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Add support for cross-referencing PDFs.

finite-set-exercises
Joshua Potter 2023-05-10 18:26:01 -06:00
parent 8c5029f8ec
commit cadb07018a
11 changed files with 89 additions and 110 deletions
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57
Bookshelf/Apostol/Chapter_1_07.tex
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@ -6,8 +6,9 @@
\graphicspath{{./images/}} \graphicspath{{./images/}}
\newcommand{\larea}[2]{\lean{../..}{Common/Real/Geometry/Area}{#1}{#2}} \externaldocument[A:]
\newcommand{\lrect}[2]{\lean{../..}{Common/Real/Geometry/Rectangle}{#1}{#2}} {../../Common/Real/Geometry/Area}
[../../Common/Real/Geometry/Area.pdf]
\begin{document} \begin{document}
@ -29,10 +30,9 @@ A set consisting of a single point.
\begin{proof} \begin{proof}
Let $S$ be a set consisting of a single point. Let $S$ be a set consisting of a single point.
By definition of a \lrect{Real.Point}{Point}, $S$ is a rectangle in which all By definition of a Point, $S$ is a rectangle in which all vertices coincide.
vertices coincide. By \nameref{A:sec:choice-scale} $S$ is measurable with area its width times
By \larea{Choice-of-Scale}{Choice of Scale}, $S$ is measurable with area its its height.
width times its height.
The width and height of $S$ is trivially zero. The width and height of $S$ is trivially zero.
Therefore $a(S) = (0)(0) = 0$. Therefore $a(S) = (0)(0) = 0$.
@ -65,7 +65,7 @@ A set consisting of a finite number of points in a plane.
By construction, $S_{k+1} = S_k \cup T$. By construction, $S_{k+1} = S_k \cup T$.
By the induction hypothesis, $S_k$ is measurable with area $0$. By the induction hypothesis, $S_k$ is measurable with area $0$.
By \nameref{sub:exercise-1a}, $T$ is measurable with area $0$. By \nameref{sub:exercise-1a}, $T$ is measurable with area $0$.
By the \larea{Additive-Property}{Additive Property}, $S_k \cup T$ is By the \nameref{A:sec:additive-property}, $S_k \cup T$ is
measurable, $S_k \cap T$ is measurable, and measurable, $S_k \cap T$ is measurable, and
\begin{align} \begin{align}
a(S_{k+1}) a(S_{k+1})
@ -111,10 +111,10 @@ The union of a finite collection of line segments in a plane.
\paragraph{Base Case}% \paragraph{Base Case}%
Consider a set $S$ consisting of a single line segment in a plane. Consider a set $S$ consisting of a single line segment in a plane.
By definition of a \lrect{Real.LineSemgnet}{Line Segment}, $S$ is a By definition of a Line Segment, $S$ is a rectangle in which one side has
rectangle in which one side has dimension $0$. dimension $0$.
By \larea{Choice-of-Scale}{Choice of Scale}, $S$ is measurable with area its By \nameref{A:sec:choice-scale}, $S$ is measurable with area its width $w$
width $w$ times its height $h$. times its height $h$.
Therefore $a(S) = wh = 0$. Therefore $a(S) = wh = 0$.
Thus $P(1)$ holds. Thus $P(1)$ holds.
@ -128,8 +128,8 @@ The union of a finite collection of line segments in a plane.
By construction, $S_{k+1} = S_k \cup T$. By construction, $S_{k+1} = S_k \cup T$.
By the induction hypothesis, $S_k$ is measurable with area $0$. By the induction hypothesis, $S_k$ is measurable with area $0$.
By the base case, $T$ is measurable with area $0$. By the base case, $T$ is measurable with area $0$.
By the \larea{Additive-Property}{Additive Property}, $S_k \cup T$ is By the \nameref{A:sec:additive-property}, $S_k \cup T$ is measurable,
measurable, $S_k \cap T$ is measurable, and $S_k \cap T$ is measurable, and
\begin{align} \begin{align}
a(S_{k+1}) a(S_{k+1})
& = a(S_k \cup T) \nonumber \\ & = a(S_k \cup T) \nonumber \\
@ -190,10 +190,10 @@ Prove that every triangular region is measurable and that its area is one half
\centering \centering
\end{figure} \end{figure}
By \larea{Choice-of-Scale}{Choice of Scale}, both $R$ and $S$ are measurable. By \nameref{A:sec:choice-scale}, both $R$ and $S$ are measurable.
By this same axiom, $a(R) = ab$ and $a(S) = ca\sin{\theta}$. By this same axiom, $a(R) = ab$ and $a(S) = ca\sin{\theta}$.
By the \larea{Additive-Property}{Additive Property}, $R \cup S$ and $R \cap S$ By the \nameref{A:sec:additive-property}, $R \cup S$ and $R \cap S$ are both
are both measurable. measurable.
$a(R \cap S) = a(T)$ and $a(R \cup S)$ can be determined by noting that $a(R \cap S) = a(T)$ and $a(R \cup S)$ can be determined by noting that
$R$'s construction implies identity $a(R) = 2a(T)$. $R$'s construction implies identity $a(R) = 2a(T)$.
Therefore Therefore
@ -206,8 +206,8 @@ Prove that every triangular region is measurable and that its area is one half
& = ab + ca\sin{\theta} - ca\sin{\theta} - a(T). & = ab + ca\sin{\theta} - ca\sin{\theta} - a(T).
\end{align*} \end{align*}
