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Is my conjecture below mathematically and linguistically correctly formulated and well formulated?

How can the conjecture be improved and shortened/simplified and made more intelligible? The conjecture should also be easily understood by non-mathematicians.

It's a new conjecture. The conjecture answers the solvability of some kinds of rational equations of transcendental functions by elementary numbers (a kind of closed-form numbers).

Conjecture:
Let
$f$ be a nonconstant elementary function,
$n\in\mathbb{N}_{\ge 1}$,
$P(x,y)\in\overline{\mathbb{Q}}[x,y]\setminus(\overline{\mathbb{Q}}[x]\cup\overline{\mathbb{Q}}[y])$ irreducible,
$Q(x,y)\in\overline{\mathbb{Q}}[x,y]$ not zero,
$p(x)\in\overline{\mathbb{Q}}[x]$,
$q(x)\in\overline{\mathbb{Q}}[x]$ not zero,
$R(x,y)=\frac{P(x,y)}{Q(x,y)\cdot q(x)^n}$,
$r(x)=\frac{p(x)}{q(x)}$ not constant
so that
$P(x,y)$ and $Q(x,y)$ coprime,
$p(x)$ and $q(x)$ coprime.
Suppose Schanuel's conjecture is true.
If a $\tilde{P}(x,y)\in\overline{\mathbb{Q}}[x,y]\setminus(\overline{\mathbb{Q}}[x]\ \cup\ \overline{\mathbb{Q}}[y])$ of $degree_x=n$ with $\frac{P(x,y)}{{q(x)}^n}=\tilde{P}(r(x),y)$ exists and $R(f(z_0),e^{r(f(z_0))})=0$ for $z_0\in\mathbb{C}$ and $r(f(z_0))\neq 0$, then $z_0$ is not an elementary number.

IV_
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  • I do not think that such a complex conjecture can be easily formulated. It is even more complicated than the Lindemann-Weierstrass-theorem. – Peter Jul 10 '20 at 12:14
  • @Peter If we wouldn't have the $Q(x,y)$ in the conjecture, we could formulate the conjecture e.g. with using the formulation "reduced rationals". – IV_ Jul 10 '20 at 12:18

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