Formal Approaches to Semantic Microvariation: Adverbial ...

(16) Theorem 2: For all s ∈ N, BCPs
QF = BCPs 1 ◦ ∃ NOM
PROOF: Let t ∈ N to show BCP QF
t = BCP 1
t ◦ ∃ NOM . Let R ∈ P(E × E) to
show that BCPt
QF (R) = BCPt GQ ◦ ∃ NOM .
Case 1: BCP QF
t
(R) = 1. Then BCPt
GQ (Ran(R)) = 1. Therefore, R ≠ Ø. That
means that for all x ∈ Ran(R), there is some e ∈ Dom(R) such that < e,x >∈ R.
Therefore, E NOM (R) = Ran(R). Since BCP GQ
t applied to a property is BCP 1
t ,
BCP 1
t (Ran(R)) = 1, BCP 1
t ◦ ∃ NOM (R) = 1. □
Case 2: BCP QF
t
(R) = 0. Therefore, BCPt
GQ (Ran(R)) = 0. Since ∃ NOM (R) is
simply Ran(R), and BCP 1
t (Ran(R)) = 0, BCP 1
t ◦ ∃ NOM (R) = 0. □
In summary, I have shown that quantification in Québec French is reducible to iterated
unary quantification; namely, I showed that the truth conditions for sentences formed
with the binary quantifier BCP QF are the same as the truth conditions for sentences
formed by existentially closing the event argument, and then applying BCP 1 .
3.2.1.1 Binary Quantification?
The formal result presented in the previous section raises the question of whether QAD
in Québec French is really binary quantification. We saw with the unreducibility proof
in chapter 2 that binary quantification is necessary to account for QAD in Standard
French, but since BCP QF is reducible to iterations of unary quantifiers, ∃ and the type
< 1 > BCP 1 , perhaps the meaning of J’ai beaucoup lu de livres is better modeled
using two different quantifiers, yielding the logical representation in (17).
(17) J’ai beaucoup lu de livres = BCP 1 s x ∃e(Reading (e,I,x) & Book(x))
An argument for a unary analysis of QAD in Québec French is that degree verbs do not
seem to block QAD in this dialect, or at least, the effect is not as strong. Cyr (1991)
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