Let $\mathcal{B}\subset\mathcal{A}$ be an inclusion of unital $C^*$-algebras. Let $\mathcal{C}$ be an unital simple subalgebra of $\mathcal{A}$ which is under the image of a faithful conditional expectation from $\mathcal{A}$, i.e., there exists a unital completely positive projection $\Phi: \mathcal{A}\to\mathcal{C}$. Moreover, $\Phi(\mathcal{B})\subsetneq \mathcal{C}$. Suppose that $I\triangleleft\mathcal{A}$ is a non-trivial two sided closed ideal in $\mathcal{A}$.
Standing assumption: $\mathcal{A}$ is generated as a $C^*$-algebra by $\mathcal{B}$ and $\mathcal{C}$.
Claim: $\mathcal{B}\subsetneq\mathcal{B}+I\subsetneq \mathcal{A}$.
Attempt: Since $I\triangleleft\mathcal{A}$ is an ideal of $\mathcal{A}$, it follows that $I+\mathcal{B}$ is a $C^*$-subalgebra of $\mathcal{A}$. Now, $\Phi(I)\triangleleft\mathcal{C}$ is an ideal of $\mathcal{C}$. Since $\mathcal{C}$ is simple, it follows that $\Phi(I)=0$ or $\Phi(I)=\mathcal{C}$. Since $\Phi$ is faithful, it must be the case that $\Phi(I)=\mathcal{C}$. Since $\Phi(\mathcal{B})\subsetneq \mathcal{C}$, it follows that $\mathcal{B}\subsetneq \mathcal{B}+I$.
Now, $I\cap\mathcal{C}$ is an ideal of $\mathcal{C}$. Since $\mathcal{C}$ is simple and $I$ is non-trivial, it must be the case that $I\cap\mathcal{C}=0$. Here, I am using the assumption that $\mathcal{A}$ is generated by $\mathcal{B}$ and $\mathcal{C}$. All that remains to be shown is that $\mathcal{B}+I\subsetneq \mathcal{A}$. I am trying to show that $\left(\mathcal{B}+I\right)\cap \mathcal{C}=\mathcal{B}\cap\mathcal{C}$. Once this is establishes, we shall have that $\mathcal{B}\cap\mathcal{C}\subset \Phi(\mathcal{B})\subsetneq\mathcal{C}$ which will ensure that $\mathcal{B}+I\subsetneq \mathcal{A}$.
I am unable to do it. I need to use that $I\cap\mathcal{C}=0$. How do I do it?
Thank you for your time.