The six exponentials theorem for ℒ_u = ℒ + ℚu + ℚū (u transcendental over the field of ℒ̃, uū algebraic): no ℚ-independent x₁, x₂ and y₁, y₂, y₃ have all xᵢyⱼ in ℒ_u

Here $\mathcal{L}$ is the set of logarithms of algebraic numbers (the $z \in \mathbb{C}$ with $e^z$ algebraic), and $\widetilde{\mathcal{L}}$ is the $\overline{\mathbb{Q}}$-vector space spanned by $1$ and $\mathcal{L}$. Let $F$ be the subfield of $\mathbb{C}$ generated by $\widetilde{\mathcal{L}}$, that is, by $\overline{\mathbb{Q}}$ and $\mathcal{L}$. Let $u \in \mathbb{C}$ be transcendental over $F$, with $u\bar u$ algebraic, and put $\mathcal{L}_u = \mathrm{span}_{\mathbb{Q}}(\mathcal{L}) + \mathrm{span}_{\mathbb{Q}}(u, \bar u)$; as $\mathcal{L}$ is a $\mathbb{Q}$-vector space, this is $\mathcal{L} + \mathbb{Q}u + \mathbb{Q}\bar u$. Then no $x_1, x_2 \in \mathbb{C}$, linearly independent over $\mathbb{Q}$, and $y_1, y_2, y_3 \in \mathbb{C}$, linearly independent over $\mathbb{Q}$, have all six products $x_iy_j$ in $\mathcal{L}_u$.

No hypothesis is made: the proof needs neither Roy's strong six exponentials theorem nor anything about Diaz's conjecture.

Proof. Both families are $\overline{\mathbb{Q}}$-independent: a relation $\sum_ig_ix_i = 0$ with algebraic $g_i$, not all zero, gives the vanishing form $\sum_ig_i(x_iy_1) = 0$ in the $\mathbb{Q}$-independent elements $x_iy_1$ of $\mathcal{L}_u$, which Baker's theorem for $\mathcal{L}_u$ (DiazModulus.logAlg_adjoin_circle_point_baker_linear_forms) forbids, and the same with the $x_1y_j$ for $y$. Next, by DiazModulus.mem_logAlg_adjoin_circle_point_iff, $\mathcal{L}_u \subseteq \widetilde{\mathcal{L}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}u^{-1}$, as $\bar u = (u\bar u)u^{-1}$, so DiazModulus.rank_one_config_inverse_adjoin with $K = \overline{\mathbb{Q}}$ and $V_0 = \widetilde{\mathcal{L}}$ puts every $x_iy_j$ in $\widetilde{\mathcal{L}} \subseteq F$. Writing $x_iy_j = c + pu + q\bar u$ with $c \in \mathcal{L}$, DiazModulus.transcendental_add_mul_add_mul_conj forces $p = q = 0$, so all six products lie in $\mathcal{L}$, against the six exponentials theorem (DiazModulus.six_exponentials). Neither Roy's theorem nor the Gel'fond–Schneider theorem is used.

Novelty. With $\mathcal{L}_u$ in place of $\mathcal{L}$, this is the six exponentials theorem (Lang 1966, Ramachandra 1968), from which it is derived together with Baker's theorem for $\mathcal{L}_u$. Not found in the sources read, and neither is a route that avoids Roy's strong six exponentials theorem; routine.