I’ve recently been learning functional programming from the ground up. My previous attempts—while not complete failures—typically resulted in me learning how to program in a functional language, but I didn’t feel like I was understanding the core philosophy of the paradigm.

Obviously, it’s not always necessary to have a deep understanding of theory before being able to successfully apply a concept. However, in the case of functional programming, it really is centered around a mathematical model of computation (and a quite beautiful one, at that). I would be so bold as to say that any self-respecting hacker should understand these fundamentals—they really do change how you think about computation, and for the better.

With that said, the following is a distillation of what I think are the building blocks behind one of the origins of functional programming: the lambda calculus. My hope is that this will provide enough of a baseline to encourage further exploration in understanding this important development in computer science.

A Simple Model of Computation

The lambda calculus is a formal system for expressing computation. The term calculus is well-chosen, as it is meant to indicate that the system is very small and simple (the original Latin meaning of calculus is small pebble).

This simplicity at first seems limiting, but it is for good reason. It allows us to easily prove properties of the system, and these properties will still be valid as we add features to the system to make it more like the programming languages we know and love.

So how simple are we talking? Well, here’s a full grammar for all lambda calculus expressions, in Backus-Naur form (don’t worry if you aren’t familiar with the notation—I’ll explain in detail below):

$$ \begin{aligned} \langle expression \rangle & ::= \langle variable \rangle \\ & | \hspace{7mm} \langle abstraction \rangle \\ & | \hspace{7mm} \langle application \rangle \\ \langle abstraction \rangle & ::= \lambda \langle variable \rangle . \langle expression \rangle \\ \langle application \rangle & ::= \langle expression \rangle \langle expression \rangle \end{aligned} $$

If you’ve ever seen the BNF grammar for any popular programming language, you’ll immediately realize from the grammar above how simple lambda calculus expressions are. In a nutshell, an expression can be one of three things: a variable name, a function declaration (called an abstraction), or a function invocation (called an application). The simplest expression is therefore:

$$ x $$

What is $x$? It’s just a variable, free to take on any value we choose (you can think of it as an input to your program).

To declare functions (also referred to as abstractions), we write a λ followed by a variable name to declare the parameter of the function, and follow it by another expression representing the body of the function. Thus a simple function could be:

$$ \lambda x . x $$

This is an example of the identity function—it takes a parameter $x$ and returns whatever argument was given for $x$. For example, applying the identity function to the value $y$, we get the original value $y$:

$$ (\lambda x . x) y \rightarrow y $$

We just saw an example of the final kind of expression: an application. In an application, the abstraction ($\lambda x . x$) is applied to the argument $y$. We can view the process of application as substituting $y$ for $x$ everywhere that $x$ appears in the body of the abstraction, removing the $\lambda$ and variable name while we’re at it. The technical terms for the left-side and right-side are rator (from operator) and rand (from operand), respectively.

One quick note that should be mentioned is that there is no requirement for a variable of an abstraction to appear in the body of the abstraction. Therefore, the following is also a valid abstraction:

$$ \lambda x . y $$

This is an example of a constant function. It essentially ignores its argument and always returns the same value $y$, regardless of the value substituted for $x$.

Functions Are Values

At first glance, it would seem that the lone $x$ expression and the identity function $\lambda x.x$ mentioned earlier are the same thing. For example, $x$ can take on any value in the former, and any value can be substituted for $x$ in the latter.

There is, however, a quite important distinction, which is why these two expressions are indeed far from the same. In the second expression, the abstraction itself is the value of the expression. Since it is not applied to any argument, no value is substituted for $x$, and so it remains as $\lambda x.x$. It cannot be simplified further, because that changes the meaning of the expression.

The concept of treating functions as values is one of the most important traits of functional programming. It allows us to create functions that can take other functions as arguments, as well as return functions as results. This flexibility gives the lambda calculus an incredible amount of expressiveness. A lot of modern programming languages have included this ability in their own design for this very reason. JavaScript, for example, frequently uses the passing of functions as values in order to supply custom callback code for asynchronous requests.

