+Variables:+ +Each variable is an expression. For any expressions M and N and variable a, the following are also expressions: + +x
,y
,z
... +
+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.
+
+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.
+
+
+
+
+
+
Declarative/functional vs Imperatival/dynamic models of computation
===================================================================
Many of you, like us, will have grown up thinking the paradigm of computation is a sequence of changes. Let go of that. It will take some care to separate the operative notion of "sequencing" here from other notions close to it, but once that's done, you'll see that languages that have no significant notions of sequencing or changes are Turing complete: they can perform any computation we know how to describe. In itself, that only puts them on equal footing with more mainstream, imperatival programming languages like C and Java and Python, which are also Turing complete. But further, the languages we want you to become familiar with can reasonably be understood to be more fundamental. They embody the elemental building blocks that computer scientists use when reasoning about and designing other languages.
Jim offered the metaphor: think of imperatival languages, which include "mutation" and "sideeffects" (we'll flesh out these keywords as we proceeed), as the pate of computation. We want to teach you about the meat and potatoes, where as it turns out there is no sequencing and no changes. There's just the evaluation or simplification of complex expressions.
+Jim offered the metaphor: think of imperatival languages, which include "mutation" and "sideeffects" (we'll flesh out these keywords as we proceeed), as the pâté of computation. We want to teach you about the meat and potatoes, where as it turns out there is no sequencing and no changes. There's just the evaluation or simplification of complex expressions.
Now, when you ask the Scheme interpreter to simplify an expression for you, that's a kind of dynamic interaction between you and the interpreter. You may wonder then why these languages should not also be understood imperatively. The difference is that in a purely declarative or functional language, there are no dynamic effects in the language itself. It's just a static semantic fact about the language that one expression reduces to another. You may have verified that fact through your dynamic interactions with the Scheme interpreter, but that's different from saying that there are dynamic effects in the language itself.
@@ 45,9 +266,9 @@ Many of the slogans and keywords we'll encounter in discussions of these issues
For example, you'll encounter the claim that declarative languages are distinguished by their **referential transparency.** What's meant by this is not always exactly the same, and as a cluster, it's related to but not the same as this means for philosophers and linguists.
The notion of "function" that we'll be working with will be one that, by default, sometimes counts as nonidentical functions that map all their inputs to the very same outputs. For example, two functions from jumbled decks of cards to sorted decks of cards may use different algorithms and hence be different functions.
+The notion of **function** that we'll be working with will be one that, by default, sometimes counts as nonidentical functions that map all their inputs to the very same outputs. For example, two functions from jumbled decks of cards to sorted decks of cards may use different algorithms and hence be different functions.
It's possible to enhance the lambda calculus so that functions do get identified when they map all the same inputs to the same outputs. This is called making the calculus **extensional**. Church called languages which didn't do this "intensional." If you try to understand this in terms of functions from worlds to extensions (an idea also associated with Church), you will hurt yourself. So too if you try to understand it in terms of mental stereotypes, another notion sometimes designated by "intension."
+It's possible to enhance the lambda calculus so that functions do get identified when they map all the same inputs to the same outputs. This is called making the calculus **extensional**. Church called languages which didn't do this **intensional**. If you try to understand that kind of "intensionality" in terms of functions from worlds to extensions (an idea also associated with Church), you may hurt yourself. So too if you try to understand it in terms of mental stereotypes, another notion sometimes designated by "intension."
It's often said that dynamic systems are distinguished because they are the ones in which **order matters**. However, there are many ways in which order can matter. If we have a trivalent boolean system, for exampleeasily had in a purely functional calculuswe might choose to give a truthtable like this for "and":
@@ 86,243 +307,473 @@ sense such that only so matters in imperatival languages.
Finally, you'll see the term **dynamic** used in a variety of ways in the literature for this course:
 * dynamic versus static typing
+* dynamic versus static typing
 * dynamic versus lexical scoping
+* dynamic versus lexical scoping
 * dynamic versus static control operators
+* dynamic versus static control operators
 * finally, we're used ourselves to talking about dynamic versus static semantics
+* finally, we're used ourselves to talking about dynamic versus static semantics
For the most part, these uses are only loosely connected to each other. We'll tend to use "imperatival" to describe the kinds of semantic properties made available in dynamic semantics, languages which have robust notions of sequencing changes, and so on.
