Cleanup
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@ -57,225 +57,7 @@
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Based on a book of the same name.
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}
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\section{Introduction}
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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 $AB$.
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Bird $AB$ is, by definition, $A$'s response to $B$.
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\vspace{2mm}
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As you wander around this forest, you quickly discover two interesting facts:
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\begin{enumerate}[itemsep = 1mm]
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\item $A$'s responds to $B$ mustn't be the same as $B$'s response to $A$.
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\item Given three birds $A$, $B$, and $C$, $(AB)C$ and $A(BC)$ are not necessarily the same bird. \\
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Bird $A(BC)$ is $A$'s response to bird $BC$, while $(AB)C$ is $AB$'s response to $C$. \\
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Thus, $ABC$ is ambiguous. Parenthesis are mandatory.
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\end{enumerate}
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\vspace{2mm}
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You also find that this forest has two laws:
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\begin{enumerate}[itemsep = 1mm]
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\item $L_1$, \textit{The Law of Composition}: \\
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For any two birds $A$ and $B$, there must be a bird $C$ so that $Cx = A(Bx)$
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\item $L_2$, \textit{The Law of the Mockingbird}: \\
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The forest must contain the Mockingbird $M$, which always satisfies $Mx = xx$. \\
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In other words, the Mockingbird's response to any bird $x$ is the same as $x$'s response to itself.
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\end{enumerate}
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\vfill
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\definition{}
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We say a bird $A$ is fond of a bird $B$ if $A$ responds to $B$ with $B$. \\
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In other words, $A$ is fond of $B$ if $AB = B$.
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\vfill
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\definition{}
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We say a bird $C$ \textit{composes} $A$ with $B$ if for any bird $x$,
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$$
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Cx = A(Bx)
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$$
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In other words, this means that $C$'s response to $x$ is the same as $A$'s response to $B$'s response to $x$.
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\vfill
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\pagebreak
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\section{To Mock a Mockingbird}
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\problem{}
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The bear, a lifelong resident of the forest, tells you that any bird $A$ is fond of at least one other bird. \\
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Complete his proof.
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\begin{alltt}
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let A \cmnt{Let A be any any bird.}
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let Cx := A(Mx) \cmnt{Define C as the composition of A and M}
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\cmnt{The rest is up to you.}
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CC = ??
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\end{alltt}
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\begin{helpbox}
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\texttt{Law:} There exists a Mockingbird, $Mx := xx$ \\
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\texttt{Def:} $A$ is fond of $B$ if $AB = B$
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\end{helpbox}
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\begin{solution}
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\begin{alltt}
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let A \cmnt{Let A be any any bird.}
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let Cx := A(Mx) \cmnt{Define C as the composition of A and M}
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CC = A(MC)
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= A(CC) \qed{}
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\end{alltt}
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\end{solution}
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\vfill
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\problem{}
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We say a bird $A$ is \textit{egocentric} if it is fond if itself.
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Show that the laws of the forest guarantee that at least one bird is egocentric.
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\begin{helpbox}
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\texttt{Law:} There exists a Mockingbird, $Mx := xx$ \\
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\texttt{Def:} $A$ is fond of $B$ if $AB = B$ \\
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\texttt{Lem:} Any bird is fond of at least one bird.
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\end{helpbox}
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\begin{solution}
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\begin{alltt}
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\cmnt{We know M is fond of at least one bird.}
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let E so that ME = E
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ME = E \cmnt{By definition of fondness}
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ME = EE \cmnt{By definition of M}
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\thus{} EE = E \qed{}
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\end{alltt}
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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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. \\
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This means that $Ax = Bx$.
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\begin{helpbox}
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\texttt{Def:} $Mx := xx$
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\end{helpbox}
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\begin{solution}
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We know that $Mx = xx$. \\
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From this definition, we see that $M$ agrees with any $x$ on $x$ itself.
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\end{solution}
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\vfill
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\problem{}
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Take two birds $A$ and $B$. Let $C$ be their composition. \\
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Show that $A$ must be agreeable if $C$ is agreeable. \\
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The bear has again given you a hint.
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\begin{alltt}
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\cmnt{Given information}
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let A, B
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let Cx := A(Bx)
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let D \cmnt{Arbitrary bird}
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let Ex := D(Bx) \cmnt{Define E as the composition of D and B}
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Cy = ??
