A rational combination lies off both rational rays
ProvedDiaz.conj_combination_off_raysLet lie off the two axes, that is and (equivalently and ), and let . Then
In particular , which is the instance of either clause.
Where this sits. This is the closing clause of Theorem 3.9 (thm:mixed-rigidity, Mixed-coordinate rigidity at a Diaz point) of the manuscript. Theorem 3.9 takes on the Diaz locus, puts , and for concludes
Its final paragraph verifies that the headline instance with really does satisfy that hypothesis: a relation would force and put on the axes. This node is exactly that verification, and it is unconditional — the deep input of Theorem 3.9 is Théorème 7.1 of Roy–Waldschmidt (1997), which is not available in this Mathlib revision and is not on the mission graph.
Proof. Off the axes, and are linearly independent over (Diaz.indep_of_not_axis: from and , a relation with gives on real parts and on imaginary parts, hence ).
- If with , then , so — contradicting .
- If with , then , so — contradicting .
Formalisation notes. The hypothesis is stated in the off-axes form , which is the weakest form and is exactly what Diaz.indep_of_not_axis consumes; a consumer working on the Diaz locus obtains it from Diaz.not_on_axes. The two ray exclusions quantify over all (not merely ), so the non-vanishing that the manuscript notes separately is the case and is not stated as a third conjunct.
Source. Carlo Perassi, Rigidity of logarithms with algebraic modulus — Around a conjecture of Diaz (private manuscript, 15 August 2026), Theorem 3.9, closing paragraph of the proof. The mathematics is his; this node records one step of it in Lean and claims no novelty of its own. Elementary; possibly known, not checked against the literature.
import Mathlib import Definitions.Def_Diaz_Closure import Definitions.Def_Diaz_Instantiation open ComplexConjugate open Diaz
theorem Diaz.conj_combination_off_rays {u : ℂ}
(h1 : conj u ≠ u) (h2 : conj u ≠ -u)
{a b : ℚ} (ha : a ≠ 0) (hb : b ≠ 0) :
(∀ c : ℚ, (a : ℂ) * u + (b : ℂ) * conj u ≠ (c : ℂ) * u) ∧
(∀ c : ℚ, (a : ℂ) * u + (b : ℂ) * conj u ≠ (c : ℂ) * conj u) := by sorry