Functions of Multiple Variables

It may seem odd that functions in the lambda calculus can only be declared with a single parameter. However, we can see that this does not sacrifice any amount of expressivity. We can define a function which takes more than one parameter to simply be a function that returns a function that takes a parameter. For example, consider the following function:

$$ \lambda x . \lambda y . (x y) $$

The expression above takes a paramater $x$ and returns a function that takes a parameter $y$, which in turn returns the result of applying $x$ to $y$. Therefore, in order to apply the function to two arguments, we would first have to apply it to a value to substitute for $x$, and then apply the result of that to a value to substitute for $y$, i.e.:

$$ ((\lambda x . \lambda y . (x y)) (\lambda z. z)) w $$

Here, we applied the value $\lambda z.z$ to the function first, then applied the result of that (notice the placement of the brackets) to the value $w$. If we evaluated this expression, it would reduce to $w$, since the function applies the first argument (the identity function) to the second, resulting in the second argument $w$ being returned.

In general, a function of $n$ variables ($n > 1$) will be represented by a function of one variable that produces a function of $n - 1$ variables. Functions of this kind are called curried functions (named after Haskell Curry), and have some nice properties that we won’t get into now.

Free and Bound Variables

We’ve talked about defining functions and applying them to arguments, but we haven’t formalized how the variables in the body of a function are substituted during an application. In order to do so, we need to talk about the concept of free and bound variables in a lambda calculus expression.

In the expression consisting of the lone variable $x$, $x$ is free to take on any value, thus $x$ is considered a free variable. However, in the body of the abstraction $\lambda x.x$, we say that $x$ is bound by the binding occurrence of $x$ as the abstraction’s parameter.

It’s still possible for the body of an abstraction to contain free variables, so long as they are not bound by any binding occurrences in the containing abstraction definitions. For example, $y$ is free in $\lambda x.y$.

We can define the set of free variables ($FV$) and bound variables ($BV$) of a lambda calculus expression as follows:

$$ \begin{align*} FV[x] & = \{x\} \\ FV[\lambda x.E] & = FV[E] - \{x\} \\ FV[MN] & = FV[M] \cup FV[N] \\ \\ BV[x] & = \emptyset \\ BV[\lambda x.E] & = BV[E] \cup \{x\} \\ BV[MN] & = BV[M] \cup BV[N] \end{align*} $$

These definitions are relatively intuitive, given what we’ve currently discussed about free and bound variables. The main idea to note is that abstractions are responsible for introducing bound variables and removing free variables.

Given these definitions of free and bound variables, we can now start reusing the same variable name in an expression (this has been avoided up until now). For example, $x$ is both free and bound in the following expression:

$$ (\lambda x.x) x $$

Why is this the case? The rightmost occurrence of $x$ is free because it is not contained in the body of any abstraction, thus it can take on any value. The middle occurrence of $x$, however, is bound by the binding occurrence of the leftmost occurrence of $x$. This value of $x$ cannot take any value; it must take the value that is substituted for the bound occurrences of $x$ in the body.


When we apply an abstraction to its argument, we replace all occurrences of the parameter of the abstraction with the argument expression. We use the following notation to represent this process:

$$ (\lambda x.B) E = [x \mapsto E]B $$

This notation simply means that all free occurrences of $x$ in $B$ are replaced with $E$.

However, we have to be careful when performing this substitution, since if we blindly replace the variable we may end up causing a variable in the expression we’re replacing it with to be unnecessarily captured by a binding occurrence of an outer abstraction. For example, $[x \mapsto y](\lambda y.xy)$ should not become $\lambda y.yy$, as that changes the meaning of the expression. When we substituted $y$ for $x$, $y$ was free, but after the substitution $y$ became bound.

