Introduction to the Lambda Calculus
We often talk about "the Lambda Calculus", as if there were just one; but in fact, there are many, many variations. The one we will start with, and will explore in some detail, is often called "the pure" or "the untyped" Lambda Calculus. Actually, there are many variations even under that heading. But all of the variations share a strong family resemblance, so what we learn now will apply to all of them.
Fussy note: calling this the "pure" Lambda Calculus is entrenched terminology, but it coheres imperfectly with other uses of "pure" we'll encounter. There are three respects in which the lambda calculus we'll be presenting might claim to deserve the name "pure": (1) it has no pre-defined constants and a very spare syntax; (2) it has no types; (3) it has no side-effects, and is insensitive to the order of evaluation.
Sense (3) corresponds most closely to the other uses of "pure" you'll see in the surrounding literature. With respect to this point, it may be true that the Lambda Calculus has no side effects. (Let's revisit that assumption at the end of term.) But as we'll see next week, it is not true that it's insensitive to the order of evaluation. So if that's what we mean by "pure", this lambda calculus isn't as pure as you might hope to get. Some typed lambda calculi will turn out to be more pure in that respect.
But this lambda calculus is at least "pure" in sense (2). At least, it doesn't explicitly talk about any types. Some prefer to say that the Lambda Calculus does have types implicitly, it's just that there's only one type, so that every expression is a member of that one type. If you say that, you have to say that functions from this type to this type also belong to this type. Which is weird... In fact, though, such types are studied, under the name "recursive types." More about these later in the seminar.
Well, at least this lambda calculus is "pure" in sense (1). As we'll see next week, though, there are some computational formal systems whose syntax is even more spare, in that they don't even have variables. So if that's what mean by "pure", again this lambda calculus isn't as pure as you might hope to get.
Still, if you say you're talking about "the pure" Lambda Calculus, or "the untyped" Lambda Calculus, or even just "the" Lambda Calculus, this is the system that people will understand you to be referring to.
Here is its syntax:
Each variable is an expression. For any variable
a and (possibly complex) expressions
N, the following are also expressions:
We'll tend to write
(λa M) as just
(\a M), so we
don't have to write out the markup code for the
λ. You can yourself write
(\a M) or
(lambda a M).
Expressions in the lambda calculus are called "terms". Here is the syntax of the lambda calculus given in the form of a context-free grammar:
T --> Var
T --> ( λ Var T)
T --> ( T T )
Var --> x
Var --> y
Var --> z
Very, very simple.
Sometimes the first two production rules are further distinguished, and those
are called more specifically "value terms". Whereas Applications (terms of the
(M N)) are terms but not value terms.
Examples of terms:
x (y x) (x x) (\x y) (\x x) (\x (\y x)) (x (\x x)) ((\x (x x)) (\x (x x)))
The lambda calculus has an associated proof theory. For now, we can regard the proof theory as having just one rule, called the rule of beta-reduction or "beta-contraction". Suppose you have some expression of the form:
((\a M) N)
This is an application whose first element is an abstract. This
compound form is called a redex, meaning it's a "beta-reducible
(\a M) is called the head of the redex;
called the argument, and
M is called the body.
The rule of beta-reduction permits a transition from that expression to the following:
What this means is just
M, with any free occurrences inside
a replaced with the term
N (--- "without capture", which
we'll explain in the advanced notes).
What is a free occurrence?
Any occurrence of a variable
ais bound in T when T has the form
An occurrence of a variable
ais bound in T when T has the form
(\b N)--- that is, with a different variable
b--- just in case that occurrence of
ais bound in the subexpression
When T has the form
(M N), any occurrences of
athat are bound in the subexpression
Mare also bound in T, and so too any occurrences of
athat are bound in the subexpression
An occurrence of a variable is free if it's not bound.
For instance consider the following term
(x (\x (\y (x (y z)))))
The first occurrence of
x in T is free. The
\x we won't regard as
containing an occurrence of
x. The next occurrence of
within a form
(\x (\y ...)) that begins with
\x, so it is bound as well. The
y is bound; and the occurrence of
z is free.
To read further:
Here's an example of beta-reduction:
((\x (y x)) z)
We'll write that like this:
((\x (y x)) z) ~~> (y z)
Different authors use different terminology and notation. Some authors use the term "contraction" for a single reduction step, and reserve the term "reduction" for the reflexive transitive closure of that, that is, for zero or more reduction steps. Informally, it seems easiest to us to say "reduction" for one or more reduction steps. So when we write:
M ~~> N
we'll mean that you can get from
N by one or more reduction
N are such that there is some common term (perhaps just one
of those two, or perhaps a third term) that they both reduce to,
we'll say that
N are beta-convertible. We'll write that
M <~~> N
More details about the notation and metatheory of the lambda calculus are in this week's advanced notes.
