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Finished TMAM hanout

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Mark
2023-03-16 19:36:13 -07:00
parent 9b55f682ce
commit 8d27bf87ce
4 changed files with 211 additions and 56 deletions
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23
Advanced/Mock a Mockingbird/main.tex
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@ -7,6 +7,9 @@
\usepackage{mathtools} % for \coloneqq
%\usepackage{lua-visual-debug}
\usepackage{censor}
\usepackage{alltt}
@ -37,10 +40,24 @@
}
% Logic block comment
\newcommand{\cmnt}[1]{
\textcolor{gray}{\# #1}
\newcommand{\cmnt}[1]{\textcolor{gray}{\# #1}}
\newcounter{allttLineCounter}
\setcounter{allttLineCounter}{0}
\newcommand{\linenoref}[1]{\colorbox{gray!30!white}{#1}}
\newcommand{\lineno}{
\stepcounter{allttLineCounter}%
\linenoref{\ifnum\value{allttLineCounter}<10 0\fi\arabic{allttLineCounter}}%
}
% Redefine alltt so it automatically
% resets allttLineCounter
\let\oldalltt\alltt
\renewenvironment{alltt}
{\setcounter{allttLineCounter}{0}\begin{oldalltt}}
{\end{oldalltt}}
\newcommand{\thus}{\(\Rightarrow\)}
\newcommand{\qed}{\(\blacksquare\)}
@ -59,5 +76,5 @@
\input{parts/00 intro}
\input{parts/01 tmam}
\input{parts/02 kestrel}
\end{document}

4
Advanced/Mock a Mockingbird/parts/00 intro.tex
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@ -38,7 +38,9 @@ We say a bird $C$ \textit{composes} $A$ with $B$ if for any bird $x$,
$$
Cx = A(Bx)
$$
In other words, this means that $C$'s response to $x$ is the same as $A$'s response to $B$'s response to $x$.
In other words, this means that $C$'s response to $x$ is the same as $A$'s response to $B$'s response to $x$. \\
Note that $C$ is exactly the kind of bird $L_1$ guarantees.
\vfill
\pagebreak

103
Advanced/Mock a Mockingbird/parts/01 tmam.tex
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@ -1,11 +1,11 @@
\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. \\
Mark 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}
let Cx = A(Mx) \cmnt{Define C as the composition of A and M}
\cmnt{The rest is up to you.}
CC = ??
@ -20,17 +20,16 @@ Complete his proof.
\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{}
\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{}
\end{alltt}
\end{solution}
\vfill
\problem{}
We say a bird $A$ is \textit{egocentric} if it is fond if itself.
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.
@ -42,12 +41,12 @@ Show that the laws of the forest guarantee that at least one bird is egocentric.
\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{}
\lineno{} \cmnt{We know M is fond of at least one bird.}
\lineno{} let E so that ME = E
\lineno{}
\lineno{} ME = E \cmnt{By definition of fondness}
\lineno{} ME = EE \cmnt{By definition of M}
\lineno{} \thus{} EE = E \qed{}
\end{alltt}
\end{solution}
@ -57,7 +56,7 @@ Show that the laws of the forest guarantee that at least one bird is egocentric.
\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 $Ax = Bx$.
In other words, $A$ is agreeable if $Ax = Bx$ for some $x$ for all $B$.
\begin{helpbox}
\texttt{Def:} $Mx := xx$
@ -73,15 +72,14 @@ This means that $Ax = Bx$.
\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.
Show that $A$ must be agreeable if $C$ is agreeable.
\begin{alltt}
\cmnt{Given information}
let A, B
let Cx := A(Bx)
let Cx = A(Bx)
let D \cmnt{Arbitrary bird}
let Ex := D(Bx) \cmnt{Define E as the composition of D and B}
let Ex = D(Bx) \cmnt{Define E as the composition of D and B}
Cy = ??
\end{alltt}
@ -93,15 +91,16 @@ The bear has again given you a hint.
\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{}
\lineno{} \cmnt{Given information}
\lineno{} let A, B
\lineno{} let Cx = A(Bx)
\lineno{}
\lineno{} let D \cmnt{Arbitrary bird}
\lineno{} let Ex = D(Bx) \cmnt{Define E as the composition of D and B}
\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{}
\end{alltt}
\end{solution}
@ -109,18 +108,18 @@ The bear has again given you a hint.
\pagebreak
\problem{}
Given three arbitrary birds $A$, $B$, and $C$, show that there exists a bird $D$ defined by $Dx = A(B(Cx))$
Given three arbitrary birds $A$, $B$, and $C$, show that there exists a bird $D$ satisfying $Dx = A(B(Cx))$
\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{}
\lineno{} let A, B, C
\lineno{}
\lineno{} \cmnt{Invoke the Law of Composition:}
\lineno{} let Q = BC
\lineno{} let D = AQ
\lineno{}
\lineno{} D = AQ
\lineno{} = A(BC) \qed{}
\end{alltt}
\end{solution}
@ -133,7 +132,7 @@ 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)
let Cx = A(Bx)
\end{alltt}
\begin{helpbox}
@ -143,16 +142,16 @@ Show that any two birds in this forest are compatible. \\
\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{}
\lineno{} let A, B
\lineno{}
\lineno{} let Cx = A(Bx) \cmnt{Composition}
\lineno{} let y = Cy \cmnt{Let C be fond of y}
\lineno{}
\lineno{} Cy = y
\lineno{} = A(By)
\lineno{}
\lineno{} let x = By \cmnt{Rename By to x}
\lineno{} Ax = y \qed{}
\end{alltt}
\end{solution}
@ -163,13 +162,13 @@ 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{}
\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

