Simplify lean link formatting.

Bookshelf/Enderton/Logic.tex

@@ -19,8 +19,8 @@
 \hyperlabel{ref:construction-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*}
       & \epsilon_i \text{ is a sentence symbol} \\
       & \epsilon_i = \mathcal{E}_\neg(\epsilon_j) \text{ for some } j < i \\
@@ -39,8 +39,9 @@ An \textbf{expression} is a finite sequence of symbols.
 \hyperlabel{ref:well-formed-formula}
 
   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
+    equations:
     \begin{align*}
       \mathcal{E}_{\neg}(\alpha)
         & = (\neg \alpha) \\
@@ -66,17 +67,13 @@ An \nameref{ref:expression} that can be built up from the sentence symbols by
 \hyperlabel{sec: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}
 
   \begin{proof}
-
     TODO
-
   \end{proof}
 
 \chapter{Sentential Logic}%
@@ -89,16 +86,13 @@ Then $x_1 = \langle y_1, \ldots, y_{k+1} \rangle$.
 \hyperlabel{sub:induction-principle-1}
 
   \begin{theorem}
-
+    If $S$ is a set of wffs containing all the sentence symbols and closed under
-If $S$ is a set of wffs containing all the sentence symbols and closed under all
+      all five formula-building operations, then $S$ is the set of \textit{all}
-  five formula-building operations, then $S$ is the set of \textit{all} wffs.
+      wffs.
-
   \end{theorem}
 
   \begin{proof}
-
     TODO
-
   \end{proof}
 
 \section{Exercises 1}%
@@ -113,9 +107,7 @@ The sentences shoudl be chosen so as to have an interesting structure, and the
     translations should each contain 15 or more symbols.
 
   \begin{answer}
-
     TODO
-
   \end{answer}
 
 \subsection{\sorry{Exercise 1.1.2}}%
@@ -124,11 +116,9 @@ The sentences shoudl be chosen so as to have an interesting structure, and the
   Show that there are no wffs of length 2, 3, or 6, but that any other positive
     length is possible.
 
-\begin{answer}
+  \begin{proof}
-
     TODO
-
+  \end{proof}
-\end{answer}
 
 \subsection{\sorry{Exercise 1.1.3}}%
 \hyperlabel{sub:exercise-1.1.3}
@@ -137,29 +127,26 @@ Let $\alpha$ be a wff; let $c$ be the number of places at which binary
     connective symbols $(\land, \lor, \Rightarrow, \Leftrightarrow)$ occur in
     $\alpha$; let $s$ be the number of places at which sentence symbols occur in
     $\alpha$.
-(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
+    $s = 2$.)
   Show by using the induction principle that $s = c + 1$.
 
-\begin{answer}
+  \begin{proof}
-
     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
     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.
 
-\begin{answer}
+  \begin{proof}
-
     TODO
-
+  \end{proof}
-\end{answer}
 
 \subsection{\sorry{Exercise 1.1.5}}%
 \hyperlabel{sub:exercise-1.1.5}
@@ -173,11 +160,9 @@ Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
   \textit{Suggestion}: Apply induction to show that the length is of the form
     $4k + 1$ and the number of sentence symbols is $k + 1$.
 
-\begin{answer}
+  \begin{proof}
-
     TODO
-
+  \end{proof}
-\end{answer}
 
 \subsection{\sorry{Exercise 1.1.6}}%
 \hyperlabel{sub:exercise-1.1.6}
@@ -189,10 +174,8 @@ Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
     \item Show that more than a quarter of the symbols are sentence symbols.
   \end{enumerate}
 
-\begin{answer}
+  \begin{proof}
-
     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

@@ -22,11 +22,11 @@ Let $(a_i)_{i \geq 0}$ be an arithmetic sequence with common difference $d$.
       \sum_{i=0}^n a_i = \frac{(n + 1)(a_0 + a_n)}{2}.
     \end{equation}
 
-\begin{proof}
-
   \code{Common/Real/Sequence/Arithmetic}
     {Real.Arithmetic.sum\_recursive\_closed}
 
+  \begin{proof}
+
     Let $(a_i)_{i \geq 0}$ be an arithmetic sequence with common difference $d$.
     By definition, for all $k \in \mathbb{N}$,
       \begin{equation}
@@ -82,11 +82,11 @@ Let $(a_i)_{i \geq 0}$ be a geometric sequence with common ratio $r \neq 1$.
       \sum_{i=0}^n a_i = \frac{a_0(1 - r^{n+1})}{1 - r}.
     \end{equation}
 
-\begin{proof}
-
   \code{Common/Real/Sequence/Geometric}
     {Real.Geometric.sum\_recursive\_closed}
 
