Simplify lean link formatting.

Bookshelf/Enderton/Logic.tex

@@ -18,41 +18,42 @@
 \section{\defined{Construction Sequence}}%
 \hyperlabel{ref:construction-sequence}
 
-A \textbf{construction sequence} is a finite sequence
+  A \textbf{construction sequence} is a finite sequence
-  $\langle \epsilon_1, \ldots, \epsilon_n \rangle$ of expressions such that for
+    $\langle \epsilon_1, \ldots, \epsilon_n \rangle$ of expressions such that
-  each $i \leq n$ we have at least one of
+    for each $i \leq n$ we have at least one of
-  \begin{align*}
+    \begin{align*}
-    & \epsilon_i \text{ is a sentence symbol} \\
+      & \epsilon_i \text{ is a sentence symbol} \\
-    & \epsilon_i = \mathcal{E}_\neg(\epsilon_j) \text{ for some } j < i \\
+      & \epsilon_i = \mathcal{E}_\neg(\epsilon_j) \text{ for some } j < i \\
-    & \epsilon_i = \mathcal{E}_\square(\epsilon_j, \epsilon_k)
+      & \epsilon_i = \mathcal{E}_\square(\epsilon_j, \epsilon_k)
-      \text{ for some } j < i, k < i
+        \text{ for some } j < i, k < i
-  \end{align*}
+    \end{align*}
-  where $\square$ is one of the binary connectives $\land$, $\lor$,
+    where $\square$ is one of the binary connectives $\land$, $\lor$,
-    $\Rightarrow$, $\Leftrightarrow$.
+      $\Rightarrow$, $\Leftrightarrow$.
 
 \section{\defined{Expression}}%
 \hyperlabel{ref:expression}
 
-An \textbf{expression} is a finite sequence of symbols.
+  An \textbf{expression} is a finite sequence of symbols.
 
 \section{\defined{Well-Formed Formula}}%
 \hyperlabel{ref:well-formed-formula}
 
-An \nameref{ref:expression} that can be built up from the sentence symbols by
+  An \nameref{ref:expression} that can be built up from the sentence symbols by
-  applying some finite number of times the \textbf{formula-building operations}
+    applying some finite number of times the
-  (on expressions) defined by the equations:
+    \textbf{formula-building operations} (on expressions) defined by the
-  \begin{align*}
+    equations:
-    \mathcal{E}_{\neg}(\alpha)
+    \begin{align*}
-      & = (\neg \alpha) \\
+      \mathcal{E}_{\neg}(\alpha)
-    \mathcal{E}_{\land}(\alpha, \beta)
+        & = (\neg \alpha) \\
-      & = (\alpha \land \beta) \\
+      \mathcal{E}_{\land}(\alpha, \beta)
-    \mathcal{E}_{\lor}(\alpha, \beta)
+        & = (\alpha \land \beta) \\
-      & = (\alpha \lor \beta) \\
+      \mathcal{E}_{\lor}(\alpha, \beta)
-    \mathcal{E}_{\Rightarrow}(\alpha, \beta)
+        & = (\alpha \lor \beta) \\
-      & = (\alpha \Rightarrow \beta) \\
+      \mathcal{E}_{\Rightarrow}(\alpha, \beta)
-    \mathcal{E}_{\Leftrightarrow}(\alpha, \beta)
+        & = (\alpha \Rightarrow \beta) \\
-      & = (\alpha \Leftrightarrow \beta)
+      \mathcal{E}_{\Leftrightarrow}(\alpha, \beta)
-  \end{align*}
+        & = (\alpha \Leftrightarrow \beta)
+    \end{align*}
 
 \endgroup
 
@@ -65,19 +66,15 @@ An \nameref{ref:expression} that can be built up from the sentence symbols by
 \section{\sorry{Lemma 0A}}%
 \hyperlabel{sec:lemma-0a}
 
