# let divide' num den = if den = 0 then None else Some (num/den);; val divide' : int -> int -> int option =as basically a function from two integers to an integer, except with this little bit of option plumbing on the side. A note on notation: Haskell uses the infix operator `>>=` to stand for `bind`. Chris really hates that symbol. Following Wadler, he prefers to use an infix five-pointed star, or on a keyboard, `*`. Jim on the other hand thinks `>>=` is what the literature uses and students won't be able to avoid it. Moreover, although ⋆ is OK (though not a convention that's been picked up), overloading the multiplication symbol invites its own confusion and Jim feels very uneasy about that. If not `>>=` then we should use some other unfamiliar infix symbol (but `>>=` already is such...) In any case, the course leaders will work this out somehow. In the meantime, as you read around, wherever you see `m >>= f`, that means `bind m f`. Also, if you ever see this notation: do x <- m f x That's a Haskell shorthand for `m >>= (\x -> f x)`, that is, `bind m f`. Similarly: do x <- m y <- n f x y is shorthand for `m >>= (\x -> n >>= (\y -> f x y))`, that is, `bind m (fun x -> bind n (fun y -> f x y))`. Those who did last week's homework may recognize this. (Note that the above "do" notation comes from Haskell. We're mentioning it here because you're likely to see it when reading about monads. It won't work in OCaml. In fact, the `<-` symbol already means something different in OCaml, having to do with mutable record fields. We'll be discussing mutation someday soon.) The Monad laws -------------- Just like good robots, monads must obey three laws designed to prevent them from hurting the people that use them or themselves. * **Left identity: unit is a left identity for the bind operation.** That is, for all `f:'a -> 'a m`, where `'a m` is a monadic object, we have `(unit x) * f == f x`. For instance, `unit` is a function of type `'a -> 'a option`, so we have

# let ( * ) m f = match m with None -> None | Some n -> f n;; val ( * ) : 'a option -> ('a -> 'b option) -> 'b option =The parentheses is the magic for telling OCaml that the function to be defined (in this case, the name of the function is `*`, pronounced "bind") is an infix operator, so we write `m * f` or `( * ) m f` instead of `* m f`. * **Associativity: bind obeys a kind of associativity**. Like this: (m * f) * g == m * (fun x -> f x * g) If you don't understand why the lambda form is necessary (the "fun x" part), you need to look again at the type of bind. Some examples of associativity in the option monad:# let unit x = Some x;; val unit : 'a -> 'a option = # unit 2;; - : int option = Some 2 # unit 2 * unit;; - : int option = Some 2 # divide 6 2;; - : int option = Some 3 # unit 2 * divide 6;; - : int option = Some 3 # divide 6 0;; - : int option = None # unit 0 * divide 6;; - : int option = None

# Some 3 * unit * unit;; - : int option = Some 3 # Some 3 * (fun x -> unit x * unit);; - : int option = Some 3 # Some 3 * divide 6 * divide 2;; - : int option = Some 1 # Some 3 * (fun x -> divide 6 x * divide 2);; - : int option = Some 1 # Some 3 * divide 2 * divide 6;; - : int option = None # Some 3 * (fun x -> divide 2 x * divide 6);; - : int option = NoneOf course, associativity must hold for arbitrary functions of type `'a -> 'a m`, where `m` is the monad type. It's easy to convince yourself that the bind operation for the option monad obeys associativity by dividing the inputs into cases: if `m` matches `None`, both computations will result in `None`; if `m` matches `Some n`, and `f n` evalutes to `None`, then both computations will again result in `None`; and if the value of `f n` matches `Some r`, then both computations will evaluate to `g r`. * **Right identity: unit is a right identity for bind.** That is, `m * unit == m` for all monad objects `m`. For instance,

# Some 3 * unit;; - : int option = Some 3 # None * unit;; - : 'a option = NoneNow, if you studied algebra, you'll remember that a *monoid* is an associative operation with a left and right identity. For instance, the natural numbers along with multiplication form a monoid with 1 serving as the left and right identity. That is, temporarily using `*` to mean arithmetic multiplication, `1 * n == n == n * 1` for all `n`, and `(a * b) * c == a * (b * c)` for all `a`, `b`, and `c`. As presented here, a monad is not exactly a monoid, because (unlike the arguments of a monoid operation) the two arguments of the bind are of different types. But if we generalize bind so that both arguments are of type `'a -> 'a m`, then we get plain identity laws and associativity laws, and the monad laws are exactly like the monoid laws (see

Extensional types Intensional types Examples ------------------------------------------------------------------- S s->t s->t John left DP s->e s->e John VP s->e->t s->(s->e)->t left Vt s->e->e->t s->(s->e)->(s->e)->t saw Vs s->t->e->t s->(s->t)->(s->e)->t thoughtThis system is modeled on the way Montague arranged his grammar. There are significant simplifications: for instance, determiner phrases are thought of as corresponding to individuals rather than to generalized quantifiers. If you're curious about the initial `s`'s in the extensional types, they're there because the behavior of these expressions depends on which world they're evaluated at. If you are in a situation in which you can hold the evaluation world constant, you can further simplify the extensional types. Usually, the dependence of the extension of an expression on the evaluation world is hidden in a superscript, or built into the lexical interpretation function. The main difference between the intensional types and the extensional types is that in the intensional types, the arguments are functions from worlds to extensions: intransitive verb phrases like "left" now take intensional concepts as arguments (type s->e) rather than plain individuals (type e), and attitude verbs like "think" now take propositions (type s->t) rather than truth values (type t). The intenstional types are more complicated than the intensional types. Wouldn't it be nice to keep the complicated types to just those attitude verbs that need to worry about intensions, and keep the rest of the grammar as extensional as possible? This desire is parallel to our earlier desire to limit the concern about division by zero to the division function, and let the other functions, like addition or multiplication, ignore division-by-zero problems as much as possible. So here's what we do: In OCaml, we'll use integers to model possible worlds: type s = int;; type e = char;; type t = bool;; Characters (characters in the computational sense, i.e., letters like `'a'` and `'b'`, not Kaplanian characters) will model individuals, and OCaml booleans will serve for truth values.