Solving for $a(T)$ gives the desired identity: $$a(T) = \frac{1}{2}ab.$$ Solving for $a(T)$ gives the desired identity: $$a(T) = \frac{1}{2}ab.$$
By \larea{Invariance-Under-Congruence}{Invariance Under Congruence}, By \nameref{A:sec:invariance-under-congruence}, $a(T') = a(T)$, concluding our
$a(T') = a(T)$, concluding our proof. proof.
\end{proof} \end{proof}
@ -233,10 +233,10 @@ Prove that every trapezoid and every parallelogram is measurable and derive the
Suppose $S$ is a right trapezoid. Suppose $S$ is a right trapezoid.
Then $S$ is the union of non-overlapping rectangle $R$ of width $b_1$ and Then $S$ is the union of non-overlapping rectangle $R$ of width $b_1$ and
height $h$ with right triangle $T$ of base $b_2 - b_1$ and height $h$. height $h$ with right triangle $T$ of base $b_2 - b_1$ and height $h$.
By \larea{Choice-of-Scale}{Choice of Scale}, $R$ is measurable. By \nameref{A:sec:choice-scale}, $R$ is measurable.
By \nameref{sec:exercise-2}, $T$ is measurable. By \nameref{sec:exercise-2}, $T$ is measurable.
By the \larea{Additive-Property}{Additive Property}, $R \cup T$ and $R \cap T$ By the \nameref{A:sec:additive-property}, $R \cup T$ and $R \cap T$ are both
are both measurable and measurable and
\begin{align*} \begin{align*}
a(S) a(S)
& = a(R \cup T) \\ & = a(R \cup T) \\
@ -256,8 +256,8 @@ Prove that every trapezoid and every parallelogram is measurable and derive the
Then $R$ has longer base edge of length $b_2 - c$. Then $R$ has longer base edge of length $b_2 - c$.
By \nameref{sec:exercise-2}, $T$ is measurable. By \nameref{sec:exercise-2}, $T$ is measurable.
By Case 1, $R$ is measurable. By Case 1, $R$ is measurable.
By the \larea{Additive-Property}{Additive Property}, $R \cup T$ and $R \cap T$ By the \nameref{A:sec:additive-property}, $R \cup T$ and $R \cap T$ are both
are both measurable and measurable and
\begin{align*} \begin{align*}
a(S) a(S)
& = a(T) + a(R) - a(R \cap T) \\ & = a(T) + a(R) - a(R \cap T) \\
@ -274,7 +274,7 @@ Prove that every trapezoid and every parallelogram is measurable and derive the
Let $c$ denote the length of base $T$. Let $c$ denote the length of base $T$.
Reflect $T$ vertically to form another right triangle, say $T'$. Reflect $T$ vertically to form another right triangle, say $T'$.
Then $T' \cup R$ is an acute trapezoid. Then $T' \cup R$ is an acute trapezoid.
By \larea{Invariance-Under-Congruence}{Invariance Under Congruence}, By \nameref{A:sec:invariance-under-congruence},
\begin{equation} \begin{equation}
\label{par:exercise-3-case-3-eq1} \label{par:exercise-3-case-3-eq1}
\tag{3.1} \tag{3.1}
@ -301,7 +301,7 @@ Prove that every trapezoid and every parallelogram is measurable and derive the
Let $c$ denote the length of base $T$. Let $c$ denote the length of base $T$.
Reflect $T$ vertically to form another right triangle, say $T'$. Reflect $T$ vertically to form another right triangle, say $T'$.
Then $T' \cup R$ is an acute trapezoid. Then $T' \cup R$ is an acute trapezoid.
By \larea{Invariance-Under-Congruence}{Invariance Under Congruence}, By \nameref{A:sec:invariance-under-congruence},
\begin{equation} \begin{equation}
\label{par:exercise-3-eq2} \label{par:exercise-3-eq2}
\tag{3.2} \tag{3.2}
@ -338,8 +338,7 @@ Prove that the formula is valid for rectangles with sides parallel to the
We assume $P$ has three non-collinear points, ruling out any instances of We assume $P$ has three non-collinear points, ruling out any instances of
points or line segments. points or line segments.
By \larea{Choice-of-Scale}{Choice of Scale}, $P$ is measurable with area By \nameref{A:sec:choice-scale}, $P$ is measurable with area $a(P) = wh$.
$a(P) = wh$.
By construction, $P$ has $I = (w - 1)(h - 1)$ interior lattice points and By construction, $P$ has $I = (w - 1)(h - 1)$ interior lattice points and
$B = 2(w + h)$ lattice points on its boundary. $B = 2(w + h)$ lattice points on its boundary.
The following shows the lattice point area formula is in agreement with The following shows the lattice point area formula is in agreement with
@ -456,8 +455,8 @@ Use induction on the number of edges to construct a proof for general polygons.
& = a(S) + (I_T + \frac{1}{2}B_T - 1) & \text{induction hypothesis} \\ & = a(S) + (I_T + \frac{1}{2}B_T - 1) & \text{induction hypothesis} \\
& = a(S) + a(T). & \text{base case} & = a(S) + a(T). & \text{base case}
\end{align*} \end{align*}
By the \larea{Additive-Property}{Additive Property}, $S \cup T$ is By the \nameref{A:sec:additive-property}, $S \cup T$ is measurable,
measurable, $S \cap T$ is measurable, and $S \cap T$ is measurable, and
\begin{align*} \begin{align*}
a(P) a(P)
& = a(S \cup T) \\ & = a(S \cup T) \\

