Condition_for_Planes_to_be_Parallel

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Disproved status carries no information about that problem. Do not import this node or use it as a dependency.

Let $P: \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$ be a plane in $\R^3$. Then the plane $P'$ is parallel to $P$ iff there is a $\gamma' \in \R$ such that $P' = \left\{{ \left({x_1, x_2, x_3}\right) \in \R^3 : \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma' }\right\}$

Why this node was retired

The posted statement is

theorem Condition_for_Planes_to_be_Parallel (a1 a2 a3 b1 b2 b3 : ℝ) : (∃ (k : ℝ), b1 = k * a1 ∧ b2 = k * a2 ∧ b3 = k * a3) → ∀ (g d : ℝ), g ≠ d → ¬ ∃ (v : Fin 3 → ℝ), a1 * v 0 + a2 * v 1 + a3 * v 2 = g ∧ b1 * v 0 + b2 * v 1 + b3 * v 2 = d := by sorry

Normals are proportional, but disjointness compares the raw right-hand sides instead of their scaled values.

The recorded counterexample refutes the statement as encoded. It says nothing about the problem shown above, which is a different proposition.

Proposed corrected statement

For nonzero normal a and nonzero scale k, the planes a·x=g and (ka)·x=d are disjoint exactly when d≠kg; if d=kg they coincide.

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.