-\begin{lemma}[0A]
+  \begin{lemma}[0A]
+    Assume that $\langle x_1, \ldots, x_m \rangle =
+      \langle y_1, \ldots, y_m, \ldots, y_{m+k} \rangle$.
+    Then $x_1 = \langle y_1, \ldots, y_{k+1} \rangle$.
+  \end{lemma}
 
-Assume that $\langle x_1, \ldots, x_m \rangle =
+  \begin{proof}
-  \langle y_1, \ldots, y_m, \ldots, y_{m+k} \rangle$.
+    TODO
-Then $x_1 = \langle y_1, \ldots, y_{k+1} \rangle$.
+  \end{proof}
-
-\end{lemma}
-
-\begin{proof}
-
-  TODO
-
-\end{proof}
 
 \chapter{Sentential Logic}%
 \hyperlabel{chap:sentential-logic}
@@ -88,18 +85,15 @@ Then $x_1 = \langle y_1, \ldots, y_{k+1} \rangle$.
 \subsection{\sorry{Induction Principle}}%
 \hyperlabel{sub:induction-principle-1}
 
-\begin{theorem}
+  \begin{theorem}
+    If $S$ is a set of wffs containing all the sentence symbols and closed under
+      all five formula-building operations, then $S$ is the set of \textit{all}
+      wffs.
+  \end{theorem}
 
-If $S$ is a set of wffs containing all the sentence symbols and closed under all
+  \begin{proof}
-  five formula-building operations, then $S$ is the set of \textit{all} wffs.
+    TODO
-
+  \end{proof}
-\end{theorem}
-
-\begin{proof}
-
-  TODO
-
-\end{proof}
 
 \section{Exercises 1}%
 \hyperlabel{sec:exercises-1}
@@ -107,92 +101,81 @@ If $S$ is a set of wffs containing all the sentence symbols and closed under all
 \subsection{\sorry{Exercise 1.1.1}}%
 \hyperlabel{sub:exercise-1.1.1}
 
-Give three sentences in English together with translations into our formal
+  Give three sentences in English together with translations into our formal
-  language.
+    language.
-The sentences shoudl be chosen so as to have an interesting structure, and the
+  The sentences shoudl be chosen so as to have an interesting structure, and the
-  translations should each contain 15 or more symbols.
+    translations should each contain 15 or more symbols.
 
-\begin{answer}
+  \begin{answer}
-
+    TODO
-  TODO
+  \end{answer}
-
-\end{answer}
 
 \subsection{\sorry{Exercise 1.1.2}}%
 \hyperlabel{sub:exercise-1.1.2}
 
-Show that there are no wffs of length 2, 3, or 6, but that any other positive
+  Show that there are no wffs of length 2, 3, or 6, but that any other positive
-  length is possible.
+    length is possible.
 
-\begin{answer}
+  \begin{proof}
-
+    TODO
-  TODO
+  \end{proof}
-
-\end{answer}
 
 \subsection{\sorry{Exercise 1.1.3}}%
 \hyperlabel{sub:exercise-1.1.3}
 
-Let $\alpha$ be a wff; let $c$ be the number of places at which binary
+  Let $\alpha$ be a wff; let $c$ be the number of places at which binary
-  connective symbols $(\land, \lor, \Rightarrow, \Leftrightarrow)$ occur in
+    connective symbols $(\land, \lor, \Rightarrow, \Leftrightarrow)$ occur in
-  $\alpha$; let $s$ be the number of places at which sentence symbols occur in
+    $\alpha$; let $s$ be the number of places at which sentence symbols occur in
-  $\alpha$.
+    $\alpha$.
-(For exmaple, if $\alpha$ is $(A \Rightarrow (\neg A))$ then $c = 1$ and $s = 2$.)
+  (For example, if $\alpha$ is $(A \Rightarrow (\neg A))$ then $c = 1$ and
-Show by using the induction principle that $s = c + 1$.
+    $s = 2$.)
+  Show by using the induction principle that $s = c + 1$.
 