In order to prevent this accidental capture of free variables, we use the following recursive definitions when performing substitution:

$$ \begin{aligned} \lbrack x \mapsto E \rbrack x & = E \\ \lbrack x \mapsto E \rbrack y & = y \\ \lbrack x \mapsto E \rbrack (MN) & = (\lbrack x \mapsto E \rbrack M)(\lbrack x \mapsto E \rbrack N) \\ \lbrack x \mapsto E \rbrack (\lambda x.P) & = \lambda x.P \\ \lbrack x \mapsto E \rbrack (\lambda y.P) & = \lambda y.(\lbrack x \mapsto E \rbrack P), y \notin FV \lbrack E \rbrack \\ \lbrack x \mapsto E \rbrack (\lambda y.P) & = \lambda z.(\lbrack x \mapsto E \rbrack \lbrack y \mapsto z \rbrack P), y \in FV \lbrack E \rbrack , z \text{ is a fresh variable} \\ \end{aligned} $$

The last definition is what prevents us from accidentally capturing free variables. If the parameter of the abstraction occurs free in the expression we’re substituting, we need to rename the variable with a fresh variable so that no accidental capture takes place.

Equivalent Expressions

In the previous example we learned that $(\lambda x.x)x$ contains both free and bound variables, despite these variables having the same name. This would perhaps make more sense if we were trying to find the free and bound variables of $(\lambda y.y)x$, as it becomes clearer that $x$ is free and $y$ is bound with respect to the abstraction it is contained within. In a way, the two expressions $(\lambda x.x)x$ and $(\lambda y.y)x$—while containing different variable names—were equivalent.

Formally, we call two expressions that are similar in this way $\alpha$-equivalent, written as:

$$ (\lambda x.x)x =_{\alpha} (\lambda y.y)x $$

To define $\alpha$-equivalence, we need to be careful. The expression $x$ and the expression $y$ on their own are not $\alpha$-equivalent to each other, as they are two free variables which individually could take on any value.

$$ x \ne_{\alpha} y $$

However, the abstractions $\lambda x.x$ and $\lambda y.y$ are $\alpha$-equivalent.

$$ \lambda x.x =_{\alpha} \lambda y.y $$

From these examples we can see that $\alpha$-equivalence is simply a way of formalizing when two lambda calculus expressions are semantically equivalent (i.e. they compute the same thing). If we can rename bound variables so that both expressions are exactly the same, while at the same time preserving the semantics of the expression, those two expressions are $\alpha$-equivalent.

Formally, we can define $\alpha$-equivalence using the following recursive definitions:

$$ \begin{aligned} x & =_{\alpha} x,& \text{ (i.e. they are the same variable)} \\ x & \ne_{\alpha} y,& \text{ (i.e. they aren't the same variable)} \\ \lambda x.E_1 & =_{\alpha} \lambda y.E_2,& \text{if } \lbrack x \mapsto z \rbrack E_1 =_{\alpha} \lbrack y \mapsto z \rbrack E_2, \text{ where } z \text{ is a fresh variable} \\ E_1 E_2 & =_{\alpha} E_3 E_4,& \text{if } E_1 =_{\alpha} E_3, E_2 =_{\alpha} E_4 \end{aligned} $$

The cases for variables and applications are pretty straightforward. Abstractions require that we rename the variable name of each abstraction to the same fresh variable (i.e. a newly-generated variable that is not used anywhere else in the expression), and if the bodies of the abstractions with the newly substituted variable are the same, then the two abstractions are the same.


Now that we’ve gotten a sense for lambda calculus expressions, the next question is: what exactly does computation in this model look like?

Substitution is a core part of the semantics of the lambda calculus. We know that the expression $(\lambda x.yx)z$ should reduce to $yz$, since $z$ is subtituted for $x$ in the abstraction. We call this substitution process $\beta$-reduction, and denote it:

$$ (\lambda x.yx) z \rightarrow_{\beta} yz $$

There may be more than one way to reduce an expression. For example, the expression $(\lambda x.(\lambda y.x)w)z$ can be reduced in two different ways:

$$ \begin{aligned} (\lambda x.(\lambda y.x)w)z & \rightarrow_{\beta} (\lambda y.z)w \\ (\lambda x.(\lambda y.x)w)z & \rightarrow_{\beta} (\lambda x.x)z \end{aligned} $$

In the first example, we reduced the outermost application by substituting $z$ for all bound occurrences of $x$. In the second example, we reduced the innermost application, substituting $w$ for all bound occurrences of $y$ (of which there were none).

