```
T ≡ (\x. x y) z ≡ (\z. z y) z
```

This:
T ~~> z y
means that T beta-reduces to `z y`. This:
M <~~> T
means that M and T are beta-convertible, that is, that there's something they both reduce to in zero or more steps.
Combinators and Combinatorial Logic
===================================
Lambda expressions that have no free variables are known as **combinators**. Here are some common ones:
> **I** is defined to be `\x x`
> **K** is defined to be `\x y. x`, That is, it throws away its second argument. So `K x` is a constant function from any (further) argument to `x`. ("K" for "constant".) Compare K to our definition of **true**.
> **get-first** was our function for extracting the first element of an ordered pair: `\fst snd. fst`. Compare this to **K** and **true** as well.
> **get-second** was our function for extracting the second element of an ordered pair: `\fst snd. snd`. Compare this to our definition of **false**.
> **ω** is defined to be: `\x. x x`
It's possible to build a logical system equally powerful as the lambda calculus (and readily intertranslatable with it) using just combinators, considered as atomic operations. Such a language doesn't have any variables in it: not just no free variables, but no variables at all.
One can do that with a very spare set of basic combinators. These days the standard base is just three combinators: K and I from above, and also one more, **S**, which behaves the same as the lambda expression `\f g x. f x (g x)`. behaves. But it's possible to be even more minimalistic, and get by with only a single combinator. (And there are different single-combinator bases you can choose.)
These systems are Turing complete. In other words: every computation we know how to describe can be represented in a logical system consisting of only a single primitive operation!
Here's more to read about combinatorial logic:
* [[!wikipedia Combinatory logic]] at Wikipedia
* [Combinatory logic](http://plato.stanford.edu/entries/logic-combinatory/) at the Stanford Encyclopedia of Philosophy
* [[!wikipedia SKI combinatory calculus]]
* [[!wikipedia B,C,K,W system]]
* [Chris Barker's Iota and Jot](http://semarch.linguistics.fas.nyu.edu/barker/Iota/)
* Jeroen Fokker, "The Systematic Construction of a One-combinator Basis for Lambda-Terms" Formal Aspects of Computing 4 (1992), pp. 776-780.
`ω`

; so this can also be written:
`ω ω`

This compound expression---the self-application of `ω`

---is named Ω. It has the form of an application of an abstract (`ω`

) to an argument (which also happens to be `ω`

), so it's a redex and can be reduced. But when we reduce it, we get `ω ω`

again. So there's no stage at which this expression has been reduced to a point where it can't be reduced any further. In other words, evaluation of this expression "never terminates." (This is the standard language, however it has the unfortunate connotation that evaluation is a process or operation that is performed in time. You shouldn't think of it like that. Evaluation of this expression "never terminates" in the way that the decimal expansion of π never terminates. These are static, atemporal facts about their mathematical properties.)
There are infinitely many formulas in the lambda calculus that have this same property. Ω is the syntactically simplest of them. In our meta-theory, it's common to assign such formulas a special value, `⊥`

, pronounced "bottom." When we get to discussing types, you'll see that this value is counted as belonging to every type. To say that a formula has the bottom value means that the computation that formula represents never terminates and so doesn't evaluate to any orthodox, computed value.
From a "Fregean" or "weak Kleene" perspective, if any component of an expression fails to be evaluable (to an orthodox, computed value), then the whole expression should be unevaluable as well.
However, in some such cases it seems *we could* sensibly carry on evaluation. For instance, consider:
```
(\x. y) (ω ω)
```

Should we count this as unevaluable, because the reduction of `(ω ω)`

never terminates? Or should we count it as evaluating to `y`?
This question highlights that there are different choices to make about how evaluation or computation proceeds. It's helpful to think of three questions in this neighborhood:
> Q1. When arguments are complex, as `(ω ω)`

is, do we reduce them before substituting them into the abstracts to which they are arguments, or later?
> Q2. Are we allowed to reduce inside abstracts? That is, can we reduce:
> (\x y. x z) (\x. x)
> only this far:
> \y. (\x. x) z
> or can we continue reducing to:
> \y. z
> Q3. Are we allowed to "eta-reduce"? That is, can we reduce expressions of the form:
> \x. M x
> where x does not occur free in `M`, to `M`?
With regard to Q3, it should be intuitively clear that `\x. M x` and `M` will behave the same with respect to any arguments they are given. It can also be proven that no other functions can behave differently with respect to them. However, the logical system you get when eta-reduction is added to the proof theory is importantly different from the one where only beta-reduction is permitted.
MORE on extensionality
If we answer Q2 by permitting reduction inside abstracts, and we also permit eta-reduction, then where neither `y` nor `z` occur in M, this:
\x y z. M y z
will eta-reduce by two steps to:
\x. M
The evaluation strategy which answers Q1 by saying "reduce arguments first" is known as **call-by-value**. The evaluation strategy which answers Q1 by saying "substitute arguments in unreduced" is known as **call-by-name** or **call-by-need** (the difference between these has to do with efficiency, not semantics).
When one has a call-by-value strategy that also permits reduction to continue inside unapplied abstracts, that's known as "applicative order" reduction. When one has a call-by-name strategy that permits reduction inside abstracts, that's known as "normal order" reduction. Consider an expression of the form:
((A B) (C D)) (E F)
Its syntax has the following tree:
((A B) (C D)) (E F)
/ \
/ \
((A B) (C D)) \
/\ (E F)
/ \ /\
/ \ E F
(A B) (C D)
/\ /\
/ \ / \
A B C D
Applicative order evaluation does what's called a "post-order traversal" of the tree: that is, we always go down when we can, first to the left, and we process a node only after processing all its children. So `(C D)` gets processed before `((A B) (C D))` does, and `(E F)` gets processed before `((A B) (C D)) (E F)` does.
Normal order evaluation, on the other hand, will substitute the expresion `(C D)` into the abstract that `(A B)` evaluates to, without first trying to compute what `(C D)` evaluates to. That computation may be done later.
With normal-order evaluation (or call-by-name more generally), if we have an expression like:
(\x. y) (C D)
the computation of `(C D)` won't ever have to be performed. Instead, `(\x. y) (C D)` reduces directly to `y`. This is so even if `(C D)` is the non-evaluable `(ω ω)`

