Session 08

Introduction to Semantics 

Session 8 

Quantification 

Cornelia Endriss 

Cognitive Science Program 

University of Osnabrück 

cendriss@uos.de

Outline for Today 

1. Homework from 2 weeks ago 

2. Quantificational DPs: Type and Lexical Semantics 

• First Try: Individuals 

• Second Try: Sets 

• Finally: Sets of Sets 

3. Quantificational Determiners 

4. Formal Properties of Quantifiers 

03.06.2008 Slides based on semantics textbook from I. Heim & A. Kratzer 

and lecture notes from M. Krifka 

2

Homework 

Homework 



3

Homework 

1. The intensifier very combines with adjectives as the following DP 

illustrates: 

the very intelligent student 

Which class of adjectives can be intensified by very? Why can 

others not be intensified by very? Figure out the semantic type of 

very and try to give a lexical semantics for it. For instance, you can 

think of a very intelligent student as intelligent compared to the 

intelligent students. 

2. H&K p.79: Exercise 1 (escalator in South College) on the definite 

determiner. 

3. H&K p.80: Exercise 2 (apples in a row) on the definite determiner. 



4

Very 

Which class of adjectives can be intensified by very? 

The so-called gradable ones. According to the classification from 

H&K’s textbook the non-intersective non-intensional ones: 

(a) the very tall boy, the very intelligent girl. 

I.e. not the intersective ones: 

• (b) #the very red ball, #the very round circle 

but watch out for metalinguistic uses 

• Note also: 

(c) the very wet coat, the very full cupboard 

those so-called absolute standard adjectives are still gradable 

and hence modifiable by very 

I.e. not the intensional ones: 

(d) #the very former president 



5

Very 

Figure out the semantic type of very and try to give a lexical 

semantics for it. For instance, you can think of a very intelligent 

student as intelligent compared to the intelligent students. 

• Type: 

• Idea: very is an adjective-duplicater. 

• Adjective doubling is common in many languages 

Lit: 

the mean mean dog = 

‘the very mean dog’ 



6

Very 

’very÷ = λR λQ . R(R(Q)) 

’mean÷ = λPλx. P(x) & the meanness of x is above the average meanness of 

the elements of {y : P(y)} 

’very÷(’mean÷) = λQ λPλx. P(x) & the meanness of x is above the average 

meanness of the elements of {y : P(y)}(λPλx. P(x) & the meanness of x is 

above the average meanness of the elements of {y : P(y)}(Q)) 

= λQ λPλx. P(x) & the meanness of x is above the average meanness of the 

elements of {y : P(y)}(λx. Q(x) & the meanness of x is above the average 

meanness of the elements of {y : Q(y)}) 

= λQ λx. Q(x) & the meanness of x is above the average meanness of the 

elements of {y : Q(y)} & the meanness of x is above the average meanness 

of the elements of {y : Q(y) & the meanness of y is above the average 

meanness of the elements of {y : Q(y)}} 



7

Very 

’very÷(’mean÷)(’dog÷) = λQ λx. Q(x) & the meanness of x is above the 

average meanness of the elements of {y : Q(y)} & the meanness of x is 

above the average meanness of the elements of {y : Q(y) & the meanness of 

y is above the average meanness of the elements of {y : Q(y)}} (λz. dog(z)) 

= λx. [λz.dog(z)(x)] & the meanness of x is above the average meanness of 

the elements of {y : λz. dog(z)(y)} & the meanness of x is above the average 

meanness of the elements of {y : [λz. dog(z)(y)] & the meanness of y is 

above the average meanness of the elements of {y : [λz. dog(z) (y)]}} 

= λx. dog(x) & the meanness of x is above the average meanness of the 

elements of {y : dog(y)} & the meanness of x is above the average 

meanness of the elements of {y : dog(y) & the meanness of y is above the 

average meanness of the elements of {y : dog(y)}} 



8

Very 

= λx. dog(x) & the meanness of x is above the average meanness of the 

elements of {y : dog(y) & the meanness of y is above the average meanness 

of the elements of {y : dog(y)}} 

= λx. dog(x) & the meanness of x is above the average meanness of a mean 

dog 



9

Very 

Note that many solutions that seem easier do not work, e.g.: 

’very÷ c 

= λPλx. x’s P-property is above s, where s is the standard 

made salient by the context of all y which are P in c. 

