [F3]. Since ${1 / 2} \subset (3 / 8, 5 / 8)$ , the algorithm returns "unique-root" by [F1], and [F2] proves that this is the only root in the box. The box therefore certifies the stated positive equilibrium.

Depends on

Used by

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Dependency tree · two levels

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