Hoeksema (1983): negative polarity and the comparative #
This file formalizes Hoeksema's semantics for the two comparative constructions of English and
Dutch. The NP-comparative taller than NP sends a generalized quantifier Q to the individuals
x for which the set of individuals shorter than x belongs to Q. This map is a Boolean
homomorphism and the only one that gets the comparative right on proper names, so it is
monotone increasing and, by Ladusaw's criterion, no environment for negative polarity items.
The S-comparative taller than S sends a set of degrees to the individuals above all of them.
It is anti-additive and hence monotone decreasing, so it licenses negative polarity items,
including the Dutch ook maar, which Zwarts's hypothesis restricts to anti-additive triggers.
The two comparatives coincide on a proper name and the singleton of its degree.
Main definitions #
npComparative: the NP-comparative as a complete lattice homomorphism (Definition 3).PreservesOrdering: a map from quantifiers to predicates respects the grading relation on proper names (Definition 4).
Main results #
PreservesOrdering.eq: two homomorphisms that preserve the ordering are equal (Fact 2).PreservesOrdering.eq_npComparative: the NP-comparative is the only one (§3.5).not_antitone_npComparative: the NP-comparative is not monotone decreasing (§3.6).gtOverSet_antitone: the S-comparative is monotone decreasing (§3.8).npComparative_individual: the two comparatives agree on proper names (§3.9).not_isAntiAdditive_notEveryDutchman: niet iedere Nederlander is not anti-additive (§4.1).
Implementation notes #
Degrees are values of a measure μ in a preorder rather than the equivalence classes of
Definition 7, and the grading relation x > y is μ y < μ x. The results on the
NP-comparative need neither of the paper's conditions on grading relations (Definitions 1
and 2). The homomorphism preserves arbitrary unions and intersections, which Fact 2 needs on
infinite domains (the paper's footnote 11). The S-comparative of Definition 8 is
(μ ⁻¹' strictUpperBounds ·), and its anti-additivity (Fact 5) is
Degree.gtOverSet_isAntiAdditive.
References #
The NP-comparative #
The NP-comparative of Definition 3 holds of x and a quantifier Q when the individuals
below x on the scale μ form a set in Q. As a preimage map it preserves intersections,
unions and complements, which is (22).
Equations
- Hoeksema1983.npComparative μ = CompleteLatticeHom.setPreimage fun (x : Entity) => μ ⁻¹' Set.Iio (μ x)
Instances For
The NP-comparative is monotone increasing (Fact 3), as every homomorphism is.
On a nonempty domain the NP-comparative is not monotone decreasing, so by Ladusaw's criterion it is not a negative polarity environment (§3.6).
Uniqueness #
A map f from quantifiers to predicates preserves the ordering (Definition 4) when a > b
holds exactly when a falls under f of the quantifier that the name of b denotes.
Equations
- Hoeksema1983.PreservesOrdering μ f = ∀ (a b : Entity), μ b < μ a ↔ a ∈ f (Quantifier.NP.individual b)
Instances For
The NP-comparative preserves the ordering, as the truth conditions of (30) require.
Two maps that preserve the ordering agree on every proper name (Fact 1).
Two homomorphisms from quantifiers to predicates that preserve the ordering are equal (Fact 2). Each is the preimage map of a function, which Fact 1 determines.
The NP-comparative is the only homomorphism from quantifiers to predicates that preserves the ordering (§3.5).
The S-comparative #
The S-comparative is monotone decreasing in its set of degrees (§3.8), since it is anti-additive (Fact 5) and anti-additive maps are monotone decreasing (Fact 4).
The NP-comparative of a proper name is the S-comparative of the singleton of its degree (§3.9), which accounts for the equivalence of I am bigger than you and I am bigger than you are in (44).
Triggers of ook maar #
The two Dutchmen of the §4.1 model, Jan and Piet.
Instances For
The quantifier niet iedere Nederlander 'not every Dutchman' holds of a property that some Dutchman lacks.
Equations
- Hoeksema1983.notEveryDutchman P = ¬Quantifier.GQ.every (fun (x : Hoeksema1983.Dutchman) => True) fun (x : Hoeksema1983.Dutchman) => x ∈ P
Instances For
In (51), Jan walks but is silent and Piet sits and talks. Not every Dutchman walks and not every Dutchman talks, yet every Dutchman walks or talks, so niet iedere Nederlander is not anti-additive and, by Zwarts's hypothesis, cannot trigger ook maar (45b).