+  \begin{proof}
+
     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}

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.
-% custom one (code).
 \newcommand\@leanlink[4]{%
-  \textcolor{blue}{\raisebox{-4.5pt}{%
+  \textcolor{BlueViolet}{\raisebox{-4.5pt}{%
-    \tikz{\draw (0, 0) node[yscale=-1,xscale=1] {\faFont};}}}%
+    \tikz{\draw (0, 0) node[yscale=-1,xscale=1] {\faFont};}}{-\;}}%
-  {-\;}\href{#1/#2.html\##3}{#4}}
+  \href{#1/#2.html\##3}{\color{BlueViolet}{#4}}}
-
 \newcommand\@codelink[4]{%
-  \textcolor{blue}{\raisebox{-4.5pt}{%
+  \textcolor{MidnightBlue}{\raisebox{-4.5pt}{%
-    \tikz{\draw (0, 0) node[] {\faCodeBranch};}}}%
+    \tikz{\draw (0, 0) node[xshift=8pt] {\faCodeBranch};}}{-\;}}%
-  {-\;}\href{#1/#2.html\##3}{#4}}
+  \href{#1/#2.html\##3}{\color{MidnightBlue}{#4}}}
 
-% Reference to an anchor of Lean documentation.
+% Reference to an anchor of generated Lean documentation.
 \newcommand\leanref[3]{%
-  \@leanlink{#1}{#2}{#3}{#3}\vspace{10pt}}
-\WithSuffix\newcommand\leanref*[3]{%
   \@leanlink{#1}{#2}{#3}{#3}}
-
 \newcommand\coderef[3]{%
-  \@codelink{#1}{#2}{#3}{#3}\vspace{10pt}}
-\WithSuffix\newcommand\coderef*[3]{%
   \@codelink{#1}{#2}{#3}{#3}}
-
-% Variants that allows customizing display text.
 \newcommand\leanpref[4]{%
-  \@leanlink{#1}{#2}{#3}{#4}\vspace{10pt}}
-\WithSuffix\newcommand\leanpref*[4]{%
   \@leanlink{#1}{#2}{#3}{#4}}
-
 \newcommand\codepref[4]{%
-  \@codelink{#1}{#2}{#3}{#4}\vspace{10pt}}
-\WithSuffix\newcommand\codepref*[4]{%
   \@codelink{#1}{#2}{#3}{#4}}
 
 % Macro to build all Lean related commands relative to a specified directory.
 \newcommand\makeleancommands[1]{%
   \newcommand\lean[2]{%
-    \leanref{#1}{##1}{##2}}
+    \noindent\leanref{#1}{##1}{##2}}
   \WithSuffix\newcommand\lean*[2]{%
-    \leanref*{#1}{##1}{##2}}
+    \vspace{6pt}\noindent\leanref{#1}{##1}{##2}}
 
   \newcommand\code[2]{%
-    \coderef{#1}{##1}{##2}}
+    \noindent\coderef{#1}{##1}{##2}}
   \WithSuffix\newcommand\code*[2]{%
-    \coderef*{#1}{##1}{##2}}
+    \vspace{6pt}\noindent\coderef{#1}{##1}{##2}}
 
   \newcommand\leanp[3]{%
-    \leanpref{#1}{##1}{##2}{##3}}
+    \noindent\leanpref{#1}{##1}{##2}{##3}}
   \WithSuffix\newcommand\leanp*[3]{%
-    \leanpref*{#1}{##1}{##2}{##3}}
+    \vspace{6pt}\noindent\leanpref{#1}{##1}{##2}{##3}}
 
   \newcommand\codep[3]{%
-    \codepref{#1}{##1}{##2}{##3}}
+    \noindent\codepref{#1}{##1}{##2}{##3}}
   \WithSuffix\newcommand\codep*[3]{%
-    \codepref*{#1}{##1}{##2}{##3}}
+    \vspace{6pt}\noindent\codepref{#1}{##1}{##2}{##3}}
 }
 
 % ========================================
@@ -124,11 +110,7 @@
 
 \newcommand\@statement[1]{%
   \linedivider*\paragraph{\normalfont\normalsize\textit{#1.}}}
-\newcommand{\statementpadding}{\ \vspace{8pt}}
-
 \newenvironment{answer}{\@statement{Answer}}{\hfill$\square$}
-\newenvironment{axiom}{\@statement{Axiom}}{\hfill$\square$}
-\newenvironment{definition}{\@statement{Definition}}{\hfill$\square$}
 \renewenvironment{proof}{\@statement{Proof}}{\hfill$\square$}
 
 \newtheorem{corollaryinner}{Corollary}