Fixed a few errors

Advanced/Lambda Calculus/parts/00 intro.tex

@@ -178,7 +178,6 @@ Reduce the following expressions. \par
 	\textbf{Solution for $(I~I)$:}\par
 	Recall that $I = \lm x.x$. First, we rewrite the left $I$ to get $(\lm x . x )~I$. \par
 	Applying this function by replacing $x$ with $I$, we get $I$:
-
 	$$
 		I ~ I = 
 		(\lm x . \tzm{b}x )~\tzm{a}I =
@@ -193,11 +192,7 @@ Reduce the following expressions. \par
 			\draw[->,gray,shorten >=5pt,shorten <=3pt]
 				(a.center) to (b.east);
 		\end{tikzpicture}
-	$$
-	\vspace{0.5mm}
-
-	So, $I~I$ reduces to itself. This makes sense, since the identity
-	function doesn't change its input!
+	$$\null
 \end{examplesolution}
 
 
@@ -214,11 +209,20 @@ Rewrite the following expressions with as few parentheses as possible, without c
 Remember that lambda calculus is left-associative.
 \vspace{2mm}
 \begin{itemize}[itemsep=2mm]
-	\item $(\lm x. (\lm y. \lm (z. ((xz)(yz)))))$
+	\item $(\lm x. (\lm y. \lm z. ((xz)(yz))))$
 	\item $((ab)(cd))((ef)(gh))$
 	\item $(\lm x. ((\lm y.(yx))(\lm v.v)z)u) (\lm w.w)$
 \end{itemize}
 
+\begin{solution}
+	$(\lm x. ((\lm y.(yx))(\lm v.v)z)u) (\lm w.w) \implies (\lm x. (\lm y.yx) (\lm v.v)~z~u) \lm w.w$
+
+	\vspace{2mm}
+
+	It's important that a function's output (everything after the dot) will continue until we hit a close-paren.
+	This is why we need the parentheses in the above example.
+\end{solution}
+
 
 \vfill
 \pagebreak
@@ -317,17 +321,18 @@ We've already seen this on the previous page: $K$ takes an input $x$ and uses it
 You can think of $K$ as a \say{factory} that constructs functions using the input we provide.
 
 \problem{}<firstcardinal>
-Let $C = \lm f. \Bigl[\lm g. \Bigl( \lm x. [~ g(f(x)) ~] \Bigr)\Bigr]$. For now, we'll call it the \say{composer.}
+Let $C = \lm f. \Bigl[\lm g. \Bigl( \lm x. [~ f(g(x)) ~] \Bigr)\Bigr]$. For now, we'll call it the \say{composer.} \par
+\note[Note]{We could also call $C$ the \say{right-associator.} Why?}
 
 \vspace{1mm}
 
-Note that $C$ has three \say{layers} of curry: it makes a function ($\lm g$) that makes another function ($\lm x$). \par
+$C$ has three \say{layers} of curry: it makes a function ($\lm g$) that makes another function ($\lm x$). \par
 If we look closely, we'll find that $C$ pretends to take three arguments.
 
 \vspace{1mm}
 
 What does $C$ do? Evaluate $(C~a~b~x)$ for arbitary expressions $a, b,$ and $x$. \par
-\hint{Place parentheses first. Remember, function application is left-associative.}
+\hint{Evaluate $(C~a)$ first. Remember, function application is left-associative.}
 
 \vfill