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