This solution does not work, because we do not know the “Pproperty”, 

we simply know the extension of P. 

Hence all solutions that assume ’very÷ to be of type 

are doomed to fail. 

Otherwise the set of very intelligent students would necessarily 

have to be the same as the set of very tall students, if the sets of 

intelligent creatures and the set of tall creatures happen to coincide 

in the world and the context under discussion. 



10

Escalator: H&K 

What does our current theory predict? 

Undefined because semantics of one 

of the daughter nodes undefinded 





’the÷(’escalator in South College÷) = 

λP∈D and there is ex. 1 x st. P(x)=1. 

(the) y such that P(y) = 1 (λx.e-i-SC(x)) 

is undefined 

because P is the empty 

set. Hence 

presupposition failure. 



11

Escalator: Russel 

What would Russel predict? 

i’John uses the escalator in South College÷ = 1 

’John uses the escalator in South College÷ = 

∃x.e-i-SC(x) ∧ ∀y[e-i-SC(y) → x=y] ∧ use(john,x) = 0 



12

Escalator: intuitions 

What is empirically correct? 

Debatable. Depends on the context. There are contexts (maybe 

contrastive ones) where one could say something like John did 

not use the escalator in South College, because there is none. 

So here it seems that the definite determiner is not 

presuppositional or at least the entire sentence does not carry 

an existence presupposition. So here the current theory would 

make false predictions. 



13

Apples: intuitions 

b 1 b 2 b 3 b 4 b 5 b 6 

(i) (ii) (iii) = undefined 

(i) the leftmost apple in the row 

(ii) the leftmost dark apple in the row 

(iii) the apple that is both leftmost in the row and dark 



14


DP 

’the÷(’leftmost apple in the row÷) = 

the unique y such that y is an apple in the row 

and y is leftmost 

the 

NP 

PM( x.x is an apple in the row, 

x.x is leftmost) = 

x.x is an apple in the row and x is leftmost 

leftmost N' 

x.x is an apple in the row 

apple in the row 

Assumption: ’leftmost÷ = λx. x is leftmost 



15


b 1 b 2 b 3 b 4 b 5 b 6 

(i) (ii) 

(iii) = undefined 

̌ 






16


the 

 

DP 

leftmost 

 

’the÷( x. x is leftmost and x is dark and x is an apple 

in the row) = the unique y such that y is an apple 

in the row and y is leftmost and y is dark 

NP 

PM( x.x is an apple in the row and x is dark, 

x.x is leftmost) = x.x is an apple in the row and 

x is dark and x is leftmost 

N' 

PM( x.x is an apple in the row, 

x.x is leftmost) = 

x.x is an apple in the row and x is dark 

dark N' 

 





17


b 1 b 2 b 3 b 4 b 5 b 6 

(i) (ii) 


̌ 

 






18

(iii) the apple that is both leftmost in 

the row and dark 

b 1 b 2 b 3 b 4 b 5 b 6 

(i) (ii) 


̌ 

 

̌ 






19

What we derive so far 

• our one-place predicate treatment of “leftmost” is 

appropriate for (i) and (iii) 

• but “the leftmost dark apple in the row” (ii) has the 

same truth-conditions as “the apple that is both 

leftmost in the row and dark” (iii) and thus goes 

against our intuitions 



20

Lexical Semantics of leftmost 

• It has to be a function of type D 

• ’leftmost÷ = λP c D . [λx c D e . P(x) and x is the leftmost 

element among the elements of {y: P(y)}] 

• “the leftmost dark apple in the row” denotes the leftmost apple 

in the row of dark apples 



21


the unique y st. y is an apple in the row and y is dark and y is the leftmost 

element among the elements of {y: y is an apple in the row and y is dark}] 