Map
===
+
+
+
+
+
+
+
+
+
+
+
+Scheme (functional part)  OCaml (functional part)  C, Java, Pasval +Scheme (imperative part) +OCaml (imperative part) 
lambda calculus +combinatorial logic  
 Turing complete   
+  more advanced type systems, such as polymorphic types +  + 
+  simplytyped lambda calculus (what linguists mostly use) +  + 
 Abstract: ( λa M )
+ In the lambda calculus: here we're on our own, we don't have predefined constants like `+` and `3` and `2` to work with. We've got to build up everything from scratch. We'll be seeing how to do that over the next weeks.
 Application: ( M N )

+ But supposing you had constructed appropriate values for `+` and `3` and `2`, you'd place them in the ellided positions in:
We'll tend to write ( λa M )
as just `( \a M )`.
+ (((\three (\two ((... three) two))) ...) ...)
+
+ In an ordinary imperatival language like C:
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.
+ int three = 3;
+ int two = 2;
+ three + two;
Examples of expressions:
+2. Mutation
 x
 (y x)
 (x x)
 (\x y)
 (\x x)
 (\x (\y x))
 (x (\x x))
 ((\x (x x)) (\x (x x)))
+ In C this looks almost the same as what we had before:
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:
+ int x = 3;
+ x = 2;
 ((\a M) N)
+ Here we first initialize `x` to hold the value 3; then we mutate `x` to hold a new value.
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**.
+ In (the imperatival part of) Scheme, this could be done as:
The rule of betareduction permits a transition from that expression to the following:
+ (let ((x (box 3)))
+ (setbox! x 2))
 M {a:=N}
+ In general, mutating operations in Scheme are named with a trailing `!`. There are other imperatival constructions, though, like `(print ...)`, that don't follow that convention.
What this means is just `M`, with any *free occurrences* inside `M` of the variable `a` replaced with the term `N`.
+ In (the imperatival part of) OCaml, this could be done as:
What is a free occurrence?
+ let x = ref 3 in
+ x := 2
> An occurrence of a variable `a` is **bound** in T if T has the form `(\a N)`.
+ Of course you don't need to remember any of this syntax. We're just illustrating it so that you see that in Scheme and OCaml it looks somewhat different than we had above. The difference is much more obvious than it is in C.
> 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`.
+ In the lambda calculus: sorry, you can't do mutation. At least, not natively. Later in the term we'll be learning how in fact, really, you can embed mutation inside the lambda calculus even though the lambda calculus has no primitive facilities for mutation.
> 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.
+3. Anonymous functions
Here's an example of betareduction:
+ Functions are "firstclass values" in the lambda calculus, in Scheme, and in OCaml. What that means is that they can be arguments to other functions. They can be the results of the application of other functions to some arguments. They can be stored in data structures. And so on.
 ((\x (y x)) z)
+ First, we'll show what "anonymous" functions look like. These are functions that have not been bound as values to any variables. That is, there are no variables whose value they are.
betareduces to:
+ In the lambda calculus:
 (y z)
+ (\x M)
We'll write that like this:
+ is always anonymous! Here `M` stands for any expression of the language, simple or complex. It's only when you do
 ((\x (y x)) z) ~~> (y z)
+ ((\y N) (\x M))
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:
+ that `(\x M)` has a "name" (it's named `y` during the evaluation of `N`).
 M ~~> N
+ In Scheme, the same thing is written:
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 (triangle..sub1) and (triangle).
+ (lambda (x) M)
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:
+ Not very different, right? For example, if `M` stands for `(+ 3 x)`, then this is an anonymous function that adds 3 to whatever argument it's given:
 M <~~> N
+ (lambda (x) (+ 3 x))
This is what plays the role of equality in the lambda calculus. Hankin uses the symbol `=` for this. So too do Hindley and Seldin.
