144 lines
9.0 KiB
TeX
144 lines
9.0 KiB
TeX
We define $W^t$ as the set of processes that are winners in round $r$ at time $t$.
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\begin{theorem}
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$\forall j_1, j_2 \text{ corrects}, W_{j_1} = W_{j_2}$, where $W_j$ is the set of processes that are winners in round $r$.
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\end{theorem}
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\begin{proof}
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\begin{align*}
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J = \{j_1, ..., j_n\} & \text{(set of all processes)} \\
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B \subseteq J, B = \{b_1, ..., b_f\} & \text{(set of faulty processes)} \\
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C \subseteq J, C = \{c_1, ..., c_{n-f}\} & \text{(set of correct processes)} \\
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\textbf{Let's assume } \exists b_1 \in B, \exists t_0 & \text{ such that } \texttt{R-Broadcast}_{b_1}(PROP, S, r, b_1) \text{ occurs} \\
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\Rightarrow\; & \exists K^{t_0} \subseteq C \text{ such that } \forall k \in K^{t_0}, \texttt{R-Delivered}_k(PROP, S, r, b_1) \text{ occurs} \\
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& \wedge |K^{t_0}| = n - 2f \\
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\Rightarrow\; & \texttt{PROVE}_k[k](<r, b_1>) \text{ is valid for all } k \in K^{t_0} \\
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\Rightarrow\; & b_1 \not\in W^{t_0} \text{ since } \texttt{PROVE}_k[k](<r, b_1>) \text{ is valid less than } n - f \text{ times} \\
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\text{in the same way,} & \\
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\textbf{Let's assume } \exists L^{t_0} \subseteq C \text{ such that } & \forall l \in L^{t_0}, \texttt{R-Broadcast}_{l}(PROP, \_, r, l) \text{ occurs} \\
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\textbf{And } \exists M^{t_0} \subseteq C \text{ such that } & \forall m \in M^{t_0}, \texttt{R-Delivered}_m(PROP, \_, r, m) \text{ occurs} \\
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& \text{with } |L^{t_0}| = n - f \text{ and } |M^{t_0}| = n - f \\
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\Rightarrow\; & \forall m, l : \exists (m, PROVE(<r, l>)) \in \texttt{READ}[m]() \\
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& \textbf{And because } |M^{t_0}| \geq n - f \\
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\Rightarrow\; & \exists O^{t_0} \subseteq M^{t_0} \text{ such that } \forall o \in O^{t_0}, W^{t_0}_o \not\ni b_1 \\
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& \exists t_1 \geq t_0 : \forall b \in B, \texttt{PROVE}_b[b](<r, b_1>) \text{ occurs} \\
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\Rightarrow\; & \textbf{at time } t_1, \forall k \in K : \exists (k, \texttt{PROVE}(<r, b_1>)) \in \texttt{READ}[k]() \\
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& \textbf{And } \forall b \in B, \exists (b, \texttt{PROVE}(<r, b_1>)) \in \texttt{READ}[b]() \\
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\Rightarrow\; & \textbf{Because } |K| + |B| = n - 2f + f = n - f \text{ the condition is satisfied} \\
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\Rightarrow\; & W^{t_1} \ni b_1 \\
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\end{align*}
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\end{proof}
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\begin{theorem}
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% $\exist j \text{ correct } W^{t_0}$
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\end{theorem}
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\begin{theorem}[Integrity]
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If a message $m$ is delivered by any process, then it was previously broadcast by some process via the \texttt{AB-broadcast} primitive.
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\end{theorem}
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\begin{proof}
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% Let $j$ be a process such that $\texttt{AB-deliver}_j(m)$ occurs.
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% \begin{align*}
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% &\texttt{AB-deliver}_j(m) & \text{(line 18)} \\
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% \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)} \\
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% \Rightarrow\; & \exists j_0,\ r_0 : m \in \textit{prop}[r_0][j_0] & \text{(line 16)} \\
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% \Rightarrow\; & \textit{prop}[r_0][j_0] = S,\ \text{with } \texttt{RB-delivered}_{j}(PROP, S, r_0, j_0) & \text{(line 22)} \\
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% \Rightarrow\; & S \text{ was sent in } \texttt{RB-cast}(PROP, S, r_0, j_0) & \text{(line 9)} \\
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% \Rightarrow\; & S = \textit{received}_{j_0} \setminus \textit{delivered}_{j_0} & \text{(line 6)} \\
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% \Rightarrow\; & m' \in \textit{received}_{j_0}\ \text{where } m' \text{ broadcast by } j_0 & \text{(line 4)} \\
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% \Rightarrow\; & \textbf{if } m = m' \\
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% & \quad \Rightarrow \texttt{RB-Broadcast}_{j_0}(m) \text{ occurred} & \text{(line 3)} \\
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% & \quad \Rightarrow \texttt{AB-Broadcast}_{j_0}(m) \text{ occurred} & \text{(line 1)} & \hspace{1em} \square \\
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% & \textbf{else: } m \in \textit{received}_{j_0} \setminus \textit{delivered}_{j_0} \\
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% & \quad \Rightarrow m \in \textit{received}_{j_0} & \text{(line 4)} \\
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% & \quad \Rightarrow \texttt{RB-delivered}_{j_0}(m) \text{ occurred} & \text{(line 3)} \\
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% & \quad \Rightarrow \exists j_1 : \texttt{RB-Broadcast}_{j_1}(m) \text{ occurred} & \text{(line 2)} \\
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% & \quad \Rightarrow \texttt{AB-Broadcast}_{j_1}(m) \text{ occurred} & \text{(line 1)} & \hspace{1em} \square
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% \end{align*}
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% Therefore, every delivered message $m$ must originate from some call to \texttt{AB-Broadcast}.