DP P x.P(x) and x is the leftmost element 

among the elements of {y: P(y)} 

( x.x is an apple in the row and x is dark) 

the NP = x.x is an apple in the row and x is dark and 

 

x is the leftmost element among the elements 

of {y: y is an apple in the row and y is dark} 

leftmost 

 

N' 

x.x is an apple in the row and x is dark 

dark 

 

N' 





22


b 1 b 2 b 3 b 4 b 5 b 6 

(ii) (ii) 


̌ 






23





24

More DPs 

Do all DPs behave like proper names or definite descriptions? 

S 

DP 

VP 

Fred 

Fred 

walks 

λx ∈ D e 

.x walks 

’Fred walks ÷ = 1 iff λx ∈D e 

. x walks(Fred) 

iff Fred ∈ walk 



25

More DPs 

Do all DPs behave like proper names or definite descriptions? 

S 

DP 

VP 

D 

NP 

the man walks 

λf∈D the unique… λx ∈ D e .x is a man λx ∈ D e .x walks 



26

More DPs 

A woman 

Somebody 

Everybody 

Few people 

Nobody 

At most two people 

– an arbitrary individual? 

– every individual? A set? 

– an arbitrary small set? 

– no individual? The empty set? 

– an arbitrary set of 

at most two people? 



28


Quantificational DPs 

as Individuals 



29

Quantifiers as Individuals? 

John lives in Osnabrück 

u John lives in Germany 

(i) 

’ John ÷ ∈ D e 

(ii) ’ lives in Osnabrück ÷ ⊆ ’ lives in Germany ÷ 

(iii) F char 

(x) = 1 iff x ∈ F 

Nobody lives in Osnabrück 

^ Nobody lives in Germany 

At most two people live in OS 

^ At most two people live in Germany 

Few people live in Osnabrück 

^ Few people live in Germany 

Intuitively invalid, but could be derived if the DPs were of type e! 



30


John is in this room 

u It’s not the case: John is outside this room 

(i) 

(ii) 


’ is in this room ÷ ∩ ’ is outside this room ÷ = ∅ 

(iii) F(x) = 1 iff x ∈ F 

A woman is in this room 

^ It’s not the case: a woman is outside this room 

Somebody is in this room 

^ It’s not the case: somebody is outside this room 

Three people are in this room 

^ It’s not the case: three people are outside this room 

Intuitively invalid, but could be derived if the DPs were of type e! 



31


Tautological (= always true) statement: 

(i) 

(ii) 

John is over 30 years old or John is under 40 years old 


’ be over 30 years old ÷ ∪ ’ be under 40 years old ÷ = D e 

(iii) F(x) = 1 iff x ∈ F 

Everybody is over 30 years old or everybody is under 40 years old 

At most three people are over 30 years old or at most three people are under 40 

years old 

Intuitively non-tautological, but would be derived as tautology if the 

DPs were of type e! 



32



as Sets 



33

Quantifiers as Sets? 

The problems indicate: DPs do (in general) not denote an individual of 

type e. So maybe they denote sets of individuals (i.e. they are of type 

)? 

’ Fred ÷ = { Fred } 

’ everything ÷ = D e 

’ nothing ÷ = ∅ 

’ at most 2 people ÷ = an arbitrary set of at most 2 people 

Compositional rules (or the semantics of verbs) has to be changed 

accordingly. Assume: 

’S ÷ = ’ DP VP ÷ = 1 

iff ’ DP ÷ ⊆ ’ VP ÷ 

Alternatively, change the semantics of the 

VP: 

’ lives in Osnabrück ÷ = 

λX ∈ Pow(D). X ⊆ {y: y lives in Osnabrück} 



34

Does this proposal yield adequate results for the following 

sentences? 

S 


DP 

N 

Fred 

VP 

V 

lives in Osnabrück 

S 

{ Fred } ⊆ {y: y lives in Osnabrück } 

iff Fred lives in Osnabrück 

DP 

Everybody 

VP 

V 

D e ⊆ {y: y lives in Osnabrück } 

iff every individual lives in Osnabrück 




35


Does this proposal yield adequate results for the following 

sentences? 

