\begin{theorem}[Integrity] If a message $m$ is delivered by any process, then it was previously broadcast by some process via the \texttt{AB-broadcast} primitive. \end{theorem} \begin{proof} Let $j$ be a process such that $\texttt{AB-deliver}_j(m)$ occurs. \begin{align*} &\texttt{AB-deliver}_j(m) & \text{(line 18)} \\ \Rightarrow\; & m \in \texttt{ordered}(T),\ \text{with } T = \bigcup_{j' \in \textit{winner}^r} \textit{prop}[r][j'] \setminus \textit{delivered} & \text{(lines 16-17)} \\ \Rightarrow\; & \exists j_0,\ r_0 : m \in \textit{prop}[r_0][j_0] & \text{(line 16)} \\ \Rightarrow\; & \textit{prop}[r_0][j_0] = S,\ \text{with } \texttt{RB-delivered}_{j}(PROP, S, r_0, j_0) & \text{(line 22)} \\ \Rightarrow\; & S \text{ was sent in } \texttt{RB-cast}(PROP, S, r_0, j_0) & \text{(line 9)} \\ \Rightarrow\; & S = \textit{received}_{j_0} \setminus \textit{delivered}_{j_0} & \text{(line 6)} \\ \Rightarrow\; & m' \in \textit{received}_{j_0}\ \text{where } m' \text{ broadcast by } j_0 & \text{(line 4)} \\ \Rightarrow\; & \textbf{if } m = m' \\ & \quad \Rightarrow \texttt{RB-Broadcast}_{j_0}(m) \text{ occurred} & \text{(line 3)} \\ & \quad \Rightarrow \texttt{AB-Broadcast}_{j_0}(m) \text{ occurred} & \text{(line 1)} & \hspace{1em} \square \\ & \textbf{else: } m \in \textit{received}_{j_0} \setminus \textit{delivered}_{j_0} \\ & \quad \Rightarrow m \in \textit{received}_{j_0} & \text{(line 4)} \\ & \quad \Rightarrow \texttt{RB-delivered}_{j_0}(m) \text{ occurred} & \text{(line 3)} \\ & \quad \Rightarrow \exists j_1 : \texttt{RB-Broadcast}_{j_1}(m) \text{ occurred} & \text{(line 2)} \\ & \quad \Rightarrow \texttt{AB-Broadcast}_{j_1}(m) \text{ occurred} & \text{(line 1)} & \hspace{1em} \square \end{align*} Therefore, every delivered message $m$ must originate from some call to \texttt{AB-Broadcast}. \end{proof} \begin{theorem}[No Duplication] No message is delivered more than once by any process. \end{theorem} \begin{proof} Assume by contradiction that a process $j$ delivers the same message $m$ more than once, i.e., \[ \texttt{AB-deliver}_j(m) \text{ occurs at least twice.} \] \begin{align*} &\texttt{AB-deliver}_j(m) \text{ occurs} & \text{(line 19)} \\ \Rightarrow\; & m \in \texttt{ordered}(T),\ \text{where } T = \bigcup_{j' \in \textit{winner}^r} \textit{prop}[r][j'] \setminus \textit{delivered} & \text{(lines 16-17)} \\ \Rightarrow\; & m \notin \textit{delivered} \text{ at that time} \\ \\ \text{However:} \\ & \texttt{delivered} \gets \texttt{delivered} \cup \{m\} & \text{(line 18)} \\ \Rightarrow\; & m \in \textit{delivered} \text{ permanently} \\ \Rightarrow\; & \text{In any future round, } m \notin T' \text{ since } T' = \bigcup_{j' \in \textit{winner}^r} \textit{prop}[r'][j'] \setminus \textit{delivered} \\ \Rightarrow\; & m \text{ will not be delivered again} \\ \Rightarrow\; & \text{Contradiction.