ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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A two-channel cycle obstructs reaction offsets

Example

Set $A = {1, 2}$ and $r = 2$ , with independently supplied data $ν_{1} = (1, - 1)$ , $d_{1} = - 2$ , $ν_{2} = (- 1, 1)$ , and $d_{2} = 1$ . No offset $b \in R^{2}$ satisfies both $b \cdot ν_{a} + d_{a} \geq 0$ .

Facts & Assumptions

[A1]

Channel data in the Example: $ν_{1} = (1, - 1)$ , $d_{1} = - 2$ , $ν_{2} = (- 1, 1)$ , and $d_{2} = 1$ .

[F1]

Feasibility of finite reaction offsets: for independently supplied real channel vectors and lower-bound data, feasibility requires every nonnegative zero-cycle $λ$ to have $\sum_{a} λ_{a} d_{a} \geq 0$ .

[F2]

Finite inequality alternative by Fourier–Motzkin elimination: dot products are finite coordinate sums, vector sums are coordinatewise, and an infeasibility certificate is a nonnegative combination with zero left coefficient and positive right side.

Verification

Given: [A1]–[F2].

1.1A1F1algebra

The multiplier $λ = (1, 1)$ is nonnegative and satisfies $λ_{1} ν_{1} + λ_{2} ν_{2} = (1, - 1) + (- 1, 1) = (0, 0)$ . But $λ_{1} d_{1} + λ_{2} d_{2} = - 2 + 1 = - 1 < 0$ , violating the necessary condition in [F1]; therefore no offset exists.

2.1A1F2step 1.1algebra∎

Directly, the two inequalities would require $b_{1} - b_{2} \geq 2$ and $- b_{1} + b_{2} \geq - 1$ , equivalently $b_{1} - b_{2} \leq 1$ . Adding the original inequalities gives $0 \geq 1$ , a contradiction. This is the positive-right-side zero-left certificate of [F2].

Depends on

Dependency tree · 2 levels

Sources

Stanford EE464, Fourier–Motzkin Elimination lecture notes