Perform same aggregation on Enderton and Sequence tex files.

Bookshelf/Apostol.tex

@@ -7,6 +7,7 @@
 \graphicspath{{./Apostol/images/}}
 
 \newcommand{\lean}[2]{\leanref{../#1.html\##2}{#2}}
+\newcommand{\leanPretty}[3]{\leanref{../#1.html\##2}{#3}}
 
 \begin{document}
 
@@ -26,7 +27,7 @@ Nonempty set $S$ has supremum $L$ if and only if set $-S$ has infimum $-L$.
 
 \begin{proof}
 
-  \lean{Chapter\_I\_03}
+  \lean{Bookshelf/Apostol/Chapter\_I\_03}
     {Apostol.Chapter\_I\_03.is\_lub\_neg\_set\_iff\_is\_glb\_set\_neg}
 
   \divider
@@ -48,7 +49,7 @@ Every nonempty set $S$ that is bounded below has a greatest lower bound; that
 
 \begin{proof}
 
-  \lean{Chapter\_I\_03}
+  \lean{Bookshelf/Apostol/Chapter\_I\_03}
     {Apostol.Chapter\_I\_03.exists\_isGLB}
 
   \divider
@@ -68,7 +69,7 @@ For every real $x$ there exists a positive integer $n$ such that $n > x$.
 
 \begin{proof}
 
-  \lean{Chapter\_I\_03}
+  \lean{Bookshelf/Apostol/Chapter\_I\_03}
     {Apostol.Chapter\_I\_03.exists\_pnat\_geq\_self}
 
   \divider
@@ -92,7 +93,7 @@ If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive
 
 \begin{proof}
 
-  \lean{Chapter\_I\_03}
+  \lean{Bookshelf/Apostol/Chapter\_I\_03}
     {Apostol.Chapter\_I\_03.exists\_pnat\_mul\_self\_geq\_of\_pos}
 
   \divider
@@ -112,7 +113,7 @@ If three real numbers $a$, $x$, and $y$ satisfy the inequalities
 
 \begin{proof}
 
-  \lean{Chapter\_I\_03}
+  \lean{Bookshelf/Apostol/Chapter\_I\_03}
     {Apostol.Chapter\_I\_03.forall\_pnat\_leq\_self\_leq\_frac\_imp\_eq}
 
   \divider
@@ -156,7 +157,7 @@ If three real numbers $a$, $x$, and $y$ satisfy the inequalities
 
 \begin{proof}
 
-  \lean{Chapter\_I\_03}
+  \lean{Bookshelf/Apostol/Chapter\_I\_03}
     {Apostol.Chapter\_I\_03.forall\_pnat\_frac\_leq\_self\_leq\_imp\_eq}
 
   \divider
@@ -204,7 +205,7 @@ If $S$ has a supremum, then for some $x$ in $S$ we have $x > \sup{S} - h$.
 
 \begin{proof}
 
-  \lean{Chapter\_I\_03}
+  \lean{Bookshelf/Apostol/Chapter\_I\_03}
     {Apostol.Chapter\_I\_03.sup\_imp\_exists\_gt\_sup\_sub\_delta}
 
   \divider
@@ -227,7 +228,7 @@ If $S$ has an infimum, then for some $x$ in $S$ we have $x < \inf{S} + h$.
 
 \begin{proof}
 
-  \lean{Chapter\_I\_03}
+  \lean{Bookshelf/Apostol/Chapter\_I\_03}
     {Apostol.Chapter\_I\_03.inf\_imp\_exists\_lt\_inf\_add\_delta}
 
   \divider
@@ -259,7 +260,7 @@ If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and
 
 \begin{proof}
 
-  \lean{Chapter\_I\_03}
+  \lean{Bookshelf/Apostol/Chapter\_I\_03}
     {Apostol.Chapter\_I\_03.sup\_minkowski\_sum\_eq\_sup\_add\_sup}
 
   \divider
@@ -328,7 +329,7 @@ If each of $A$ and $B$ has an infimum, then $C$ has an infimum, and
 
 \begin{proof}
 
-  \lean{Chapter\_I\_03}
+  \lean{Bookshelf/Apostol/Chapter\_I\_03}
     {Apostol.Chapter\_I\_03.inf\_minkowski\_sum\_eq\_inf\_add\_inf}
 
