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Co-authored-by: Mark <mark@betalupi.com>
Co-committed-by: Mark <mark@betalupi.com>
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2025-01-22 12:28:44 -08:00
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src/Advanced/Mock a Mockingbird/main.tex Executable file
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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\usepackage{mathtools} % for \coloneqq
%\usepackage{lua-visual-debug}
\usepackage{censor}
\usepackage{alltt}
\newenvironment{helpbox}[1][0.5]{
\begin{center}
\begin{tcolorbox}[
colback=white!90!black,
colframe=white!90!black,
coltitle=black,
center title,
width = #1\textwidth,
leftrule = 0mm,
rightrule = 0mm,
toprule = 0mm,
bottomrule = 0mm,
left = 1mm,
right = 1mm,
top = 1mm,
bottom = 1mm,
toptitle = 1mm,
lefttitle = 1mm,
titlerule = 1pt,
title={\textbf{Things you will need:}}
]
}{
\end{tcolorbox}
\end{center}
}
% Logic block comment
\newcommand{\cmnt}[1]{\textcolor{gray}{\# #1}}
\newcounter{allttLineCounter}
\setcounter{allttLineCounter}{0}
\newcommand{\linenoref}[1]{\colorbox{gray!30!white}{#1}}
\newcommand{\lineno}{
\stepcounter{allttLineCounter}%
\linenoref{\ifnum\value{allttLineCounter}<10 0\fi\arabic{allttLineCounter}}%
}
% Redefine alltt so it automatically
% resets allttLineCounter
\let\oldalltt\alltt
\renewenvironment{alltt}
{\setcounter{allttLineCounter}{0}\begin{oldalltt}}
{\end{oldalltt}}
\newcommand{\thus}{\(\Rightarrow\)}
\newcommand{\qed}{\(\blacksquare\)}
\uptitlel{Advanced 2}
\uptitler{\smallurl{}}
\title{To Mock a Mockingbird}
\subtitle{
Prepared by Mark on \today \\
Based on a book of the same name.
}
\begin{document}
\maketitle
\input{parts/00 intro}
\input{parts/01 tmam}
\input{parts/02 kestrel}
\end{document}

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src/Advanced/Mock a Mockingbird/meta.toml Normal file
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[metadata]
title = "Mock a Mockingbird"
[publish]
handout = true
solutions = true

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src/Advanced/Mock a Mockingbird/parts/00 intro.tex Normal file
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\section{Introduction}
A certain enchanted forest is inhabited by talking birds. Each of these birds has a name, and will respond whenever it hears the name of another. Suppose you are exploring this forest and come across the bird $A$. You call the name of bird $B$. $A$ hears you and responds with the name of some other bird, which we will designate $AB$.
Bird $AB$ is, by definition, $A$'s response to $B$.
\vspace{2mm}
As you wander around this forest, you quickly discover two interesting facts:
\begin{enumerate}[itemsep = 1mm]
\item $A$'s response to $B$ mustn't be the same as $B$'s response to $A$.
\item Given three birds $A$, $B$, and $C$, $(AB)C$ and $A(BC)$ are not necessarily the same bird. \\
Bird $A(BC)$ is $A$'s response to bird $BC$, while $(AB)C$ is $AB$'s response to $C$. \\
Thus, $ABC$ is ambiguous. Parenthesis are mandatory.
\end{enumerate}
\vspace{2mm}
You also find that this forest has two laws:
\begin{enumerate}[itemsep = 1mm]
\item \textit{The Law of Composition}: \\
For any two birds $A$ and $B$, there must be a bird $C$ so that $Cx = A(Bx)$
\item \textit{The Law of the Mockingbird}: \\
The forest must contain the Mockingbird $M$, which always satisfies $Mx = xx$. \\
In other words, the Mockingbird's response to any bird $x$ is the same as $x$'s response to itself.
\end{enumerate}
\vfill
\definition{}
We say a bird $A$ is fond of a bird $B$ if $A$ responds to $B$ with $B$. \\
In other words, $A$ is fond of $B$ if $AB = B$.
\vfill
\definition{}
We say a bird $C$ \textit{composes} $A$ with $B$ if for any bird $x$,
$$
Cx = A(Bx)
$$
In other words, this means that $C$'s response to $x$ is the same as $A$'s response to $B$'s response to $x$. \\
Note that $C$ is exactly the kind of bird $L_1$ guarantees.
\vfill
\pagebreak

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src/Advanced/Mock a Mockingbird/parts/01 tmam.tex Normal file
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\section{To Mock a Mockingbird}
\problem{}
Mark tells you that any bird $A$ is fond of at least one other bird. \\
Complete his proof.
\begin{alltt}
let A \cmnt{Let A be any any bird.}
let Cx = A(Mx) \cmnt{Define C as the composition of A and M}
\cmnt{The rest is up to you.}
CC = ??