24
Bookshelf/Apostol/Chapter_1_11.tex
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@ -2,11 +2,9 @@
\input{../../preamble} \input{../../preamble}
\newcommand{\link}[1]{\lean{../..} \newcommand{\lean}[1]{\href
{Bookshelf/Apostol/Chapter\_1\_11} % Location {./Chapter\_1\_11.html\#Apostol.Chapter\_1\_11.#1}
{Apostol.Chapter\_1\_11.#1} % Fragment {Apostol.Chapter\_1\_11.#1}}
{Chapter\_1\_11.#1} % Presentation
}
\begin{document} \begin{document}
@ -24,7 +22,7 @@ $\floor{x + n} = \floor{x} + n$ for every integer $n$.
\begin{proof} \begin{proof}
\link{exercise\_4a} \lean{exercise\_4a}
\end{proof} \end{proof}
@ -42,8 +40,8 @@ $\floor{-x} =
\ % Force space prior to *Proof.* \ % Force space prior to *Proof.*
\begin{enumerate}[(a)] \begin{enumerate}[(a)]
\item \link{exercise\_4b\_1} \item \lean{exercise\_4b\_1}
\item \link{exercise\_4b\_2} \item \lean{exercise\_4b\_2}
\end{enumerate} \end{enumerate}
\end{proof} \end{proof}
@ -55,7 +53,7 @@ $\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$.
\begin{proof} \begin{proof}
\link{exercise\_4c} \lean{exercise\_4c}
\end{proof} \end{proof}
@ -66,7 +64,7 @@ $\floor{2x} = \floor{x} + \floor{x + \frac{1}{2}}.$
\begin{proof} \begin{proof}
\link{exercise\_4d} \lean{exercise\_4d}
\end{proof} \end{proof}
@ -77,7 +75,7 @@ $\floor{3x} = \floor{x} + \floor{x + \frac{1}{3}} + \floor{x + \frac{2}{3}}.$
\begin{proof} \begin{proof}
\link{exercise\_4e} \lean{exercise\_4e}
\end{proof} \end{proof}
@ -92,7 +90,7 @@ State and prove such a generalization.
\begin{proof} \begin{proof}
\link{exercise\_5} \lean{exercise\_5}
\divider \divider
@ -223,7 +221,7 @@ Now apply Exercises 4(a) and (b) to the bracket on the right.
\begin{proof} \begin{proof}
\link{exercise\_7b} \lean{exercise\_7b}
\end{proof} \end{proof}

