From a7dfe2728a3595deac14d512e9df02fc1ffcc623 Mon Sep 17 00:00:00 2001
From: Mark <mark@betalupi.com>
Date: Tue, 17 Oct 2023 09:15:31 -0700
Subject: [PATCH] Improved intro

---
 Advanced/Lambda Calculus/parts/00 intro.tex | 230 ++++++++++++++------
 1 file changed, 158 insertions(+), 72 deletions(-)

diff --git a/Advanced/Lambda Calculus/parts/00 intro.tex b/Advanced/Lambda Calculus/parts/00 intro.tex
index 10b74b0..97a1f3d 100755
--- a/Advanced/Lambda Calculus/parts/00 intro.tex	
+++ b/Advanced/Lambda Calculus/parts/00 intro.tex	
@@ -1,18 +1,82 @@
-\section{Definitions}
+\section{Introduction}
+\textit{Lambda calculus} is a model of computation, much like the Turing machine.
+As we're about to see, it works in a fundamentally different way, which has a few 
+practical applications we'll discuss at the end of class.
 
-\generic{$\lm$ Notation:}
-$\lm$ notation is used to define functions, and looks like $(\lm ~ \text{input} ~ . ~ \text{output})$. \par
-Consider the following statement:
-$$
-	I = \lm a . a
-$$
-This tells us that $I$ is a function that takes its input, $a$, to itself. We'll call this the \textit{identity function}.
+\vspace{2mm}
 
-To apply functions, put them next to their inputs. We'll omit the usual parentheses to save space.
+
+A lambda function starts with a lambda ($\lm$), followed by the names of any inputs used in the expression,
+followed by the function's output.
+
+For example, $\lm x . x + 3$ is the function $f(x) = x + 3$ written in lambda notation.
+
+\vspace{4mm}
+Let's disect $\lm x . x + 3$ piece by piece:
+
+\begin{itemize}[itemsep=3mm]
+	\item \say{$\lm$} tells us that this is the beginning of an expression. \par
+	$\lm$ here doesn't have a special value or definition; \par
+	it's just a symbol that tells us \say{this is the start of a function.}
+	
+
+	\item \say{$\lm x$} says that the variable $x$ is \say{bound} to the function (i.e, it is used for input).\par
+	Whenever we see $x$ in the function's output, we'll replace it with the input of the same name.
+
+	\vspace{1mm}
+
+	This is a lot like normal function notation: In $f(x) = x + 3$, $(x)$ is \say{bound}
+	to $f$, and we replace every $x$ we see with our input when evaluating.
+
+
+
+	\item The dot tells us that what follows is the output of this expression. \par
+	This is much like $=$ in our usual function notation: \par
+	The symbols after $=$ in $f(x) = x + 3$ tell us how to compute the output of this function.
+\end{itemize}
+
+\problem{}
+Rewrite the following functions using this notation:
+\begin{itemize}
+	\item $f(x) = 7x + 4$
+	\item $f(x) = x^2 + 2x + 1$
+\end{itemize}
+
+\vfill
+\pagebreak
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+To evaluate $\lm x . x + 3$, we need to input a value:
 $$
-(I~\star) =
-(\lm \tzm{b}a. a)~\tzmr{a}\star =
-\star
+	(\lm x . x + 3)~5
+$$
+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.
+
+
+\vspace{2mm}
+
+
+We evaluate this by removing the \say{$\lm$} prefix and substituting $3$ for $x$ wheverever it appears:
+$$
+(\lm x . \tzm{b}x + 3)~\tzmr{a}5 =
+5 + 3 = 8
 \begin{tikzpicture}[
 	overlay,
 	remember picture,
@@ -24,10 +88,89 @@ $$
 		(a.center) to (b.center);
 \end{tikzpicture}
 $$
-Functions are left-associative: If $A$ and $B$ are functions, $(A~B~\star)$ is equivalent to $((A~B)~\star)$.
-As usual, we'll use parentheses to group terms if we want to override this order: $(A~(B~\star)) \neq (A~B~\star)$ \par
-In this handout, all types of parentheses ( $(), [~],$ etc ) are equivalent.
 