\end{alltt}
\begin{helpbox}
\texttt{Law:} There exists a Mockingbird, $Mx := xx$ \\
\texttt{Def:} $A$ is fond of $B$ if $AB = B$
\end{helpbox}
\begin{solution}
\begin{alltt}
\lineno{} let A \cmnt{Let A be any any bird.}
\lineno{} let Cx = A(Mx) \cmnt{Define C as the composition of A and M}
\lineno{} CC = A(MC)
\lineno{} = A(CC) \qed{}
\end{alltt}
\end{solution}
\vfill
\problem{}
We say a bird $A$ is \textit{egocentric} if it is fond of itself. \\
Show that the laws of the forest guarantee that at least one bird is egocentric.
\begin{helpbox}
\texttt{Law:} There exists a Mockingbird, $Mx := xx$ \\
\texttt{Def:} $A$ is fond of $B$ if $AB = B$ \\
\texttt{Lem:} Any bird is fond of at least one bird.
\end{helpbox}
\begin{solution}
\begin{alltt}
\lineno{} \cmnt{We know M is fond of at least one bird.}
\lineno{} let E so that ME = E
\lineno{}
\lineno{} ME = E \cmnt{By definition of fondness}
\lineno{} ME = EE \cmnt{By definition of M}
\lineno{} \thus{} EE = E \qed{}
\end{alltt}
\end{solution}
\vfill
\pagebreak
\definition{}
We say a bird $A$ is \textit{agreeable} if for all birds $B$, there is at least one bird $x$ on which $A$ and $B$ agree. \\
In other words, $A$ is agreeable if given any $B$, we can find a bird $x$ satisfying $Ax = Bx$.
\problem{}
Is the Mockingbird agreeable?
\begin{solution}
We know that $Mx = xx$. \\
From this definition, we see that $M$ agrees with any $x$ on $x$ itself.
\end{solution}
\vfill
\problem{}
Take two birds $A$ and $B$. Let $C$ be their composition. \\
Show that if $C$ is agreeable, $A$ is agreeable.
\begin{alltt}
\cmnt{Given information}
let A, B
let Cx = A(Bx)
let D \cmnt{Arbitrary bird}
let Ex = D(Bx) \cmnt{Define E as the composition of D and B}
Cy = ??
\end{alltt}
\begin{helpbox}[0.65]
\texttt{Def:} $A$ is agreeable if $Ax = Bx$ for all $B$ with some $x$. \\
\texttt{Law:} For any $A, B$, there is C defined by Cx = A(Bx)
\end{helpbox}
\begin{solution}
\begin{alltt}
\lineno{} \cmnt{Given information}
\lineno{} let A, B
\lineno{} let Cx = A(Bx)
\lineno{}
\lineno{} let D \cmnt{Arbitrary bird}
\lineno{} let Ex = D(Bx) \cmnt{Define E as the composition of D and B}
\lineno{} let y so that Cy = Ey \cmnt{Such a y must exist because C is agreeable}
\lineno{}
\lineno{} A(By) = Ey
\lineno{} = D(By) \qed{}
\end{alltt}
\end{solution}
\vfill
\pagebreak
\problem{}
Given three arbitrary birds $A$, $B$, and $C$, show that there exists a bird $D$ satisfying $Dx = A(B(Cx))$
\begin{solution}
\begin{alltt}
\lineno{} let A, B, C
\lineno{}
\lineno{} \cmnt{Invoke the Law of Composition:}
\lineno{} let Qx = B(Cx)
\lineno{} let Dx = A(Qx)
\lineno{}
\lineno{} Dx = A(Qx)
\lineno{} = A(B(Cx)) \qed{}
\end{alltt}
\end{solution}
\vfill
\definition{}
We say two birds $A$ and $B$ are \textit{compatible} if there are birds $x$ and $y$ so that $Ax = y$ and $By = x$. \\
Note that $x$ and $y$ may be the same bird. \\
\problem{}
Show that any two birds in this forest are compatible. \\
\begin{alltt}
let A, B
let Cx = A(Bx)
\end{alltt}
\begin{helpbox}
\texttt{Law:} Law of composition \\
\texttt{Lem:} Any bird is fond of at least one bird.
\end{helpbox}
\begin{solution}
\begin{alltt}
\lineno{} let A, B
\lineno{}
\lineno{} let Cx = A(Bx) \cmnt{Composition}
\lineno{} let y = Cy \cmnt{Let C be fond of y}
\lineno{}
\lineno{} Cy = y
\lineno{} = A(By)
\lineno{}
\lineno{} let x = By \cmnt{Rename By to x}
\lineno{} Ax = y \qed{}
\end{alltt}
\end{solution}
\vfill
\problem{}
Show that any bird that is fond of at least one bird is compatible with itself.
\begin{solution}
\begin{alltt}
\lineno{} let A
\lineno{} let x so that Ax = x \cmnt{A is fond of at least one other bird}
\lineno{} Ax = x \qed{}
\end{alltt}
That's it.
