Cleanup

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

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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