The grammar we gave for the lambda calculus leads to some verbosity. There are several informal conventions in widespread use, which enable the language to be written more compactly. (If you like, you could instead articulate a formal grammar which incorporates these additional conventions. Instead of showing it to you, we'll leave it as an exercise for those so inclined.)
Parentheses Outermost parentheses around applications can be dropped. Moreover, applications will associate to the left, so
M N P will be understood as
((M N) P). Finally, you can drop parentheses around abstracts, but not when they're part of an application. So you can abbreviate:
(\x (x y))
\x (x y)
but you should include the parentheses in:
(\x (x y)) z
z (\x (x y))
Dot notation Dot means "Insert a left parenthesis here, and the matching right parenthesis as far to the right as possible without creating unbalanced parentheses---so long as doing so would enclose an application or abstract not already wrapped in parentheses." Thus:
\x (\y (x y))
can be abbreviated as:
\x (\y. x y)
and that as:
\x. \y. x y
\x. \y. (x y) x
\x (\y ((x y) x))
This on the other hand:
(\x. \y. (x y)) x
((\x (\y (x y))) x)
The outermost parentheses were added because we have an application.
(\x (\y. ...)) because of the rule for dots. We didn't
insert any parentheses around the inner body of
\y. (x y) because they were
already there. That is, in expressions of the form
\y. (...), the dot abbreviates
nothing. It's harmless to write such a dot, though, and it can be conceptually
helpful especially in light of the next convention.
Similarly, we permit
\x. x, which is shorthand for
\x x, not for
\x (x), which
our syntax forbids. (The lambda evaluator however tolerates such expressions.)
Merging lambdas An expression of the form
(\x (\y M)), or equivalently,
(\x. \y. M), can be abbreviated as:
(\x y. M)
(\x (\y (\z M))) can be abbreviated as:
(\x y z. M)
Lambda terms represent functions
Let's pause and consider the following fundamental question: what is a
function? One popular answer is that a function can be represented by
a set of ordered pairs. This set is called the graph of the
function. If the ordered pair
(a, b) is a member of the graph of
that means that
f maps the argument
a to the value
f: a ↦
f (a) == b.
In order to count as a function (rather than as merely a more general relation), we require that the graph not contain two (distinct) ordered pairs with the same first element. That is, in order to count as a proper function, each argument must correspond to a unique result.
The lambda calculus seems to be wonderfully well-suited for representing functions. In fact, the untyped lambda calculus is Turing Complete (see Turing completeness): all (recursively computable) functions can be represented by lambda terms. Which, by most people's lights, means that all functions we can "effectively decide" --- that is, always apply in a mechanical way without requiring any ingenuity or insight, and be guaranteed of a correct answer after some finite number of steps --- can be represented by lambda terms. As we'll see, though, it will be fun (that is, not straightforward) unpacking how these things can be so "represented."
For some lambda terms, it is easy to see what function they represent:
(\x x)represents the identity function: given any argument
M, this function simply returns
((\x x) M) ~~> M.
(\x (x x))duplicates its argument (applies it to itself):
((\x (x x)) M) ~~> (M M)
(\x (\y (y x)))reorders its two arguments:
(((\x (\y (y x))) M) N) ~~> (N M)
(\x (\y x))throws away its second argument:
(((\x (\y x)) M) N) ~~> M
and so on. In order to get an intuitive feel for the power of the lambda calculus, note that duplicating, reordering, and deleting elements is all that it takes to simulate the behavior of a general word processing program. That means that if we had a big enough lambda term, it could take a representation of Emma as input and produce Hamlet as a result.
Some of these functions are so useful that we'll give them special
names. In particular, we'll call the identity function
I, and the function
(\x (\y x)) K (for "konstant":
K x is
a "constant function" that accepts any second argument
y and ignores
it, always returning
It is easy to see that distinct lambda expressions can represent the same function, considered as a mapping from input to outputs. Obviously:
both represent the same function, the identity function. However, we said in the advanced notes that we would be regarding these expressions as synactically equivalent, so they aren't yet really examples of distinct lambda expressions representing a single function. However, all three of these are distinct lambda expressions:
(\y x. y x) (\z z) (\x. (\z z) x) (\z z)
yet when applied to any argument
M, all of these will always return
M. So they
have the same extension. It's also true, though you may not yet be in a
position to see, that no other function can differentiate between them when
they're supplied as an argument to it. However, these expressions are all
The first two expressions are (beta-)convertible: in particular the first reduces to the second via a single instance of beta reduction. So they can be regarded as proof-theoretically equivalent even though they're not syntactically identical. However, the proof theory we've given so far doesn't permit you to reduce the second expression to the third. So these lambda expressions are non-equivalent.