DP 

Nobody 

S 

VP 

V 

∅ ⊆ {y: y lives in Osnabrück } 

always true! 


DP 

At most 2 people 

S 

VP 

V 

an arbitrary set of at most 2 people 

⊆ {y: y lives in Osnabrück } 

also always true! 

live in Osnabrück 



36


What about unintuitive inferences? 

Nobody lives in Osnabrück ^ Nobody lives in Germany 

Still predicted! 

∅ ⊆ {y: y lives in Osnabrück } ⊆ {y: y lives in Germany } 



37



as Sets of Sets 



38

Solution: Change Type of DPs 

So far: 

VP (DP ) 

S 

 

DP 

 

VP 

 

… how about: 

DP (VP ) 

S 

 

DP 

 

VP 

 



39

What does it mean to change the type of DPs 

from to ? 

What sort of things are of the type ? 

functions are properties of properties, 

i.e. 2nd order properties 

For instance: 

At least 2 people walk d 

Everybody walks d 

Properties of Properties 

the property of walking is a property 

that at least 2 people have. 

the property of walking is a property 

that everybody has. 

The walk property is in the set of properties that 2 people 

have/everybody has. 



40

Functional View vs. Set View 

A Generalized Quantifier (GQ) is a function from functions to truth 

values: 

’ nothing ÷ = λf ∈ D 

. there is no x∈D e 

such that f(x) = 1 

’ something ÷ = λf ∈ D 

. there is some x∈D e 

such that f(x) = 1 

’ everything ÷ = λf ∈ D 

. for all x∈D e 

: f(x) = 1 

A different view of the same thing: A GQ is a set of sets 

’ nothing ÷ = { X ∈ Pow(D) : X = ∅ } 

’ something ÷ = { X ∈ Pow(D) : X ≠ ∅ } 

’everything ÷ = { X ∈ Pow(D) : X = D } 

Definitions for truth and falsity have to be amended accordingly: 

’ S ÷ = ’ DP VP ÷ = 1 iff ’ VP ÷ ∈ ’ DP ÷ 



41

Quantifiers as Sets of Sets 

GQs as sets of sets: 

’ no woman ÷ = { X ∈ Pow(D) : woman ∩ X = ∅ } 

’ nobody ÷ = { X ∈ Pow(D) : people ∩ X = ∅ } 

’ a man ÷ = { X ∈ Pow(D) : man ∩ X ≠ ∅ } 

’ somebody ÷ = { X ∈ Pow(D) : people ∩ X ≠ ∅ } 

’ every dog ÷ = { X ∈ Pow(D) : dog ⊆ X } 

’ everybody ÷ = { X ∈ Pow(D) : people ⊆ X } 

’ at least 2 people÷ = { X ∈ Pow(D) : |X ∩ people| ≥ 2 } 

’ at most 2 people÷ = { X ∈ Pow(D) : |X ∩ people| ≤ 2 } 



42

Problematic Cases Solved 

At most 2 people live in Osnabrück. 

’ live in Osnabrück ÷ 

’ at most 2 people ÷ 

= { X ∈ Pow(D) : |X ∩ people| ≤ 2 } 

Only true iff at most 2 people 

and no more live in Osnabrück. 



43

Problematic Cases Solved 

Nobody lives in Osnabrück. 


’ Nobody ÷ 

= { X ∈ Pow(D) : X ∩ people = ∅ } 

Not tautological, but only true iff 

nobody lives in Osnabrück. 



