Suppose that there is a transitive closure set $\mathscr{S}$ of clients, at index $n$. Then there is some total message sequence $T$ of length $n$ such that every client $C$ in $\mathscr{S}$ sees a partial sequence $P_C$ consistent with $T$. \end{theorem}\r
\r
\begin{proof}\r
-The definition of consistency of $P_C$ with $T$ is that every message $p \in P_C$ with index $i(p) \le n$ is equal to the message in that slot in $T$. Let $C_1$ be some client in the transitive closure set, with partial message sequence $P_{C_1}$, and let $u$ be some message with $i(u) > i$ that $C_1$ shares with another client. Then let $T$ be the portion of the path of $u$ ending at index $i$ and $t$ be the message at that index. Clearly, by Lemma 1, $P_{C_1}$ is consistent with $T$. We will show that, for every other client $D$ with partial sequence $P_D$, $P_D$ has some message whose path includes $t$. Because $D$ is in the transitive closure, there is a sequence of edges from $C_1$ to $D$. Call this $\mathscr{C} = (C_1, C_2, ..., D)$. \r
+ \r
-We subsequently prove by induction that $D$ has a message whose path includes $t$.\r
+The definition of consistency of $P_C$ with $T$ is that every message $p \in P_C$ with index $i(p) \le n$ is equal to the message in that slot in $T$. Let $C_1$ be some client in the transitive closure set, with partial message sequence $P_{C_1}$, and let $u$ be some message with $i(u) > n$ that $C_1$ shares with another client. Then let $T$ be the portion of the path of $u$ ending at index $n$ and $t$ be the message at that index. Clearly, by Lemma 1, $P_{C_1}$ is consistent with $T$. We will show that, for every other client $D$ with partial sequence $P_D$, $P_D$ has some message whose path includes $t$. Because $D$ is in the transitive closure, there is a sequence of clients $\mathscr{C} = (C_1, C_2, ..., D)$ from $C_1$ to $D$, where each shares an edge with the next.\r
+\r
+We prove by induction that $P_D$ has a message whose path includes $t$.\r
\begin{itemize}\r
-\item For the base case, $P_{C_1}$ includes $u$, whose path includes $t$. \r
-\item For the inductive step, suppose $P_{C_k}$ has a message $w$ with a path that includes $t$, and shares message $x$ with $P_{C_{k+1}}$ such that $i(x) > i$. If $i(w) = i(x)$, then $w = x$. If $i(w) < i(x)$, then, by Lemma 1, $w$ is in the path of $x$. If $i(w) > i(x)$, $x$ is in the path of $w$; note again that its index is greater than $i$. In any case, $t$ is in the path of $u_k+1$.\r
-\item Let $w$ the message of $D$ whose path includes $t$. By Lemma 1, every message in $P_D$ with index smaller than $i(w)$ is in the path of $w$. Since $t$ is in the path of $w$, every message in $P_D$ with smaller index than $i(t)$ is in $T$. Therefore, $P_D$ is consistent with $T$.\r
+\item Base case: $P_{C_1}$ includes $u$, whose path includes $t$.\r
+\r
+\item Inductive step: Each client in $\mathscr{C}$ has a partial message sequence with a message that includes $t$ if the previous client does. Suppose $P_{C_k}$ has a message $w$ with a path that includes $t$, and shares message $x$ with $P_{C_{k+1}}$ such that $i(x) > n$. By Lemma 1, $w$ or $x$, whichever has the least sequence number, is in the path of the other, and therefore by Lemma 2, $t$ is in the path of $x$.\r
+\r
+\item Let $z$ be the message of $D$ whose path includes $t$. By Lemma 1, every message in $P_D$ with index smaller than $i(w)$ is in the path of $z$. Since $t$ is in the path of $z$, every message in $P_D$ with smaller index than $i(t)$ is in $T$. Therefore, $P_D$ is consistent with $T$.\r
++\r
\end{itemize}\r
\end{proof}\r
\r
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