\end{solution}
\vfill
\pagebreak

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src/Advanced/Mock a Mockingbird/parts/02 kestrel.tex Normal file
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\section{The Curious Kestrel}
\definition{}
Recall that a bird is \textit{egocentric} if it is fond of itself. \\
A bird is \textit{hopelessly egocentric} if $Bx = B$ for all birds $x$.
\definition{}
More generally, we say that a bird $A$ is \textit{fixated} on a bird $B$ if $Ax = B$ for all $x$. \\
Convince yourself that a hopelessly egocentric bird is fixated on itself.
\problem{}
Say $A$ is fixated on $B$. Is $A$ fond of $B$?
\begin{solution}
Yes! See the following proof.
\begin{alltt}
\lineno{} let A
\lineno{} let B so that Ax = B
\lineno{} \thus{} AB = B \qed{}
\end{alltt}
\end{solution}
\vfill
\definition{}
The \textit{Kestrel} $K$ is defined by the following relationship:
$$
(Kx)y = x~~~\forall x, y
$$
In other words, this means that for every bird $x$, the bird $Kx$ is fixated on $x$.
\problem{}
Show that an egocentric Kestrel is hopelessly egocentric.
\begin{solution}
\begin{alltt}
\lineno{} KK = K
\lineno{} \thus{} (KK)y = K \cmnt{By definition of the Kestrel}
\lineno{} \thus{} Ky = K \qed{} \cmnt{By 01}
\end{alltt}
\end{solution}
\vfill
\pagebreak
\problem{}
Assume the forest contains a Kestrel. \\
Given the Law of Composition and the Law of the Mockingbird, show that at least one bird is hopelessly egocentric.
\begin{helpbox}[0.75]
\texttt{Def:} $K$ is defined by $(Kx)y = x$ \\
\texttt{Def:} $A$ is fond of $B$ if $AB = B$ \\
\texttt{???:} You'll need one more result from the previous section. Good luck!
\end{helpbox}
\begin{solution}
The final piece is a lemma we proved earlier: \\
Any bird is fond of at least one bird
\begin{alltt}
\lineno{} let A so that KA = A \cmnt{Any bird is fond of at least one bird}
\lineno{} (KA)y = y \cmnt{By definition of the kestrel}
\lineno{} \thus{} Ay = A \qed{} \cmnt{By 01}
\end{alltt}
\end{solution}
\vfill
\problem{Kestrel Left-Cancellation}<leftcancel>
In general, $Ax = Ay$ does not imply $x = y$. However, this is true if $A$ is $K$. \\
Show that $Kx = Ky \implies x = y$.
\begin{alltt}
\cmnt{This is a hint.}
let x, y so that Kx = Ky
\end{alltt}
\begin{solution}
\begin{alltt}
\lineno{} let x, y so that Kx = Ky
\lineno{} let z
\lineno{}
\lineno{} (Kx)z = (Ky)z \cmnt{By 01}
\lineno{}
\lineno{} \cmnt{By the definition of K}
\lineno{} (Kx)z = x
\lineno{} (Ky)z = y
\lineno{}
\lineno{} \thus{} x = (Kx)z = (Ky)z = y \qed{}
\end{alltt}
\end{solution}
\vfill
\pagebreak
\problem{}
Show that if $K$ is fond of $Kx$, $K$ is fond of $x$.
\begin{solution}
\begin{alltt}
\lineno{} let x so that K(Kx) = Kx
\lineno{} (K(Kx))y = (Kx)y
\lineno{} = Kx \cmnt{By definition of K}
\lineno{} x = Kx \cmnt{By 03 and definition of K}
\end{alltt}
\end{solution}
\vfill
\problem{}
An egocentric Kestrel must be extremely lonely. Why is this?
\begin{solution}
If a Kestrel is egocentric, it must be the only bird in the forest!
\begin{alltt}
\lineno{} \cmnt{Given}
\lineno{} Kx = K for some x
\lineno{} \cmnt{We have shown that an egocentric kestrel is hopelessly egocentric}
\lineno{} Kx = K for all x
\lineno{}
\lineno{} let x, y
\lineno{} Kx = K
\lineno{} Ky = K
\lineno{} Kx = Ky
\lineno{} x = y for all x, y \cmnt{By \ref{leftcancel}}
\lineno{} x = y = K \qed{} \cmnt{By 10, and since K exists}
\end{alltt}
\end{solution}
\vfill
\pagebreak