Proposition 1

In an FBAS $(V, Q)$, the union of two quorums is a quorum.

Proof

Let $U_1, U_2$ be two quorums. Let $v \in U_1 \cup U_2$. Then $v \in U_i$ for $i = 1$ or $i = 2$. Then $q \subset U_i$ for some $q \in Q(v)$. Therefore, $q \subset U_1 \cup U_2$, so $U_1 \cup U_2$ is indeed a quorum.

Proposition 2

In an FBAS $(V, Q)$, the intersection of two quorums is not necessarily a quorum.

Proof

Let $V = \{ v_1, \cdots, v_4 \}$ and

• $Q(v_1) = \{ \{ v_1, v_2, v_3 \}, \{ v_1, v_2, v_4 \}, \{ v_1, v_3, v_4 \} \}$,
• $\vdots$
• $Q(v_4) = \{ \{ v_1, v_2, v_4 \}, \{ v_1, v_3, v_4 \}, \{ v_2, v_3, v_4 \} \}$.

In other words, $Q(v_i) = \{ \{ v_i, v_j, v_k \} \mid i \ne j, j \ne k, i \ne k \}$.

Then $U_1 = \{ v_1, v_2, v_3 \}$ is a quorum, and $U_2 = \{ v_2, v_3, v_4 \}$ is a quorum. However, $U_1 \cap U_2 = \{ v_2, v_3 \}$ is not a quorum because the size of any quorum slice is 3.

Proposition 3

In an FBAS $(V, Q)$, $V$ is a quorum.

Proof

For any $v \in V$, for any $q \in Q(v)$, $q \subset V$. Therefore, $V$ is indeed a quorum.