+\begin{lem} If there are two packets $t$ and $u$, with $i(t) <= i(u)$, such that $t$ is in the path of $u$, then for any message $p$ with $i(p) <= i(t)$, iff $p$ is in the path of $t$, it is in the path of $u$. \end{lem}\r
+\r
+\begin{proof}\r
+If $i(t) = i(u)$ or $i(p) = i(t)$, then we are done, because the two relevant messages are the same.\r
+\r
+Reverse direction: The definition of $t$ being in the path of $u$ is the existence of a message sequence $(..., t, ..., u)$ such that each message except $u$ is the parent of the succeeding message. The path of $u$ must contain some message with index $i(p)$; because $p$ is in the path of $u$, this message is $p$ itself. The path of $t$ is then the prefix of this path ending at $t$, which clearly contains $p$.\r
+\r
+Forward direction: The path of $t$ is a substring of the path of $u$, so if the path of $t$ contains $p$, so does the path of $u$.\r
+\end{proof}\r
+\r