handouts/Advanced/Mock a Mockingbird/parts/01 tmam.tex

\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