+\problem{}
+Evaluate the following:
+\begin{itemize}[itemsep = 2mm]
+	\item $(\lm x. 2x + 1)~4$
+	\item $(\lm x. x^2 + 2x + 1)~3$
+	\item $(\lm x. (\lm y. 9y) x + 3)~2$ \par
+	\hint{
+		This function has a function inside, but the evaluation process doesn't change. \\
+		Replace all $x$ with 2 and evaluate again.
+	}
+\end{itemize}
+
+
+\vfill
+
+As we saw above, we denote function application by simply putting functions next to their inputs. \par
+If we want to apply $f$ to $5$, we write \say{$f~5$}, without any parentheses around the function's argument.
+
+\vspace{4mm}
+
+You may have noticed that we've been using arithmetic in the last few problems.
+This isn't fully correct: addition is not defined in lambda calculus. In fact, nothing is defined:
+not even numbers!
+
+In lambda calculus, we have only one kind of object: the function.
+The only action we have is function application, which works by just like the examples above.
+
+\vspace{2mm}
+
+Don't worry if this sounds confusing, we'll see a few examples soon.
+
+\pagebreak
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+\definition{}
+The first \say{pure} functions we'll define are $I$ and $M$:
+\begin{itemize}
+	\item $I = \lm x . x$
+	\item $M = \lm x . x x$
+\end{itemize}
+
+Note that $I$ and $M$ don't have a meaning on their own. They are not formal functions. \par
+Rather, it's notation that says \say{write $\lm x.x$ whenever you see $I$.}
+
+
+
+
+\problem{}
+Reduce the following expressions.
+\begin{itemize}[itemsep=2mm]
+	\item $I~I$
+	\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}
+
+
+\vfill
+
+In lambda calculus, functions are left-associative: \par
+$(f~g~h)$ is equivalent to $((f~g)~h)$, not $(f~(g~h))$
+
+As usual, we use parentheses to group terms if we want to override this order: $(f~(g~h)) \neq ((f~g)~h)$ \par
+In this handout, all types of parentheses ( $(), [~],$ etc ) are equivalent.
 
 \problem{}
 Rewrite the following expressions with as few parentheses as possible, without changing their meaning or structure.
@@ -53,63 +196,6 @@ Remember that lambda calculus is left-associative.
 
 
 
-\generic{$\beta$-Reduction:}
-
-$\beta$-reduction is a fancy name for \say{simplifying an expression.} We've already done it once above,
-while evaluating $(I \star)$. Let's make another function:
-$$
-	M = \lm f . f f
-$$
-The function $M$ simply repeats its input. What is $(M~I)$?
-$$
-	(M~I) =
-	((\lm \tzm{b}f.f f)~\tzmr{a}I) =
-	(I~I) =
-	((\lm \tzm{d}a.a)~\tzmr{c}I) =
-	I
-	\begin{tikzpicture}[
-		overlay,
-		remember picture,
-		out=225,
-		in=315,
-		distance=0.5cm
-	]
-		\draw[->,gray,shorten >=5pt,shorten <=3pt]
-			(a.center) to (b.center);
-		\draw[->,gray,shorten >=5pt,shorten <=3pt]
-			(c.center) to (d.center);
-	\end{tikzpicture}
-$$
-We cannot go any further, so we stop. Our expression is now in \textit{$\beta$-normal} or \say{reduced} form.
-
-
-
-\problem{}
-Reduce the following expressions:
-\begin{itemize}
-	\item $I~I$
-	\item $I~I~I$
-	\item $(\lm a .(a~a~a)) ~ I$
-	\item $(\lm a . (\lm b . a)) ~ M ~ I$
-\end{itemize}
-
-\vfill
-\pagebreak
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
 
 \generic{Currying:}