Semantics Boot Camp

Elizabeth Coppock

Compilation of handouts from NASSLLI 2012, Austin Texas(revised December 15, 2012)

Contents

1 Semantics 2

2 Predicate calculus 42.1 Syntax of Predicate Calculus . . . . . . . . . . . . . . . . . . . . . . 4

2.1.1 Atomic symbols . . . . . . . . . . . . . . . . . . . . . . . . . 42.1.2 Syntactic composition rules . . . . . . . . . . . . . . . . . . 5

2.2 Semantics of Predicate Calculus . . . . . . . . . . . . . . . . . . . .62.2.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2.2 Ordered pairs and relations . . . . . . . . . . . . . . . . . . . 72.2.3 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2.4 Set theory as meta-language . . . . . . . . . . . . . . . . . . 92.2.5 Models and interpretation functions . . . . . . . . . . . . . .92.2.6 Interpretation rules . . . . . . . . . . . . . . . . . . . . . . . 10

3 English as a formal language 113.1 A fragment of English with set denotations . . . . . . . . . . . .. . 11

3.1.1 Models and the Lexicon . . . . . . . . . . . . . . . . . . . . 113.1.2 Composition rules . . . . . . . . . . . . . . . . . . . . . . . . 13

3.2 A fragment of English satisfying Frege’s conjecture . . .. . . . . . 143.3 Transitive and ditransitive verbs . . . . . . . . . . . . . . . . . .. . . 19

4 Fun with Functional Application 204.1 Rick Perry is conservative . . . . . . . . . . . . . . . . . . . . . . . . 204.2 Rick Perry is in Texas . . . . . . . . . . . . . . . . . . . . . . . . . . 214.3 Rick Perry is proud of Texas . . . . . . . . . . . . . . . . . . . . . . . 224.4 Rick Perry is a Republican . . . . . . . . . . . . . . . . . . . . . . . . 22

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5 Predicate Modification 235.1 Rick Perry is a conservative Republican . . . . . . . . . . . . . .. . 235.2 Austin is a city in Texas . . . . . . . . . . . . . . . . . . . . . . . . . 24

6 The definite article 256.1 The negative square root of 4 . . . . . . . . . . . . . . . . . . . . . . 256.2 Top-down evaluation . . . . . . . . . . . . . . . . . . . . . . . . . . . 266.3 Bottom-up style . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

7 Predicate calculus: now with variables! 32

8 Relative clauses 34

9 Quantifiers 389.1 Typee? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399.2 Solution: Generalized quantifiers . . . . . . . . . . . . . . . . . .. . 41

10 The problem of quantifiers in object position 4210.1 The problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4210.2 Anin situ approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4210.3 A Quantifier Raising approach . . . . . . . . . . . . . . . . . . . . . 4410.4 Arguments in favor of the movement approach . . . . . . . . . .. . 47

11 Free and Bound Variable Pronouns 4811.1 Toward a unified theory of anaphora . . . . . . . . . . . . . . . . . .4811.2 Assignments as part of the context . . . . . . . . . . . . . . . . . .. 52

12 Our fragment of English so far 5312.1 Composition Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5312.2 Additional principles . . . . . . . . . . . . . . . . . . . . . . . . . . .5412.3 Lexical items . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

1 Semantics

Semantics:The study of meaning. What ismeaning? How can you tell whethersomebody or somethingunderstands? Does Google understand language?

An argument that it does not: Google can’t doinferences:

(1) Everyone who smokes gets cancer→ No heavy smoker avoids cancer

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(cf. “No alleged smoker avoids cancer”...)

A hallmark of a system or agent that understands language / grasps meaning is thatit can do these kinds of inferences.

Another way to put it: A good theory of meaning should be able to account for theconditions under which one sentence implies another sentence.

Some different kinds of inferences:

1. Entailment (domain of semantics): A entails B if and only if whenever A istrue, B is true too. E.g.Obama was born in 1961entailsObama was born inthe 1960s.

2. Presupposition(semantics/pragmatics): A presupposes B if and only if anutterance of A takes B for granted. E.g.Sue has finally stopped smokingpresupposes that Sue smoked in the recent past.

3. Conversational Implicature (pragmatics): A conversationally implicates Bif and only if a hearer can infer B from an utterance of A by making use of theassumption that the speaker is being cooperative. E.g.Some of the studentspassedconversationally implicates that not all of the students passed.

