first term in the RHS of (\[eqn:set1\_III\]). Note that, $X_3$ is required in this condition due to the dependance on $M_2$. Similarly, the condition of the first term in the RHS of (\[eqn:set1\_II\]) enables Rx2 to decode $M_1$. Moreover, $(M_1,M_2)$ can be decoded at Rx3 under (\[eqn:set1\_IV\]) and the second terms in the RHS of (\[eqn:set1\_II\]) and (\[eqn:set1\_III\]). Therefore, ${\mathcal{C}}_1$ gives the capacity region for a compound three-user MAC with common information, where $R_1$ and $R_2$ are the common rates of Tx1-Tx3 and Tx2-Tx3, respectively, $R_3$ is the private rate for Tx3, and the private rates for Tx1 and Tx2 are zero.

\[remark:cap\_set1\_II\] If we omit the second pair, i.e., $X_2=Y_2=\emptyset$ and $R_2=0$, the model reduces to a two-user C-IFC and ${\mathcal{C}}_1$ coincides with the capacity region of the strong interference channel with unidirectional cooperation, which was characterized in [@MariYatKra07].

First, we provide a useful lemma which we need in the proof of the converse part for Theorem \[thm:cap\_set1\].

\[lemma:cond\_str\^n\] If (\[eqn:set1\_I\])-(\[eqn:set1\_III\]) hold for all distribution $p(.)\in{\mathcal{P}}$, then we have:

[rcl]{} I(X\_3\^n;Y\_3\^n|X\_1\^n,X\_2\^n,U)&&I(X\_3\^n;Y\_1\^n|X\_1\^n,X\_2\^n,U)\[eqn:set1\_I\^n1\]\
I(X\_3\^n;Y\_3\^n|X\_1\^n,X\_2\^n,U)&&I(X\_3\^n;Y\_2\^n|X\_1\^n,X\_2\^n,U)\[eqn:set1\_I\^n2\]\
I(X\_1\^n,X\_3\^n;Y\_1\^n|X\_2\^n,U)&&I(X\_1\^n;Y\_2\^n|X\_2\^n,U)\[eqn:set1\_II\^n\]\
I(X\_2\^n,X\_3\^n;Y\_2\^n|X\_1\^n,U)&&I(X\_2\^n;Y\_1\^n|X\_1\^n,U).\[eqn:set1\_III\^n\]

Proof relies on the result in [@KorMar77 Proposition 1] and follows the same lines as in [@MariYatKra07 Lemma 5] and [@CosElg79 Lemma].

Considering (\[eqn:set1\_II\])-(\[eqn:set1\_IV\]), the 