Generalized quantifier

In linguistic semantics, a generalized quantifier is an expression that denotes a property of a property, also called a higher-order property. This is the standard semantics assigned to quantified noun phrases, also called determiner phrases, in short: DP. The DP every boy below says of a property X that the set of entities that are boys is a subset of the set of entities that have property X. So the following sentence says that the set of boys is a subset of the set of sleepers.
 * Every boy sleeps.
 * $$\{x\,|\, x \mbox{ is a boy}\} \subseteq \{x\,|\,x \mbox{ sleeps}\}$$

This treatment of quantifiers has been essential in achieving a compositional semantics for sentences containing quantifiers.

Type theory
A version of type theory is often used to make the semantics of different kinds of expressions explicit. The standard construction defines the set of types recursively as follows:
 * 1) e and t are types.
 * 2) If a and b are both types, then so is $$\langle a,b\rangle$$
 * 3) Nothing is a type, except what can be constructed on the basis of lines 1 and 2 above.

Given this definition, we have the simple types e and t, but also a countable infinity of complex types, some of which include:
 * $$\langle e,t\rangle;\qquad \langle t,t\rangle;\qquad \langle\langle e,t\rangle, t\rangle; \qquad\langle e,\langle e,t\rangle\rangle; \qquad \langle\langle e,t\rangle,\langle \langle e, t\rangle, t\rangle\rangle;\qquad \ldots$$


 * Expressions of type e denote elements of the universe of discourse, the set of entities the discourse is about. This set is usually written as $$D_e$$. Examples of type e expressions include John and he.
 * Expressions of type t denote a truth value, usually rendered as the set$$\{0,1\}$$, where 0 stands for "false" and 1 stands for "true".  Examples of expressions that are sometimes said to be of type t are sentences or propositions.
 * Expressions of type $$\langle e,t\rangle$$ denote functions from the set of entities to the set of truth values. This set of functions is rendered as $$D_t^{D_e}$$. Such functions are characteristic functions of sets.  They map every individual that is an element of the set to "true", and everything else to "false."  It is common to say that they denote sets rather than characteristic functions, although, strictly speaking, the latter is more accurate. Examples of expressions of this type are predicates, nouns and some kinds of adjectives.
 * In general, expressions of complex types $$\langle a,b\rangle$$ denote functions from the set of entities of type $$a$$ to the set of entities of type $$b$$, a construct we can write as follows: $$D_b^{D_a}$$.

We can now assign types to the words in our sentence above (Every boy sleeps) as follows.
 * Type(boy)=$$\langle e,t\rangle$$
 * Type(sleeps)=$$\langle e,t\rangle$$
 * Type(every)= $$\langle\langle e,t\rangle,\langle \langle e, t\rangle, t\rangle\rangle$$

Thus, every denotes a function from a set to a function from a set to a truth value. Put differently, it denotes a function from a set to a set of sets. It is that function which for any two sets A,B, every(A)(B)= 1 if and only if $$A\subseteq B$$.

The typed lambda calculus
A useful way to write complex functions is the lambda calculus. For example, one can write the meaning of sleeps as the following lambda expression, which is a function from an individual x to the proposition that x sleeps.
 * $$\lambda x. sleep(x)$$

Such lambda terms are functions whose domain is what precedes the period, and whose range are the type of thing that follows the period. If x is a variable that ranges over elements of $$D_e$$, then the following lambda term denotes the identity function on individuals:
 * $$\lambda x.x$$

We can now write the meaning of every with the following lambda term, where X,Y are variables of type $$\langle e,t\rangle$$:


 * $$\lambda X.\lambda Y. X\subseteq Y$$

If we abbreviate the meaning of boy and sleeps as "B" and "S", respectively, we have that the sentence every boy sleeps now means the following:
 * $$(\lambda X.\lambda Y. X\subseteq Y)(B)(S)$$ — β-reduction
 * $$(\lambda Y. B \subseteq Y)(S)$$ — β-reduction
 * $$B\subseteq S$$

The expression every is a determiner. Combined with a noun, it yields a generalized quantifier of type $$\langle\langle e,t\rangle,t\rangle$$.

Monotone increasing GQs
A generalized quantifier GQ is said to be monotone increasing, also called upward entailing, just in case, for any two sets X and Y the following holds:
 * if $$X\subseteq Y$$, then GQ(X) entails GQ(Y).

The GQ every boy is monotone increasing. For example, the set of things that run fast is a subset of the set of things that run. Therefore, the first sentence below entails the second:
 * 1) Every boy runs fast.
 * 2) Every boy runs.

Monotone decreasing GQs
A GQ is said to me monotone decreasing, also called downward entailing just in case, for any two sets X and Y, the following holds:
 * If $$X\subseteq Y$$, then GQ(Y) entails GQ(X).