-\begin{answer}
+  \begin{proof}
-
+    TODO
-  TODO
+  \end{proof}
-
-\end{answer}
 
 \subsection{\sorry{Exercise 1.1.4}}%
 \hyperlabel{sub:exercise-1.1.4}
 
-Assume we have a construction sequence ending in $\phi$, where $\phi$ does not
+  Assume we have a construction sequence ending in $\phi$, where $\phi$ does not
-  contain the symbol $A_4$.
+    contain the symbol $A_4$.
-Suppose we delete all the expressions in the construction sequence that contain
+  Suppose we delete all the expressions in the construction sequence that
-  $A_4$.
+    contain $A_4$.
-Show that the result is still a legal construction sequence.
+  Show that the result is still a legal construction sequence.
 
-\begin{answer}
+  \begin{proof}
-
+    TODO
-  TODO
+  \end{proof}
-
-\end{answer}
 
 \subsection{\sorry{Exercise 1.1.5}}%
 \hyperlabel{sub:exercise-1.1.5}
 
-Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
+  Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
-\begin{enumerate}[(a)]
+  \begin{enumerate}[(a)]
-  \item Show that the length of $\alpha$ (i.e., the number of symbols in the
+    \item Show that the length of $\alpha$ (i.e., the number of symbols in the
-    string) is odd.
+      string) is odd.
-  \item Show that more than a quarter of the symbols are sentence symbols.
+    \item Show that more than a quarter of the symbols are sentence symbols.
-\end{enumerate}
+  \end{enumerate}
-\textit{Suggestion}: Apply induction to show that the length is of the form
+  \textit{Suggestion}: Apply induction to show that the length is of the form
-  $4k + 1$ and the number of sentence symbols is $k + 1$.
+    $4k + 1$ and the number of sentence symbols is $k + 1$.
 
-\begin{answer}
+  \begin{proof}
-
+    TODO
-  TODO
+  \end{proof}
-
-\end{answer}
 
 \subsection{\sorry{Exercise 1.1.6}}%
 \hyperlabel{sub:exercise-1.1.6}
 
-Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
+  Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
-\begin{enumerate}[(a)]
+  \begin{enumerate}[(a)]
-  \item Show that the length of $\alpha$ (i.e., the number of symbols in the
+    \item Show that the length of $\alpha$ (i.e., the number of symbols in the
-    string) is odd.
+      string) is odd.
-  \item Show that more than a quarter of the symbols are sentence symbols.
+    \item Show that more than a quarter of the symbols are sentence symbols.
-\end{enumerate}
+  \end{enumerate}
 
-\begin{answer}
+  \begin{proof}
-
+    TODO
-  TODO
+  \end{proof}
-
-\end{answer}
 
 \end{document}

Bookshelf/Enderton/Set/Chapter_2.lean

@@ -38,6 +38,153 @@ theorem commutative_law_ii (A B : Set α)
     exact and_comm
   _ = B ∩ A := rfl
 