   \divider
@@ -398,7 +399,7 @@ Given two nonempty subsets $S$ and $T$ of $\mathbb{R}$ such that $$s \leq t$$
 
 \begin{proof}
 
-  \lean{Chapter\_I\_03}
+  \lean{Bookshelf/Apostol/Chapter\_I\_03}
     {Apostol.Chapter\_I\_03.forall\_mem\_le\_forall\_mem\_imp\_sup\_le\_inf}
 
   \divider
@@ -436,7 +437,8 @@ For each set $S$ in $\mathscr{M}$, we have $a(S) \geq 0$.
 
 \begin{axiom}
 
-  \lean{Common/Real/Geometry/Area}{Nonnegative-Property}
+  \leanPretty{Common/Real/Geometry/Area}{Nonnegative-Property}
+    {Nonnegative Property}
 
 \end{axiom}
 
@@ -448,7 +450,8 @@ If $S$ and $T$ are in $\mathscr{M}$, then $S \cup T$ and $S \cap T$ are in
 
 \begin{axiom}
 
-  \lean{Common/Real/Geometry/Area}{Additive-Property}
+  \leanPretty{Common/Real/Geometry/Area}{Additive-Property}
+    {Additive Property}
 
 \end{axiom}
 
@@ -460,7 +463,8 @@ If $S$ and $T$ are in $\mathscr{M}$ with $S \subseteq T$, then $T - S$ is in
 
 \begin{axiom}
 
-  \lean{Common/Real/Geometry/Area}{Difference-Property}
+  \leanPretty{Common/Real/Geometry/Area}{Difference-Property}
+    {Difference Property}
 
 \end{axiom}
 
@@ -472,7 +476,8 @@ If a set $S$ is in $\mathscr{M}$ and if $T$ is congruent to $S$, then $T$ is
 
 \begin{axiom}
 
-  \lean{Common/Real/Geometry/Area}{Invariance-Under-Congruence}
+  \leanPretty{Common/Real/Geometry/Area}{Invariance-Under-Congruence}
+    {Invariance Under Congruence}
 
 \end{axiom}
 
@@ -484,7 +489,8 @@ If the edges of $R$ have lengths $h$ and $k$, then $a(R) = hk$.
 
 \begin{axiom}
 
-  \lean{Common/Real/Geometry/Area}{Choice-of-Scale}
+  \leanPretty{Common/Real/Geometry/Area}{Choice-of-Scale}
+    {Choice of Scale}
 
 \end{axiom}
 
@@ -503,7 +509,8 @@ If there is one and only one number $c$ which satisfies the inequalities
 
 \begin{axiom}
 
-  \lean{Common/Real/Geometry/Area}{Exhaustion-Property}
+  \leanPretty{Common/Real/Geometry/Area}{Exhaustion-Property}
+    {Exhaustion Property}
 
 \end{axiom}
 

Bookshelf/Enderton.tex

@@ -0,0 +1,36 @@
+\documentclass{report}
+
+\input{../preamble}
+
+\newcommand{\lean}[2]{\leanref{../#1.html\##2}{#2}}
+\newcommand{\leanPretty}[3]{\leanref{../#1.html\##2}{#3}}
+
+\begin{document}
+
+\header
+  {A Mathematical Introduction to Logic}
+  {Herbert B. Enderton}
+
+\tableofcontents
+
+% Sets first chapter to `0` to match Enderton book.
+\setcounter{chapter}{0}
+\addtocounter{chapter}{-1}
+
+\chapter{Useful Facts About Sets}%
+\label{chap:useful-facts-about-sets}
+
+\section{\unverified{Lemma 0A}}%
+\label{sec:lemma-0a}
+
+Assume that $\langle x_1, \ldots, x_m \rangle =
+  \langle y_1, \ldots, y_m, \ldots, y_{m+k} \rangle$.
+Then $x_1 = \langle y_1, \ldots, y_{k+1} \rangle$.
+
+\begin{proof}
+
+  \lean{Bookshelf/Enderton/Chapter\_0}{Enderton.Chapter\_0.lemma\_0a}
+
+\end{proof}
+
+\end{document}