+ Scheme uses a lot of parentheses, and they are always significant, never optional. In `(+ 3 x)` the parentheses mean "apply the function `+` to the arguments `3` and `x`. In `(lambda (x) ...)` the parentheses have a different meaning: they mark where the anonymous function you're defining begins and ends, and so on. As you'll see, parentheses have yet further roles in Scheme. I know it's confusing.
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 (three bars) 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:
+ In OCaml, we write our anonymous function like this:
> T is defined to be `(M N)`.
+ fun x > (3 + x)
We'll regard the following two expressions:
+ or:
 (\x x y)
+ fun x > (( + ) 3 x)
 (\z z y)
+ In OCaml, parentheses only serve a grouping function and they often can be omitted. Or more could be added. For instance, we could equally well say:
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.
+ fun x > ( + ) 3 x
Note that neither of those expressions are identical to:
+ or:
 (\x x w)
+ (fun x > (( + ) (3) (x)))
because here it's a free variable that's been changed. Nor are they identical to:
+ As we saw above, parentheses can often be omitted in the lambda calculus too. But not in Scheme. Every parentheses has a specific role.
 (\y y y)
+4. Supplying an argument to an anonymous function
because here the second occurrence of `y` is no longer free.
+ Just because the functions we built aren't named doesn't mean we can't do anything with them. We can give them arguments. For example, in Scheme we can say:
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.
+ ((lambda (x) (+ 3 x)) 2)
+ The outermost parentheses here mean "apply the function `(lambda (x) (+ 3 x))` to the argument `2`.
Shorthand

+ In OCaml:
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.)
+ (fun x > ( + ) 3 x) 2
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:
+5. Binding variables to values with "let"
 (\x (\y (xy)))
+ Let's go back and reconsider this Scheme expression:
can be abbreviated as:
+ (let* ((three 3))
+ (let ((two 2))
+ (+ three two)))
 (\x (\y. x y))
+ Scheme also has a simple `let` (without the `*`), and it permits you to group several variable bindings together in a single `let` or `let*`statement, like this:
and:
+ (let* ((three 3) (two 2))
+ (+ three two))
 (\x \y. (z y) z)
+ Often you'll get the same results whether you use `let*` or `let`. However, there are cases where it makes a difference, and in those cases, `let*` behaves more like you'd expect. So you should just get into the habit of consistently using that. It's also good discipline for this seminar, especially while you're learning, to write things out the longer way, like this:
would abbreviate:
+ (let* ((three 3))
+ (let ((two 2))
+ (+ three two)))
 (\x \y ((z y) z))
+ However, here you've got the double parentheses in `(let* ((three 3)) ...)`. They're doubled because the syntax permits more assignments than just the assignment of the value `3` to the variable `three`. Myself I tend to use `[` and `]` for the outer of these parentheses: `(let* [(three 3)] ...)`. Scheme can be configured to parse `[...]` as if they're just more `(...)`.
This on the other hand:
+ Someone asked in seminar if the `3` could be replaced by a more complex expression. The answer is "yes". You could also write:
 ((\x \y. (z y) z)
+ (let* [(three (+ 1 2))]
+ (let [(two 2)]
+ (+ three two)))
+
+ The question also came up whether the `(+ 1 2)` computation would be performed before or after it was bound to the variable `three`. That's a terrific question. Let's say this: both strategies could be reasonable designs for a language. We are going to discuss this carefully in coming weeks. In fact Scheme and OCaml make the same design choice. But you should think of the underlying form of the `let`statement as not settling this by itself.
would abbreviate:
+ Repeating our starting point for reference:
 ((\x (\y (z y))) z)
+ (let* [(three 3)]
+ (let [(two 2)]
+ (+ three two)))
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:
+ Recall in OCaml this same computation was written:
 (\x x y)
+ let three = 3 in
+ let two = 2 in
+ ( + ) three two
as:
+6. Binding with "let" is the same as supplying an argument to a lambda
 \x. x y
+ The preceding expression in Scheme is exactly equivalent to:
but you should include the parentheses in:
+ (((lambda (three) (lambda (two) (+ three two))) 3) 2)
 (\x. x y) z
+ The preceding expression in OCaml is exactly equivalent to:
and:
+ (fun three > (fun two > ( + ) three two)) 3 2
 z (\x. x y)
+ Read this several times until you understand it.