+/-! #### Associative Laws
+
+For any sets `A`, `B`, and `C`,
+```
+A ∪ (B ∪ C) = (A ∪ B) ∪ C
+A ∩ (B ∩ C) = (A ∩ B) ∩ C
+```
+-/
+
+#check Set.union_assoc
+
+theorem associative_law_i (A B C : Set α)
+  : A ∪ (B ∪ C) = (A ∪ B) ∪ C := calc A ∪ (B ∪ C)
+  _ = { x | x ∈ A ∨ x ∈ B ∪ C } := rfl
+  _ = { x | x ∈ A ∨ (x ∈ B ∨ x ∈ C) } := rfl
+  _ = { x | (x ∈ A ∨ x ∈ B) ∨ x ∈ C } := by
+    ext _
+    simp only [Set.mem_setOf_eq]
+    rw [← or_assoc]
+  _ = { x | x ∈ A ∪ B ∨ x ∈ C } := rfl
+  _ = (A ∪ B) ∪ C := rfl
+
+#check Set.inter_assoc
+
+theorem associative_law_ii (A B C : Set α)
+  : A ∩ (B ∩ C) = (A ∩ B) ∩ C := calc A ∩ (B ∩ C)
+  _ = { x | x ∈ A ∧ (x ∈ B ∩ C) } := rfl
+  _ = { x | x ∈ A ∧ (x ∈ B ∧ x ∈ C) } := rfl
+  _ = { x | (x ∈ A ∧ x ∈ B) ∧ x ∈ C } := by
+    ext _
+    simp only [Set.mem_setOf_eq]
+    rw [← and_assoc]
+  _ = { x | x ∈ A ∩ B ∧ x ∈ C } := rfl
+  _ = (A ∩ B) ∩ C := rfl
+
+/-! #### Distributive Laws
+
+For any sets `A`, `B`, and `C`,
+```
+A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
+A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
+```
+-/
+
+#check Set.inter_distrib_left
+
+theorem distributive_law_i (A B C : Set α)
+  : A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) := calc A ∩ (B ∪ C)
+  _ = { x | x ∈ A ∧ x ∈ B ∪ C } := rfl
+  _ = { x | x ∈ A ∧ (x ∈ B ∨ x ∈ C) } := rfl
+  _ = { x | (x ∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x ∈ C) } := by
+    ext _
+    exact and_or_left
+  _ = { x | x ∈ A ∩ B ∨ x ∈ A ∩ C } := rfl
+  _ = (A ∩ B) ∪ (A ∩ C) := rfl
+
+#check Set.union_distrib_left
+
+theorem distributive_law_ii (A B C : Set α)
+  : A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) := calc A ∪ (B ∩ C)
+  _ = { x | x ∈ A ∨ x ∈ B ∩ C } := rfl
+  _ = { x | x ∈ A ∨ (x ∈ B ∧ x ∈ C) } := rfl
+  _ = { x | (x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ C) } := by
+    ext _
+    exact or_and_left
+  _ = { x | x ∈ A ∪ B ∧ x ∈ A ∪ C } := rfl
+  _ = (A ∪ B) ∩ (A ∪ C) := rfl
+
+/-! #### De Morgan's Laws
+
+For any sets `A`, `B`, and `C`,
+```
+C - (A ∪ B) = (C - A) ∩ (C - B)
+C - (A ∩ B) = (C - A) ∪ (C - B)
+```
+-/
+
+#check Set.diff_inter_diff
+
+theorem de_morgans_law_i (A B C : Set α)
+  : C \ (A ∪ B) = (C \ A) ∩ (C \ B) := calc C \ (A ∪ B)
+  _ = { x | x ∈ C ∧ x ∉ A ∪ B } := rfl
+  _ = { x | x ∈ C ∧ ¬(x ∈ A ∨ x ∈ B) } := rfl
+  _ = { x | x ∈ C ∧ (x ∉ A ∧ x ∉ B) } := by
+    ext _
+    simp only [Set.mem_setOf_eq]
+    rw [not_or_de_morgan]
+  _ = { x | (x ∈ C ∧ x ∉ A) ∧ (x ∈ C ∧ x ∉ B) } := by
+    ext _
+    exact and_and_left
+  _ = { x | x ∈ C \ A ∧ x ∈ C \ B } := rfl
+  _ = (C \ A) ∩ (C \ B) := rfl
+
+#check Set.diff_inter
+
+theorem de_morgans_law_ii (A B C : Set α)
+  : C \ (A ∩ B) = (C \ A) ∪ (C \ B) := calc C \ (A ∩ B)
+  _ = { x | x ∈ C ∧ x ∉ A ∩ B } := rfl
+  _ = { x | x ∈ C ∧ ¬(x ∈ A ∧ x ∈ B) } := rfl
+  _ = { x | x ∈ C ∧ (x ∉ A ∨ x ∉ B) } := by
+    ext _
+    simp only [Set.mem_setOf_eq]
+    rw [not_and_de_morgan]
+  _ = { x | (x ∈ C ∧ x ∉ A) ∨ (x ∈ C ∧ x ∉ B) } := by
+    ext _
+    exact and_or_left
+  _ = { x | x ∈ C \ A ∨ x ∈ C \ B } := rfl
+  _ = (C \ A) ∪ (C \ B) := rfl
+
+/-! #### Identities Involving ∅
+
+For any set `A`,
+```
+A ∪ ∅ = A
+A ∩ ∅ = ∅
+A ∩ (C - A) = ∅
+```
+-/
+
+#check Set.union_empty
+
+theorem emptyset_identity_i (A : Set α)
+  : A ∪ ∅ = A := calc A ∪ ∅
+  _ = { x | x ∈ A ∨ x ∈ ∅ } := rfl
+  _ = { x | x ∈ A ∨ False } := rfl
+  _ = { x | x ∈ A } := by simp
+  _ = A := rfl
+
+#check Set.inter_empty
+
+theorem emptyset_identity_ii (A : Set α)
+  : A ∩ ∅ = ∅ := calc A ∩ ∅
+  _ = { x | x ∈ A ∧ x ∈ ∅ } := rfl
+  _ = { x | x ∈ A ∧ False } := rfl
+  _ = { x | False } := by simp
+  _ = ∅ := rfl
+
+#check Set.inter_diff_self
+
+theorem emptyset_identity_iii (A C : Set α)
+  : A ∩ (C \ A) = ∅ := calc A ∩ (C \ A)
+  _ = { x | x ∈ A ∧ x ∈ C \ A } := rfl
+  _ = { x | x ∈ A ∧ (x ∈ C ∧ x ∉ A) } := rfl
+  _ = { x | x ∈ C ∧ False } := by simp
+  _ = { x | False } := by simp
+  _ = ∅ := rfl
+
 /-- #### Exercise 2.1
 
