quantum mechanics - von Neumann Entropy of a joint state

Definition 1 The von Neumann entropy of a density matrix is given by $$S(\rho) := - \mathrm{Tr}[\rho \ln \rho] = H[\lambda (\rho)] $$ where $H[\lambda (\rho)] $ is the Shannon entropy of the set of probabilities $\lambda (\rho) $ (which are eigenvalues of the density operator $\rho$).

Definition 2 If a system is prepared in the ensemble $\{ p_j, \rho_j \} $ then we define the Holevo $\chi$ quantity for the ensemble by $$\chi := S(\rho) - \sum_j p_j S(\rho_j) $$

Short Question : Let $A$ and $B$ be two quantum systems in a state of the form $$\rho^{(AB)} = \sum_i q_i |a_{i}^{(A)} \rangle \langle a_{i}^{(A)} | \otimes \rho_{i}^{(B)} $$ where the states $|a_{i}^{(A)} \rangle$ are orthogonal. What property of the von Neumann entropy implies that the von Neumann entropy of the joint state is $$S(\rho^{(AB)}) = H(\vec{q}) + \sum_i q_i S(\rho_{i}^{(B)})? $$

Proposal: I'm pretty sure it makes use of the following properties of von Neumann entropy: $$S(\rho_{A} \otimes \rho_{B}) = S(\rho_{A}) + S(\rho_{B})$$ and if $\rho_{A} = \sum_{x} p_x | \phi_x \rangle \langle \phi_x|$ then $$S(\rho_{A} \otimes \rho_{B}) = H(X) + S(\rho_{B})$$ where $X = \{| \phi_x \rangle, p_x \}$.

Thanks for any assistance.

Answer

$$ \rho^{(AB)} = \sum_i q_i |a_i^{(A)}\rangle \langle a_i^{(A)}| \otimes \rho_i^{(B)} \tag 1 $$ Note that subsystem $A$ and $B$ are separable, let $$ \rho^{(B)}_i = \sum_j \lambda^j_i |{b_i^j}^{(B)}\rangle \langle {b_i^j}^{(B)}|. \tag 2 $$ Substitute equation $(2)$ into $(1)$: $$ \eqalignno{ S(\rho^{(AB)}) &= -\sum_{ij} q_i \lambda_i^j \ln(q_i \lambda_i^j) \\ &= -\sum_{ij} q_i \lambda_i^j (\ln q_i + \ln \lambda_i^j) \\ &= -\sum_i q_i \ln q_i - \sum_i q_i \sum_j \lambda_i^j \ln \lambda_i^j \\ &= H(\vec q) + \sum_i q_i S(\rho_i^{(B)}). &(3) } $$