Let $a,b,c,d \in \mathbb{C}[x,y]$. Assume that:
(1) Each two of $\{a,b,c,d\}$ are algebraically independent. In other words, the Jacobian of each two of $\{a,b,c,d\}$ is non-zero.
(2) $a,b$ are symmetric with respect to the involution $\beta: (x,y) \mapsto (x,-y)$ and $c,d$ are skew-symmetric w.r.t. $\beta$.
Is it true that $\mathbb{C}[a,b,c,d]=\mathbb{C}[x,y]$? Or at least (if this helps) that $\mathbb{C}[a,b,c,d,y]=\mathbb{C}[x,y]$?
I wonder if this can be proved or there exists a counterexample?
Of course, for general four elements not satisfying conditions (1) and (2), there exist counterexamples, such as: $\mathbb{C}[x^2,x^3,y^2,y^3]$.
A counterexample, satisfying condition (2) but not condition (1): $\mathbb{C}[x^2,x^3,y^3,y^5] \subsetneq \mathbb{C}[x,y]$ and $\mathbb{C}[x^2,x^3,y^3,y^5,y] \subsetneq \mathbb{C}[x,y]$.
A counterexample, satisfying condition (1) but not condition (2): $\mathbb{C}[x^2,x^2+y, y^2,x^2+y^2]$.
Edit: After writing this question I have noticed that $\mathbb{C}[x^2,x^2+y^2,y^3,xy]$ is a counterexample. So now I ask:
What additional conditions guarantee that $\mathbb{C}[a,b,c,d]=\mathbb{C}[x,y]$? Perhaps $\operatorname{Jac}(a+c,b+d) \in \mathbb{C}^{\times}$ or $\operatorname{Jac}(a+d,b+c) \in \mathbb{C}^{\times}$?
Remark: I thought considering $\mathbb{C}(x)[y]$ and $\mathbb{C}(y)[x]$, but not sure if and how this may help. For $\mathbb{C}[x^2,x^2+y^2,y^3,xy]$, we have: $\mathbb{C}(x)[x^2,x^2+y^2,y^3,xy]=\mathbb{C}(x)[y]$ and $\mathbb{C}(y)[x^2,x^2+y^2,y^3,xy]=\mathbb{C}(y)[x]$.
Now also asked in MO.
Thank you very much!
What if we also assume (3), which says that $\mathbb{C}(a,b,c)=\mathbb{C}(x,y)$ the field generated by $a,b,c$ (without $d$) equals the field of fractions of $\mathbb{C}[x,y]$? Could one find a counterexample? Observe that $\mathbb{C}[x^2,x^2+y^2,y^3,xy]$ does not satisfy the new condition (3) (not with $a,b,c$ and not with $a,b,d$). Indeed: $\mathbb{C}(x^2,x^2+y^2,y^3)=\mathbb{C}(x^2,y) \subsetneq \mathbb{C}(x,y)$