From a5411ad5a5692f1b90f4a269bafe4db98384b75b Mon Sep 17 00:00:00 2001 From: Mark Date: Sun, 15 Feb 2026 10:31:45 -0800 Subject: [PATCH] Mockingbird edits --- src/Advanced/Mock a Mockingbird/main.tex | 1 + .../Mock a Mockingbird/parts/01 tmam.tex | 38 ++++--- .../Mock a Mockingbird/parts/02 kestrel.tex | 10 +- .../Mock a Mockingbird/parts/03 bonus.tex | 102 ++++++++++++++++++ 4 files changed, 126 insertions(+), 25 deletions(-) create mode 100644 src/Advanced/Mock a Mockingbird/parts/03 bonus.tex diff --git a/src/Advanced/Mock a Mockingbird/main.tex b/src/Advanced/Mock a Mockingbird/main.tex index 1904d20..f5ef898 100755 --- a/src/Advanced/Mock a Mockingbird/main.tex +++ b/src/Advanced/Mock a Mockingbird/main.tex @@ -81,5 +81,6 @@ \input{parts/00 intro} \input{parts/01 tmam} \input{parts/02 kestrel} + \input{parts/03 bonus} \end{document} \ No newline at end of file diff --git a/src/Advanced/Mock a Mockingbird/parts/01 tmam.tex b/src/Advanced/Mock a Mockingbird/parts/01 tmam.tex index 55a8af1..c568588 100644 --- a/src/Advanced/Mock a Mockingbird/parts/01 tmam.tex +++ b/src/Advanced/Mock a Mockingbird/parts/01 tmam.tex @@ -23,7 +23,7 @@ Complete his proof. \lineno{} let A \cmnt{Let A be any any bird.} \lineno{} let Cx = A(Mx) \cmnt{Define C as the composition of A and M} \lineno{} CC = A(MC) - \lineno{} = A(CC) \qed{} + \lineno{} = A(CC) \end{alltt} \end{solution} @@ -46,7 +46,7 @@ Show that the laws of the forest guarantee that at least one bird is egocentric. \lineno{} \lineno{} ME = E \cmnt{By definition of fondness} \lineno{} ME = EE \cmnt{By definition of M} - \lineno{} \thus{} EE = E \qed{} + \lineno{} \thus{} EE = E \end{alltt} \end{solution} @@ -99,7 +99,7 @@ Show that if $C$ is agreeable, $A$ is agreeable. \lineno{} let y so that Cy = Ey \cmnt{Such a y must exist because C is agreeable} \lineno{} \lineno{} A(By) = Ey - \lineno{} = D(By) \qed{} + \lineno{} = D(By) \end{alltt} \end{solution} @@ -129,6 +129,20 @@ Given three arbitrary birds $A$, $B$, and $C$, show that there exists a bird $D$ 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. \\ + +\problem{} +Show that any bird that is fond of at least one bird is compatible with itself. + +\begin{solution} + \begin{alltt} + \lineno{} let A + \lineno{} let x so that Ax = x \cmnt{A is fond of at least one other bird} + \lineno{} Ax = x + \end{alltt} +\end{solution} + +\vfill + \problem{} Show that any two birds in this forest are compatible. \\ \begin{alltt} @@ -144,7 +158,6 @@ Show that any two birds in this forest are compatible. \\ \begin{solution} \begin{alltt} \lineno{} let A, B - \lineno{} \lineno{} let Cx = A(Bx) \cmnt{Composition} \lineno{} let y = Cy \cmnt{Let C be fond of y} \lineno{} @@ -152,24 +165,9 @@ Show that any two birds in this forest are compatible. \\ \lineno{} = A(By) \lineno{} \lineno{} let x = By \cmnt{Rename By to x} - \lineno{} Ax = y \qed{} + \lineno{} Ax = y \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} - \lineno{} let A - \lineno{} let x so that Ax = x \cmnt{A is fond of at least one other bird} - \lineno{} Ax = x \qed{} - \end{alltt} - - That's it. -\end{solution} - \vfill \pagebreak \ No newline at end of file diff --git a/src/Advanced/Mock a Mockingbird/parts/02 kestrel.tex b/src/Advanced/Mock a Mockingbird/parts/02 kestrel.tex index 293b508..0f6c027 100644 --- a/src/Advanced/Mock a Mockingbird/parts/02 kestrel.tex +++ b/src/Advanced/Mock a Mockingbird/parts/02 kestrel.tex @@ -18,7 +18,7 @@ Say $A$ is fixated on $B$. Is $A$ fond of $B$? \begin{alltt} \lineno{} let A \lineno{} let B so that Ax = B - \lineno{} \thus{} AB = B \qed{} + \lineno{} \thus{} AB = B \end{alltt} \end{solution} \vfill @@ -39,7 +39,7 @@ Show that an egocentric Kestrel is hopelessly egocentric. \begin{alltt} \lineno{} KK = K \lineno{} \thus{} (KK)y = K \cmnt{By definition of the Kestrel} - \lineno{} \thus{} Ky = K \qed{} \cmnt{By 01} + \lineno{} \thus{} Ky = K \cmnt{By 01} \end{alltt} \end{solution} @@ -64,7 +64,7 @@ Given