More edits

Advanced/Lambda Calculus/parts/00 intro.tex

@@ -67,7 +67,7 @@ $$
 $$
 This is very similar to the usual way we call functions: we usually write $f(5)$. \par
 Above, we define our function $f$ \say{in-line} using lambda notation, \par
-and we ommit the parentheses around 5 for the sake of simpler notation.
+and we omit the parentheses around 5 for the sake of simpler notation.
 
 
 \vspace{2mm}
@@ -147,8 +147,13 @@ The first \say{pure} functions we'll define are $I$ and $M$:
 	\item $I = \lm x . x$
 	\item $M = \lm x . x x$
 \end{itemize}
+Both $I$ and $M$ take one function ($x$) as an input. \par
+$I$ does nothing, it just returns $x$. \par
+$M$ is a bit more interesting: it applies the function $x$ on a copy of itself.
 
-Note that $I$ and $M$ don't have a meaning on their own. They are not formal functions. \par
+\vspace{4mm}
+
+Also, note that $I$ and $M$ don't have a meaning on their own. They are not formal functions. \par
 Rather, they are abbreviations that say \say{write $\lm x.x$ whenever you see $I$.}
 
 
@@ -163,11 +168,38 @@ Reduce the following expressions. \par
 
 \begin{itemize}[itemsep=2mm]
 	\item $I~I$
+	\item $M~I$
 	\item $(I~I)~I$
 	\item $\Bigl(~ \lm a .(a~(a~a)) ~\Bigr) ~ I$
 	\item $\Bigl(~(\lm a . (\lm b . a)) ~ M~\Bigr) ~ I$
 \end{itemize}
 
+\begin{examplesolution}
+	\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 $I$:
+
+	$$
+		I ~ I = 
+		(\lm x . \tzm{b}x )~\tzm{a}I =
+		I
+		\begin{tikzpicture}[
+			overlay,
+			remember picture,
+			out=-90,
+			in=-90,
+			distance=0.5cm
+		]
+			\draw[->,gray,shorten >=5pt,shorten <=3pt]
+				(a.center) to (b.east);
+		\end{tikzpicture}
+	$$
+	\vspace{0.5mm}
+
+	So, $I~I$ reduces to itself. This makes sense, since the identity
+	function doesn't change its input!
+\end{examplesolution}
+
 
 \vfill
 
@@ -204,8 +236,8 @@ Remember that lambda calculus is left-associative.
 
 \generic{Currying:}
 
-In lambda calculus, functions are only allowed to take one argument.
-However, we can emulate multivariable functions through \textit{currying}\footnotemark{}\hspace{-1ex}.
+In lambda calculus, functions are only allowed to take one argument. \par
+If we want multivariable functions, we'll have to emulate them through \textit{currying}\footnotemark{}\hspace{-1ex}.
 
 \footnotetext{After Haskell Brooks Curry\footnotemark{}\hspace{-1ex}, a logician that contributed to the theory of functional computation.}
 \footnotetext{
@@ -216,9 +248,9 @@ However, we can emulate multivariable functions through \textit{currying}\footno
 \vspace{1ex}
 
 The idea behind currying is fairly simple: we make functions that return functions. \par
-Consider the expression $A$ below. As before, it takes one input $f$ and returns the function $\lm a. f(f(a))$ (underlined).
+Consider the expression $A$ below. It takes one input $f$ and returns the function $\lm a. f~(f~a)$ (underlined).
 $$
-A = \lm f .(\tzm{a} ~ \lm a . f(f(a)) ~ \tzm{b})
+A = \lm f .(\tzm{a} ~ \lm a . f~(f~a) ~ \tzm{b})
 \begin{tikzpicture}[
 	overlay,
 	remember picture
@@ -230,11 +262,11 @@ A = \lm f .(\tzm{a} ~ \lm a . f(f(a)) ~ \tzm{b})
 \end{tikzpicture}
 $$
 When we evaluate $A$ with one input, it constructs a new function with the argument we give it. \par
-For example, let's apply $A$ to an arbirary function $N$:
+For example, let's apply $A$ to an arbirary function $n$:
 $$
-	A~N =
-	\lm \tzm{b}f.(~\lm a.f(f(a))~)~\tzmr{a}N =
-	\lm a.N(N(a))
+	A~n =
+	\lm f.(~\lm a.\tzm{b}f~(\tzm{c}f~a)~)~\tzmr{a}n =
+	\lm a.n~(n~a)
 	\begin{tikzpicture}[
 		overlay,
 		remember picture,
@@ -244,13 +276,15 @@ $$
 	]
 		\draw[->,gray,shorten >=5pt,shorten <=3pt]
 			(a.center) to (b.center);
+		\draw[->,gray,shorten >=5pt,shorten <=3pt]
+			(a.center) to (c.center);
 	\end{tikzpicture}
 $$
-Above, $A$ replaced every $f$ in its definition with an $N$. \par
+Above, $A$ replaced every $f$ in its definition with an $n$. \par
 You can think of $A$ as a \say{factory} that constructs functions using the inputs we provide.
 
