-Clearly $\mathsf{C}$ will throw an error if $\mathsf{s_t = s_u}$. So \r
-$\mathsf{s_t < s_u}$. Additionally, if $\mathsf{C}$ receives $\mathsf{u}$ before \r
-$\mathsf{t}$, this will cause it to throw an error, so $\mathsf{t}$ is received \r
-before $\mathsf{u}$.\r
+By contradiction, we will prove that if $\mathsf{t}$ is not in the path of \r
+$\mathsf{u}$, then it is impossible for client $\mathsf{C}$ to receive both\r
+messages without throwing any errors. Clearly $\mathsf{C}$ will throw an error \r
+if $\mathsf{s_t = s_u}$. So $\mathsf{s_t < s_u}$. Additionally, if $\mathsf{C}$ \r
+receives $\mathsf{u}$ before $\mathsf{t}$, this will cause it to throw an \r
+error, so $\mathsf{t}$ is received before $\mathsf{u}$.\r