Entanglement breaking channels

Question

I can't get the proof of Theorem 2 of this. So we have a superoperator as $\Phi\in End(H_B)$ defined using a POVM $\{R_i\}_{i=1}^k$ meaning, every $R_i$ is a positive operator and $\sum_{i=1}^kR_i=I_{H_B}$ where $I_{H_B}$ is the identity on $H_B$, and a set of arbitrary states on $H_B$, say $\{F_i\}_{i=1}^k$, state means a normalized positive operator and normalized means trace equal to one $$\Phi(\rho):=\sum_{i=1}^k(tr(R_i\rho))R_i$$ Where $tr$ is the usual trace. Now we want to prove that $\Phi$ is entanglement breaking, so I have to show for every state $\sigma\in End(H_A\otimes H_B)$, outcome of $(I_{H_A}\otimes\Phi)(\sigma)$ is separable. $$\sigma\in End(H_A\otimes H_B)\longrightarrow \sigma=\sum_{i=1}^l\lambda_i\sigma_A^{(i)}\sigma_B^{(i)}$$ Now; $$\begin{array}{ll} I_{H_A}\otimes\Phi(\sigma) & =(I_{H_A}\otimes\Phi)(\sum_{i=1}^l\lambda_i\sigma_A^{(i)}\sigma_B^{(i)})\\ & =\sum_{i=1}^l\lambda_iI_{H_A}(\sigma_A^{(i)})\otimes\Phi(\sigma_B^{(i)})\\ & =\sum_{i=1}^l\lambda_i\sigma_A^{(i)}\otimes\sum_{j=1}^k(tr(F_j\sigma_B^{(i)}))R_j\end{array}$$ I can't continue from here, how can I get the convex combination?