Proposition

Let $U$ be a quorum in FBAS $\ev{V, Q}$, let $B \subset V$ be a set of nodes, and let $U’ = U \setminus B$. If $U’ \ne \emptyset$, then $U’$ is a quorum in $\ev{V, Q}^B$.

Solution

Since $U’ \ne \emptyset$, it suffices to show that $\forall v \in U’, \exists q \in Q^B(v), q \subset U’$. Let $v \in U’$. Then $v \in U$. Since $U$ is a quorum in $\ev{V, Q}$, we can find $q \in Q(v)$ such that $q \subset U$. Then $q’ = q \setminus B \in Q^B(v)$, and $q’ = q \setminus B \subset U \setminus B = U’$. Therefore, $U’$ is a quorum in $\ev{V, Q}^B$.