-Basics of Lambda Calculus
-=========================
+## Syntax of 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 that we will explore in some detail, is the "pure"
-Lambda Calculus. And actually, there are many variations even in the
-pure Lambda Calculus. But all of the variations share a strong family
+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.
-The lambda calculus we'll be focusing on for the first part of the
-course has no types. Some prefer to say it does have types, 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
+> Fussy note: calling this/these 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
+this 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 this
+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:
<blockquote>
<strong>Abstract</strong>: <code>(λa M)</code>
</blockquote>
-We'll tend to write <code>(λa M)</code> as just `(\a M)`, so we don't have to write out the markup code for the <code>λ</code>. You can yourself write <code>(λa M)</code> or `(\a M)` or `(lambda a M)`.
+We'll tend to write <code>(λa M)</code> as just `(\a M)`, so we
+don't have to write out the markup code for the
+<code>λ</code>. You can yourself write <code>(λa
+M)</code> or `(\a M)` or `(lambda a M)`.
<blockquote>
<strong>Application</strong>: <code>(M N)</code>
</blockquote>
+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 --> ( lambda 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
+form `(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)))
-Examples of expressions:
- x
- (y x)
- (x x)
- (\x y)
- (\x x)
- (\x (\y x))
- (x (\x x))
- ((\x (x x)) (\x (x x)))
+## Beta-Reduction ##
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)
+ ((\a M) N)
-that is, an application of an abstract to some other expression. This compound form is called a **redex**, meaning it's a "beta-reducible expression." `(\a M)` is called the **head** of the redex; `N` is called the **argument**, and `M` is called the **body**.
+This is an application whose first element is an abstract. This
+compound form is called a **redex**, meaning it's a "beta-reducible
+expression." `(\a M)` is called the **head** of the redex; `N` is
+called the **argument**, and `M` is called the **body**.
The rule of beta-reduction permits a transition from that expression to the following:
- M [a:=N]
+ M [a <-- N]
-What this means is just `M`, with any *free occurrences* inside `M` of the variable `a` replaced with the term `N`.
+What this means is just `M`, with any *free occurrences* inside `M` of
+the variable `a` replaced with the term `N` (--- "without capture", which
+we'll explain in the [[advanced notes|FIXME]]).
What is a free occurrence?
-> An occurrence of a variable `a` is **bound** in T if T has the form `(\a N)`.
+> Any occurrence of a variable `a` is **bound** in T when T has the form `(\a N)`.
-> If T has the form `(M N)`, any occurrences of `a` that are bound in `M` are also bound in T, and so too any occurrences of `a` that are bound in `N`.
+> An occurrence of a variable `a` is **bound** in T when T has the form `(\b
+N)` --- that is, with a *different* variable `b` --- just in case that
+occurrence of `a` is bound in the subexpression `N`.
-> An occurrence of a variable is **free** if it's not bound.
+> When T has the form `(M N)`, any occurrences of `a` that are bound in the
+subexpression `M` are also bound in T, and so too any occurrences of `a` that
+are bound in the subexpression `N`.
-For instance:
+> An occurrence of a variable is **free** if it's not bound.
+For instance consider the following term `T`:
-> T is defined to be `(x (\x (\y (x (y z)))))`
+ (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 `x` occurs within a form that begins with `\x`, so it is bound as well. The occurrence of `y` is bound; and the occurrence of `z` is free.
+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 `x` occurs
+within a form `(\x (\y ...))` that begins with `\x`, so it is bound as well. The
+occurrence of `y` is bound; and the occurrence of `z` is free.
To read further:
-* [[!wikipedia Free variables and bound variables]]
+* [[!wikipedia Free variables and bound variables]]
Here's an example of beta-reduction:
- ((\x (y x)) z)
+ ((\x (y x)) z)
beta-reduces to:
- (y z)
+ (y z)
We'll write that like this:
- ((\x (y x)) z) ~~> (y z)
-
-Different authors use different notations. 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 M to N by one or more reduction steps. Hankin uses the symbol <code><big><big>→</big></big></code> for one-step contraction, and the symbol <code><big><big>↠</big></big></code> for zero-or-more step reduction. Hindley and Seldin use <code><big><big><big>⊳</big></big></big><sub>1</sub></code> and <code><big><big><big>⊳</big></big></big></code>.