the Law of Composition and the Law of the Mockingbird, show that at least \begin{alltt} \lineno{} let A so that KA = A \cmnt{Any bird is fond of at least one bird} \lineno{} (KA)y = y \cmnt{By definition of the kestrel} - \lineno{} \thus{} Ay = A \qed{} \cmnt{By 01} + \lineno{} \thus{} Ay = A \cmnt{By 01} \end{alltt} \end{solution} @@ -90,7 +90,7 @@ Show that $Kx = Ky \implies x = y$. \lineno{} (Kx)z = x \lineno{} (Ky)z = y \lineno{} - \lineno{} \thus{} x = (Kx)z = (Ky)z = y \qed{} + \lineno{} \thus{} x = (Kx)z = (Ky)z = y \end{alltt} \end{solution} @@ -128,7 +128,7 @@ An egocentric Kestrel must be extremely lonely. Why is this? \lineno{} Ky = K \lineno{} Kx = Ky \lineno{} x = y for all x, y \cmnt{By \ref{leftcancel}} - \lineno{} x = y = K \qed{} \cmnt{By 10, and since K exists} + \lineno{} x = y = K \cmnt{By 10, and since K exists} \end{alltt} \end{solution} diff --git a/src/Advanced/Mock a Mockingbird/parts/03 bonus.tex b/src/Advanced/Mock a Mockingbird/parts/03 bonus.tex new file mode 100644 index 0000000..73b3f02 --- /dev/null +++ b/src/Advanced/Mock a Mockingbird/parts/03 bonus.tex @@ -0,0 +1,102 @@ +\section{Bonus Problems} + +\definition{} +The identity bird has sometimes been maligned, owing to +the fact that whatever bird x you call to $I$, all $I$ does is to echo +$x$ back to you. + +\vspace{2mm} + +Superficially, the bird $I$ appears to have no intelligence or imagination; all it can do is repeat what it hears. +For this reason, in the past, thoughtless students of ornithology +referred to it as the idiot bird. However, a more profound or- +nithologist once studied the situation in great depth and dis- +covered that the identity bird is in fact highly intelligent! The +real reason for its apparently unimaginative behavior is that it +has an unusually large heart and hence is fond of every bird! +When you call $x$ to $I$, the reason it responds by calling back $x$ +is not that it can't think of anything else; it's just that it wants +you to know that it is fond of $x$! + +\vspace{2mm} + +Since an identity bird is fond of every bird, then it is also +fond of itself, so every identity bird is egocentric. However, +its egocentricity doesn't mean that it is any more fond of itself +than of any other bird!. + + +\problem{} +The laws of the forest no longer apply. + +Suppose we are told that the forest contains an identity bird +$I$ and that $I$ is agreeable. \ +Does it follow that every bird must be fond of at least one bird? + +\vfill + + +\problem{} +Suppose we are told that there is an identity bird $I$ and that +every bird is fond of at least one bird. \ +Does it necessarily follow that $I$ is agreeable? + +\vfill +\pagebreak + + +\problem{} +Suppose we are told that there is an identity bird $I$, but we are +not told whether $I$ is agreeable or not. + +However, we are told that every pair of birds is compatible. \ +Which of the following conclusiens can be validly drawn? + +\begin{itemize} + \item Every bird is fond of at least one bird + \item $I$ is agreeable. +\end{itemize} + +\vfill + +\problem{} +The identity bird $I$, though egocentric, is in general not hope- +lessly egocentric. Indeed, if there were a hopelessly egocentric +identity bird, the situation would be quite sad. Why? + +\vfill + +\definition{} +A bird $L$ is called a lark if the following +holds for any birds $x$ and $y$: + +\[ + (Lx)y = x(yy) +\] + +\problem{} +Prove that if the forest contains a lark $L$ and an identity bird +$I$, then it must also contain a mockingbird $M$. + +\vfill +\pagebreak + + +\problem{} +Why is a hopelessly egocentric lark unusually attractive? + +\vfill + + +\problem{} +Assuming that no bird can be both a lark and a kestrel---as +any ornithologist knows!---prove that it is impossible for a +lark to be fond of a kestrel. + +\vfill + +\problem{} +It might happen, however, that a kestrel is fond of a lark. \par +Show that in this case, \textit{every} bird is fond of the lark. + +\vfill