handouts/Advanced/Mock a Mockingbird/parts/00 intro.tex

\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 $B$.

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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}

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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}

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\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 $AB = B$.

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\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.


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