 Assume that `A` is the set of integers divisible by `4`. Similarly assume that

Bookshelf/Enderton/Set/Chapter_3.lean

@@ -19,7 +19,7 @@ namespace Enderton.Set.Chapter_3
 
 If `x ∈ C` and `y ∈ C`, then `⟨x, y⟩ ∈ 𝒫 𝒫 C`.
 -/
-theorem theorem_3b {C : Set α} (hx : x ∈ C) (hy : y ∈ C)
+theorem lemma_3b {C : Set α} (hx : x ∈ C) (hy : y ∈ C)
   : OrderedPair x y ∈ 𝒫 𝒫 C := by
   have hxs : {x} ⊆ C := Set.singleton_subset_iff.mpr hx
   have hxys : {x, y} ⊆ C := Set.mem_mem_imp_pair_subset hx hy

Common/Real/Sequence.tex

@@ -15,118 +15,118 @@
 \subsection{\verified{Arithmetic Series}}%
 \hyperlabel{sub:sum-arithmetic-series}
 
-Let $(a_i)_{i \geq 0}$ be an arithmetic sequence with common difference $d$.
+  Let $(a_i)_{i \geq 0}$ be an arithmetic sequence with common difference $d$.
-  Then for some $n \in \mathbb{N}$,
+    Then for some $n \in \mathbb{N}$,
-  \begin{equation}
+    \begin{equation}
-    \hyperlabel{sub:sum-arithmetic-series-eq1}
+      \hyperlabel{sub:sum-arithmetic-series-eq1}
-    \sum_{i=0}^n a_i = \frac{(n + 1)(a_0 + a_n)}{2}.
+      \sum_{i=0}^n a_i = \frac{(n + 1)(a_0 + a_n)}{2}.
-  \end{equation}
+    \end{equation}
-
-\begin{proof}
 
   \code{Common/Real/Sequence/Arithmetic}
     {Real.Arithmetic.sum\_recursive\_closed}
 
-  Let $(a_i)_{i \geq 0}$ be an arithmetic sequence with common difference $d$.
+  \begin{proof}
-  By definition, for all $k \in \mathbb{N}$,
-    \begin{equation}
-      \hyperlabel{sub:sum-arithmetic-series-eq2}
-      a_k = (a_0 + kd).
-    \end{equation}
-  Define predicate $P(n)$ as "identity \eqref{sub:sum-arithmetic-series-eq1}
-    holds for value $n$."
-  We use induction to prove $P(n)$ holds for all $n \geq 0$.
 