If an expression can be simplified in this manner, we say that expression is $\beta$-reducible. If it can’t, we say that the expression is in $\beta$-normal form (or just normal form).

We can now think of computation as simply repeatedly reducing an expression to its normal form. However, does every expression have a normal form? Considering that one can write a program with an infinite loop, it seems like not all expressions should have a normal form (i.e. computation does not terminate for all expressions), and indeed this is the case. The simplest example is the $\Omega$ expression:

$$ (\lambda x.xx)(\lambda x.xx) \rightarrow_{\beta} (\lambda x.xx)(\lambda x.xx) $$

In this case, we say that the expression diverges under $\beta$-reduction.

One important result with regards to normal forms is the Church-Rosser theorem, which states that any lambda calculus expression has at most one normal form. In other words, an expression will either diverge forever or will eventually reduce to a normal form.

Finding the Answer

A problem with arbitrarily reducing an expression is that while a normal form may exist, we may perform a reduction which leads us down a path that results in divergence. For example consider the following two reductions of the same expression:

$$ \begin{aligned} (\lambda y.z) ((\lambda x.xx)(\lambda x.xx)) & \rightarrow_{\beta} (\lambda y.z) ((\lambda x.xx)(\lambda x.xx)) \\ (\lambda y.z) ((\lambda x.xx)(\lambda x.xx)) & \rightarrow_{\beta} z \end{aligned} $$

In the first case, we reduced the innermost application, $(\lambda x.xx)(\lambda x.xx)$, which reduced to the same value, and thus resulted in no change. In the second case, we reduced the outermost application of the abstraction $\lambda y.z$ to its argument, which simplified to the value $z$, since $y$ is not bound anywhere in the body of the abstraction.

Therefore, it is important that we choose our $\beta$-reductions carefully. Luckily, reducing the leftmost, outermost application will always find a normal form, if it exists. This reduction strategy is called Normal Order Reduction (NOR). A series of normal order reductions are shown below:

$$ \begin{aligned} ((\lambda x.(\lambda y.x (\lambda z.yz)r))w)q & \rightarrow_{\beta} (\lambda y.w (\lambda z.yz)r)q \\ & \rightarrow_{\beta} w(\lambda z.qz)r \\ & \rightarrow_{\beta} w q r \end{aligned} $$

While probably the most intuitive of the reduction strategies, NOR is rarely used in actual programming languages, because it requires us to rewrite abstractions during the process of computation. This can get expensive, so NOR tends to be avoided for practical use.

Normal Order Evaluation (NOE) is another reduction strategy that is very similar to NOR, except it forbids reduction within the body of abstractions. It is also known as call by name, as arguments to a function are not evaluated beforehand, but are rather substituted directly into the body and left to be evaluated when the entire function is evaluated later. The Haskell language uses a modified version of call by name, call by need, which is where if the function argument is evaluated, its value is stored for subsequent uses (resulting in less repeated computation if the parameter is used multiple times in the abstraction).

Applicative Order Reduction (AOR) reduces the leftmost, innermost application, and Applicative Order Evaluation (AOE) does not reduce applications in the body of abstractions. AOE is also known as call by value, and is used in functional programming languages like Scheme and ML, as well as most imperative programming languages.

Here’s a concrete example of an expression where reduction by NOR, AOR, and AOE all result in a different initial reduction, but whose repeated application leads to the same normal form. Reduction by NOR and NOE are the same in this list, as by definition NOE will either be the same or will stop earlier due to its restrictions on not simplifying applications within abstractions.