-
-When M and N are such that there's some P that M reduces to by zero or more steps, and that N also reduces to by zero or more steps, then we say that M and N are **beta-convertible**. We'll write that like this:
-
- M <~~> N
-
-This is what plays the role of equality in the lambda calculus. Hankin uses the symbol `=` for this. So too do Hindley and Seldin. Personally, I keep confusing that with the relation to be described next, so let's use this notation instead. Note that `M <~~> N` doesn't mean that each of `M` and `N` are reducible to each other; that only holds when `M` and `N` are the same expression. (Or, with our convention of only saying "reducible" for one or more reduction steps, it never holds.)
-
-In the metatheory, it's also sometimes useful to talk about formulas that are syntactically equivalent *before any reductions take place*. Hankin uses the symbol <code>≡</code> for this. So too do Hindley and Seldin. We'll use that too, and will avoid using `=` when discussing the metatheory. Instead we'll use `<~~>` as we said above. When we want to introduce a stipulative definition, we'll write it out longhand, as in:
+ ((\x (y x)) z) ~~> (y z)
-> T is defined to be `(M N)`.
+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:
-We'll regard the following two expressions:
+ M ~~> N
- (\x (x y))
+we'll mean that you can get from M to N by one or more reduction
+steps.
- (\z (z y))
+When `M` and `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 `M` and `N` are **beta-convertible**. We'll write that
+like this:
-as syntactically equivalent, since they only involve a typographic change of a bound variable. Read Hankin section 2.3 for discussion of different attitudes one can take about this.
+ M <~~> N
-Note that neither of those expressions are identical to:
+More details about the notation and metatheory of
+the lambda calculus here:
- (\x (x w))
+* [[topics/_week2_lambda_calculus_fine_points.mdwn]]
-because here it's a free variable that's been changed. Nor are they identical to:
- (\y (y y))
+## Shorthand ##
-because here the second occurrence of `y` is no longer free.
-
-There is plenty of discussion of this, and the fine points of how substitution works, in Hankin and in various of the tutorials we've linked to about the lambda calculus. We expect you have a good intuitive understanding of what to do already, though, even if you're not able to articulate it rigorously.
-
-* [More discussion in week 2 notes](/week2/#index1h1)
-
-
-Shorthand
----------
-
-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.)
+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))
as:
- \x (x y)
+ \x (x y)
but you should include the parentheses in:
- (\x (x y)) z
+ (\x (x y)) z
and:
- z (\x (x y))
+ z (\x (x y))
-**Dot notation** Dot means "put a left paren here, and put the right
+**Dot notation** Dot means "assume a left paren here, and the matching right
paren as far the right as possible without creating unbalanced
parentheses". So:
- \x (\y (x y))
+ \x (\y (x y))
can be abbreviated as:
- \x (\y. x y)
+ \x (\y. x y)
and that as:
- \x. \y. x y
+ \x. \y. x y
This:
- \x. \y. (x y) x
+ \x. \y. (x y) x
abbreviates:
- \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
abbreviates:
- ((\x (\y (x y))) x)
+ ((\x (\y (x y))) x)
+
+We didn't have to insert any parentheses around the inner body of `\y. (x y)` because they were already there.
**Merging lambdas** An expression of the form `(\x (\y M))`, or equivalently, `(\x. \y. M)`, can be abbreviated as:
- (\x y. M)
+ (\x y. M)
Similarly, `(\x (\y (\z M)))` can be abbreviated as:
- (\x y z. M)
+ (\x y z. M)
+
+## Lambda terms represent functions ##
-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 `f`,
+that means that `f` maps the argument `a` to the value `b`. In
+symbols, `f: a` ↦ `b`, or `f (a) == b`.
