Each variable is an expression. For any expressions M and N and variable a, the following are also expressions:Variables:`x`

,`y`

,`z`

...

We'll tend to writeAbstract:`(λa M)`

`(λa M)`

as just `(\a M)`, so we don't have to write out the markup code for the `λ`

. You can yourself write `(λa M)`

or `(\a M)` or `(lambda a M)`.
Examples of expressions: 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) 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**. The rule of beta-reduction permits a transition from that expression to the following: 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 is a free occurrence? > An occurrence of a variable `a` is **bound** in T if 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 is **free** if it's not bound. For instance: > T is defined to be `(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. To read further: * [[!wikipedia Free variables and bound variables]] Here's an example of beta-reduction: ((\x (y x)) z) beta-reduces to: (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 symbolApplication:`(M N)`

`→`

for one-step contraction, and the symbol `↠`

for zero-or-more step reduction. Hindley and Seldin use `⊳`_{1}

and `⊳`

.
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 `≡`

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:
> T is defined to be `(M N)`.
We'll regard the following two expressions:
(\x (x y))
(\z (z y))
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.
Note that neither of those expressions are identical to:
(\x (x w))
because here it's a free variable that's been changed. Nor are they identical to:
(\y (y y))
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.)
**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))
as:
\x (x y)
but you should include the parentheses in:
(\x (x y)) z
and:
z (\x (x y))
**Dot notation** Dot means "put a left paren here, and put the right
paren as far the right as possible without creating unbalanced
parentheses". So:
\x (\y (x y))
can be abbreviated as:
\x (\y. x y)
and that as:
\x. \y. x y
This:
\x. \y. (x y) x
abbreviates:
\x (\y ((x y) x))
This on the other hand:
(\x. \y. (x y)) x
abbreviates:
((\x (\y (x y))) x)
**Merging lambdas** An expression of the form `(\x (\y M))`, or equivalently, `(\x. \y. M)`, can be abbreviated as:
(\x y. M)
Similarly, `(\x (\y (\z M)))` can be abbreviated as:
(\x y z. M)
Lambda terms represent functions
--------------------------------
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:
> `(\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 (\y x))` throws away its second argument:
`(((\x (\y x)) M) N) ~~> M`
and so on.
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)
and:
(\z z)
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:
(\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 *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.
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).
Booleans and pairs
==================
Our definition of these is reviewed in [[Assignment1]].
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.
There's also a (slow, bare-bones, but perfectly adequate) version of Scheme available for online use at