Merging lambdas: an expression of the form `(\x (\y M))`, or equivalently, `(\x. \y. M)`, can be abbreviated as:
+7. Functions can also be bound to variables (and hence, cease being "anonymous").
 (\x y. M)
+ In Scheme:
Similarly, `(\x (\y (\z M)))` can be abbreviated as:
+ (let* [(bar (lambda (x) B))] M)
 (\x y z. 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)`.
+ Similarly, in OCaml:
Lambda terms represent functions

+ let bar = fun x > B in
+ M
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:
+ This in Scheme:
(\x x) represents the identity function: given any argument M, this function
simply returns M: ((\x x) M) ~~> M.
+ (let* [(bar (lambda (x) B))] (bar A))
(\x (x x)) duplicates its argument:
((\x (x x)) M) ~~> (M M)
+ as we've said, means the same as:
(\x (\y x)) throws away its second argument:
(((\x (\y x)) M) N) ~~> M
+ ((lambda (bar) (bar A)) (lambda (x) B))
and so on.
+ which, as we'll see, is equivalent to:
It is easy to see that distinct lambda expressions can represent the same
function, considered as a mapping from input to outputs. Obviously:
+ ((lambda (x) B) A)
 (\x x)
+ and that means the same as:
and:
+ (let* [(x A)] B)
 (\z z)
+ in other words: evaluate `B` with `x` assigned to the value `A`.
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:
+ Similarly, this in OCaml:
 (\y x. y x) (\z z)
+ let bar = fun x > B in
+ bar A
 (\x. (\z z) x)
+ is equivalent to:
 (\z z)
+ (fun x > B) A
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 argument can differentiate between them when they're supplied as an argument to it. However, these expressions are all syntactically distinct.
+ and that means the same as:
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.
+ let x = A in
+ B
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 seminar (and further discussion is best pursued in person).
+8. Pushing a "let"binding from now until the end
+ What if you want to do something like this, in Scheme?
+ (let* [(x A)] ... for the rest of the file or interactive session ...)
+ or this, in OCaml:
+ let x = A in
+ ... for the rest of the file or interactive session ...
Booleans and pairs
==================
+ Scheme and OCaml have syntactic shorthands for doing this. In Scheme it's written like this:
Our definition of these is reviewed in [[Assignment1]].
+ (define x A)
+ ... rest of the file or interactive session ...
+
+ In OCaml it's written like this:
+
+ 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.)
+
+
+9. Some shorthand
+
+ OCaml permits you to abbreviate:
+
+ let bar = fun x > B in
+ M
+
+ as:
+
+ let bar x = B in
+ M
+
+ It also permits you to abbreviate:
+
+ let bar = fun x > B;;
+
+ as:
+
+ let bar x = B;;
+
+ Similarly, Scheme permits you to abbreviate:
+
+ (define bar (lambda (x) B))
+
+ as:
+
+ (define (bar x) B)
+
+ and this is the form you'll most often see Scheme definitions written in.
+
+ However, conceptually you should think backwards through the abbreviations and equivalences we've just presented.
+
+ (define (bar x) B)
+
+ just means:
+
+ (define bar (lambda (x) B))
+
+ which just means:
+
+ (let* [(bar (lambda (x) B))] ... rest of the file or interactive session ...)
+
+ which just means:
+
+ (lambda (bar) ... rest of the file or interactive session ...) (lambda (x) B)
+
+ or in other words, interpret the rest of the file or interactive session with `bar` assigned the function `(lambda (x) B)`.
+
+
+10. Shadowing
+
+ You can override a binding with a more inner binding to the same variable. For instance the following expression in OCaml:
+
+ let x = 3 in
+ let x = 2 in
+ x
+
+ will evaluate to 2, not to 3. It's easy to be lulled into thinking this is the same as what happens when we say in C:
+
+ int x = 3;
+ x = 2;
+
+ but it's not the same! In the latter case we have mutation, in the former case we don't. You will learn to recognize the difference as we proceed.
