\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.

\vspace{2mm}


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:
$$
	(\lm x . x + 3)~5
$$
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 omit 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:
$$
(\lm x . \tzm{b}x + 3)~\tzmr{a}5 =
5 + 3 = 8
\begin{tikzpicture}[
	overlay,
	remember picture,
	out=225,
	in=315,
	distance=0.5cm
]
	\draw[->,gray,shorten >=5pt,shorten <=3pt]
		(a.center) to (b.center);
\end{tikzpicture}
$$

\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}
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.

\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 $I$.}




\problem{}
Reduce the following expressions. \par
\hint{
	Of course, your final result will be a function. \\
	Functions are the only objects we have!
}

\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

In lambda calculus, functions are left-associative: \par
$(f~g~h)$ means $((f~g)~h)$, 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.
Remember that lambda calculus is left-associative.
\vspace{2mm}
\begin{itemize}[itemsep=2mm]
	\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}


\vfill
\pagebreak












\generic{Currying:}

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{
	There are three programming languages named after him: Haskell, Brook, and Curry. \par
	Two of these are functional, and one is an oddball GPU language last released in 2007.
}

\vspace{1ex}

The idea behind currying is fairly simple: we make functions that return functions. \par
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})
\begin{tikzpicture}[
	overlay,
	remember picture
]
	\node[below = 0.8mm] at (a.center) (aa) {};
	\node[below = 0.8mm] at (b.center) (bb) {};

	\path[draw = gray] (aa) to (bb);
\end{tikzpicture}
$$
When we evaluate $A$ with one input, it constructs a new function with the argument we give it. \par
For example, let's apply $A$ to an arbirary function $n$:
$$
	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,
		out=225,
		in=315,
		distance=0.5cm
	]
		\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
You can think of $A$ as a \say{factory} that constructs functions using the inputs we provide.

\problem{}<firstcardinal>
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.}

\vfill

\problem{}
Using the definition of $C$ above, evaluate $C~M~I~\star$ \par
Then, evaluate $C~I~M~I$ \par
\note[Note]{$\star$ represents an arbitrary expression. Treat it like an unknown variable.}

\vfill

As we saw above, currying allows us to create multivariable functions by nesting single-variable functions.
You may have notice that curried expressions can get very long. We'll use a bit of shorthand to make them more palatable:
If we have an expression with repeated function definitions, we'll combine their arguments under one $\lm$.

\vspace{1ex}

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.

\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 will cause errors.

\pagebreak


























\definition{Equivalence}
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

\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

	\vspace{2mm}

	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.

\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.

	\vspace{2mm}

	Variables inside functions are \say{scoped.} We must take care to keep separate variables separate.

	For example, take the functions \par
	$A = \lm a b . a$ \par
	$B = \lm b . b$

	\vspace{2ex}

	We could say that $(A~B) = \lm b . (\lm b . b)$, and therefore
	$$
	((A~B)~I)
	= (~ (\lm \tzm{b}b . (\lm b . b))~\tzmr{a}I ~)
	= \lm I . I
	\begin{tikzpicture}[
		overlay,
		remember picture,
		out=225,
		in=315,
		distance=0.5cm
	]
		\draw[->,gray,shorten >=5pt,shorten <=3pt]
			(a.center) to (b.center);
	\end{tikzpicture}
	$$

	Which is, of course, incorrect. $\lm I . I$ is not a valid 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 $Q = \lm abc.b$. 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 $(Q~a_1~c_1~b_1)$:
	\begin{align*}
		Q = (\lm abc.b) &= (\lm a.\lm b.\lm c.b) \\
		(\lm a.\lm b.\lm c.b)~a_1 &= (\lm b.\lm c.b) \\
		(\lm b.\lm c.b)~c_1 &= (\lm c.c_1) \\
		(\lm c.c_1)~b_1 &= c_1
	\end{align*}
\end{solution}

\vfill

\problem{}
Reduce $((\lm a.a)~\lm bc.b)~d~\lm eg.g$

\begin{solution}
	$((\lm a.a)~\lm bc.b)~d~\lm eg.g$ \\
	$= (\lm bc.b)~d~\lm eg.g$ \\
	$= (\lm c.d)~\lm eg.g$ \\
	$= d$
\end{solution}

\vfill
\pagebreak