Quantificational determiners

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We’ve now talked about quantifiers—expressions of syntactic category \((s/(np\backslash s))\) (at least, when they occur in subject position).

Let’s begin analyzing the quantificational determiners that quantifiers arise from. Quantificational determiners are expressions like every, some, and no—they combine with a noun (e.g.,dog), or a modified noun (e.g., furry dog), to their right, in order to produce a quantifier. Hence, the syntactic category of a quantificational determiner—at least, when it gives rise to a quantifier that can occur in subject position—should be \(((s/(np\backslash s))/n)\).

Moreover, because quantificational determiners have this syntactic category, we can also determine their semantic type to be \(((e → t) → ((e → t) → t))\). The noun (or modified noun) they combine with has semantic type \((e → t)\), as does the verb phrase they combine with next. The main function of a quantificational determiner is therefore to say something about the relation between the meaning of the noun and the meaning of the verb phrase; for example, maybe everything in the set characterized by the noun has to also be in the set characterized by the verb phrase, as in the case of the determiner every; or maybe, the two sets have to overlap—i.e., share some elements—as in the case of some; or not overlap at all, as in the case of no.

Every

Recall that we provided the “lexical entry” for every dog given in (1).

1. \(⟨\textit{every dog}, (λf.\ct{dog}_{SET} ⊆ f_{SET})⟩ ⊢ (s/(np\backslash s))\)

To get a lexical entry for every, i.e., on its own, we need only (a) adjust the syntactic category, and (b) abstract over the meaning of the noun, so that it is determined by the argument of a function. That is, we should substitute a variable where \(\ct{dog}\) went in the meaning of every dog, and then add a ‘\(λ\)’ to the meaning representation which binds that variable. The goal here is just to allow the function denoted by every to take an additional argument of type \((e → t)\)—one which the meaning of dog can get slotted in for when every combines with dog.

2. \(⟨\textit{every}, (λf.(λg.f_{SET} ⊆ g_{SET}))⟩ ⊢ ((s/(np\backslash s))/n)\)

Now we can see that every takes two characteristic functions—\(f\) for the noun, and \(g\) for the verb phrase—and it says that the set characterized by \(f\) must be a subset of the set characterized by \(g\). Thus every effectively says that everything in the noun set must also be in the verb phrase set.

No

We can use the same strategy to give a meaning to the quantificational determiner no. Recall the “lexical entry” we provided for no dogs.

3. \(⟨\textit{no dogs}, (λf.\ct{dog}_{SET} ∩ f_{SET} = \varnothing)⟩ ⊢ (s/(np\backslash s))\)

In this case, if we change the syntactic category and abstract over the noun phrase meaning, we end up with the result in (4).

4. \(⟨\textit{no}, (λf.(λg.f_{SET} ∩ g_{SET} = \varnothing)⟩ ⊢ ((s/(np\backslash s))/n)\)

Thus once no combines with the noun to its right, and then the verb phrase, it gives you a sentence whose value is \(\true\) just in case the sets characterized by the noun and verb phrase, respectively, have an empty intersection.

Some

We assigned some dog the “lexical entry” in (5).

5. \(⟨\textit{some dog}, (λf.\ct{dog}_{SET} ∩ f_{SET} ≠ \varnothing)⟩ ⊢ (s/(np\backslash s))\)

Consistent with this analysis, we should assign some the lexical entry in (6).

6. \(⟨\textit{some}, (λf.(λg.f_{SET} ∩ g_{SET} ≠ \varnothing)⟩ ⊢ ((s/(np\backslash s))/n)\)

Three

Assuming we analyze three dogs as in (7),

7. \(⟨\textit{three dogs}, (λf.|\ct{dog}_{SET} ∩ f_{SET}| ≥ 3)⟩ ⊢ (s/(np\backslash s))\)

we should assign three the lexical entry in (8).

8. \(⟨\textit{three}, (λf.(λg.|f_{SET} ∩ g_{SET}| ≥ 3))⟩ ⊢ ((s/(np\backslash s))/n)\)

Example derivation

Using lexical entries like the above, we can derive sentences such as (9).

9. Some dog slept.

Here’s a derivation for this sentence:

\[\begin{prooftree} \AxiomC{\(⟨\textit{some}, (λf.(λg.f_{SET} ∩ g_{SET} ≠ \varnothing))⟩ ⊢ ((s/(np\backslash s))/n)\)} \AxiomC{\(⟨\textit{dog}, (λx.\ct{dog}(x))⟩ ⊢ n\)} \RightLabel{\(/\)}\BinaryInfC{\(⟨\textit{some dog}, (λf.(λg.f_{SET} ∩ g_{SET} ≠ \varnothing))((λx.\ct{dog}(x)))⟩ ⊢ (s/(np\backslash s))\)} \RightLabel{\(/\)}\AxiomC{\(⟨\textit{slept}, (λx.\ct{sleep}(x))⟩ ⊢ (np\backslash s)\)} \BinaryInfC{\(⟨\textit{some dog slept}, (λf.(λg.f_{SET} ∩ g_{SET} ≠ \varnothing))((λx.\ct{dog}(x)))((λx.\ct{sleep}(x)))⟩ ⊢ s\)} \end{prooftree}\]

Evaluating the resulting meaning representation, we get:

\[(λf.(λg.f_{SET} ∩ g_{SET} ≠ \varnothing))((λx.\ct{dog}(x)))((λx.\ct{sleep}(x)))\] \[⇒ (λg.(λx.\ct{dog}(x))_{SET} ∩ g_{SET} ≠ \varnothing)((λx.\ct{sleep}(x)))\] \[⇒ (λx.\ct{dog}(x))_{SET} ∩ (λx.\ct{sleep}(x))_{SET} ≠ \varnothing\]

We can see that (9) is analyzed as true (i.e., as denoting the value \(\true\)) just in case intersecting the set of dogs with the set of sleeping things results in a set with at least one entity in it. Some dog has to have slept.