Bethe-vector eigenvalue conjecture for transfer matrices of the super Yangian

Let $M$ be a finite-dimensional irreducible $\mathrm{Y}(\mathfrak{gl}_{m|n})$-module of highest $\ell$-weight $\boldsymbol\zeta=(\zeta_i(u))_{i\in\bar I}^s$. Let $\boldsymbol l=(l_i)_{i\in I}$ be a sequence of non-negative integers, let $\boldsymbol t=(t_j^{(i)})$ with $i\in I$ and $j=1,\ldots,l_i$, define

$$y_i(u)=\prod_{j=1}^{l_i}(u-t_j^{(i)}),$$

and set $y_0(u)=y_{m+n}(u)=1$. Suppose that $\boldsymbol t$ satisfies the Bethe ansatz equations

$$\frac{\zeta_i(t_j^{(i)})}{\zeta_{i+1}(t_j^{(i)})}\frac{y_{i-1}(t_j^{(i)}+s_i)}{y_{i-1}(t_j^{(i)})}\frac{y_i(t_j^{(i)}-s_i)}{y_i(t_j^{(i)}+s_{i+1})}\frac{y_{i+1}(t_j^{(i)})}{y_{i+1}(t_j^{(i)}-s_{i+1})}=1,$$

for $i\in I$ and $j=1,\ldots,l_i$. Bethe-vector eigenvalue conjecture. Then

$$\mathfrak D(u,\tau;q)\,\mathbb B_{\boldsymbol l}(\boldsymbol t)=\mathfrak D(u,\tau,\boldsymbol\zeta,\boldsymbol y;q)\,\mathbb B_{\boldsymbol l}(\boldsymbol t).$$

Here $\mathbb B_{\boldsymbol l}(\boldsymbol t)$ is the Bethe vector and $\mathfrak D(u,\tau,\boldsymbol\zeta,\boldsymbol y;q)$ is the rational difference operator defined in the source. The conjecture predicts that a Bethe vector satisfying the Bethe ansatz equations is a simultaneous eigenvector for the transfer-matrix difference operator, with the displayed rational difference operator as eigenvalue. The source does not indicate whether this conjecture has been resolved.