For u transcendental over the field of ℒ̃ with uū algebraic: if ℚ[S] has transcendence degree ≤ 1 and contains c + pu + qū with c ∈ ℒ and (p, q) ≠ (0, 0), its only element of ℒ is 0

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 $\mathcal{L} + \mathbb{Q}u + \mathbb{Q}\bar u$. Let $S \subseteq \mathbb{C}$ be any set such that $\mathbb{Q}[S]$, the $\mathbb{Q}$-subalgebra of $\mathbb{C}$ generated by $S$, has transcendence degree at most $1$ over $\mathbb{Q}$. Let $c \in \mathcal{L}$ and $p, q \in \mathbb{Q}$, not both zero, with $c + pu + q\bar u \in \mathbb{Q}[S]$. If $w \in \mathcal{L}$ lies in $\mathbb{Q}[S]$, then $w = 0$.

Proof. Algebraic numbers, rationals and $\mathcal{L}$ lie in $F$, so DiazModulus.transcendental_add_mul_add_mul_conj makes $e = c + pu + q\bar u$ transcendental over $F$, hence over $\mathbb{Q}$. In a ring of transcendence degree at most one every element is algebraic over $\mathbb{Q}(e)$, since otherwise it would form with $e$ an algebraically independent pair, two elements against the bound; so $w$ is algebraic over $\mathbb{Q}(e)$. As $w \in F$ and $e$ is transcendental over $F$, DiazModulus.isAlgebraic_of_mem_of_isAlgebraic_adjoin_transcendental makes $w$ algebraic, and Hermite–Lindemann (DiazModulus.hermite_lindemann_holds), with $e^w$ algebraic, forces $w = 0$.

Novelty. Not found in the sources read; routine.