The first step towards the decode-forward lower bound for the relay channel is the multihop lower bound. In this scheme the direct link between the sender and the receiver is not used and all the information flows first through the relay.
The relay channel is defined by $(\mathcal{X}_1 \times \mathcal{X}_2, p(y_2, y_3 | x_1, x_2), \mathcal{Y}_2, \mathcal{Y}_3)$ where $\mathcal{X}_1$ is the codeword transmitted by the transmitter, $\mathcal{X}_2$ the codeword transmitted by the relay, $\mathcal{Y}_2$ the output channel received by the relay and $\mathcal{Y}_3$ the output channel received by the receiver
The encoding scheme uses the block markov coding, which basically consits on sending $b-1$ messages over $b$ $n$-tranmission blocks. At each block j, the sender send the message $m_j$ while the relay sends $m_{j-1}$.
Regarding the codebook generation, we generate a codebook for each block,
\begin{equation*} \mathcal{C}_j = \{(x_1^n(m_j), x_2^n(m_{j-1})): m_{j-1}, m_{j} \in [1 : 2^{nR}]\},\;\; j\in [1:b] \end{equation*}
where $x_1(m_j)$ is generated according to $\prod_{i=1}^{n} p(x_{1i})$ and $x_2(m_{j-1})$ according to $\prod_{i=1}^{n} p(x_{2i})$.
The sender transmits $x_1^n(m_j)$ from codebook $\mathcal{C}_j$, being $m_j$ the message to be sent in block $j$.
The problem comes with the relay encoding scheme, which is the following: At the end of block $j$, the relay finds the unique message $\tilde{m}_j$ such that $(x_1^n(\tilde{m}_j), x_2^n(\tilde{m}_{j-1}), y_2^n(j)) \in \mathcal{T}_{\epsilon}^{(n)}$. I don't understand why we check for this triplet to be jointly typical. $x_2^n(\tilde{m}_{j-1})$ is independent from $x_1^n(\tilde{m}_j)$ and $y_2^n(j)$.
In the same way, i don't understand why the probability of the following event goes to 0 with n (suppose that the sender transmitted $m_j = 1$),
\begin{equation*} \tilde{\mathcal{E}}_1(j) = \{(X_1^n(1), X_2^n(\tilde{M}_{j-1}), Y_2^n(j)) \not\in \mathcal{T}_{\epsilon}^{(n)}\} \end{equation*}
I am following the "Network information theory" book by Abbas El Gamal and Young-Han Kim. What I have explained above comes from chapter 16 pag 387 from this book. Thank you very much in advance!