-We will show that $\mathsf{C}$ sees $\mathsf{r_1}$. Assume not. Then $\mathsf{(r_2, ..., u)}$ must have at least $\mathsf{{max_g}_C} >= 2$ messages for $\mathsf{r_1}$ to fall off the end of the queue. Consider the sender of $\mathsf{r_3}$ and call it $\mathsf{H}$. $\mathsf{H \neq J}$ by Proposition 3 and the existence of $\mathsf{m}$. Since $\mathsf{H \neq J}$, then by Proposition 3 it could not also have sent a message in $\mathsf{(l_2,..., u}$. Therefore, $\mathsf{s_{u_H} < s_q + 2 = s_{t_H}}$, so upon receipt of $\mathsf{u}$, $\mathsf{C}$ will throw an error by the decrease in a last sequence number similar to Case 1, a contradiction.\r
-\r
-Now that we know that $\mathsf{C}$ sees $\mathsf{r_1}$, note that C receives $\mathsf{u}$ immediately after $\mathsf{t}$ by Proposition 4. Therefore, \r
+We will show that $\mathsf{C}$ sees $\mathsf{r_1}$. Assume not. Then $\mathsf{(r_2, ..., u)}$ \r
+must have at least $\mathsf{{max_g}_C} \geq 2$ messages for $\mathsf{r_1}$ to fall off the \r
+end of the queue. Consider the sender of $\mathsf{r_3}$ and call it $\mathsf{H}$. \r
+$\mathsf{H \neq J}$ by Proposition \ref{prop:bothmessages} and the existence of $\mathsf{m}$. \r
+Since $\mathsf{H \neq J}$, then by Proposition \ref{prop:bothmessages} it could not also \r
+have sent a message in $\mathsf{(l_2,..., u)}$. Therefore, $\mathsf{s_{u_H} < s_q + 2 = s_{t_H}}$, \r
+so upon receipt of $\mathsf{u}$, $\mathsf{C}$ will throw an error by the decrease in a \r
+last sequence number similar to Case 1, a contradiction.\r
+\r
+Now that we know that $\mathsf{C}$ sees $\mathsf{r_1}$, note that C receives $\mathsf{u}$ \r
+immediately after $\mathsf{t}$ by Proposition \ref{prop:seqnumb}. Therefore, \r