probability $\prod\limits_{i=1}^np(x_{3,i}|x_{1,i},x_{2,i})$. Index them as $x_{3}^n(m_{3},m_2,m_1)$ where $m_{3}\in[1,2^{nR_{3}}]$.

***Encoding**:* In order to transmit the message $(m_{1},m_{2},m_3)$, Txu sends $x_{u}^n(m_{u})$ for $u\in\{1,2\}$ and Tx3 sends $x_{3}^n(m_{3},m_2,m_1)$.

***Decoding**:*

*Rx1:* After receiving $y_1^n$, Rx1 looks for a unique index $\hat{m}_{1}$ and some $(\hat{m}_{2},\hat{m}_{3})$ such that,

[c]{} (y\_1\^n,x\_[1]{}\^n(\_[1]{}),x\_[2]{}\^n(\_[2]{}),x\_[3]{}\^n(\_[3]{},\_2,\_1))A\_\^n(Y\_1,X\_1,X\_2,X\_3).

Using the packing lemma [@ElgKim11 P. 45] (or [@CovTho06 Theorem 15.2.3]), for large enough $n$, with arbitrarily high probability $\hat{m}_{1}=m_{1}$ if

[rcl]{} R\_[1]{}+R\_[3]{}&& I(X\_1,X\_3;Y\_1|X\_2)\[eqn:ach1\_rec1\_I\]\
R\_[1]{}+R\_[2]{}+R\_[3]{}&& I(X\_1,X\_2,X\_3;Y\_1).\[eqn:ach1\_rec1\_II\]

*Rx2:* The decoding process at Rx2 is similar to Rx1. Therefore, based on packing lemma [@ElgKim11 P. 45], the decoding error at Rx2 can be made sufficiently small if

[rcl]{} R\_[2]{}+R\_[3]{}&& I(X\_2,X\_3;Y\_2|X\_1)\[eqn:ach1\_rec2\_I\]\
R\_[1]{}+R\_[2]{}+R\_[3]{}&& I(X\_1,X\_2,X\_3;Y\_2).\[eqn:ach1\_rec2\_II\]

*Rx3:* After receiving $y_3^n$, Rx3 finds a unique index $\hat{\hat{m}}_{3}$ and some pair $(\hat{\hat{m}}_{1},\hat{\hat{m}}_{2})$ such that,

[c]{} (y\_3\^n,x\_[1]{}\^n(\_[1]{}),x\_[2]{}\^n(\_[2]{}),x\_[3]{}\^n(\_[3]{},\_2,\_1))A\_\^n(Y\_3,X\_1,X\_2,X\_3).

Using packing lemma [@ElgKim11 P. 45], With arbitrary high probability $\hat{\hat{m}}_{3}=m_3$, if $n$ is large enough and

[rcl]{} R\_[3]{}&& I(X\_3;Y\_3|X\_1,X\_2) \[eqn:ach1\_rec3\_I\]\
R\_[1]{}+R\_[3]{}&& I(X\_1,X\_3;Y\_3|X\_2)\[eqn:ach1\_rec3\_II\]\
R\_[2]{}+R\_[3]{}&& I(X\_2,X\_3;Y\_3 |X\_1)\[eqn:ach1\_rec3\_III\]\
R\_[1]{}+R\_[2]{}+R\_[3]{}&& I(X\_1,X\_2,X\_3;Y\_1).\[eqn:ach1\_rec3\_IV\]

This completes the proof.

The *codebook generation* and *encoding* parts of the proof follow the same lines as in Theorem \[thm:ach1\]. Therefore, we only describe the decoding part. Similar to Theorem \[thm:ach1\], all three receivers decode $m_1$, $m_2$ and $m_3$. However, here Rx1 (or Rx2) first decodes $m_2$ (or $m_1$). 