.
@@ 66,7 +280,7 @@ It's often said that dynamic systems are distinguished because they are the ones
false and * = false
false and false = false
And then we'd notice that ` * and false` has a different intepretation than `false and *`. (The same phenomenon is already present with the mateial conditional in bivalent logics; but seeing that a nonsymmetric semantics for `and` is available even for functional languages is instructive.)
+And then we'd notice that `* and false` has a different intepretation than `false and *`. (The same phenomenon is already present with the material conditional in bivalent logics; but seeing that a nonsymmetric semantics for `and` is available even for functional languages is instructive.)
Another way in which order can matter that's present even in functional languages is that the interpretation of some complex expressions can depend on the order in which subexpressions are evaluated. Evaluated in one order, the computations might never terminate (and so semantically we interpret them as having "the bottom value"we'll discuss this). Evaluated in another order, they might have a perfectly mundane value. Here's an example, though we'll reserve discussion of it until later:
@@ 132,6 +346,11 @@ Rosetta Stone
Here's how it looks to say the same thing in various of these languages.
+The following site may be useful; it lets you run a Scheme interpreter inside your web browser:
+
+* [Try Scheme in your web browser](http://tryscheme.sourceforge.net/)
+
+
1. Binding suitable values to the variables `three` and `two`, and adding them.
In Scheme:
@@ 309,7 +528,7 @@ Here's how it looks to say the same thing in various of these languages.
(let* [(bar (lambda (x) B))] M)
 then wherever `bar` occurs in `M` (and isn't rebound by a more local "let" or "lambda"), it will be interpreted as the function `(lambda (x) B)`.
+ then wherever `bar` occurs in `M` (and isn't rebound by a more local `let` or `lambda`), it will be interpreted as the function `(lambda (x) B)`.
Similarly, in OCaml:
@@ 369,8 +588,7 @@ Here's how it looks to say the same thing in various of these languages.
let x = A;;
... rest of the file or interactive session ...
 It's easy to be lulled into thinking this is a kind of imperative construction. *But it's not!* It's really just a shorthand for the compound "let"expressions we've already been looking at, taking the maximum syntactically permissible scope. (Compare the "dot" convention in the lambda calculus, discussed above.)

+ It's easy to be lulled into thinking this is a kind of imperative construction. *But it's not!* It's really just a shorthand for the compound `let`expressions we've already been looking at, taking the maximum syntactically permissible scope. (Compare the "dot" convention in the lambda calculus, discussed above.)
9. Some shorthand
@@ 442,9 +660,8 @@ Here's how it looks to say the same thing in various of these languages.
and there's no more mutation going on there than there is in:
  ∀x. (F x or ∀x (not (F x)))

+ ∀x. (F x or ∀x (not (F x)))
+
When a previouslybound variable is rebound in the way we see here, that's called **shadowing**: the outer binding is shadowed during the scope of the inner binding.
@@ 521,7 +738,7 @@ Or even:
(define foo B)
(foo 2)
don't involve any changes or sequencing in the sense we're trying to identify. As we said, these programs are just syntactic variants of (single) compound syntactic structures involving "let"s and "lambda"s.
+don't involve any changes or sequencing in the sense we're trying to identify. As we said, these programs are just syntactic variants of (single) compound syntactic structures involving `let`s and `lambda`s.
Since Scheme and OCaml also do permit imperatival constructions, they do have syntax for genuine sequencing. In Scheme it looks like this:
@@ 561,245 +778,4 @@ We'll discuss this more as the seminar proceeds.
Basics of Lambda Calculus
=========================

The lambda calculus we'll be focusing on for the first part of the course has no types. (Some prefer to say it instead has a single typebut if you say that, you have to say that functions from this type to this type also belong to this type. Which is weird.)

Here is its syntax:


Variables: x
, y
, z
...


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


Abstract: (λa M)


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 (λa M)
or `(\a M)` or `(lambda a M)`.


Application: (M N)


Some authors reserve the term "term" for just variables and abstracts. We won't participate in that convention; we'll probably just say "term" and "expression" indiscriminately for expressions of any of these three forms.

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 **betareduction** or "betacontraction". 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 "betareducible expression." `(\a M)` is called the **head** of the redex; `N` is called the **argument**, and `M` is called the **body**.

The rule of betareduction 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 being 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.

Here's an example of betareduction:

 ((\x (y x)) z)

betareduces 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 symbol →
for onestep contraction, and the symbol ↠
for zeroormore 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 **betaconvertible**. 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 metatheory for the lambda calculus. 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.


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.)


**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:

 (\x (\y. (z y) z))

would abbreviate:

 (\x (\y ((z y) z)))

This on the other hand:

 (\x (\y. z y) z)

would abbreviate:

 (\x (\y (z y)) z)

**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)

**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


All (recursively computable) functions can be represented by lambda
terms (the untyped lambda calculus is Turing complete). 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 prooftheoretically 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 nonequivalent.

There's an extension of the prooftheory 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 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.





1. Declarative vs imperatival models of computation.
2. Variety of ways in which "order can matter."
3. Variety of meanings for "dynamic."
4. Schoenfinkel, Curry, Church: a brief history
5. Functions as "firstclass values"
6. "Curried" functions

1. Beta reduction
1. Encoding pairs (and triples and ...)
1. Encoding booleans