} \end{align*} Therefore, no message can be delivered more than once by the same process. $\square$ \end{proof} \begin{theorem}[Validity] If a correct process invokes $\texttt{AB-Broadcast}_j(m)$, then all correct processes eventually deliver $m$. \end{theorem} \begin{proof} Let $j$ be a correct process such that $\texttt{AB-Broadcast}_j(m)$ occurs (line 5). \begin{align*} &\texttt{AB-Broadcast}_j(m) & \text{(line 1)}\\ \Rightarrow\; & \texttt{RB-Broadcast}_j(m) \text{ occurs} & \text{(line 2)} \\ \Rightarrow\; & \forall j_0 : \texttt{RB-delivered}_{j_0}(m) & \text{(line 3)} \\ \Rightarrow\; & m \in \textit{received}_{j_0} & \text{(line 4)} \\ \Rightarrow\; & \textbf{if } m \in \texttt{delivered}_{j_0} \\ & \quad \Rightarrow \textit{delivered}_{j_0} \gets textit{delivered}_{j_0} \cup \{m\} & \text{(line 18)} \\ & \quad \Rightarrow \texttt{AB-delivered}_{j_0}(m) & \text{(line 19)} & \hspace{1em} \square \\ & \textbf{else } m \notin \textit{delivered}_{j_0} : \\ & \quad \Rightarrow m \in S_{j_0}\ \text{since } S_{j_0} = \textit{receieved}_{j_0} \setminus \textit{delivered}_{j_0} & \text{(line 6)} \\ & \quad \Rightarrow \exists r : \texttt{RB-cast}_{j_0}(texttt{PROP}, S_{j_0}, r, j_0) & \text{(line 9)} \\ & \quad \quad \Rightarrow \forall j_1 : \texttt{RB-Deliver}_{j_1}(\texttt{PROP}, S_{j_0}, r, j_0)\ \text{occurs} & \text{(line 21)} \\ & \quad \quad \Rightarrow \textit{prop}[r][j_0] = S_{j_0} & \text{(line 22)} \\ & \quad \Rightarrow \exists j_2 \in j_0 : \texttt{PROVE}_{j_2}(r)\ \text{is valid} & \text{(line 10)} \\ & \quad \Rightarrow j_2 \in textit{winner}^r & \text{(line 14)} \\ & \quad \Rightarrow T_{j_0} \ni \textit{prop}[r][j_2] \setminus \textit{delivered}_{j_0} & \text{(line 16)} \\ & \quad \Rightarrow T_{j_0} \ni S_{j_2} \setminus \textit{delivered}_{j_0} \ni m & \text{(line 16)} \\ & \quad \Rightarrow \texttt{AB-deliver}_{j_0}(m) & \text{(line 19)} & \hspace{1em} \square \\ \end{align*} \end{proof} \begin{theorem}[Total Order] If two correct processes deliver two messages $m_1$ and $m_2$, then they deliver them in the same order. \end{theorem} \begin{proof} \begin{align*} & \forall j_0 : \texttt{AB-Deliver}_{j_0}(m_0) \wedge \texttt{AB-Deliver}_{j_0}(m_1) & \text{(line 19)} \\ \Rightarrow\; & \exists r_0, r_1 : m_0 \in \texttt{ordered}(T^{r_0}) \wedge m_1 \in \texttt{ordered}(T^{r_1}) & \text{(line 17)} \\ \Rightarrow\; & T^{r_0} = \bigcup_{j' \in \textit{winner}^{r_0}} \textit{prop}[r_0][j'] \setminus \textit{delivered}\ \wedge \\ & T^{r_1} = \bigcup_{j' \in \textit{winner}^{r_1}} \textit{prop}[r_1][j'] \setminus \textit{delivered} & \text{(line 16)} \\ \Rightarrow\; & \textbf{if } r_0 = r_1 \\ & \quad \Rightarrow T^{r_0} = T^{r_1} \\ & \quad \Rightarrow m_0, m_1 \in \texttt{ordered}(T^{r_0})\ \text{since \texttt{ordered} is deterministic} \\ & \quad \Rightarrow \textbf{if } m_0 < m_1 : \\ & \quad \quad \Rightarrow \texttt{AB-Deliver}_{j_0}(m_0) < \texttt{AB-Deliver}_{j_0}(m_1) & & \hspace{1em} \square\\ & \textbf{else if } r_0 < r_1 \\ & \quad \Rightarrow \forall m \in T^{r_0}, \forall m' \in T^{r_1} : \texttt{AB-Deliver}(m) < \texttt{AB-Deliver}(m') & & \hspace{1em} \square\\ \end{align*} Therefore, for all correct processes, messages are delivered in the same total order. \end{proof}