Mockingbird edits

src/Advanced/Introduction to Quantum/src/parts/01 bits.tex

@@ -230,7 +230,7 @@ The \textit{tensor product} of two vectors is defined as follows:
 That is, we take our first vector, multiply the second
 vector by each of its components, and stack the result.
 You could think of this as a generalization of scalar
-mulitiplication, where scalar mulitiplication is a
+multiplication, where scalar multiplication is a
 tensor product with a vector in $\mathbb{R}^1$:
 \begin{equation*}
 	a

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}

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

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}
 

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

src/Advanced/Retrograde Analysis/parts/03 medium.tex

@@ -251,7 +251,7 @@ What is it, and what is its color? \par
 	\textbf{Part 4:}
 
 	The promoted black bishop on H2 must have been promoted on G1. The pawn which was promoted must have come from G7,
-	since neither of the pawns from F6 or H6 could make a capture to get to the G-file (all six missing white pieces have been accouted for).
+	since neither of the pawns from F6 or H6 could make a capture to get to the G-file (all six missing white pieces have been accounted for).
 	The Pawn from E7 has promoted to the bishop on A2.
 
 	What happened was this: the white pawn from G2 made its capture on H3 while the pawn on G3 was still on H2. This allowed the black pawn

src/Intermediate/Instant Insanity/main.tex

@@ -331,7 +331,7 @@
 	representing all four cubes. \\
 
 	\begin{center} \begin{small}
-	\begin{tikzpicture} \label{pic:II_comfiguration}
+	\begin{tikzpicture} \label{pic:II_configuration}
 		\filldraw [blue] (0,5) -- (1,5) -- (1,6) --
 		(0,6) -- (0,5);
 		\draw [line width = 1.5pt] (0,5) --