An example of a monotone decreasing GQ is no boy. For this GQ we have that the first sentence below entails the second. The lambda term for the determiner no is the following. It says that the two sets have an empty intersection.
 * 1) No boy runs.
 * 2) No boy runs fast.
 * $$\lambda X.\lambda Y. X\cap Y= \emptyset$$

Monotone decreasing GQs are among the expressions that can license a negative polarity item, such as any. Monotone increasing GQs do not license negative polarity items.
 * 1) Good: No boy has any money.
 * Bad: *Every boy has any money.

Non-monotone GQs
A GQ is said to be non-monotone if it is neither monotone increasing nor monotone decreasing. An example of such a GQ is exactly three boys. Neither of the following two sentences entail the other. The first sentence doesn't entail the second. The fact that the number of students that ran is exactly three doesn't entail that each of these students ran fast, so the number of students that did that can be smaller than 3. Conversely, the second sentence doesn't entail the first. The sentence exactly three students ran fast can be true, even though the number of students who merely ran (i.e. not so fast) is greater than 3.
 * 1) Exactly three students ran.
 * 2) Exactly three students ran fast.

The lambda term for the (complex) determiner exactly three is the following. It says that the cardinality of the intersection between the two sets equals 3.
 * $$\lambda X.\lambda Y. |X\cap Y|=3$$

This would represent the first assertion above, where X means the set of people that are students, and Y the set of people that ran. The second sentence would be written similarly using an apparently equivalent lambda-expression
 * $$\lambda X.\lambda Z. |X\cap Z|=3$$

where Z means the set of people that ran fast.

Although we have Y included in Z neither of the two assertions can be entailed from the other, because there's no free variable in the first expression that could be be interpreted as having the same meaning in the second instance of the same lambda-expression, because they are bound to distinct sets (the set of people that ran, and the distinct set of people thatran fast could be both bound separately and indepdantly to the first of second lambda-expression without implying anything about the entailment of either one of the two generated instances in the second one).

For such entailment to be possible, the rebindings of quantified (non-free) variables into effective sets (free variables) must be done identically in all instances. If this is not the case, the quantified (non-free) variables must be renamed so that one create a new assertion stating the entailment of one lamda-expression into the other, with those explicit bindings such as:
 * $$\lambda X.\lambda Y.\lambda Z. |X\cap Y|=3 \subseteq |X\cap Z|=3$$
 * $$\lambda X.\lambda Y.\lambda Z. |X\cap Z|=3 \subseteq |X\cap Y|=3$$

which can't be entailed from any of the two above statements because it attempts to bind the same non-free variable X to both statements. The only correct way to explicit the combination of the two statements above as a single assertion is to rename all quantified (non-free) variables so that they don't conflict within the two instances:
 * $$\lambda X_0.\lambda Y_0. |X_0\cap Y_0|=3$$ and
 * $$\lambda X_1.\lambda Y_1. |X_1\cap Y_1|=3$$

and then create the compound assertion:
 * $$(\lambda X_0.\lambda Y_0. |X_0\cap Y_0|=3) \and (\lambda X_1.\lambda Y_1. |X_1\cap Y_1|=3)$$

which can now safely be reordered with all lambda quantifiers factorized at the begining (because these quantifiers are commutative and associative providing that there's no modified binding to their existing variables):
 * $$\lambda X_0.\lambda Y_0.\lambda X_1.\lambda Y_1. (|X_0\cap Y_0|=3) \and (|X_1\cap Y_1|=3)$$

and then bind their associated sets according to each instance:
 * $$\lambda X.\lambda Y.\lambda X.\lambda Z. (|X\cap Y|=3) \and (|X\cap Z|=3)$$

which finally reduces as:
 * $$\lambda X.\lambda Y.\lambda Z. (|X\cap Y|=3) \and (|X\cap Z|=3)$$

because two equal quantifiers reduce to a single equivalent one. As there's no derivation between this last statement and the first two statements, even if we also try to include that fact that we also know that Y includes Z (a third statement to the first two ones):
 * $$\lambda X.\lambda Y.\lambda Z. (|X\cap Y|=3) \and (|X\cap Z|=3) \and (Y \subseteq Z)$$

Conservativity
A determiner D is said to be conservative if the following equivalence holds:
 * $$D(A)(B)\leftrightarrow D(A)(A\cap B)$$

For example, the following two sentences are equivalent.
 * 1) Every boy sleeps.
 * 2) Every boy is a boy who sleeps.

It has been proposed that all natural language determiners (i.e. in every language) are conservative (Barwise and Cooper 1981). The expression only is not conservative. The following two sentences are not equivalent. But it is, in fact not common to analyze only as a determiner. Rather, it is standardly treated as a focus-sensitive adverb.
 * 1) Only boys sleep.
 * 2) Only boys are boys who sleep.