30
Bookshelf/Apostol/Chapter_I_03.tex
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@ -2,11 +2,9 @@
\input{../../preamble} \input{../../preamble}
\newcommand{\link}[1]{\lean{../..} \newcommand{\lean}[1]{\href
{Bookshelf/Apostol/Chapter\_I\_03} % Location {./Chapter\_I\_03.html\#Apostol.Chapter\_I\_03.#1}
{Apostol.Chapter\_I\_03.#1} % Fragment {Apostol.Chapter\_I\_03.#1}}
{Chapter\_I\_03.#1} % Presentation
}
\begin{document} \begin{document}
@ -19,7 +17,7 @@ Nonempty set $S$ has supremum $L$ if and only if set $-S$ has infimum $-L$.
\begin{proof} \begin{proof}
\link{is\_lub\_neg\_set\_iff\_is\_glb\_set\_neg} \lean{is\_lub\_neg\_set\_iff\_is\_glb\_set\_neg}
\divider \divider
@ -40,7 +38,7 @@ Every nonempty set $S$ that is bounded below has a greatest lower bound; that
\begin{proof} \begin{proof}
\link{exists\_isGLB} \lean{exists\_isGLB}
\divider \divider
@ -59,7 +57,7 @@ For every real $x$ there exists a positive integer $n$ such that $n > x$.
\begin{proof} \begin{proof}
\link{exists\_pnat\_geq\_self} \lean{exists\_pnat\_geq\_self}
\divider \divider
@ -82,7 +80,7 @@ If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive
\begin{proof} \begin{proof}
\link{exists\_pnat\_mul\_self\_geq\_of\_pos} \lean{exists\_pnat\_mul\_self\_geq\_of\_pos}
\divider \divider
@ -101,7 +99,7 @@ If three real numbers $a$, $x$, and $y$ satisfy the inequalities
\begin{proof} \begin{proof}
\link{forall\_pnat\_leq\_self\_leq\_frac\_imp\_eq} \lean{forall\_pnat\_leq\_self\_leq\_frac\_imp\_eq}
\divider \divider
@ -144,7 +142,7 @@ If three real numbers $a$, $x$, and $y$ satisfy the inequalities
\begin{proof} \begin{proof}
\link{forall\_pnat\_frac\_leq\_self\_leq\_imp\_eq} \lean{forall\_pnat\_frac\_leq\_self\_leq\_imp\_eq}
\divider \divider
@ -191,7 +189,7 @@ If $S$ has a supremum, then for some $x$ in $S$ we have $x > \sup{S} - h$.
\begin{proof} \begin{proof}
\link{sup\_imp\_exists\_gt\_sup\_sub\_delta} \lean{sup\_imp\_exists\_gt\_sup\_sub\_delta}
\divider \divider
@ -213,7 +211,7 @@ If $S$ has an infimum, then for some $x$ in $S$ we have $x < \inf{S} + h$.
\begin{proof} \begin{proof}
\link{inf\_imp\_exists\_lt\_inf\_add\_delta} \lean{inf\_imp\_exists\_lt\_inf\_add\_delta}
\divider \divider
@ -244,7 +242,7 @@ If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and
\begin{proof} \begin{proof}
\link{sup\_minkowski\_sum\_eq\_sup\_add\_sup} \lean{sup\_minkowski\_sum\_eq\_sup\_add\_sup}
\divider \divider
@ -312,7 +310,7 @@ If each of $A$ and $B$ has an infimum, then $C$ has an infimum, and
\begin{proof} \begin{proof}
\link{inf\_minkowski\_sum\_eq\_inf\_add\_inf} \lean{inf\_minkowski\_sum\_eq\_inf\_add\_inf}
\divider \divider
@ -381,7 +379,7 @@ Given two nonempty subsets $S$ and $T$ of $\mathbb{R}$ such that $$s \leq t$$
\begin{proof} \begin{proof}
\link{forall\_mem\_le\_forall\_mem\_imp\_sup\_le\_inf} \lean{forall\_mem\_le\_forall\_mem\_imp\_sup\_le\_inf}
\divider \divider

