Started Mockingbird handout

Advanced/Mock a Mockingbird/main.tex

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+% use [nosolutions] flag to hide solutions.
+% use [solutions] flag to show solutions.
+\documentclass[
+	solutions,
+	singlenumbering
+]{../../resources/ormc_handout}
+
+\usepackage{mathtools} % for \coloneqq
+
+\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}
+}
+
+
+\newcommand{\thus}{\(\Rightarrow\)}
+\newcommand{\qed}{\(\blacksquare\)}
+
+
+\begin{document}
+
+	\maketitle
+		<Advanced 2>
+		<Fall 2022>
+		{To Mock a Mockingbird}
+		{
+			Prepared by Mark on \today \\
+			Based on a book of the same name.
+		}
+
+	\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 responds 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 $L_1$, \textit{The Law of Composition}: \\
+		For any two birds $A$ and $B$, there must be a bird $C$ so that $Cx = A(Bx)$
+		\item $L_2$, \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 $x$.
+
+	\vfill
+	\pagebreak
+
+	\section{To Mock a Mockingbird}
+
+	\problem{}
+	The bear, a lifelong resident of the forest, 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}
+			let A           \cmnt{Let A be any any bird.}
+			let Cx := A(Mx) \cmnt{Define C as the composition of A and M}
+			CC = A(MC)
+			   = A(CC)  \qed{}
+		\end{alltt}
+	\end{solution}
+
+	\vfill
+	\problem{}
+	We say a bird $A$ is \textit{egocentric} if it is fond if 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}
+			\cmnt{We know M is fond of at least one bird.}
+			let E so that ME = E
+
+			ME = E     \cmnt{By definition of fondness}
+			ME = EE    \cmnt{By definition of M}
+			\thus{} EE = E  \qed{}
+		\end{alltt}
+	\end{solution}
+
+	\vfill
+	\pagebreak
+
+
+	\problem{}
+	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. \\
+	This means that $Ax = Bx$.
+
+	\begin{helpbox}
+		\texttt{Def:} $Mx := xx$
+	\end{helpbox}
+
+	\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 $A$ must be agreeable if $C$ is agreeable. \\
+	The bear has again given you a hint.
+	\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}
+			\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 = Ey         \cmnt{For some y, because C is agreeable}
+			\thus{} A(By) = Ey
+			\thus{} A(By) = D(By)  \qed{}
+		\end{alltt}
+	\end{solution}
+
+	\vfill
+	\pagebreak
+
+	\problem{}
+	Given three arbitrary birds $A$, $B$, and $C$, show that there exists a bird $D$ defined by $Dx = A(B(Cx))$
+
+	\begin{solution}
+		\begin{alltt}
+			let A, B, C
+
+			\cmnt{Invoke the Law of Composition:}
+			let Q := BC
+			let D := AQ
+
+			D = AQ
+			  = A(BC)  \qed{}
+		\end{alltt}
+	\end{solution}
+
+	\vfill
+
+
+	\problem{}
+	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. \\
+	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}
+			let A, B
+
+			let Cx := A(Bx)  \cmnt{Composition}
+			let y := Cy      \cmnt{Let C be fond of y}
+
+			Cy = y
+			\thus{} A(By) = y
+
+			let x := By  \cmnt{Rename By to x}
+			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}
+			let A
+			let x so that Ax := x
+			Ax = x  \qed{}
+		\end{alltt}
+		That's it.
+	\end{solution}
+
+
+	\vfill
+	\pagebreak
+
+\end{document}