Then, it jointly decodes $m_1$ (or $m_2$) and $m_3$.

***Decoding**:*

*Rx1:* After receiving $y_1^n$, Rx1 first finds a unique index $\hat{m}_{2}$ such that,

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

Applying packing lemma [@ElgKim11 P. 45], with arbitrary high probability $\hat{m}_{2}=m_{2}$, if $n$ is large enough and

[rcl]{} R\_[2]{}&& I(X\_2;Y\_1). \[eqn:ach2\_rec1\_I\]

Then, it looks for a unique index $\hat{m}_{1}$ and some $\hat{m}_{3}$ such that,

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

For large enough $n$, with arbitrarily high probability $\hat{m}_{1}=m_{1}$ if (\[eqn:ach1\_rec1\_I\]) holds.

*Rx2:* Rx2 proceeds similarly. This step can be accomplished with sufficiently small probability of error for large enough $n$, if (\[eqn:ach1\_rec2\_I\]) holds and

[rcl]{} R\_[1]{}&& I(X\_1;Y\_2).\[eqn:ach2\_rec2\_I\]

The decoding procedure at Rx3 is similar to Theorem \[thm:ach1\] and the error in this receiver can be bounded, if (\[eqn:ach1\_rec3\_I\])-(\[eqn:ach1\_rec3\_IV\]) hold. This completes the proof.

Proof of the Converse Part for Theorem \[thm:Gaus\_cap\_set1\] {#app:thm:Gaus_converse}
==============================================================

For any rate triple $(R_1,R_2,R_3)\in{\mathcal{C}}$, Rx1 is able to decode $m_1$ reliably. Assume that Rx1 knows $X_2$ by a genie. Obviously, the genie aided channel has a larger capacity region than ${\mathcal{C}}$. Now, Rx1 knows $X_1$ from $m_1$ and $X_2$ from genie. Then, Rx1 is able to construct

[rcl]{} \_3&=&(Y\_1-h\_[11]{}X\_1-h\_[12]{}X\_2)+h\_[13]{}X\_1+h\_[23]{}X\_2\
&=&h\_[13]{}X\_1+h\_[23]{}X\_2+h\_[33]{}X\_3+Z\_1

If condition (\[eqn:set1\_I\_Gaus\]) holds, then $\tilde{Y}_3$ is a less noisy version of $Y_3$. Since 