type 'a intension = s -> 'a;; let unit x (w:s) = x;; let ann = unit 'a';; let bill = unit 'b';; let cam = unit 'c';;In our monad, the intension of an extensional type `'a` is `s -> 'a`, a function from worlds to extensions. Our unit will be the constant function (an instance of the K combinator) that returns the same individual at each world. Then `ann = unit 'a'` is a rigid designator: a constant function from worlds to individuals that returns `'a'` no matter which world is used as an argument. Let's test compliance with the left identity law:

# let bind m f (w:s) = f (m w) w;; val bind : (s -> 'a) -> ('a -> s -> 'b) -> s -> 'b =We'll assume that this and the other laws always hold. We now build up some extensional meanings: let left w x = match (w,x) with (2,'c') -> false | _ -> true;; This function says that everyone always left, except for Cam in world 2 (i.e., `left 2 'c' == false`). Then the way to evaluate an extensional sentence is to determine the extension of the verb phrase, and then apply that extension to the extension of the subject:# bind (unit 'a') unit 1;; - : char = 'a'

let extapp fn arg w = fn w (arg w);; extapp left ann 1;; # - : bool = true extapp left cam 2;; # - : bool = false`extapp` stands for "extensional function application". So Ann left in world 1, but Cam didn't leave in world 2. A transitive predicate: let saw w x y = (w < 2) && (y < x);; extapp (extapp saw bill) ann 1;; (* true *) extapp (extapp saw bill) ann 2;; (* false *) In world 1, Ann saw Bill and Cam, and Bill saw Cam. No one saw anyone in world two. Good. Now for intensions: let intapp fn arg w = fn w arg;; The only difference between intensional application and extensional application is that we don't feed the evaluation world to the argument. (See Montague's rules of (intensional) functional application, T4 -- T10.) In other words, instead of taking an extension as an argument, Montague's predicates take a full-blown intension. But for so-called extensional predicates like "left" and "saw", the extra power is not used. We'd like to define intensional versions of these predicates that depend only on their extensional essence. Just as we used bind to define a version of addition that interacted with the option monad, we now use bind to intensionalize an extensional verb:

let lift pred w arg = bind arg (fun x w -> pred w x) w;; intapp (lift left) ann 1;; (* true: Ann still left in world 1 *) intapp (lift left) cam 2;; (* false: Cam still didn't leave in world 2 *)Because `bind` unwraps the intensionality of the argument, when the lifted "left" receives an individual concept (e.g., `unit 'a'`) as argument, it's the extension of the individual concept (i.e., `'a'`) that gets fed to the basic extensional version of "left". (For those of you who know Montague's PTQ, this use of bind captures Montague's third meaning postulate.) Likewise for extensional transitive predicates like "saw":

let lift2 pred w arg1 arg2 = bind arg1 (fun x -> bind arg2 (fun y w -> pred w x y)) w;; intapp (intapp (lift2 saw) bill) ann 1;; (* true: Ann saw Bill in world 1 *) intapp (intapp (lift2 saw) bill) ann 2;; (* false: No one saw anyone in world 2 *)Crucially, an intensional predicate does not use `bind` to consume its arguments. Attitude verbs like "thought" are intensional with respect to their sentential complement, but extensional with respect to their subject (as Montague noticed, almost all verbs in English are extensional with respect to their subject; a possible exception is "appear"):

let think (w:s) (p:s->t) (x:e) = match (x, p 2) with ('a', false) -> false | _ -> p w;;Ann disbelieves any proposition that is false in world 2. Apparently, she firmly believes we're in world 2. Everyone else believes a proposition iff that proposition is true in the world of evaluation.

intapp (lift (intapp think (intapp (lift left) (unit 'b')))) (unit 'a') 1;; (* true *)So in world 1, Ann thinks that Bill left (because in world 2, Bill did leave). The `lift` is there because "think Bill left" is extensional wrt its subject. The important bit is that "think" takes the intension of "Bill left" as its first argument.

intapp (lift (intapp think (intapp (lift left) (unit 'c')))) (unit 'a') 1;; (* false *)But even in world 1, Ann doesn't believe that Cam left (even though he did: `intapp (lift left) cam 1 == true`). Ann's thoughts are hung up on what is happening in world 2, where Cam doesn't leave. *Small project*: add intersective ("red") and non-intersective adjectives ("good") to the fragment. The intersective adjectives will be extensional with respect to the nominal they combine with (using bind), and the non-intersective adjectives will take intensional arguments. Finally, note that within an intensional grammar, extensional funtion application is essentially just bind:

# let swap f x y = f y x;; # bind cam (swap left) 2;; - : bool = false