Can a fully general adaptive strategy attain a larger Stein exponent than every nested-adaptive strategy for discriminating two quantum superchannels? Fix finite-dimensional physical realizations

 \begin{equation}
 \Theta_i(\mathcal N)
 =\mathcal D_i\circ(\mathcal N\otimes\operatorname{id}_{S_i})
 \circ\mathcal E_i,
 \qquad i\in\{1,2\}.
\tag{1}
\end{equation} 
The systems $S_i$ in Eq. (1) are internal memories. A nested strategy recursively places complete uses of $\Theta_i$ inside one another. A fully general strategy may interleave the preprocessing and postprocessing components of different uses in any causally valid order, provided each $\mathcal E_i$ precedes its matched $\mathcal D_i$ and the tester cannot access the internal memory $S_i$. Define the vanishing-type-I-error Stein exponent for a strategy class $\mathsf S$ by

 \begin{equation}
 \zeta_{\mathsf S}(\Theta_1\|\Theta_2)
 :=\lim_{\varepsilon\downarrow0}\liminf_{n\to\infty}
 -\frac1n\log_2
 \inf_{\substack{P\in\mathsf S_n:\alpha_n(P)\leq\varepsilon}}
 \beta_n(P).
\tag{2}
\end{equation} 
The definition in Eq. (2) uses the type-I and type-II errors $\alpha_n(P)$ and $\beta_n(P)$ of protocol $P$. Does there exist a pair of realizations for which

 \begin{equation}
 \zeta_{\mathrm{fg}}(\Theta_1\|\Theta_2)
 >\zeta_{\mathrm{nest}}(\Theta_1\|\Theta_2),
\tag{3}
\end{equation} 
where $\mathrm{fg}$ denotes fully general strategies? If not, prove equality in Eq. (3) with $>$ replaced by $=$ for all superchannel pairs.
