proof for bound (\[eq:Ra\_lossless\]) follows immediately since $A^{n}$ is selected only as a function of message $M_{b}$. As for the other two bounds, namely (\[eq:R1+Rb\_lossless\])-(\[eq:R2+Rb\_lossless\]), the proof of the converse can be established following cut-set arguments and using the point-to-point result of [@Permuter]. For achievability, we use the code structure proposed in [@Permuter] along with rate splitting. Specifically, Node 1 first maps sequence $X^{n}$ into the action sequence $A^{n}$. This mapping requires a codebook of rate $I(X;A)$. This rate has to be conveyed over link $R_{b}$ by the definition of the problem and is thus received by both Node 2 and Node 3. Given the so obtained sequence $A^{n}$, communicating $X$ losslessly to Node 2 requires rate $H(X|A,Y)$. We split this rate into two rates $r_{1b}$ and $r_{1d}$, such that the message corresponding to the first rate is carried over the broadcast link of rate $R_{b}$ and the second on the direct link of rate $R_{1}$. Note that Node 2 can thus recover sequence $X$ losslessly. The rate $H(X|A,Z)$ which is required to send $X$ losslessly to Node 3, is then split into two parts, of rates $r_{2b}$ and $r_{2d}$. The message corresponding to the rate $r_{2b}$ is sent to 