$$ \begin{aligned} (\lambda x.(\lambda y.y)x)((\lambda w.w)z) & \overset{NOR}{\rightarrow} (\lambda y.y)((\lambda w.w)z) \\ & \overset{NOR}{\rightarrow} (\lambda w.w)z \\ & \overset{NOR}{\rightarrow} z \\ (\lambda x.(\lambda y.y)x)((\lambda w.w)z) & \overset{NOE}{\rightarrow} (\lambda y.y)((\lambda w.w)z) \\ & \overset{NOE}{\rightarrow} (\lambda w.w)z \\ & \overset{NOE}{\rightarrow} z \\ (\lambda x.(\lambda y.y)x)((\lambda w.w)z) & \overset{AOR}{\rightarrow} (\lambda x.x)((\lambda w.w)z) \\ & \overset{AOR}{\rightarrow} (\lambda x.x)z \\ & \overset{AOR}{\rightarrow} z \\ (\lambda x.(\lambda y.y)x)((\lambda w.w)z) & \overset{AOE}{\rightarrow} (\lambda x.(\lambda y.y)x)z \\ & \overset{AOE}{\rightarrow} (\lambda y.y)z \\ & \overset{AOE}{\rightarrow} z \end{aligned} $$

The main takeaway from all this is that different methods of computation can result in the same final answer (normal form). Computation in different programming languages varies based on the evaluation strategy used, and each strategy has different advantages and disadvantages that these languages utilize.

How Do You Program in the Lambda Calculus?

If you’ve made it this far, you’re probably wondering where all of this is going. Given how apparently restrictive the lambda calculus is, how does one actually write any non-trivial program with it? Where are the if statements, the arrays, or even the ability to perform arithmetic?

Remember that the lambda calculus was designed to be incredibly small and simple so that it was easy to prove various properties about it. In order to be able to apply these properties to higher-level languages, we now have to add the features of these languages by using the basic building blocks of the lambda calculus.

Rest assured that being able to add arithmetic and other high-level language features is possible. Church encoding provides a means of including higher-level operators and data into the lambda calculus.

To give you a taste of what is involved with these additions, we’ll show how to add the concept of true and false along with if statements to the lambda calculus. This specific Church encoding technique is known as Church booleans.

To keep us from repeating ourselves to much, we’re going to abuse the notation slightly and introduce abbreviations to make expressions easier to read. For example, the identity function will be represented as:

$$ \textbf{id} = \lambda x.x $$

To represent the Boolean true and false values, we consider how we could design them so that they could be used by if. if statements can be thought of as functions that take 3 parameters: a Boolean expression $B$ to evaluate, an expression $T$ to return if $B$ is true, and an expression $F$ to return if $B$ is false.

Consider then, the following definitions:

$$ \begin{aligned} & \textbf{true} = \lambda x. \lambda y. x \\ & \textbf{false} = \lambda x. \lambda y. y \\ & \textbf{if } B \textbf { then } T \textbf{ else } F = B T F \end{aligned} $$

Here, we represent true as a function that returns the first of two parameters, and false as a function that returns the second of two parameters. This works beautifully with an if, as then we simply need to apply the two expressions $T$ and $F$ from the if statement to the Boolean expression $B$, and the definitions of true and false will ensure that the correct expression is returned.

Using this same idea, we can also define and, or, and not.

\begin{aligned} & A \textbf{ and } B = A \text{ } B \textbf{ false} \\ & A \textbf{ or } B = A \textbf{ true } B \\ & \textbf{not } A = A \textbf{ false } \textbf{true} \end{aligned}

Representing true and false as functions becomes really powerful, allowing us to define Boolean operators by simply applying the arguments to those operators to each other.

There’s So Much More

We’ve really just scratched the surface of what is possible with the lambda calculus, but hopefully it was deep enough that it’s easy to see the power that this simple model of computation contains, and also where various features of functional programming languages came from.

If you are interested in learning more, the next step is to start adding types to the calculus. So far we’ve been dealing with an untyped lambda calculus, where no rules are in place to ensure that the right kind of arguments are given to a function—we just assumed that anyone writing a program in this model wouldn’t make such a mistake. The simply typed lambda calculus is the first step into dealing with types in a model of computation, and will eventually lead you to other calculi such as the polymorphic lambda calculus.

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Shane da Silva



Shane da  Silva

Coding by the woods

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