44

Solving the Inference Problem 

At most 2 people live in Osnabrück ^ 

At most 2 people live in Germany 


’ at most 2 people ÷ 

’ live in Germany ÷ 



45

Solving the Inference Problem 

Somebody is in this room 

^ It is not the case: somebody is outside this room 

’ be in this room ÷ 

’ somebody ÷ 

’ be outside this room ÷ 



46


Quantificational 

Determiners 



47

Quantificational Determiners 

Given that quantifying DPs have lexical entries like these: 

’ nothing÷ = λf ∈ D . there is no x∈D e such that f(x) = 1 

’ no woman ÷ = λf ∈ D . there is no x∈D e such that x is a woman 

& f(x) = 1 

What are the lexical entries for determiners like no, a, every? 

S 

DP 

VP 

D 

no 

NP 

man sleeps 



48

Quantificational Determiners: 

Functional View 

’ no ÷ = λf ∈ D . [λg ∈ D . . there is no x ∈ D e such that 

f(x) = 1 & g(x) = 1] 

’ a ÷ = λf ∈ D . [λg ∈ D . there is some x ∈ D e such that 

f(x) = 1 & g(x) = 1] 

’ every ÷ = λf ∈ D . [λg ∈ D . for all x ∈ D e such that 

f(x) = 1, g(x) = 1] 



49

Derivation of Truth Conditions 

Computing the truth conditions for A man left 

S 

there is some x ∈ D e such that x is a man and x left 

DP > 

λg ∈ D . there is some x ∈ D e such that x is a man & g(x) = 1 

D 

a 

λf ∈ D . [λg ∈ D . 

there is some x ∈ D e 

such that f(x) = 1 & g(x) = 1] 

NP 

man 

Py ∈ D e . y is a man 



VP 

left 

Pz ∈ D e . z left 

50

Quantif. Ds: Relational View 

Recall the ‘set of sets’ view on generalized quantifiers, e.g. 

’ something ÷ = { X ∈ Pow(D) : X ≠ ∅ } 

The same works for GQs corresponding to DPs in general: 

’ no woman ÷ = { X ∈ Pow(D) : woman ∩ X = ∅ } 

’ a man ÷ = { X ∈ Pow(D) : man ∩ X ≠ ∅ } 

’ every dog ÷ = { X ∈ Pow(D) : dog ⊆ X } 

Under this view, determiners denote relations between sets: 

’ no ÷ = { ∈ Pow(D) % Pow(D): A ∩ B = ∅ } 

’ a ÷ = { ∈ Pow(D) % Pow(D): A ∩ B ≠ ∅ } 

’ every ÷ = { ∈ Pow(D) % Pow(D): A ⊆ B } 



51

Functional and Relational View 

Truth conditions under the relational and the functional view: 

’ a man left ÷ = 1 b relational view 

iff ∈ ’ a ÷ 

iff ∈ { ∈ Pow(D) % Pow(D): A ∩ B ≠ ∅ } 

iff man ∩ left ≠ ∅ 

iff there is some x ∈ D e 

such that x ∈ man and x ∈ left 

iff there is some x ∈ D e 

such that x is a man and x left 

iff λg ∈ D 

[there is some x ∈ D e 

such that 

x is a man & g(x) = 1](left) 

iff λf ∈ D 

[λg ∈ D 

[there is some x ∈ D e 

such that 

f(x) = 1 & g(x) = 1](man)(left) 

iff (’ a ÷(’ man ÷))(’ left ÷) = 1 



b functional view 

52

Notations 

• People usually refer to a quantificational DP such as every 

woman/someone/most horses (of type ) as 

Generalized Quantifier (GQ) or simply as quantifier. 

• The quantificational determiner (of type 

) is sometimes also called a quantifier. 

• We say that in 

Every horse sleeps 

horse is the restrictor of the quantificational determiner 

every and sleeps is the nucleus or (nuclear) scope. 



53


Formal Properties of 

Quantifiers 



54

A determiner D can be classified according to certain formal properties of its 

meaning under the relational view ’ D ÷. 

Symmetry 

A determiner D is called symmetric 

iff for all sets A and B: if ∈ ’ D ÷ then ∈ ’ D ÷ 

For instance: is a symmetric? 