-  \paragraph{Base Case}%
+    Let $(a_i)_{i \geq 0}$ be an arithmetic sequence with common difference $d$.
+    By definition, for all $k \in \mathbb{N}$,
+      \begin{equation}
+        \hyperlabel{sub:sum-arithmetic-series-eq2}
+        a_k = (a_0 + kd).
+      \end{equation}
+    Define predicate $P(n)$ as "identity \eqref{sub:sum-arithmetic-series-eq1}
+      holds for value $n$."
+    We use induction to prove $P(n)$ holds for all $n \geq 0$.
 
-    Let $k = 0$.
+    \paragraph{Base Case}%
-    Then $$\sum_{i=0}^k a_i = a_0 = \frac{2a_0}{2} =
-      \frac{(k + 1)(a_0 + a_k)}{2}.$$
-    Therefore $P(0)$ holds.
 
-  \paragraph{Induction Step}%
+      Let $k = 0$.
+      Then $$\sum_{i=0}^k a_i = a_0 = \frac{2a_0}{2} =
+        \frac{(k + 1)(a_0 + a_k)}{2}.$$
+      Therefore $P(0)$ holds.
 
-    Assume induction hypothesis $P(k)$ holds for some $k \geq 0$.
+    \paragraph{Induction Step}%
-    Then
-      \begin{align*}
-        \sum_{i=0}^{k+1} a_i
-          & = \sum_{i=0}^k a_i + a_{k+1} \\
-          & = \frac{(k + 1)(a_0 + a_k)}{2} + a_{k+1}
-            & \text{induction hypothesis} \\
-          & = \frac{(k + 1)(a_0 + (a_0 + kd))}{2} + (a_0 + (k + 1)d)
-            & \eqref{sub:sum-arithmetic-series-eq2} \\
-          & = \frac{(k + 1)(2a_0 + kd)}{2} + (a_0 + (k + 1)d) \\
-          & = \frac{(k + 1)(2a_0 + kd) + 2a_0 + 2(k + 1)d}{2} \\
-          & = \frac{2ka_0 + k^2d + 4a_0 + kd + 2kd + 2d}{2} \\
-          & = \frac{(k + 2)(2a_0 + kd + d)}{2} \\
-          & = \frac{(k + 2)(a_0 + a_0 + (k + 1)d)}{2} \\
-          & = \frac{(k + 2)(a_0 + a_{k+1})}{2}
-            & \eqref{sub:sum-arithmetic-series-eq2} \\
-          & = \frac{((k + 1) + 1)(a_0 + a_{k+1})}{2}.
-      \end{align*}
-    Thus $P(k)$ implies $P(k + 1)$ holds true.
 
-  \paragraph{Conclusion}%
+      Assume induction hypothesis $P(k)$ holds for some $k \geq 0$.
+      Then
+        \begin{align*}
+          \sum_{i=0}^{k+1} a_i
+            & = \sum_{i=0}^k a_i + a_{k+1} \\
+            & = \frac{(k + 1)(a_0 + a_k)}{2} + a_{k+1}
+              & \text{induction hypothesis} \\
+            & = \frac{(k + 1)(a_0 + (a_0 + kd))}{2} + (a_0 + (k + 1)d)
+              & \eqref{sub:sum-arithmetic-series-eq2} \\
+            & = \frac{(k + 1)(2a_0 + kd)}{2} + (a_0 + (k + 1)d) \\
+            & = \frac{(k + 1)(2a_0 + kd) + 2a_0 + 2(k + 1)d}{2} \\
+            & = \frac{2ka_0 + k^2d + 4a_0 + kd + 2kd + 2d}{2} \\
+            & = \frac{(k + 2)(2a_0 + kd + d)}{2} \\
+            & = \frac{(k + 2)(a_0 + a_0 + (k + 1)d)}{2} \\
+            & = \frac{(k + 2)(a_0 + a_{k+1})}{2}
+              & \eqref{sub:sum-arithmetic-series-eq2} \\
+            & = \frac{((k + 1) + 1)(a_0 + a_{k+1})}{2}.
+        \end{align*}
+      Thus $P(k)$ implies $P(k + 1)$ holds true.
 