Bookshelf/Enderton/Chapter_0.tex

@@ -1,26 +0,0 @@
-\documentclass{article}
-
-\input{../../preamble}
-
-\newcommand{\lean}[1]{\leanref
-  {./Chapter\_0.html\#Enderton.Chapter\_0.#1}
-  {Enderton.Chapter\_0.#1}}
-
-\begin{document}
-
-\header{Useful Facts About Sets}{Herbert B. Enderton}
-
-\section*{\unverified{Lemma 0A}}%
-\label{sec:lemma-0a}
-
-Assume that $\langle x_1, \ldots, x_m \rangle =
-  \langle y_1, \ldots, y_m, \ldots, y_{m+k} \rangle$.
-Then $x_1 = \langle y_1, \ldots, y_{k+1} \rangle$.
-
-\begin{proof}
-
-  \lean{lemma\_0a}
-
-\end{proof}
-
-\end{document}

Common/Real/Sequence.tex

@@ -0,0 +1,44 @@
+\documentclass{article}
+
+\input{../../preamble}
+
+\newcommand{\lean}[2]{\leanref{../../#1.html\##2}{#2}}
+
+\begin{document}
+
+\header{Sequences}{}
+
+\tableofcontents
+
+\section{Summations}%
+\label{sec:summations}
+
+\subsection{\unverified{Arithmetic Series}}%
+\label{sub:sum-arithmetic-series}
+
+Let $(a_i)_{i \geq 0}$ be an arithmetic sequence with common difference $d$.
+  Then for some $n \in \mathbb{N}$,
+  $$\sum_{i=0}^n a_i = \frac{(n + 1)(a_0 + a_n)}{2}.$$
+
+\begin{proof}
+
+  \lean{Common/Real/Sequence/Arithmetic}
+    {Real.Arithmetic.sum\_recursive\_closed}
+
+\end{proof}
+
+\subsection{\unverified{Geometric Series}}%
+\label{sub:sum-geometric-series}
+
+Let $(a_i)_{i \geq 0}$ be a geometric sequence with common ratio $r \neq 1$.
+  Then for some $n \in \mathbb{N}$,
+  $$\sum_{i=0}^n a_i = \frac{a_0(1 - r^{n+1})}{1 - r}.$$
+
+\begin{proof}
+
+  \lean{Common/Real/Sequence/Geometric}
+    {Real.Geometric.sum\_recursive\_closed}
+
+\end{proof}
+
+\end{document}

Common/Real/Sequence/Arithmetic.tex

@@ -1,24 +0,0 @@
-\documentclass{article}
-
-\input{../../../preamble}
-
-\newcommand{\lean}[1]{\leanref
-  {./Arithmetic.html\#Real.Arithmetic.#1}
-  {Real.Arithmetic.#1}}
-
-\begin{document}
-
-\section{\unverified{Sum of Arithmetic Series}}%
-\label{sec:sum-arithmetic-series}
-
-Let $(a_i)_{i \geq 0}$ be an arithmetic sequence with common difference $d$.
-Then for some $n \in \mathbb{N}$,
-$$\sum_{i=0}^n a_i = \frac{(n + 1)(a_0 + a_n)}{2}.$$
-
-\begin{proof}
-
-  \lean{sum\_recursive\_closed}
-
-\end{proof}
-
-\end{document}

Common/Real/Sequence/Geometric.tex

@@ -1,24 +0,0 @@
-\documentclass{article}
-
-\input{../../../preamble}
-
-\newcommand{\lean}[1]{\leanref
-  {./Geometric.html\#Real.Geometric.#1}
-  {Real.Geometric.#1}}
-
-\begin{document}
-
-\section{\unverified{Sum of Geometric Series}}%
-\label{sec:sum-geometric-series}
-
-Let $(a_i)_{i \geq 0}$ be a geometric sequence with common ratio $r \neq 1$.
-Then for some $n \in \mathbb{N}$,
-$$\sum_{i=0}^n a_i = \frac{a_0(1 - r^{n+1})}{1 - r}.$$
-
-\begin{proof}
-
-  \lean{sum\_recursive\_closed}
-
-\end{proof}
-
-\end{document}