137
Advanced/Mock a Mockingbird/parts/02 kestrel.tex Normal file
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@ -0,0 +1,137 @@
\section{The Curious Kestrel}
\definition{}
Recall that a bird is \textit{egocenteric} if it is fond of itself. \\
A bird is \textit{hopelessly egocentric} if $Bx = B$ for all birds $x$.
\definition{}
More generally, we say that a bird $A$ is \textit{fixated} on a bird $B$ if $Ax = B$ for all $x$. \\
Convince yourself that a hopelessly egocentric bird is fixated on itself.
\problem{}
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{} \thus{} AB = B \qed{}
\end{alltt}
\end{solution}
\vfill
\definition{}
The \textit{Kestrel} $K$ is defined by the following relationship:
$$
(Kx)y = x
$$
In other words, this means that for every bird $x$, the bird $Kx$ is fixated on $x$.
\problem{}
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{}
\end{alltt}
\end{solution}
\vfill
\pagebreak
\problem{}
Assume the forest contains a Kestrel. \\
Given $L_1$ and $L_2$, show that at least one bird is hopelessly egocentric.
\begin{helpbox}[0.75]
\texttt{Def:} $K$ is defined by $(Kx)y = x$ \\
\texttt{Def:} $A$ is fond of $B$ if $AB = B$ \\
\texttt{???:} You'll need one more result from the previous section. Good luck!
\end{helpbox}
\begin{solution}
The final piece is a lemma we proved earler: \\
Any bird is fond of at least one bird
\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}
\end{alltt}
\end{solution}
\vfill
\problem{Kestrel Left-Cancellation}<leftcancel>
In general, $Ax = Ay$ does not imply $x = y$. However, this is true if $A$ is $K$. \\
Show that $Kx = Ky \implies x = y$.
\begin{alltt}
\cmnt{This is a hint.}
let x, y so that Kx = Ky
\end{alltt}
\begin{solution}
\begin{alltt}
\lineno{} let x, y so that Kx = Ky
\lineno{} let z
\lineno{}
\lineno{} (Kx)z = (Ky)z \cmnt{By 01}
\lineno{}
\lineno{} \cmnt{By the definition of K}
\lineno{} (Kx)z = x
\lineno{} (Ky)z = y
\lineno{}
\lineno{} \thus{} x = (Kx)z = (Ky)z = y \qed{}
\end{alltt}
\end{solution}
\vfill
\pagebreak
\problem{}
Show that if $K$ is fond of $Kx$, $K$ is fond of $x$.
\begin{solution}
\begin{alltt}
\lineno{} let x so that K(Kx) = Kx
\lineno{} (K(Kx))y = (Kx)y
\lineno{} = Kx \cmnt{By definition of K}
\lineno{} x = Kx \cmnt{By 03 and definition of K}
\end{alltt}
\end{solution}
\vfill
\problem{}
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:
\begin{alltt}
\lineno{} \cmnt{Given}
\lineno{} Kx = K for some x
\lineno{} \cmnt{We have shown that an egocentric kestrel is hopelessly egocentric}
\lineno{} Kx = K for all x
\lineno{}
\lineno{} let x, y
\lineno{} Kx = K
\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}
\end{alltt}
\end{solution}
\vfill
\pagebreak