-    By mathematical induction, it follows for all $n \geq 0$, $P(n)$ is true.
+    \paragraph{Conclusion}%
 
-\end{proof}
+      By mathematical induction, it follows for all $n \geq 0$, $P(n)$ is true.
+
+  \end{proof}
 
 \subsection{\verified{Geometric Series}}%
 \hyperlabel{sub:sum-geometric-series}
 
-Let $(a_i)_{i \geq 0}$ be a geometric sequence with common ratio $r \neq 1$.
+  Let $(a_i)_{i \geq 0}$ be a geometric sequence with common ratio $r \neq 1$.
-  Then for some $n \in \mathbb{N}$,
+    Then for some $n \in \mathbb{N}$,
-  \begin{equation}
+    \begin{equation}
-    \hyperlabel{sub:sum-geometric-series-eq1}
+      \hyperlabel{sub:sum-geometric-series-eq1}
-    \sum_{i=0}^n a_i = \frac{a_0(1 - r^{n+1})}{1 - r}.
+      \sum_{i=0}^n a_i = \frac{a_0(1 - r^{n+1})}{1 - r}.
-  \end{equation}
+    \end{equation}
-
-\begin{proof}
 
   \code{Common/Real/Sequence/Geometric}
     {Real.Geometric.sum\_recursive\_closed}
 
-  Let $(a_i)_{i \geq 0}$ be a geometric sequence with common ratio $r \neq 1$.
+  \begin{proof}
-  By definition, for all $k \in \mathbb{N}$,
-    \begin{equation}
-      \hyperlabel{sub:sum-geometric-series-eq2}
-      a_k = a_0r^k.
-    \end{equation}
-  Define predicate $P(n)$ as "identity \eqref{sub:sum-geometric-series-eq1}
-    holds for value $n$."
-  We use induction to prove $P(n)$ holds for all $n \geq 0$.
 
-  \paragraph{Base Case}%
+    Let $(a_i)_{i \geq 0}$ be a geometric sequence with common ratio $r \neq 1$.
+    By definition, for all $k \in \mathbb{N}$,
+      \begin{equation}
+        \hyperlabel{sub:sum-geometric-series-eq2}
+        a_k = a_0r^k.
+      \end{equation}
+    Define predicate $P(n)$ as "identity \eqref{sub:sum-geometric-series-eq1}
+      holds for value $n$."
+    We use induction to prove $P(n)$ holds for all $n \geq 0$.
 
-    Let $k = 0$.
+    \paragraph{Base Case}%
-    Then $$\sum_{i=0}^k a_i = a_0 = \frac{a_0(1 - r)}{1 - r} =
-      \frac{a_0(1 - r^{k+1})}{1 - r}$$
-    Therefore $P(0)$ holds.
 
-  \paragraph{Induction Step}%
+      Let $k = 0$.
+      Then $$\sum_{i=0}^k a_i = a_0 = \frac{a_0(1 - r)}{1 - r} =
+        \frac{a_0(1 - r^{k+1})}{1 - r}$$
+      Therefore $P(0)$ holds.
 