!
Call-by-name evaluation is often called "lazy." Call-by-value evaluation is also often called "eager" or "strict". Some authors say these terms all have subtly different technical meanings, but I haven't been able to figure out what it is. Perhaps the technical meaning of "strict" is what I above called the "Fregean" or "weak Kleene" perspective: if any argument of a function is non-evaluable or non-normalizing, so too is the application of the function to that argument.
Most programming languages, including Scheme and OCaml, use the call-by-value evaluation strategy. (But they don't permit evaluation to continue inside an unappplied function.) There are techniques for making them model call-by-name evaluation, when necessary. But by default, arguments will always be evaluated before being bound to the parameters (the `\x`s) of a function.
For languages like Scheme that permit functions to take more than one argument at a time, a further question arises: whether the multiple arguments are evaluated left-to-right, or right-to-left, or nothing is guaranteed about what order they are evaluated in. Different languages make different choices about this.
Some functional programming languages, such as Haskell, use the call-by-name evaluation strategy.
The lambda calculus can be evaluated either way. You have to decide what the rules shall be.
As we'll see in several weeks, there are techniques for *forcing* call-by-value evaluation of a computation, and also techniques for forcing call-by-name evaluation. If you liked, you could even have a nested hierarchy, where blocks at each level were forced to be evaluated in alternating ways.
Call-by-value and call-by-name have different pros and cons.
One important advantage of normal-order evaluation in particular is that it can compute orthodox values for:
Indeed, it's provable that if there's *any* reduction path that delivers a value for a given expression, the normal-order evalutation strategy will terminate with that value. An expression is said to be in **normal form** when it's not possible to perform any more reductions. (EVEN INSIDE ABSTRACTS?) There's a sense in which you *can't get anything more out of*`(\x. y) (ω ω)`

`\z. (\x. y) (ω ω)`

`ω ω`

, but it's not in normal form because it still has the form of a redex.
A computational system is said to be **confluent**, or to have the **Church-Rosser** or **diamond** property, if, whenever there are multiple possible evaluation paths, those that terminate always terminate in the same value. In such a system, the choice of which sub-expressions to evaluate first will only matter if some of them but not others might lead down a non-terminating path.
The untyped lambda calculus is confluent. So long as a computation terminates, it always terminates in the same way. It doesn't matter which order the sub-expressions are evaluated in.
A computational system is said to be **strongly normalizing** if every permitted evaluation path is guaranteed to terminate. The untyped lambda calculus is not strongly normalizing: `ω ω`

doesn't terminate by any evaluation path; and `(\x. y) (ω ω)`

terminates only by some evaluation paths but not by others.
But the untyped lambda calculus enjoys some compensation for this weakness. It's Turing complete! It can represent any computation we know how to describe. (That's the cash value of being Turing complete, not the rigorous definition. There is a rigrous definition. However, we don't know how to rigorously define "any computation we know how to describe.") And in fact, it's been proven that you can't have both. If a computational system is Turing complete, it cannot be strongly normalizing.
A computational system is said to be **weakly normalizing** if there's always guaranteed to be *at least one* evaluation path that terminates. The untyped lambda calculus is not weakly normalizing either, as we've seen.
The *typed* lambda calculus that linguists traditionally work with, on the other hand, is strongly normalizing. (And as a result, is not Turing complete.) It has expressive power (concerning types) that the untyped lambda calculus lacks, but it is also unable to represent some (terminating!) computations that the untyped lambda calculus can represent.
Other more-powerful type systems we'll look at in the course will also fail to be Turing complete, though they will turn out to be pretty powerful.
Further reading:
* [[!wikipedia Evaluation strategy]]
* [[!wikipedia Eager evaluation]]
* [[!wikipedia Lazy evaluation]]
* [[!wikipedia Strict programming language]]* [[!wikipedia Church-Rosser theorem]] * [[!wikipedia Normalization property]] * [[!wikipedia Turing completeness]] Decidability ============ The question whether two formulas are syntactically equal is "decidable": we can construct a computation that's guaranteed to always give us the answer. What about the question whether two formulas are convertible? Well, to answer that, we just need to reduce them to normal form, if possible, and check whether the results are syntactically equal. The crux is that "if possible." Some computations can't be reduced to normal form. Their evaluation paths never terminate. And if we just kept trying blindly to reduce them, our computation of what they're convertible to would also never terminate. So it'd be handy to have some way to check in advance whether a formula has a normal form: whether there's any evaluation path for it that terminates. Is it possible to do that? Sure, sometimes. For instance, check whether the formula is syntactically equal to Ω. If it is, it never terminates. But is there any method for doing this in general---for telling, of any given computation, whether that computation would terminate? Unfortunately, there is not. Church proved this in 1936; Turing also essentially proved it at the same time. Geoff Pullum gives a very reader-friendly outline of the proofs here: * [Scooping the Loop Snooper](http://www.cl.cam.ac.uk/teaching/0910/CompTheory/scooping.pdf), a proof of the undecidability of the halting problem in the style of Dr Seuss by Geoffrey K. Pullum ##[[Lists and Numbers]]##