diff --git a/inputs/lecture_06.tex b/inputs/lecture_06.tex
index 4e68c1f..a2adaad 100644
--- a/inputs/lecture_06.tex
+++ b/inputs/lecture_06.tex
@@ -172,21 +172,31 @@ In order to prove \autoref{thm2}, we need the following:
 
 \begin{proof}
     By SLLN, $\frac{S_n}{n} \xrightarrow{a.s.} m$ as $n \to \infty$.
-    Note that $N_t \uparrow \infty$ a.s.~as $t \to \infty (\ast\ast)$, since
-    $\{N_t \ge  n\} = \{X_1 + \ldots+ X_n \le t\}$ thus $N_t \uparrow \infty$ as $t \uparrow \infty$.
+    Note that
+    \begin{equation}
+        N_t \uparrow \infty \text{ a.s.~as } t \to \infty, \label{eqn:renewalnt}
+    \end{equation}
+    since $\{N_t \ge  n\} = \{X_1 + \ldots+ X_n \le t\}$.
 
     \begin{claim}
-        $\bP[\frac{S_n}{n} \xrightarrow{n \to  \infty} m , N_t \xrightarrow{t \to  \infty}  \infty] = 1$.
+        $\bP[\frac{S_n}{n} \xrightarrow{n \to  \infty} m 
+            \land N_t \xrightarrow{t \to  \infty}  \infty] = 1$.
     \end{claim}
     \begin{subproof}
-    Let $A \coloneqq  \{\omega: \frac{S_n(\omega)}{n} \xrightarrow{n \to  \infty} m\}$ and $B \coloneqq \{\omega : N_t(\omega \xrightarrow{t \to  \infty} \infty\}$.
-    By the SLLN, we have $\bP(A^C) = 0$ and $\ast\ast \implies \bP(B^C) = 0$.
+    Let $A \coloneqq  \{\omega: \frac{S_n(\omega)}{n} \xrightarrow{n \to  \infty} m\}$
+    and $B \coloneqq \{\omega : N_t(\omega) \xrightarrow{t \to  \infty} \infty\}$.
+    By the SLLN we have $\bP(A^C) = 0$
+    and by \eqref{eqn:renewalnt} it holds that $\bP(B^C) = 0$.
     \end{subproof}
 
-    Equivalently, $\bP\left[ \frac{S_{N_t}}{N_t} \xrightarrow{t \to \infty} m, \frac{S_{N_t + 1}}{N_t + 1} \xrightarrow{t \to  \infty} m \right] = 1$.
+    Equivalently,
+    $\bP\left[ \frac{S_{N_t}}{N_t} \xrightarrow{t \to \infty} m
+        \land \frac{S_{N_t + 1}}{N_t + 1} \xrightarrow{t \to  \infty} m \right] = 1$.
 
-    By definition, we have $S_{N_t} \le  t \le  S_{N_t + t}$.
-    Then $\frac{S_{N_t}}{N_t} \le  \frac{t}{N_t} \le S_{N_t + 1}{N_t} \le  \frac{S_{N_t + 1}}{N_t + 1} \cdot  \frac{N_t + 1}{N_t}$.
+    By definition, we have $S_{N_t} \le  t \le  S_{N_t + 1}$.
+    Thus
+    \[\frac{S_{N_t}}{N_t} \le \frac{t}{N_t} \le \frac {S_{N_t + 1}}{N_t}
+    \le  \frac{S_{N_t + 1}}{N_t + 1} \cdot  \frac{N_t + 1}{N_t}.\]
     Hence $\frac{t}{N_t} \to  m$.
 
 \end{proof}