+
+ The OCaml expression just means:
+
+ (fun x > ((fun x > x) 2) 3)
+
+ and there's no more mutation going on there than there is in:
+
+ + ∀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.
+
+
+Some more comparisons between Scheme and OCaml
+
+
+11. Simple predefined values
+
+ Numbers in Scheme: `2`, `3`
+ In OCaml: `2`, `3`
+
+ Booleans in Scheme: `#t`, `#f`
+ In OCaml: `true`, `false`
+
+ The eighth letter in the Latin alphabet, in Scheme: `#\h`
+ In OCaml: `'h'`
+
+12. Compound values
+
+ These are values which are built up out of (zero or more) simple values.
+
+ Ordered pairs in Scheme: `'(2 . 3)`
+ In OCaml: `(2, 3)`
+
+ Lists in Scheme: `'(2 3)`
+ In OCaml: `[2; 3]`
+ We'll be explaining the difference between pairs and lists next week.
+
+ The empty list, in Scheme: `'()`
+ In OCaml: `[]`
+
+ The string consisting just of the eighth letter of the Latin alphabet, in Scheme: `"h"`
+ In OCaml: `"h"`
+
+ A longer string, in Scheme: `"horse"`
+ In OCaml: `"horse"`
+
+ A shorter string, in Scheme: `""`
+ In OCaml: `""`
+
+13. Function application
+
+ Binary functions in OCaml: `foo 2 3`
+
+ Or: `( + ) 2 3`
+
+ These are the same as: `((foo 2) 3)`. In other words, functions in OCaml are "curried". `foo 2` returns a `2`fooer, which waits for an argument like `3` and then foos `2` to it. `( + ) 2` returns a `2`adder, which waits for an argument like `3` and then adds `2` to it.
+
+ In Scheme, on the other hand, there's a difference between `((foo 2) 3)` and `(foo 2 3)`. Scheme distinguishes between unary functions that return unary functions and binary functions. For our seminar purposes, it will be easiest if you confine yourself to unary functions in Scheme as much as possible.
+
+ Additionally, as said above, Scheme is very sensitive to parentheses and whenever you want a function applied to any number of arguments, you need to wrap the function and its arguments in a parentheses.
+
+
+What "sequencing" is and isn't
+
+
+We mentioned before the idea that computation is a sequencing of some changes. I said we'd be discussing (fragments of, and in some cases, entire) languages that have no native notion of change.
+
+Neither do they have any useful notion of sequencing. But what this would be takes some care to identify.
+
+First off, the mere concatenation of expressions isn't what we mean by sequencing. Concatenation of expressions is how you build syntactically complex expressions out of simpler ones. The complex expressions often express a computation where a function is applied to one (or more) arguments,
+
+Second, the kind of rebinding we called "shadowing" doesn't involve any changes or sequencing. All the precedence facts about that kind of rebinding are just consequences of the compound syntactic structures in which it occurs.
+
+Third, the kinds of bindings we see in:
+
+ (define foo A)
+ (foo 2)
+
+Or even:
+
+ (define foo A)
+ (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.
+
+Since Scheme and OCaml also do permit imperatival constructions, they do have syntax for genuine sequencing. In Scheme it looks like this:
+
+ (begin A B C)
+
+In OCaml it looks like this:
+
+ begin A; B; C end
+
+Or this:
+
+ (A; B; C)
+
+In the presence of imperatival elements, sequencing order is very relevant. For example, these will behave differently:
+
+ (begin (print "under") (print "water"))
+
+ (begin (print "water") (print "under"))
+
+And so too these:
+
+ begin x := 3; x := 2; x end
+
+ begin x := 2; x := 3; x end
+
+However, if A and B are purely functional, nonimperatival expressions, then:
+
+ begin A; B; C end
+just evaluates to C (so long as A and B evaluate to something at all). So:
+ begin A; B; C end
+contributes no more to a larger context in which it's embedded than C does. This is the sense in which functional languages have no serious notion of sequencing.
+We'll discuss this more as the seminar proceeds.
1. Declarative vs imperatival models of computation.