Formal Approaches to Semantic Microvariation: Adverbial ...

(90) Lemma 2: BCPm,n SF ≠ BCPm
GQ
◦ BCP GQ
n
PROOF: Let R ∈ P(E × E) such that | Dom(R) |=| Ran(R) |> m,n. Furthermore,
let R be a bijection.
Since | Dom(R) |=| Ran(R) |> m,n, BCPm,n(R) SF = 1. But, since R is a bijection,
every element in its range is related to only one element in the domain
by R. By assumption, n > 1, so BCPn
GQ (R) = Ø. Since BCPm
GQ is positive,
BCPm
GQ ◦ BCPn
GQ (R) = 0. □
The proof of Theorem 1 (88) follows immediately from Lemma 1 (89), Lemma 2
(90), and Keenan’s Reducibility Equivalence (84). □
BCP SF is unreducible because it is true of relations in which there are many events
with few or even a single participant in each event, provided that the total number
of participants is large enough to count as beaucoup. Since BCP GQ ◦ BCP GQ is the
iteration of two unary quantifiers, it builds in a scope dependency between the two
BCP GQ s. BCP GQ ◦ BCP GQ is only true of relations in which there are many events
with many participants. In QAD sentences; however, there is no such dependency.
This is why QAD in Standard French must be modeled with polyadic quantifiers.
2.2.2 A Compositional Analysis
I now present a compositional analysis of Quantification at a Distance in Standard
French. As shown by the unreducibility proof, the de phrases cannot be interpreted as
regular quantified noun phrases. I therefore propose that the intuition that de is a semantically
‘deficient’ determiner presented in the incorporation analysis is right, and,
following Heyd & Mathieu, I assume that de phrases denote bare properties. However,
in contrast to the incorporation analysis, I propose that combining the verb and
the de phrase does not existentially close the direct object. Instead, I propose that de
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