From 48ac646ee452c915eb20884c8af9ec0455aefee6 Mon Sep 17 00:00:00 2001 From: Mark Date: Tue, 28 Mar 2023 22:16:39 -0700 Subject: [PATCH] Mockingbird edits --- Advanced/Mock a Mockingbird/parts/00 intro.tex | 4 ++-- Advanced/Mock a Mockingbird/parts/01 tmam.tex | 15 +++++++-------- Advanced/Mock a Mockingbird/parts/02 kestrel.tex | 13 ++++++------- 3 files changed, 15 insertions(+), 17 deletions(-) diff --git a/Advanced/Mock a Mockingbird/parts/00 intro.tex b/Advanced/Mock a Mockingbird/parts/00 intro.tex index 9eef4df..c0267d7 100644 --- a/Advanced/Mock a Mockingbird/parts/00 intro.tex +++ b/Advanced/Mock a Mockingbird/parts/00 intro.tex @@ -18,9 +18,9 @@ As you wander around this forest, you quickly discover two interesting facts: You also find that this forest has two laws: \begin{enumerate}[itemsep = 1mm] - \item $L_1$, \textit{The Law of Composition}: \\ + \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 $L_2$, \textit{The Law of the Mockingbird}: \\ + \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} diff --git a/Advanced/Mock a Mockingbird/parts/01 tmam.tex b/Advanced/Mock a Mockingbird/parts/01 tmam.tex index c23d6fa..16fee99 100644 --- a/Advanced/Mock a Mockingbird/parts/01 tmam.tex +++ b/Advanced/Mock a Mockingbird/parts/01 tmam.tex @@ -54,13 +54,12 @@ Show that the laws of the forest guarantee that at least one bird is egocentric. \pagebreak -\problem{} +\definition{} 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. \\ In other words, $A$ is agreeable if $Ax = Bx$ for some $x$ for all $B$. -\begin{helpbox} - \texttt{Def:} $Mx := xx$ -\end{helpbox} +\problem{} +Is the Mockingbird agreeable? \begin{solution} We know that $Mx = xx$. \\ @@ -115,11 +114,11 @@ Given three arbitrary birds $A$, $B$, and $C$, show that there exists a bird $D$ \lineno{} let A, B, C \lineno{} \lineno{} \cmnt{Invoke the Law of Composition:} - \lineno{} let Q = BC - \lineno{} let D = AQ + \lineno{} let Qx = B(Cx) + \lineno{} let Dx = A(Qx) \lineno{} - \lineno{} D = AQ - \lineno{} = A(BC) \qed{} + \lineno{} Dx = A(Qx) + \lineno{} = A(B(Cx)) \qed{} \end{alltt} \end{solution} diff --git a/Advanced/Mock a Mockingbird/parts/02 kestrel.tex b/Advanced/Mock a Mockingbird/parts/02 kestrel.tex index 66018eb..b28b349 100644 --- a/Advanced/Mock a Mockingbird/parts/02 kestrel.tex +++ b/Advanced/Mock a Mockingbird/parts/02 kestrel.tex @@ -16,9 +16,8 @@ Say $A$ is fixated on $B$. Is $A$ fond of $B$? \begin{solution} Yes! See the following proof. \begin{alltt} - \lineno{} let B - \lineno{} let B - \lineno{} let A so that Ax = B + \lineno{} let A + \lineno{} let B so that Ax = B \lineno{} \thus{} AB = B \qed{} \end{alltt} \end{solution} @@ -39,8 +38,8 @@ Show that an egocenteric Kestrel is hopelessly egocentric \begin{solution} \begin{alltt} \lineno{} KK = K - \lineno{} \thus{} (KK)y = K - \lineno{} \thus{} Ky = K \qed{} + \lineno{} \thus{} (KK)y = K \cmnt{By definition of the Kestrel} + \lineno{} \thus{} Ky = K \qed{} \cmnt{By 01} \end{alltt} \end{solution} @@ -50,7 +49,7 @@ Show that an egocenteric Kestrel is hopelessly egocentric \problem{} Assume the forest contains a Kestrel. \\ -Given $L_1$ and $L_2$, show that at least one bird is hopelessly egocentric. +Given the Law of Composition and the Law of the Mockingbird, show that at least one bird is hopelessly egocentric. \begin{helpbox}[0.75] \texttt{Def:} $K$ is defined by $(Kx)y = x$ \\ @@ -116,7 +115,7 @@ Show that if $K$ is fond of $Kx$, $K$ is fond of $x$. An egocentric Kestrel must be extremely lonely. Why is this? \begin{solution} - If a Kestrel is egocenteric, it must be the only bird in the forest: + If a Kestrel is egocenteric, it must be the only bird in the forest! \begin{alltt} \lineno{} \cmnt{Given}