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\end{proof}
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\begin{theorem}[No Duplication]
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No message is delivered more than once by any process.
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\end{theorem}
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\begin{proof}
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% Assume by contradiction that a process $j$ delivers the same message $m$ more than once, i.e.,
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% \[
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% \texttt{AB-deliver}_j(m) \text{ occurs at least twice.}
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% \]
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% \begin{align*}
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% &\texttt{AB-deliver}_j(m) \text{ occurs} & \text{(line 19)} \\
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% \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)} \\
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% \Rightarrow\; & m \notin \textit{delivered} \text{ at that time} \\
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% \\
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% \text{However:} \\
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% & \texttt{delivered} \gets \texttt{delivered} \cup \{m\} & \text{(line 18)} \\
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% \Rightarrow\; & m \in \textit{delivered} \text{ permanently} \\
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% \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} \\
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% \Rightarrow\; & m \text{ will not be delivered again} \\
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% \Rightarrow\; & \text{Contradiction.}
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% \end{align*}
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% Therefore, no message can be delivered more than once by the same process. $\square$
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\end{proof}
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\begin{theorem}[Validity]
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If a correct process invokes $\texttt{AB-Broadcast}_j(m)$, then all correct processes eventually deliver $m$.
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\end{theorem}
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\begin{proof}
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% Let $j$ be a correct process such that $\texttt{AB-Broadcast}_j(m)$ occurs (line 5).
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% \begin{align*}
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% &\texttt{AB-Broadcast}_j(m) & \text{(line 1)}\\
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% \Rightarrow\; & \texttt{RB-Broadcast}_j(m) \text{ occurs} & \text{(line 2)} \\
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% \Rightarrow\; & \forall j_0 : \texttt{RB-delivered}_{j_0}(m) & \text{(line 3)} \\
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% \Rightarrow\; & m \in \textit{received}_{j_0} & \text{(line 4)} \\
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% \Rightarrow\; & \textbf{if } m \in \texttt{delivered}_{j_0} \\
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% & \quad \Rightarrow \textit{delivered}_{j_0} \gets textit{delivered}_{j_0} \cup \{m\} & \text{(line 18)} \\
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% & \quad \Rightarrow \texttt{AB-delivered}_{j_0}(m) & \text{(line 19)} & \hspace{1em} \square \\
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% & \textbf{else } m \notin \textit{delivered}_{j_0} : \\
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% & \quad \Rightarrow m \in S_{j_0}\ \text{since } S_{j_0} = \textit{receieved}_{j_0} \setminus \textit{delivered}_{j_0} & \text{(line 6)} \\
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% & \quad \Rightarrow \exists r : \texttt{RB-cast}_{j_0}(texttt{PROP}, S_{j_0}, r, j_0) & \text{(line 9)} \\
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% & \quad \quad \Rightarrow \forall j_1 : \texttt{RB-Deliver}_{j_1}(\texttt{PROP}, S_{j_0}, r, j_0)\ \text{occurs} & \text{(line 21)} \\
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% & \quad \quad \Rightarrow \textit{prop}[r][j_0] = S_{j_0} & \text{(line 22)} \\
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% & \quad \Rightarrow \exists j_2 \in j_0 : \texttt{PROVE}_{j_2}(r)\ \text{is valid} & \text{(line 10)} \\
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% & \quad \Rightarrow j_2 \in textit{winner}^r & \text{(line 14)} \\
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% & \quad \Rightarrow T_{j_0} \ni \textit{prop}[r][j_2] \setminus \textit{delivered}_{j_0} & \text{(line 16)} \\
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% & \quad \Rightarrow T_{j_0} \ni S_{j_2} \setminus \textit{delivered}_{j_0} \ni m & \text{(line 16)} \\
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% & \quad \Rightarrow \texttt{AB-deliver}_{j_0}(m) & \text{(line 19)} & \hspace{1em} \square \\
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% \end{align*}
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\end{proof}
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\begin{theorem}[Total Order]
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If two correct processes deliver two messages $m_1$ and $m_2$, then they deliver them in the same order.
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\end{theorem}
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\begin{proof}
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% \begin{align*}
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% & \forall j_0 : \texttt{AB-Deliver}_{j_0}(m_0) \wedge \texttt{AB-Deliver}_{j_0}(m_1) & \text{(line 19)} \\
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% \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)} \\
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% \Rightarrow\; & T^{r_0} = \bigcup_{j' \in \textit{winner}^{r_0}} \textit{prop}[r_0][j'] \setminus \textit{delivered}\ \wedge \\
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% & T^{r_1} = \bigcup_{j' \in \textit{winner}^{r_1}} \textit{prop}[r_1][j'] \setminus \textit{delivered} & \text{(line 16)} \\
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% \Rightarrow\; & \textbf{if } r_0 = r_1 \\
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% & \quad \Rightarrow T^{r_0} = T^{r_1} \\
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% & \quad \Rightarrow m_0, m_1 \in \texttt{ordered}(T^{r_0})\ \text{since \texttt{ordered} is deterministic} \\
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% & \quad \Rightarrow \textbf{if } m_0 < m_1 : \\
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% & \quad \quad \Rightarrow \texttt{AB-Deliver}_{j_0}(m_0) < \texttt{AB-Deliver}_{j_0}(m_1) & & \hspace{1em} \square\\
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% & \textbf{else if } r_0 < r_1 \\
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% & \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\\
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% \end{align*}
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% Therefore, for all correct processes, messages are delivered in the same total order.
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\end{proof}
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