diff --git a/others/appendix/list/list-en.tex b/others/appendix/list/list-en.tex
index 44413bf96..e7662d9c5 100644
--- a/others/appendix/list/list-en.tex
+++ b/others/appendix/list/list-en.tex
@@ -101,16 +101,14 @@ \subsection{Access}
 
 \section{Basic operations}
 \index{List!length}
-From the definition, we can count the length recursively. the length of the empty list is 0, or it is 1 plus the length of the sub-list.
+We may count the length of a list recursively: the length of the empty list is 0 or it is 1 plus the length of the sub-list.
 
-\be
-\begin{array}{rcl}
-length\ [\ ] & = & 0 \\
-length\ (x \cons xs) & = & 1 + length\ xs
-\end{array}
-\ee
+\begin{lalign}
+  length\ [\ ] & = 0 \\
+  length\ (x \cons xs) & = 1 + length\ xs
+\end{lalign}
 
-We traverse the list to count the length, the performance is bound to $O(n)$, where $n$ is the number of elements. We use $|X|$ as the length of $X$ when the context is clear. To avoid repeatedly counting, we can persist the length in a variable, and update it when mutate (add or delete). Below is the iterative length counting:
+As we traverse the list to count the length, the performance is bound to $O(n)$, where $n$ is the number of elements. Denote the length of the list $X$ by $|X|$ when the context is clear. To avoid repeatedly counting, we may persist the length in a variable and update it when mutate (add or delete). Below is the iterative length counting:
 
 \begin{algorithmic}[1]
 \Function{Length}{X}
@@ -130,7 +128,7 @@ \subsection{index}
 \be
 getAt\ i\ (x \cons xs) = \begin{cases}
   i = 0: & x \\
-  i \neq 0: & getAt\ (i - 1)\ xs \\
+  i \neq 0: & getAt\ (i - 1)\ xs
 \end{cases}
 \ee
 
diff --git a/others/preface/preface-zh-cn.tex b/others/preface/preface-zh-cn.tex
index 4a59989a2..e8f6efeb8 100644
--- a/others/preface/preface-zh-cn.tex
+++ b/others/preface/preface-zh-cn.tex
@@ -91,7 +91,7 @@ \subsection*{改进}
 对$n$个自然数$x_1, x_2, \dotsc , x_n$，如果存在小于$n$的可用数，必然存在某个$x_i$不在区间$[0, n)$\footnote{左开右闭区间$[a, b)$包括$a$但不包括$b$。}内。否则这些数一定是$0, 1, \dotsc, n - 1$的某个排列，这种情况下，最小的可用自然数是$n$。
 
 \be
-\textit{minfree}(x_1, x_2, ..., x_n) \leq n
+\textit{minfree}(x_1, x_2, \dotsc, x_n) \leq n
 \label{eq:min-free}
 \ee