Informal check via natural language: 

Symmetry 

a German won an academy award 

an academy award winner was German 

and hence 

for particular A (being German) and B (being AA winner) 

Formal proof (for general A and B): 

suppose A and B are sets such that ∈ ’ a ÷ 

hence A ∩ B ≠ ∅ and likewise B ∩ A ≠ ∅ 

therefore ∈ ’ a ÷ 



55

Symmetry 

Another example: Is every symmetric? 

Informal check via natural language: 

every German nominee won an AA 

every AA winner was a German nominee 

for particular A (being German nominee) and B (being AA winner) 

From the first sentence one cannot conclude to the second 

Formal proof (for general A and B): 

suppose A and B are sets such that ∈ ’ every ÷ 

hence A ⊆ B, 

but then B ⊆ A only if A = B, 

therefore v ’ every ÷ unless A = B 

Conclusion: a is symmetric, but every is not. 



56

Monotonicity 

Monotonicity 

Different DPs allow for different inference patterns. 

(a) A bachelor smoked cigars u A man smoked cigars 

(b) A bachelor smoked cigars u A bachelor smoked 

(c) A bachelor smoked cigars ^ A young bachelor smoked cigars 

(d) A bachelor smoked cigars ^ A bachelor smoked Cuban cigars 

(a) Every bachelor smoked cigars ^ Every man smoked cigars 

(b) Every bachelor smoked cigars u Every bachelor smoked 

(c) Every bachelor smoked cigars u Every young bachelor smoked cigars 

(d) Every bachelor smoked cigars ^ Every bachelor smoked Cuban cigars 



57

Monotonicity 

Monotonicity 

(a) No bachelor smoked cigars ^ No man smoked cigars 

(b) No bachelor smoked cigars ^ No bachelor smoked 

(c) No bachelor smoked cigars u No young bachelor smoked cigars 

(d) No bachelor smoked cigars u No bachelor smoked Cuban cigars 

Important Observation: 

In the (a) cases, the first argument of the determiner was enlarged: 

bachelor ⊆ man 

In the (b) cases, the second argument of the determiner was enlarged: 

smoke cigars ⊆ smoke 

In the (c) cases, the first argument of the determiner was reduced: 

bachelor r young bachelor 

In the (d) cases, the second argument of the determiner was reduced: 

smoke cigars r smoke Cuban cigars 



58

Monotonicity 

Monotonicity 

Determiners are classified according to the inference patterns they 

allow. 

Let A, B, C be sets such that A ⊆ B . Then a determiner D is called 

left upward monotone 

if ∈ ’ D ÷ then ∈ ’ D ÷ 

left downward monotone 

if ∈ ’ D ÷ then ∈ ’ D ÷ 

right upward monotone 

if ∈ ’ D ÷ then ∈ ’ D ÷ 

right downward monotone 

if ∈ ’ D ÷ then ∈ ’ D ÷ 

(a) cases 

(b) cases 

(c) cases 

(d) cases 



59

Monotonicity 

Monotonicity 

The examples above illustrate: 

a is left upward and right upward monotone 

every is left downward and right upward monotone 

no is left downward and right downward monotone 

Determiners that are neither upward nor downward monotone 

in one argument are called non-monotone in that argument. 



60

Homework 

To be handed in by Saturday next week (June 14th). 

– Read Chapters 6 and 7 of Heim & Kratzer’s textbook and go through the 

lecture notes again. 

– H&K p. 198: Exercise (calculating truth-conditions of (7’)). 

– It is shown in the textbook (pp.159-160) that the Aristotelian 

understanding of the determiners all, some, no differs from the "modern" 

understanding. Peter Strawson (cf. p. 161 of the textbook) proposed a 

view of the semantics of these determiners that is intended to save the 

Aristotelian interpretation. 

(a) 

(b) 

(c) 

(d) 

Describe the Aristotelian and the modern semantics of the 

determiners all, some, no (using material from the textbook). 

Show where precisely there is a conflict. 

Describe Strawson's idea for solving the conflict. 

Give your own assessment of Strawson's position and discuss any 

consequences you find interesting. 



61

Thank you! 

Thank you! 



62

Session 08

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