PLIN3004/PLING218 Advanced Semantic Theory
1
Assignment 8
Monotonicity
Assume the following lexical entries:
(1)
a.
b.
c.
d.
e.
f.
vawa, M “ rλg P Dxe,t y .rλ f P Dxe,t y . 1 iff setp f q X setpgq ‰ Hss
vnowa, M “ rλg P Dxe,t y .rλ f P Dxe,t y . 1 iff setp f q X setpgq “ Hss
vat most fivewa, M “ rλg P Dxe,t y .rλ f P Dxe,t y . 1 iff |setp f q X setpgq| ď 5ss
vexactly twowa, M “ rλg P Dxe,t y .rλ f P Dxe,t y . 1 iff |setp f q X setpgq| “ 2ss
vten or morewa, M “ rλg P Dxe,t y .rλ f P Dxe,t y . 1 iff |setp f q X setpgq| ě 10ss
|setp f q X setpgq|
vMore than 20% of thewa, M “ rλg P Dxe,t y .rλ f P Dxe,t y . 1 iff
ą 0.2ss
|setpgq|
For each of the these quantifiers, indicate its monotonicity properties by completing the following
table. The choices are Up (Ò), Down (Ó) or Non-monotonic (N). Motivate your answers with
data showing entailment patterns (no need for formal proofs, which are left for the optional
exercise).
every
no
a
at most five
exactly two
ten or more
more than 20% of the
2
Left
Ó
Right
Ò
Anti-additivity
Here’s a formal property similar to monotonicity:
(2)
A function Q P Dxet ,xet ,t yy is right anti-additive iff the following holds for any functions
f , g, g 1 P Dxe,t y :
Qp f qpgq “ 1 and Qp f qpg 1 q “ 1 iff Qp f qprλ x P De . 1 iff gpxq “ 1 or g 1 pxq “ 1sq “ 1
(3)
A function Q P Dxet ,xet ,t yy is left anti-additive iff the following holds for any functions
f , f 1 , g P Dxe,t y :
Qp f qpgq “ 1 and Qp f 1 qpgq “ 1 iff Qprλ x P De . 1 iff f pxq “ 1 or f 1 pxq “ 1sqpgq “ 1
In terms of sets, setprλ x P De . 1 iff f pxq “ 1 or f 1 pxq “ 1sq “ setp f q Y setp f 1 q. And
in English, if f and f 1 are the denotations of verbs, e.g. ‘drinks’ and ‘smokes’, rλ x P
De . 1 iff f pxq “ 1 or f 1 pxq “ 1s is the VP-disjunction ‘drinks or smokes’. Similarly for NPs,
e.g. if f and g are the denotations of ‘student’ and ‘professor’, rλ x P De . 1 iff f pxq “ 1 or f 1 pxq “ 1s
is the denotation of ‘student or professor’.
veverywa, M is left anti-additive, as illustrated by the equivalence of (4a) and (4b):
1
(4)
a.
b.
Every student drinks and every professor drinks.
Every student or professor drinks.
As you can see here, anti-additivity allows you to turn a sentence-level conjunction into a phrasal
level disjunction.
veverywa, M is not right anti-additive, as shown by the non-equivalence of (5a) and (5b).
(5)
a.
b.
Every student drinks and every student smokes.
Every student drinks or smokes.
Some notes:
• When checking left anti-additivity, keep the VP constant.
additivity, keep the NP constant.
When checking right anti-
• As in the case of monotonicity, you need a proof to show that something is left/right antiadditive, because it is required that the equivalence holds for any three functions of type
xe, ty, while to show that something is not left/right anti-additive, one example is enough.
For each of the following determiners, indicate whether it is left and/or right anti-additive, by
filling the table with Yes or No. Motivate your answers with pairs of examples (that are either
equivalent or not equivalent, like (4) and (5)). Proofs are not necessary here.
every
no
a
at most five
exactly two
ten or more
more than 20% of the
Left
Yes
Right
No
(Some scholars claim that there is a class of NPIs called strong NPIs (e.g. ‘at all’) which are
only licensed by anti-additivity. If this claim is right, anti-additivity is another formal property
relevant for linguistic phenomena.)
3
(Optional) Proofs
Give proofs to your answers above, whenever appropriate (i.e. when the answer was Ò or Ó in
Exercise 1 and when the answer was Yes in Exercise 2).
See the lecture notes for examples proofs of monotonicity.
Here’s a proof for the left anti-additivity of veverywa, M . The goal is to show the equivalence of
‘veverywa, M p f qpgq “ 1 and veverywa, M p f 1 qpgq “ 1’ and ‘veverywa, M prλ x P De . 1 iff f pxq “
1 or f 1 pxq “ 1sqpgq “ 1. You can do this by showing that entailment holds in both directions.
Proof. pñq Take three arbitrary functions f , f 1 , g P Dxe,t y , and assume that veverywa, M p f qpgq “
1 and veverywa, M p f 1 qpgq “ 1. Then we have setp f q Ď setpgq and setp f 1 q Ď setpgq. This means
every member of set(f) is a member of set(g) and every member of set(f’) is a member of
set(g). So this implies that psetp f q Y setp f 1 qq Ď setpgq. Since setp f q Y setp f 1 q “ setprλ x P
2
De . 1 iff f pxq “ 1 or f 1 pxq “ 1sq, this means veverywa, M prλ x P De . 1 iff f pxq “ 1 or f 1 pxq “
1sqpgq “ 1.
pðq Take three arbitrary functions f , f 1 , g P Dxe,t y , and assume that veverywa, M prλ x P
De . 1 iff f pxq “ 1 or f 1 pxq “ 1sqpgq “ 1 is the case. Recall that veverywa, M is downward monotonic. Given that setp f q Ď psetp f q Y setp f 1 qq, veverywa, M prλ x P De . 1 iff f pxq “ 1 or f 1 pxq “
1sqpgq “ 1 entails veverywa, M p f qpgq “ 1. Similarly, because setp f 1 q Ď psetp f q Y setp f 1 qq,
veverywa, M prλ x P De . 1 iff f pxq “ 1 or f 1 pxq “ 1sqpgq “ 1 entails veverywa, M p f 1 qpgq “ 1.
3