At a candidate u with u, w₁…w_m algebraically independent over Q̄, every 2×2 configuration in Q̄ + Q̄u + Q̄ū + ΣQ̄w_k has, in every row and every column, an entry outside the Q̄-span of the logarithms

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 $\mathcal{L}$. Let $u$ be a candidate for Diaz's conjecture (DiazModulus.IsCandidate) with $u, w_1, \dots, w_m$ algebraically independent over $\overline{\mathbb{Q}}$. For every $2 \times 2$ configuration $x, y$ over $\overline{\mathbb{Q}}$ in $\overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u + \overline{\mathbb{Q}}w_1 + \dots + \overline{\mathbb{Q}}w_m$, each row and each column of $(x_iy_j)$ contains an entry outside $\overline{\mathbb{Q}}\,\mathcal{L}$. In particular the matrix is never a matrix of logarithms of algebraic numbers, so the four exponentials theorem can never be applied to it. It concerns candidates, so it is vacuous if Diaz's conjecture holds.

Proof. $u\bar u = |u|^2$ is algebraic, so DiazModulus.circle_point_config_constant_matrix_det_ne_zero with $K = \overline{\mathbb{Q}}$ gives an invertible $c$ with $x_iy_j - c_{ij} \in \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u$; every row and column of $c$ has a non-zero entry. Since $u, \bar u \in \mathcal{L}$ (DiazModulus.logAlg_conj_stable), $x_iy_j \in \overline{\mathbb{Q}}\,\mathcal{L}$ would put the algebraic number $c_{ij}$ in $\overline{\mathbb{Q}}\,\mathcal{L}$, and DiazModulus.algebraic_mem_span_logAlg_eq_zero would give $c_{ij} = 0$.

Novelty. Not found in the sources read; it is short: the invertible constant matrix (DiazModulus.circle_point_config_constant_matrix_det_ne_zero) plus Baker.