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\end{alltt}
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\begin{helpbox}[0.65]
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\texttt{Def:} $A$ is agreeable if $Ax = Bx$ for all $B$ with some $x$. \\
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\texttt{Law:} For any $A, B$, there is C defined by Cx = A(Bx)
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\end{helpbox}
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\begin{solution}
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\begin{alltt}
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\cmnt{Given information}
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let A, B
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let Cx := A(Bx)
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let D \cmnt{Arbitrary bird}
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let Ex := D(Bx) \cmnt{Define E as the composition of D and B}
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Cy = Ey \cmnt{For some y, because C is agreeable}
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\thus{} A(By) = Ey
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\thus{} A(By) = D(By) \qed{}
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\end{alltt}
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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Given three arbitrary birds $A$, $B$, and $C$, show that there exists a bird $D$ defined by $Dx = A(B(Cx))$
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\begin{solution}
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\begin{alltt}
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let A, B, C
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\cmnt{Invoke the Law of Composition:}
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let Q := BC
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let D := AQ
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D = AQ
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= A(BC) \qed{}
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\end{alltt}
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\end{solution}
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\vfill
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\problem{}
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We say two birds $A$ and $B$ are \textit{compatible} if there are birds $x$ and $y$ so that $Ax = y$ and $By = x$. \\
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Note that $x$ and $y$ may be the same bird. \\
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Show that any two birds in this forest are compatible. \\
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\begin{alltt}
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let A, B
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let Cx := A(Bx)
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\end{alltt}
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\begin{helpbox}
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\texttt{Law:} Law of composition \\
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\texttt{Lem:} Any bird is fond of at least one bird.
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\end{helpbox}
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\begin{solution}
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\begin{alltt}
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let A, B
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let Cx := A(Bx) \cmnt{Composition}
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let y := Cy \cmnt{Let C be fond of y}
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Cy = y
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\thus{} A(By) = y
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let x := By \cmnt{Rename By to x}
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Ax = y \qed{}
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\end{alltt}
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\end{solution}
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\vfill
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\problem{}
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Show that any bird that is fond of at least one bird is compatible with itself.
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\begin{solution}
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\begin{alltt}
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let A
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let x so that Ax := x
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Ax = x \qed{}
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\end{alltt}
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That's it.
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\end{solution}
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\vfill
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\pagebreak
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\input{parts/00 intro}
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\input{parts/01 tmam}
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\end{document}
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44
Advanced/Mock a Mockingbird/parts/00 intro.tex
Normal file
44
Advanced/Mock a Mockingbird/parts/00 intro.tex
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@ -0,0 +1,44 @@
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\section{Introduction}
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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 $AB$.
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Bird $AB$ is, by definition, $A$'s response to $B$.
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\vspace{2mm}
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As you wander around this forest, you quickly discover two interesting facts:
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\begin{enumerate}[itemsep = 1mm]
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\item $A$'s responds to $B$ mustn't be the same as $B$'s response to $A$.
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\item Given three birds $A$, $B$, and $C$, $(AB)C$ and $A(BC)$ are not necessarily the same bird. \\
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Bird $A(BC)$ is $A$'s response to bird $BC$, while $(AB)C$ is $AB$'s response to $C$. \\
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Thus, $ABC$ is ambiguous. Parenthesis are mandatory.
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\end{enumerate}
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\vspace{2mm}
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You also find that this forest has two laws:
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\begin{enumerate}[itemsep = 1mm]
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\item $L_1$, \textit{The Law of Composition}: \\
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For any two birds $A$ and $B$, there must be a bird $C$ so that $Cx = A(Bx)$
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\item $L_2$, \textit{The Law of the Mockingbird}: \\
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The forest must contain the Mockingbird $M$, which always satisfies $Mx = xx$. \\
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In other words, the Mockingbird's response to any bird $x$ is the same as $x$'s response to itself.
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\end{enumerate}
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\vfill
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\definition{}
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We say a bird $A$ is fond of a bird $B$ if $A$ responds to $B$ with $B$. \\
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In other words, $A$ is fond of $B$ if $AB = B$.
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\vfill
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\definition{}
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We say a bird $C$ \textit{composes} $A$ with $B$ if for any bird $x$,
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$$
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Cx = A(Bx)
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$$
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In other words, this means that $C$'s response to $x$ is the same as $A$'s response to $B$'s response to $x$.
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\vfill
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\pagebreak
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175
Advanced/Mock a Mockingbird/parts/01 tmam.tex
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175
Advanced/Mock a Mockingbird/parts/01 tmam.tex
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@ -0,0 +1,175 @@
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\section{To Mock a Mockingbird}
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\problem{}
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The bear, a lifelong resident of the forest, tells you that any bird $A$ is fond of at least one other bird. \\
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Complete his proof.
