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222
Advanced/Mock a Mockingbird/main.tex
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Based on a book of the same name. Based on a book of the same name.
} }
\section{Introduction} \input{parts/00 intro}
\input{parts/01 tmam}
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} \end{document}

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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 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

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Advanced/Mock a Mockingbird/parts/01 tmam.tex Normal file
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\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