10
Bookshelf/Enderton/Chapter_0.tex
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@ -2,11 +2,9 @@
\input{../../preamble} \input{../../preamble}
\newcommand{\link}[1]{\lean{../..} \newcommand{\lean}[1]{\href
{Bookshelf/Enderton/Chapter\_0} % Location {./Chapter\_0.html\#Enderton.Chapter\_0.#1}
{Enderton.Chapter\_0.#1} % Fragment {Enderton.Chapter\_0.#1}}
{Chapter\_0.#1} % Presentation
}
\begin{document} \begin{document}
@ -21,7 +19,7 @@ Then $x_1 = \langle y_1, \ldots, y_{k+1} \rangle$.
\begin{proof} \begin{proof}
\link{lemma\_0a} \lean{lemma\_0a}
\end{proof} \end{proof}

20
Common/Real/Geometry/Area.tex
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@ -1,12 +1,8 @@
\documentclass{article} \documentclass{article}
\input{../../preamble} \input{../../../preamble}
\newcommand{\link}[2]{\lean{../..} \newcommand{\lean}[2]{\href{./Area.html\##1}{#2}}
{Common/Real/Geometry/Area} % Location
{#1} % Fragment
{#2} % Presentation
}
\begin{document} \begin{document}
@ -23,7 +19,7 @@ For each set $S$ in $\mathscr{M}$, we have $a(S) \geq 0$.
\begin{axiom} \begin{axiom}
\link{Nonnegative-Property}{Nonnegative Property} \lean{Nonnegative-Property}{Nonnegative Property}
\end{axiom} \end{axiom}
@ -35,7 +31,7 @@ If $S$ and $T$ are in $\mathscr{M}$, then $S \cup T$ and $S \cap T$ are in
\begin{axiom} \begin{axiom}
\link{Additive-Property}{Additive Property} \lean{Additive-Property}{Additive Property}
\end{axiom} \end{axiom}
@ -47,7 +43,7 @@ If $S$ and $T$ are in $\mathscr{M}$ with $S \subseteq T$, then $T - S$ is in
\begin{axiom} \begin{axiom}
\link{Difference-Property}{Difference Property} \lean{Difference-Property}{Difference Property}
\end{axiom} \end{axiom}
@ -59,7 +55,7 @@ If a set $S$ is in $\mathscr{M}$ and if $T$ is congruent to $S$, then $T$ is
\begin{axiom} \begin{axiom}
\link{Invariant-Under-Congruence}{Invariance Under Congruence} \lean{Invariant-Under-Congruence}{Invariance Under Congruence}
\end{axiom} \end{axiom}
@ -71,7 +67,7 @@ If the edges of $R$ have lengths $h$ and $k$, then $a(R) = hk$.
\begin{axiom} \begin{axiom}
\link{Choice-of-Scale}{Choice of Scale} \lean{Choice-of-Scale}{Choice of Scale}
\end{axiom} \end{axiom}
@ -90,7 +86,7 @@ If there is one and only one number $c$ which satisfies the inequalities
\begin{axiom} \begin{axiom}
\link{Exhaustion-Property}{Exhaustion Property} \lean{Exhaustion-Property}{Exhaustion Property}
\end{axiom} \end{axiom}

10
Common/Real/Sequence/Arithmetic.tex
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@ -2,11 +2,9 @@
\input{../../../preamble} \input{../../../preamble}
\newcommand{\link}[1]{\lean{../../..} \newcommand{\lean}[1]{\href
{Common/Real/Sequence/Arithmetic} % Location {./Arithmetic.html\#Real.Arithmetic.#1}
{Real.Arithmetic.#1} % Fragment {Real.Arithmetic.#1}}
{Real.Arithmetic.#1} % Presentation
}
\begin{document} \begin{document}
@ -19,7 +17,7 @@ $$\sum_{i=0}^n a_i = \frac{(n + 1)(a_0 + a_n)}{2}.$$
\begin{proof} \begin{proof}
\link{sum\_recursive\_closed} \lean{sum\_recursive\_closed}
\end{proof} \end{proof}

10
Common/Real/Sequence/Geometric.tex
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@ -2,11 +2,9 @@
\input{../../../preamble} \input{../../../preamble}
\newcommand{\link}[1]{\lean{../../..} \newcommand{\lean}[1]{\href
{Common/Real/Sequence/Geometric} % Location {./Geometric.html\#Real.Geometric.#1}
{Real.Geometric.#1} % Fragment {Real.Geometric.#1}}
{Real.Geometric.#1} % Presentation
}
\begin{document} \begin{document}
@ -19,7 +17,7 @@ $$\sum_{i=0}^n a_i = \frac{a_0(1 - r^{n+1})}{1 - r}.$$
\begin{proof} \begin{proof}
\link{sum\_recursive\_closed} \lean{sum\_recursive\_closed}
\end{proof} \end{proof}