-    Assume induction hypothesis $P(k)$ holds for some $k \geq 0$.
+    \paragraph{Induction Step}%
-    Then
-      \begin{align*}
-        \sum_{i=0}^{k+1} a_i
-          & = \sum_{i=0}^k a_i + a_{k+1} \\
-          & = \frac{a_0(1 - r^{k+1})}{1 - r} + a_{k + 1}
-            & \text{induction hypothesis} \\
-          & = \frac{a_0(1 - r^{k+1})}{1 - r} + a_0r^{k + 1}
-            & \eqref{sub:sum-geometric-series-eq2} \\
-          & = \frac{a_0(1 - r^{k+1}) + a_0r^{k+1}(1 - r)}{1 - r} \\
-          & = \frac{a_0(1 - r^{k+1} + r^{k+1}(1 - r))}{1 - r} \\
-          & = \frac{a_0(1 - r^{k+1} + r^{k+1} - r^{k+2})}{1 - r} \\
-          & = \frac{a_0(1 - r^{k+2})}{1 - r} \\
-          & = \frac{a_0(1 - r^{(k + 1) + 1})}{1 - r}.
-      \end{align*}
-    Thus $P(k)$ implies $P(k + 1)$ holds true.
 
-  \paragraph{Conclusion}%
+      Assume induction hypothesis $P(k)$ holds for some $k \geq 0$.
+      Then
+        \begin{align*}
+          \sum_{i=0}^{k+1} a_i
+            & = \sum_{i=0}^k a_i + a_{k+1} \\
+            & = \frac{a_0(1 - r^{k+1})}{1 - r} + a_{k + 1}
+              & \text{induction hypothesis} \\
+            & = \frac{a_0(1 - r^{k+1})}{1 - r} + a_0r^{k + 1}
+              & \eqref{sub:sum-geometric-series-eq2} \\
+            & = \frac{a_0(1 - r^{k+1}) + a_0r^{k+1}(1 - r)}{1 - r} \\
+            & = \frac{a_0(1 - r^{k+1} + r^{k+1}(1 - r))}{1 - r} \\
+            & = \frac{a_0(1 - r^{k+1} + r^{k+1} - r^{k+2})}{1 - r} \\
+            & = \frac{a_0(1 - r^{k+2})}{1 - r} \\
+            & = \frac{a_0(1 - r^{(k + 1) + 1})}{1 - r}.
+        \end{align*}
+      Thus $P(k)$ implies $P(k + 1)$ holds true.
 
-    By mathematical induction, it follows for all $n \geq 0$, $P(n)$ is true.
+    \paragraph{Conclusion}%
 
-\end{proof}
+      By mathematical induction, it follows for all $n \geq 0$, $P(n)$ is true.
+
+  \end{proof}
 
 \end{document}

preamble.tex

@@ -44,61 +44,47 @@
   \label{#1}%
   \hypertarget{#1}{}}
 
-% Denote whether we are working with a standard/Mathlib statement (lean) or a
+% Denote if working with a predefined statement/theorem or a custom one.
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+  \textcolor{BlueViolet}{\raisebox{-4.5pt}{%
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+  \href{#1/#2.html\##3}{\color{BlueViolet}{#4}}}
-
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+  \textcolor{MidnightBlue}{\raisebox{-4.5pt}{%
-    \tikz{\draw (0, 0) node[] {\faCodeBranch};}}}%
+    \tikz{\draw (0, 0) node[xshift=8pt] {\faCodeBranch};}}{-\;}}%
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+  \href{#1/#2.html\##3}{\color{MidnightBlue}{#4}}}
 
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+% Reference to an anchor of generated Lean documentation.
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-    \leanref{#1}{##1}{##2}}
+    \noindent\leanref{#1}{##1}{##2}}
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+    \vspace{6pt}\noindent\leanref{#1}{##1}{##2}}
 
   \newcommand\code[2]{%
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+    \vspace{6pt}\noindent\codepref{#1}{##1}{##2}{##3}}
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 % ========================================
@@ -124,11 +110,7 @@
 
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-\newcommand{\statementpadding}{\ \vspace{8pt}}
-
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-\newenvironment{axiom}{\@statement{Axiom}}{\hfill$\square$}
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 \newtheorem{corollaryinner}{Corollary}