the lower bound  and using the fact that as $n \rightarrow \infty$, the summation converges to the mean values, $$\begin{aligned}
&\frac{1}{n}H({\mathsf{\kappa}}|\rvye^n) \\
 &= I(\rvu;\rvyr)+ H(\rvye|\rvu, \rvs)- H(\rvye)+ H(\rvs|\rvu)\!-
\! H(\rvs|\rvu,\rvye) \!- \!o_n(1)\!\\
 &= I(\rvu;\rvyr)-I(\rvye;\rvs|\rvu) - I(\rvye;\rvu)
 +I(\rvye;\rvs|\rvu)-o_n(1)\\
 &= I(\rvu;\rvyr) - I(\rvye;\rvu) - o_n(1)\end{aligned}$$as required.

Proof of Theorem \[thm:ubNoDisc\]
---------------------------------

A sequence of length-$n$ code satisfies: $$\begin{aligned}
&\frac{1}{n}H({\mathsf{\kappa}}|\rvyr^n) \le \eps_n~\label{eq:fano}\\
&\frac{1}{n}H({\mathsf{\kappa}}|\rvye^n) \ge \frac{1}{n}H({\mathsf{\kappa}})-
\eps_n~\label{eq:secrecy}\end{aligned}$$ where  follows from the Fano’s inequality since the receiver is able to recover the secret-key ${\mathsf{\kappa}}$ given $\rvyr^n$ and  is a consequence of the secrecy constraint. Furthermore, note that ${\mathsf{\kappa}}\rightarrow (\rvx^n,\rvs^n) \rightarrow
(\rvyr^n,\rvye^n)$ holds as the encoder generates the secret key ${\mathsf{\kappa}}$. Thus we can bound the rate $R = \frac{1}{n}H({\mathsf{\kappa}})$ as below: $$\begin{aligned}
nR &\le I({\mathsf{\kappa}};\rvyr^n|\rvye^n) + 2n\eps_n \notag\\
&\le I({\mathsf{\kappa}},\rvs^n,\rvx^n;\rvyr^n|\rvye^n) + 2n\eps_n \notag\\
&\le I(\rvs^n,\rvx^n;\rvyr^n|\rvye^n) + H({\mathsf{\kappa}}|\rvs^n,\rvx^n) + 2n\eps_n \notag\\
&= I(\rvs^n,\rvx^n;\rvyr^n|\rvye^n) + 3n\eps_n \label{eq:Fano_k}\\
&\le \sum_{i=1}^n I(\rvs_i,\rvx_i;\rvy_{r,i}|\rvy_{e,i}) + 3n\eps_n \label{eq:memChannel}\\
&\le nI(\rvx,\rvs;\rvyr|\rvye) + 3n\eps_n\end{aligned}$$ where  follows from the Fano Inequality because ${\mathsf{\kappa}}$ can be obtained from $(\rvx^n,\rvs^n)$,  from from the fact that the channel is memoryless and the last step follows from the concavity of the conditional entropy term $I(\rvx,\rvs;\rvyr|\rvye)$ in the input distribution $p_{\rvx,\rvs}$ (see e.g., [@khistiTchamWornell:07]).

Finally since the secret-key capacity only 