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\begin{alltt}
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let A \cmnt{Let A be any any bird.}
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let Cx := A(Mx) \cmnt{Define C as the composition of A and M}
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\cmnt{The rest is up to you.}
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CC = ??
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\end{alltt}
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\begin{helpbox}
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\texttt{Law:} There exists a Mockingbird, $Mx := xx$ \\
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\texttt{Def:} $A$ is fond of $B$ if $AB = B$
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\end{helpbox}
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\begin{solution}
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\begin{alltt}
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let A \cmnt{Let A be any any bird.}
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let Cx := A(Mx) \cmnt{Define C as the composition of A and M}
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CC = A(MC)
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= A(CC) \qed{}
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\end{alltt}
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\end{solution}
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\vfill
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\problem{}
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We say a bird $A$ is \textit{egocentric} if it is fond if itself.
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Show that the laws of the forest guarantee that at least one bird is egocentric.
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\begin{helpbox}
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\texttt{Law:} There exists a Mockingbird, $Mx := xx$ \\
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\texttt{Def:} $A$ is fond of $B$ if $AB = B$ \\
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\texttt{Lem:} Any bird is fond of at least one bird.
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\end{helpbox}
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\begin{solution}
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\begin{alltt}
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\cmnt{We know M is fond of at least one bird.}
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let E so that ME = E
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ME = E \cmnt{By definition of fondness}
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ME = EE \cmnt{By definition of M}
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\thus{} EE = E \qed{}
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\end{alltt}
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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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. \\
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This means that $Ax = Bx$.
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\begin{helpbox}
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\texttt{Def:} $Mx := xx$
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\end{helpbox}
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\begin{solution}
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We know that $Mx = xx$. \\
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From this definition, we see that $M$ agrees with any $x$ on $x$ itself.
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\end{solution}
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\vfill
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\problem{}
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Take two birds $A$ and $B$. Let $C$ be their composition. \\
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Show that $A$ must be agreeable if $C$ is agreeable. \\
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The bear has again given you a hint.
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\begin{alltt}
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\cmnt{Given information}
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let A, B
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let Cx := A(Bx)
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let D \cmnt{Arbitrary bird}
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let Ex := D(Bx) \cmnt{Define E as the composition of D and B}
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Cy = ??
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\end{alltt}
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\begin{helpbox}[0.65]
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\texttt{Def:} $A$ is agreeable if $Ax = Bx$ for all $B$ with some $x$. \\
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\texttt{Law:} For any $A, B$, there is C defined by Cx = A(Bx)
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\end{helpbox}
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\begin{solution}
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\begin{alltt}
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\cmnt{Given information}
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let A, B
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let Cx := A(Bx)
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let D \cmnt{Arbitrary bird}
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let Ex := D(Bx) \cmnt{Define E as the composition of D and B}
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Cy = Ey \cmnt{For some y, because C is agreeable}
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\thus{} A(By) = Ey
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\thus{} A(By) = D(By) \qed{}
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\end{alltt}
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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Given three arbitrary birds $A$, $B$, and $C$, show that there exists a bird $D$ defined by $Dx = A(B(Cx))$
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\begin{solution}
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\begin{alltt}
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let A, B, C
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\cmnt{Invoke the Law of Composition:}
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let Q := BC
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let D := AQ
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D = AQ
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= A(BC) \qed{}
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\end{alltt}
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\end{solution}
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\vfill
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\problem{}
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We say two birds $A$ and $B$ are \textit{compatible} if there are birds $x$ and $y$ so that $Ax = y$ and $By = x$. \\
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Note that $x$ and $y$ may be the same bird. \\
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Show that any two birds in this forest are compatible. \\
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\begin{alltt}
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let A, B
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let Cx := A(Bx)
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\end{alltt}
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\begin{helpbox}
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\texttt{Law:} Law of composition \\
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\texttt{Lem:} Any bird is fond of at least one bird.
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\end{helpbox}
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\begin{solution}
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\begin{alltt}
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let A, B
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let Cx := A(Bx) \cmnt{Composition}
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let y := Cy \cmnt{Let C be fond of y}
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Cy = y
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\thus{} A(By) = y
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let x := By \cmnt{Rename By to x}
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Ax = y \qed{}
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\end{alltt}
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\end{solution}
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\vfill
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\problem{}
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Show that any bird that is fond of at least one bird is compatible with itself.
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\begin{solution}
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\begin{alltt}
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let A
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let x so that Ax := x
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Ax = x \qed{}
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\end{alltt}
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That's it.
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\end{solution}
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\vfill
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\pagebreak
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