Condition_for_Straight_Lines_in_Plane_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 $L: \alpha_1 x + \alpha_2 y = \beta$ be a straight line in $\R^2$. Then the straight line $L'$ is parallel to $L$ iff there is a $\beta' \in \R^2$ such that L' = \set {\tuple {x, y} \in \R^2: \alpha_1 x + \alpha_2 y = \beta'}

Why this node was retired

The posted statement is

theorem Condition_for_Straight_Lines_in_Plane_to_be_Parallel (a1 a2 b1 b2 beta gamma : ℝ) (h : a1 * b2 = a2 * b1) (hne : a1 ≠ 0 ∨ a2 ≠ 0) : ∀ (x y : ℝ), a1 * x + a2 * y = beta → b1 * x + b2 * y = gamma → False ∨ (∀ x y, a1 * x + a2 * y = beta ↔ b1 * x + b2 * y = gamma) := by sorry

The second coefficient vector may be zero, so the second equation can define the whole plane instead of a line.

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

Proposed corrected statement

Require both normal vectors to be nonzero. Under zero determinant, the corresponding lines are either coincident or disjoint, with equality of scaled constants determining coincidence.

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.