Quantum discord of a tripartite system A:BC

I know that the quantum discord of a bipartite system can be determined as:

$${D_A}({\rho _{AB}}) = I({\rho _{AB}}) - {J_A}({\rho _{AB}}),$$ The subscript of $$A$$ denotes that the measurement has been performed on the subsystem $$A$$. The quantum mutual information is defined by: $$I({\rho _{AB}}) = S({\rho _A}) + S({\rho _B}) - S({\rho _{AB}}),$$ and the classical correlation: $${J_A}({\rho _{AB}}) = S({\rho _B}) - \mathop {\min }\limits_{\{ \Pi _I^A\} } \sum\limits_i {{p_i}S({\rho _{\left. B \right|i}})} ,$$ $$S({\rho _{\left. B \right|i}}) = {1 \over {{p_i}}}t{r_A}\left( {\Pi _i^A \otimes {I_B}} \right){\rho _{AB}}\left( {\Pi _i^A \otimes {I_B}} \right)$$

The question is, in the case of a tripartite state, how can someone compute the quantum discord between $$A$$ and $$BC$$ as a whole? i.e., $${D_A}({\rho _{A:BC}})$$ when the measurement is performed on the subsystem $$A$$? What will $${D_A}({\rho _{A:BC}})$$ and $${J_A}({\rho _{A:BC}})$$ look like??

• Why is the distinction between B and C important in this context? Couldn't you simply relabel BC as B'? Do I miss something? Aug 14 '21 at 7:56
• @M.Stern Thank you for your reply. In fact, I do not know if this procedure is possible. Do I consider this an answer to my question? Aug 14 '21 at 9:50
• In that case maybe try out if it works for you to consider BC as one system and if it does you can answer your own question ;-) Aug 15 '21 at 17:25

$$D_{A}(A:BC)=I(A:BC) - J_{A}(A:BC)$$where $$J_{A}(A:BC)=max_{\{\Pi_{i}^{A}\}}(S(BC)-\sum_{i}p_{i}S(BC_{i}))$$ In this case, $$S(BC_{i})=Tr_{A}((\Pi_{i}^{A}\otimes I_{BC})\rho_{ABC}(\Pi_{i}^{A}\otimes I_{BC}))/(Tr((\Pi_{i}^{A}\otimes I_{BC})\rho_{ABC}(\Pi_{i}^{A}\otimes I_{BC})))$$
Anytime you wish to calculate the discord, the remaining conditional states will always be the residual states obtained after applying one of the projectors in the set $$\{\Pi_{i}\}$$ and then tracing out over said subsystem.