15
README.md
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@ -30,11 +30,10 @@ project.
A color/symbol code is used on generated PDF headers to indicate their status: A color/symbol code is used on generated PDF headers to indicate their status:
* Teal coloring (with a checkmark) indicates the corresponding proof is * Teal coloring indicates the corresponding proof is complete. That is, the
complete. That is, the proof has been written in TeX and also formally proof has been written in TeX and also formally verified in Lean.
verified in Lean. * Magenta coloring indicates the corresponding proof is in progress. That is, a
* Magenta coloring (with a spinner) indicates the corresponding proof is in proof in both TeX and Lean have not yet been finished, but is actively being
progress. That is, a proof in both TeX and Lean have not yet been finished, worked on.
but is actively being worked on. * Red coloring indicates the formal Lean proof has not yet been started. It may
* Red coloring (with a warning) indicates the formal Lean proof has not yet been or may not also indicate the TeX proof has been written.
started. It may or may not also indicate the TeX proof has been written.

4
lake-manifest.json
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@ -16,9 +16,9 @@
{"git": {"git":
{"url": "https://github.com/jrpotter/bookshelf-docgen.git", {"url": "https://github.com/jrpotter/bookshelf-docgen.git",
"subDir?": null, "subDir?": null,
"rev": "80a4dc8d508161c859dbbb455f3855e051a28890", "rev": "3f941dc8a814321498082da49f4a8430bbfbbb6c",
"name": "doc-gen4", "name": "doc-gen4",
"inputRev?": "80a4dc8d508161c859dbbb455f3855e051a28890"}}, "inputRev?": "3f941dc8a814321498082da49f4a8430bbfbbb6c"}},
{"git": {"git":
{"url": "https://github.com/mhuisi/lean4-cli", {"url": "https://github.com/mhuisi/lean4-cli",
"subDir?": null, "subDir?": null,

2
lakefile.lean
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@ -12,7 +12,7 @@ require std4 from git
"6006307d2ceb8743fea7e00ba0036af8654d0347" "6006307d2ceb8743fea7e00ba0036af8654d0347"
require «doc-gen4» from git require «doc-gen4» from git
"https://github.com/jrpotter/bookshelf-docgen.git" @ "https://github.com/jrpotter/bookshelf-docgen.git" @
"80a4dc8d508161c859dbbb455f3855e051a28890" "3f941dc8a814321498082da49f4a8430bbfbbb6c"
@[default_target] @[default_target]
lean_lib «Bookshelf» { lean_lib «Bookshelf» {

17
preamble.tex
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@ -3,23 +3,18 @@
\usepackage{environ} \usepackage{environ}
\usepackage{fancybox} \usepackage{fancybox}
\usepackage{fontawesome5} \usepackage{fontawesome5}
\usepackage{hyperref}
\usepackage{mathrsfs} \usepackage{mathrsfs}
\usepackage{soul, xcolor} \usepackage{soul}
\usepackage{xcolor}
% `hyperref` comes after `xr-hyper`.
\usepackage{xr-hyper}
\usepackage{hyperref}
% ======================================== % ========================================
% Linking % Linking
% ======================================== % ========================================
\hypersetup{colorlinks=true, urlcolor=blue} \hypersetup{colorlinks=true, urlcolor=blue}
% The first argument refers to a relative path upward from a current file to
% the root of the workspace (i.e. where this `preamble.tex` file is located).
% #1 - Path to root
% #2 - Location
% #3 - Fragment
% #4 - Presentation
\newcommand{\lean}[4]{\href{#1/#2.html\##3}{#4}}
\newcommand{\hyperlabel}[1]{% \newcommand{\hyperlabel}[1]{%
\label{#1}% \label{#1}%
\hypertarget{#1}{}} \hypertarget{#1}{}}
@ -56,7 +51,7 @@
\DeclareRobustCommand{\verified}[1]{% \DeclareRobustCommand{\verified}[1]{%
\texorpdfstring{\color{teal}#1\ \faCheckCircle}{#1}} \texorpdfstring{\color{teal}#1\ \faCheckCircle}{#1}}
\DeclareRobustCommand{\proceeding}[1]{% \DeclareRobustCommand{\proceeding}[1]{%
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\DeclareRobustCommand{\unverified}[1]{% \DeclareRobustCommand{\unverified}[1]{%
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