In other words, we have here different (non-equivalent) lambda terms with the same extension. This introduces some tension between the popular way of thinking of functions as represented by (identical to?) their graphs or extensions, and the idea that lambda terms express functions. Well, perhaps the lambda terms are just a finer-grained way of expressing functions than extensions are?
As we'll see, though, there are even further sources of tension between the idea of functions-as-extensions and the idea of functions embodied in the lambda calculus.
One reason is that that general mathematical conception permits many uncomputable functions, but the lambda calculus can't express those.
More problematically, lambda terms express "functions" that can take themselves as arguments. If we wanted to represent that set-theoretically, and identified functions with their extensions, then we'd have to have some extension that contained (an ordered pair containing) itself as a member. Which we're not allowed to do in mainstream set-theory. But in the lambda calculus this is permitted and common --- and in fact will turn out to be indispensable.
Here are some simple examples. We can apply the identity function to itself:
(\x x) (\x x)
This is a redex that reduces to the identity function (of course).
We can apply the K function to another argument and itself:
K z K
(\x (\y x)) z (\x (\y x))
That reduces to just
As we'll see in coming weeks, some lambda terms will turn out to be impossible to associate with any extension. This is related to the previous point.
In fact it does turn out to be possible to represent the Lambda Calculus set-theoretically. But not in the straightforward way that identifies functions with their graphs. For years, it wasn't known whether it would be possible to do this. But then Dana Scott figured out how to do it in late 1969, that is, he formulated the first "denotational semantics" for this Lambda Calculus. Scott himself had expected that this wouldn't be possible to do. He argued for its unlikelihood in a paper he wrote only a month before the discovery.
(You can find an exposition of Scott's semantics in Chapters 15 and 16 of the Hindley & Seldin book we recommended, and an exposition of some simpler models that have since been discovered in Chapter 5 of the Hankin book, and Section 16F of Hindley & Seldin. But realistically, you really ought to wait a while and get more familiar with these systems before you'll be at all ready to engage with those ideas.)
We've characterized the rule of beta-reduction as a kind of "proof theory" for the Lambda Calculus. In fact, the usual proof theory is expressed in terms of convertibility rather than in terms of reduction; but it's natural to understand reduction as being the conceptually more fundamental notion. And there are some proof theories for this system that are expressed in terms of reduction.
To use a linguistic analogy, you can think of what we're calling "proof theory" as a kind of syntactic transformation. The sentences John saw Mary and Mary was seen by John are not syntactically identical, yet (on some theories) one can be derived from the other. The key element in the analogy is that this kind of syntactic derivation is supposed to preserve meaning, so that the two sentences mean (roughly) the same thing.
There's an extension of the proof-theory we've presented so far which does permit a further move of just that sort. It would permit us to also count these functions:
(\x. (\z z) x) (\z z)
as equivalent. This additional move is called eta-reduction. It's
crucial to eta-reduction that the outermost variable binding in the
abstract we begin with (here,
\x) be of a variable that occurs free
exactly once in the body of that abstract, and that that free occurrence be the rightmost outermost constituent.
(\x (\y (y x))
can't be eta-reduced, because the rightmost outermost constituent is not
(\y (y x).
In the extended proof theory/theories we get be permitting eta-reduction/conversion as well as beta-reduction, all computable functions with the same extension do turn out to be equivalent, that is, convertible.
However, we still shouldn't assume we're working with functions traditionally conceived as just sets of ordered pairs, for the other reasons sketched above.
The analogy with
In our basic functional programming language, we used
expressions to assign values to variables. For instance,
let x match 2 in (x, x)
evaluates to the ordered pair
(2, 2). It may be helpful to think of
a redex in the lambda calculus as a particular sort of
((\x BODY) ARG)
is analogous to
let x match ARG in BODY
This analogy should be treated with caution. For one thing, our
allowed us to define recursive functions in which the name
x appeared within
ARG, but we wanted it there to be bound to the very same
ARG. We're not
ready to deal with recursive functions in the lambda calculus yet.
For another, we defined
let in terms of values: we said that the variable
x was bound to whatever value
ARG evaluated to. At this point, it's
not clear what the values are in the lambda calculus. We only have expressions.
(Though there was a suggestive remark earlier about some of the expressions but
not others being called "value terms".)
So perhaps we should translate
((\x BODY) ARG) in purely syntactic terms,
x be replaced by
Most problematically, in the lambda
calculus, an abstract such as
(\x (x x)) is perfectly well-formed
and coherent, but it is not possible to write a
let expression that
does not have an
ARG. That would be like:
let x match missing
in x x
Nevertheless, the correspondence is close enough that it can guide our intuition.