ℒ itself has all the Theorem D properties, and Diaz's conjecture holds exactly when no non-zero element of ℒ has algebraic modulus

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 $L = \mathrm{span}_{\mathbb{Q}}(\mathcal{L})$ and $\widetilde L = \mathrm{span}_{\overline{\mathbb{Q}}}(\{1\} \cup L)$. Then Diaz's conjecture (DiazModulus.DiazModulusConjecture: $e^u$ is transcendental for every $u \neq 0$ with $|u|$ algebraic) holds if and only if no $u \in L$ with $u \neq 0$ has $|u|$ algebraic, and $L$ has the following properties:

• (conjugation) $\bar z \in L$ for every $z \in L$;
• (Baker) if $n \geq 0$, $l_1, \dots, l_n \in L$ are linearly independent over $\mathbb{Q}$, and $b_0, b_1, \dots, b_n$ are algebraic numbers, not all zero, then $b_0 + \sum_{i=1}^{n} b_il_i \neq 0$;
• (Hermite–Lindemann) every algebraic $z \in L$ is $0$;
• (Gel'fond–Schneider) if $l_0 \neq 0$ and $\beta l_0$ both lie in $L$ and $\beta$ is algebraic, then $\beta \in \mathbb{Q}$;
• (Roy, relative to $\widetilde{\mathcal{L}}$) if no $x_1, x_2$, linearly independent over $\overline{\mathbb{Q}}$, and $y_1, y_2, y_3$, linearly independent over $\overline{\mathbb{Q}}$, have all six products $x_iy_j$ in $\widetilde{\mathcal{L}}$, then no such families have all six products in $\widetilde L$;
• (six exponentials) 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 $L$;
• (four exponentials in transcendence degree one) if $\ell_{11}, \ell_{12}, \ell_{21}, \ell_{22} \in L$ are non-zero, $\ell_{11}\ell_{22} = \ell_{12}\ell_{21}$, and the $\mathbb{Q}$-algebra they generate has transcendence degree at most $1$ over $\mathbb{Q}$, then there are $a, b \in \mathbb{Q}$, not both zero, with $a\ell_{11} + b\ell_{21} = a\ell_{12} + b\ell_{22} = 0$ (the rows of the matrix $(\ell_{ij})$ are linearly dependent over $\mathbb{Q}$), or with $a\ell_{11} + b\ell_{12} = a\ell_{21} + b\ell_{22} = 0$ (its columns are).

Since $\mathcal{L}$ is a $\mathbb{Q}$-vector space, $L = \mathcal{L}$ and $\widetilde L = \widetilde{\mathcal{L}}$. The Roy clause is an implication: Roy's strong six exponentials theorem for $\widetilde{\mathcal{L}}$ (Roy 1992, Cor. 2) is its hypothesis, not a hypothesis of the node, so the statement is unconditional.

What it means. DiazModulus.span_logAlg_theorem_D_properties shows that $\mathcal{L}$ itself has these properties and that Diaz's conjecture says exactly that no non-zero element of $\mathcal{L}$ has algebraic modulus, while DiazModulus.theorem_D_barrier gives a $\mathbb{Q}$-space $L' \supseteq \mathcal{L}$ with the same properties that does contain such an element. So the listed properties, as axioms about $(\mathcal{L}, \widetilde{\mathcal{L}})$ (with Roy's theorem for $\widetilde{\mathcal{L}}$ supplying the hypothesis of the Roy clause), do not prove Diaz's conjecture: an argument from them alone would apply to $L'$ and its $\overline{\mathbb{Q}}$-span with $1$ as well. Here a candidate for $L'$ means a non-zero $u \in L'$ with $|u|$ algebraic; it is not DiazModulus.IsCandidate, which asks for $e^u$ to be algebraic. The barrier is a model-existence statement about membership only: the model has no exponential function, so it says nothing about arguments through the values $e^{x_iy_j}$, such as Waldschmidt's theorem of 1973, or through Roy's rank bounds for larger matrices.

Proof. The $\mathbb{Q}$-span of $\mathcal{L}$ is $\mathcal{L}$, by DiazModulus.mem_logAlg_adjoin_circle_point_iff at $u = 0$, so membership in $L$ means $e^z$ algebraic, and Diaz's conjecture is the contrapositive of the statement that no non-zero $u \in L$ has $|u|$ algebraic. The clauses are then the classical theorems read off their nodes: conjugation is DiazModulus.logAlg_conj_stable, Baker is Schanuel.baker_linear_forms_in_logarithms, Hermite–Lindemann is DiazModulus.hermite_lindemann_holds, Gel'fond–Schneider is Schanuel.gelfond_schneider (a non-rational $\beta$ would make $e^{\beta l_0}$ transcendental), six exponentials is DiazModulus.six_exponentials, and four exponentials is DiazModulus.four_exponentials_trdeg_one, verbatim. The Roy clause holds because $\widetilde L$ is contained in $\widetilde{\mathcal{L}}$.

Novelty. Not found in the sources read as a bundle; routine. The clauses are classical theorems about $\mathcal{L}$: Baker (1966), Hermite–Lindemann, Gel'fond–Schneider (1934), the six exponentials theorem (Lang 1966, Ramachandra 1968) and the four exponentials theorem in transcendence degree one (Waldschmidt 1973, Brownawell 1974); the Roy clause is relative, so Roy's theorem (1992, Cor. 2) is not needed.