Finished TMAM hanout
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137
Advanced/Mock a Mockingbird/parts/02 kestrel.tex
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137
Advanced/Mock a Mockingbird/parts/02 kestrel.tex
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\section{The Curious Kestrel}
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\definition{}
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Recall that a bird is \textit{egocenteric} if it is fond of itself. \\
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A bird is \textit{hopelessly egocentric} if $Bx = B$ for all birds $x$.
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\definition{}
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More generally, we say that a bird $A$ is \textit{fixated} on a bird $B$ if $Ax = B$ for all $x$. \\
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Convince yourself that a hopelessly egocentric bird is fixated on itself.
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\problem{}
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Say $A$ is fixated on $B$. Is $A$ fond of $B$?
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\begin{solution}
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Yes! See the following proof.
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\begin{alltt}
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\lineno{} let B
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\lineno{} let B
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\lineno{} let A so that Ax = B
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\lineno{} \thus{} AB = B \qed{}
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\end{alltt}
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\end{solution}
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\vfill
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\definition{}
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The \textit{Kestrel} $K$ is defined by the following relationship:
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$$
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(Kx)y = x
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$$
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In other words, this means that for every bird $x$, the bird $Kx$ is fixated on $x$.
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\problem{}
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Show that an egocenteric Kestrel is hopelessly egocentric
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\begin{solution}
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\begin{alltt}
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\lineno{} KK = K
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\lineno{} \thus{} (KK)y = K
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\lineno{} \thus{} Ky = K \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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Assume the forest contains a Kestrel. \\
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Given $L_1$ and $L_2$, show that at least one bird is hopelessly egocentric.
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\begin{helpbox}[0.75]
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\texttt{Def:} $K$ is defined by $(Kx)y = x$ \\
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\texttt{Def:} $A$ is fond of $B$ if $AB = B$ \\
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\texttt{???:} You'll need one more result from the previous section. Good luck!
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\end{helpbox}
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\begin{solution}
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The final piece is a lemma we proved earler: \\
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Any bird is fond of at least one bird
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\begin{alltt}
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\lineno{} let A so that KA = A \cmnt{Any bird is fond of at least one bird}
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\lineno{} (KA)y = y \cmnt{By definition of the kestrel}
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\lineno{} \thus{} Ay = A \qed{} \cmnt{By 01}
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\end{alltt}
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\end{solution}
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\vfill
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\problem{Kestrel Left-Cancellation}<leftcancel>
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In general, $Ax = Ay$ does not imply $x = y$. However, this is true if $A$ is $K$. \\
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Show that $Kx = Ky \implies x = y$.
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\begin{alltt}
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\cmnt{This is a hint.}
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let x, y so that Kx = Ky
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\end{alltt}
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\begin{solution}
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\begin{alltt}
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\lineno{} let x, y so that Kx = Ky
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\lineno{} let z
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\lineno{}
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\lineno{} (Kx)z = (Ky)z \cmnt{By 01}
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\lineno{}
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\lineno{} \cmnt{By the definition of K}
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\lineno{} (Kx)z = x
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\lineno{} (Ky)z = y
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\lineno{}
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\lineno{} \thus{} x = (Kx)z = (Ky)z = y \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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Show that if $K$ is fond of $Kx$, $K$ is fond of $x$.
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\begin{solution}
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\begin{alltt}
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\lineno{} let x so that K(Kx) = Kx
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\lineno{} (K(Kx))y = (Kx)y
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\lineno{} = Kx \cmnt{By definition of K}
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\lineno{} x = Kx \cmnt{By 03 and definition of K}
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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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An egocentric Kestrel must be extremely lonely. Why is this?
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\begin{solution}
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If a Kestrel is egocenteric, it must be the only bird in the forest:
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\begin{alltt}
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\lineno{} \cmnt{Given}
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\lineno{} Kx = K for some x
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\lineno{} \cmnt{We have shown that an egocentric kestrel is hopelessly egocentric}
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\lineno{} Kx = K for all x
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\lineno{}
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\lineno{} let x, y
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\lineno{} Kx = K
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\lineno{} Ky = K
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\lineno{} Kx = Ky
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\lineno{} x = y for all x, y \cmnt{By \ref{leftcancel}}
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\lineno{} x = y = K \qed{} \cmnt{By 10, and since K exists}
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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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