r
/
bookshelf
1
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

Simplify lean link formatting.

finite-set-exercises
Joshua Potter 2023-08-09 07:39:41 -06:00
parent e8aa984b98
commit d89200fd9d
7 changed files with 7804 additions and 8426 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

1014
Bookshelf/Apostol.tex
View File

File diff suppressed because it is too large Load Diff

119
Bookshelf/Enderton/Logic.tex
View File

@ -18,9 +18,9 @@
\section{\defined{Construction Sequence}}%
\hyperlabel{ref:construction-sequence}
A \textbf{construction sequence} is a finite sequence
$\langle \epsilon_1, \ldots, \epsilon_n \rangle$ of expressions such that for
each $i \leq n$ we have at least one of
A \textbf{construction sequence} is a finite sequence
$\langle \epsilon_1, \ldots, \epsilon_n \rangle$ of expressions such that
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 \\
@ -33,14 +33,15 @@ A \textbf{construction sequence} is a finite sequence
\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
applying some finite number of times the \textbf{formula-building operations}
(on expressions) defined by the equations:
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} (on expressions) defined by the
equations:
\begin{align*}
\mathcal{E}_{\neg}(\alpha)
& = (\neg \alpha) \\
@ -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]
Assume that $\langle x_1, \ldots, x_m \rangle =
\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}
Then $x_1 = \langle y_1, \ldots, y_{k+1} \rangle$.
\end{lemma}
\begin{proof}
TODO
\end{proof}
\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}
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}
\begin{proof}
\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}
\begin{proof}
TODO
\end{proof}
\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.
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.
\begin{answer}
\begin{answer}
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.
\begin{answer}
\begin{proof}
TODO
\end{answer}
\end{proof}
\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
$\alpha$; 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 $s = 2$.)
Show by using the induction principle that $s = c + 1$.
\begin{answer}
(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{proof}
TODO
\end{answer}
\end{proof}
\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 $A_4$.
Suppose we delete all the expressions in the construction sequence that contain
$A_4$.
Show that the result is still a legal construction sequence.
\begin{answer}
Suppose we delete all the expressions in the construction sequence that
contain $A_4$.
Show that the result is still a legal construction sequence.
\begin{proof}
TODO
\end{answer}
\end{proof}
\subsection{\sorry{Exercise 1.1.5}}%
\hyperlabel{sub:exercise-1.1.5}
Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
\begin{enumerate}[(a)]
Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
\begin{enumerate}[(a)]
\item Show that the length of $\alpha$ (i.e., the number of symbols in the
string) is odd.
\item Show that more than a quarter of the symbols are sentence symbols.
\end{enumerate}
\textit{Suggestion}: Apply induction to show that the length is of the form
\end{enumerate}
\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{answer}
\end{proof}
\subsection{\sorry{Exercise 1.1.6}}%
\hyperlabel{sub:exercise-1.1.6}
Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
\begin{enumerate}[(a)]
Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
\begin{enumerate}[(a)]
\item Show that the length of $\alpha$ (i.e., the number of symbols in the
string) is odd.
\item Show that more than a quarter of the symbols are sentence symbols.
\end{enumerate}
\begin{answer}
\end{enumerate}
\begin{proof}
TODO
\end{answer}
\end{proof}
\end{document}

3072
Bookshelf/Enderton/Set.tex
View File

File diff suppressed because it is too large Load Diff

147
Bookshelf/Enderton/Set/Chapter_2.lean
View File

@ -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

2
Bookshelf/Enderton/Set/Chapter_3.lean
View File

@ -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

16
Common/Real/Sequence.tex
View File

@ -15,18 +15,18 @@
\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 $n \in \mathbb{N}$,
\begin{equation}
\hyperlabel{sub:sum-arithmetic-series-eq1}
\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}
@ -70,23 +70,23 @@ Let $(a_i)_{i \geq 0}$ be an arithmetic sequence with common difference $d$.
By mathematical induction, it follows for all $n \geq 0$, $P(n)$ is true.
\end{proof}
\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 $n \in \mathbb{N}$,
\begin{equation}
\hyperlabel{sub:sum-geometric-series-eq1}
\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}
@ -127,6 +127,6 @@ Let $(a_i)_{i \geq 0}$ be a geometric sequence with common ratio $r \neq 1$.
By mathematical induction, it follows for all $n \geq 0$, $P(n)$ is true.
\end{proof}
\end{proof}
\end{document}

50
preamble.tex
View File

@ -44,61 +44,47 @@
\label{#1}%
\hypertarget{#1}{}}
% Denote whether we are working with a standard/Mathlib statement (lean) or a
% custom one (code).
% Denote if working with a predefined statement/theorem or a custom one.
\newcommand\@leanlink[4]{%
\textcolor{blue}{\raisebox{-4.5pt}{%
\tikz{\draw (0, 0) node[yscale=-1,xscale=1] {\faFont};}}}%
{-\;}\href{#1/#2.html\##3}{#4}}
\textcolor{BlueViolet}{\raisebox{-4.5pt}{%
\tikz{\draw (0, 0) node[yscale=-1,xscale=1] {\faFont};}}{-\;}}%
\href{#1/#2.html\##3}{\color{BlueViolet}{#4}}}
\newcommand\@codelink[4]{%
\textcolor{blue}{\raisebox{-4.5pt}{%
\tikz{\draw (0, 0) node[] {\faCodeBranch};}}}%
{-\;}\href{#1/#2.html\##3}{#4}}
\textcolor{MidnightBlue}{\raisebox{-4.5pt}{%
\tikz{\draw (0, 0) node[xshift=8pt] {\faCodeBranch};}}{-\;}}%
\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}