 \problem{}<firstcardinal>
-Let $C = \lm f. (\lm g. \lm x. (g(f(x))))$. What does $B$ do? \par
+Let $C = \lm f. \Bigl[\lm g. \Bigl( \lm x. [~ g(f(x)) ~] \Bigr)\Bigr]$. What does $C$ do? \par
 Evaluate $(C~a~b~x)$ for arbitary expressions $a$ and $b$. \par
 \hint{Place parentheses first. Remember, function application is left-associative.}
 
@@ -269,16 +303,16 @@ If we have an expression with repeated function definitions, we'll combine their
 
 \vspace{1ex}
 
-For example, $A = \lm f . (\lm a . f(f(a)))$ will become $A = \lm fa . f(f(a))$
+For example, $A = \lm f .[ \lm a . f(f(a))]$ will become $A = \lm fa . f(f(a))$
 
 \problem{}
-Rewrite $C = \lm f. (\lm g. \lm x. (g(f(x))))$ from \ref{firstcardinal} using this shorthand.
+Rewrite $C = \lm f.\lm g.\lm x. (g(f(x)))$ from \ref{firstcardinal} using this shorthand.
 
 \vspace{25mm}
 
 Remember that this is only notation. \textbf{Curried functions are not multivariable functions, they are simply shorthand!}
 Any function presented with this notation must still be evaluated one variable at a time, just like an un-curried function.
-Substituting all curried variables at once may cause errors, as we'll see below.
+Substituting all curried variables at once will cause errors.
 
 \pagebreak
 
@@ -311,7 +345,7 @@ Substituting all curried variables at once may cause errors, as we'll see below.
 We say two functions are \textit{equivalent} if they differ only by the names of their variables:
 $I = \lm a.a = \lm b.b = \lm \heartsuit . \heartsuit = ...$ \par
 
-\vspace{2mm}
+\begin{instructornote}
 
 	The idea behind this is very similar to the idea behind \say{equivalent groups} in group theory: \par
 	we do not care which symbols a certain group or function uses, we care about their \textit{structure}. \par
@@ -321,10 +355,12 @@ we do not care which symbols a certain group or function uses, we care about the
 	If we have two groups with different elements with the same multiplication table, we look at them as identical groups.
 	The same is true of lambda functions: two lambda functions with different variable names that behave in the same way are identical.
 
-
-\vspace{4mm}
+\end{instructornote}
 
 
+
+%\iftrue
+\iffalse
 	\generic{$\alpha$-Conversion:}
 	There is one more operation we need to discuss. Those of you that have worked with any programming language may
 	find that this section sounds like something you've seen before.
@@ -357,17 +393,22 @@ $$
 	$$
 
 	Which is, of course, incorrect. $\lm I . I$ is not a valid function.
-Both $A$ and $B$ use the input $b$. However, each $b$ is \say{bound} to a different function:
+	This problem arises because both $A$ and $B$ use the input $b$.
+	However, each $b$ is \say{bound} to a different function:
 	One $b$ is bound to $A$, and the other to $B$. They are therefore distinct.
 
 	\vspace{2ex}
 
 	Let's rewrite $B$ as $\lm b_1 . b_1$ and try again: $(A~B) = \lm b . ( \lm b_1 . b_1) = \lm bb_1 . b_1$ \par
 	Now, we correctly find that $(A~B~I) = (\lm bc . c)~I = \lm c . c = B = I$.
+\fi
 
 \problem{}
-Let $Q = \lm abc.b$ \par
-Reduce $(Q~a~c~b)$.
+Let $Q = \lm abc.b$. Reduce $(Q~a~c~b)$. \par
+\hint{
+	You may want to rename a few variables. \\
+	The $a,b,c$ in $Q$ are different than the $a,b,c$ in the expression!	
+}
 
 \begin{solution}
 	I'll rewrite $(Q~a~c~b)$ as $(Q~a_1~c_1~b_1)$: