At a candidate u, with Roy's strong six exponentials theorem as hypothesis: for distinct non-zero algebraic a₁, a₂, u/(u² − a₁) and u/(u² − a₂) are not both in ℒ̃

Here $\mathcal{L}$ is the set of logarithms of algebraic numbers, and $\widetilde{\mathcal{L}}$ is the $\overline{\mathbb{Q}}$-vector space spanned by $1$ and $\mathcal{L}$. Roy's strong six exponentials theorem is carried as the hypothesis hSSE: if $x_1, x_2$ are $\overline{\mathbb{Q}}$-linearly independent and so are $y_1, y_2, y_3$, then one of the six products $x_iy_j$ is not in $\widetilde{\mathcal{L}}$.

Let $u$ be a candidate for Diaz's conjecture (DiazModulus.IsCandidate) and $a_1 \neq a_2$ non-zero algebraic numbers. Then $u/(u^2 - a_1)$ and $u/(u^2 - a_2)$ are not both in $\widetilde{\mathcal{L}}$. It concerns candidates, so it is vacuous if Diaz's conjecture holds.

Proof. $\rho = |u|^2$ is algebraic and $u \notin \overline{\mathbb{Q}}$ (DiazModulus.hermite_lindemann_holds); $u, \bar u \in \mathcal{L}$ (DiazModulus.logAlg_conj_stable) and $1 \in \widetilde{\mathcal{L}}$. If $z_1, z_2 \in \widetilde{\mathcal{L}}$, then $W \subseteq \widetilde{\mathcal{L}}$, and the configuration of DiazModulus.circle_point_two_pole_extension_carries_two_by_three_configuration contradicts hSSE.

Novelty. Not asserted: it is one substitution in Diaz (2007, Th. 7(2)), that is Fischler (2001, Lemma 6.1), with ratio $u^2$ and first term $b$; Diaz (2004, Th. 2) at $x = (u, \bar u)$ also gives it after a case split.