-The untyped lambda calculus is Turing complete: all (recursively computable) functions can be represented by lambda terms. For some lambda terms, it is easy to see what function they represent:
+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.
-> `(\x x)` represents the identity function: given any argument `M`, this function
+The lambda calculus seems to be wonderfully well-suited for
+representing functions. In fact, the untyped
+lambda calculus is Turing Complete (see [[!wikipedia 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, compute in a mechanical way without requiring any ingenuity or insight ---
+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 `M`: `((\x x) M) ~~> M`.
-> `(\x (x x))` duplicates its argument:
-`((\x (x x)) M) ~~> (M M)`
+> `(\x (x x))` duplicates its argument (applies it to itself):
+`((\x (x x)) M) ~~> (M M)` <!-- **M** or ω; W is \uv.uvv -->
+
+> `(\x (\y (y x)))` reorders its two arguments:
+`(((\x (\y (y x))) M) N) ~~> (N M)` <!-- **T**; C is \uvx.uxv -->
-> `(\x (\y x))` throws away its second argument:
-`(((\x (\y x)) M) N) ~~> M`
+> `(\x (\y x))` throws away its second argument:
+`(((\x (\y x)) M) N) ~~> M` <!-- **K** -->
-and so on.
+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 `(\x x)`
+**I**, and the function `(\x (\y x))` **K** (for "konstant": <code><b>K</b> x</code> is
+a "constant function" that accepts any second argument `y` and ignores
+it, always returning `x`).
It is easy to see that distinct lambda expressions can represent the same
function, considered as a mapping from input to outputs. Obviously:
- (\x x)
+ (\x x)
and:
- (\z z)
+ (\z z)
+
+both represent the same function, the identity function. However, we said
+(FIXME 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
+syntactically distinct.
+
+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.
+
+
+1. One reason is that that general
+mathematical conception permits many *uncomputable* functions, but the
+lambda calculus can't express those.
+
+2. More problematically, some 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:
+
+ > <code><b>K</b> z <b>K</b></code>
+
+ That is:
+
+ (\x (\y x)) z (\x (\y x))
+
+ That reduces to just `z`.
+
+3. 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
+[[!wikipedia 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 it be in the
+rightmost position.
-both represent the same function, the identity function. However, we said above 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:
+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.
- (\y x. y x) (\z z)
+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.
- (\x. (\z z) x)
- (\z z)
+## The analogy with `let` ##
-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 syntactically distinct.
+In our basic functional programming language, we used `let`
+expressions to assign values to variables. For instance,
-The first two expressions are *convertible*: in particular the first reduces to the second. 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.
+ let x match 2
+ in (x, x)
-There's an extension of the proof-theory we've presented so far which does permit this further move. And in that extended proof theory, all computable functions with the same extension do turn out to be equivalent (convertible). However, at that point, we still won't be working with the traditional mathematical notion of a function as a set of ordered pairs. One reason is that the latter but not the former permits many uncomputable functions. A second reason is that the latter but not the former prohibits functions from applying to themselves. We discussed this some at the end of Monday's meeting (and further discussion is best pursued in person).
+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 `let`
+construction.
+ ((\x BODY) ARG) is analogous to
+ let x match ARG
+ in BODY
-Booleans and pairs
-==================
+This analogy should be treated with caution. For one thing, our `letrec`
+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.
-Our definition of these is reviewed in [[Assignment1]].
+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,
+like: "let `x` be replaced by `ARG` in `BODY`"?
-It's possible to do the assignment without using a Scheme interpreter, however
-you should take this opportunity to [get Scheme installed on your
-computer](/how_to_get_the_programming_languages_running_on_your_computer), and
-[get started learning Scheme](/learning_scheme). It will help you test out
-proposed answers to the assignment.
+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`
-There's also a (slow, bare-bones, but perfectly adequate) version of Scheme available for online use at <http://tryscheme.sourceforge.net/>.
+Nevertheless, the correspondence is close enough that it can guide our
+intuition.