The primary responsibility for a theory ofsemanticsis to account for the conditionsunder which one sentenceentailsanother sentence.

Other nice things a theory of semantics could do: Account forpresuppositions,contradictions, equivalences, semantic ill-formedness,distribution patterns.

Strategy: Assigntruth conditions to sentences. The truth conditions are the con-ditions under which the sentence is true. Knowing the meaning of a sentence doesnot require knowing whether the sentence is in fact true; it only requires being ableto discriminate between situations in which the sentence istrue and situations inwhich the sentence is false.

(Cf. Heim & Kratzer’s bold first sentence: “To know the meaning of a sentence isto know its truth conditions.”)

The strategy of assigning truth conditions will allow us to account for entailments.If the circumstances under which A is true include the circumstances under whichB is true, then A entails B.

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How to assign truth conditions to sentences of natural languages like English?Montague’s idea: Let’s pretend that English is a formal language.

I reject the contention that an important theoretical difference ex-ists between formal and natural languages. ... In the present paperI shall accordingly present a precise treatment, culminating in a the-ory of truth, of a formal language that I believe may reasonably beregarded as a fragment of ordinary English. ... The treatment givenhere will be found to resemble the usual syntax and model theory (orsemantics) [due to Tarski] of the predicate calculus, but leans ratherheavily on the intuitive aspects of certain recent developments in in-tensional logic [due to Montague himself]. (Montague 1970b, p.188in Montague 1974)

2 Predicate calculus

Predicate calculus is a logic. Logics are formal languages,and as such they havesyntax and semantics.

Syntax: specifies which expressions of the logic are well-formed, and what theirsyntactic categories are.

Semantics: specifies which objects the expression correspond to, and what theirsemantic categories are.

2.1 Syntax of Predicate Calculus

2.1.1 Atomic symbols

Formulas are built up from atomic symbols drawn from the following syntacticcategories:

• individual constants: JOHN, MARY, TEXAS, 4...

• variables: x, y, z, x′, y′, z′, ...

• predicate constants

– unary predicate constants: EVEN, ODD, SLEEPY,..

– binary predicate constants: LOVE, OWN, >, ...

– ...

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• function constants:

– unary function constants: MOTHER, ABSOLUTE VALUE , ...

– binary function constants: DISTANCE, +, −,...

– ...

• logical connectives: ∧,∨,→,↔,¬• quantifiers: ∀,∃

I write constants inSMALL CAPS and variables initalics. But I will ignore quan-tifiers and variables for now (they will be discussed later).

2.1.2 Syntactic composition rules

How to build complex expressions (Greek letters are metalanguage variables):

• If π is an-ary predicate andα1...αn are terms, thenπ(α1,...αn) is an atomicformula.

– If π is a unary predicate andα is a term, thenπ(α) is an atomic formula.

– If π is a binary predicate andα1 andα2 are terms, thenπ(α1, α2) is anatomic formula.

• If α1...αn are terms, andγ is a function constant with arityn, thenγ(α1, ..., αn)is a term.

• If φ is a formula, then¬φ is a formula.

• If φ is a formula andψ is a formula, then[φ ∧ ψ] is a formula, and so are[φ ∨ ψ], [φ → ψ], and[φ↔ ψ].Expression Syntactic categoryJOHN,MARY (individual) constantx variableHAPPY, EVEN unary predicate constantLOVE, > binary predicate constantLOVE(JOHN,MARY) (atomic) formulaHAPPY(x) (atomic) formulax > 1 (atomic) formulaMOTHER unary function constantMOTHER(JOHN) term

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2.2 Semantics of Predicate Calculus

Each expression belongs to a certain semantic type. The types of our predicatecalculus are:individuals, sets, relations, functions, andtruth values.

2.2.1 Sets

Set. An abstract collection of distinct objects which are calledthe membersorelementsof that set. Elements may be concrete (like the beige 1992 Toyota CorollaI sold in 2008, David Beaver, or your computer) or abstract (like the number 2, theEnglish phoneme /p/, or the set of all Swedish soccer players). The elements of aset are not ordered, and there may be infinitely many of them ornone at all.

You can specify the members of a set in two ways:

1. By listing the elements, e.g.:{Marge,Homer,Bart,Lisa,Maggie}2. By description, e.g:{x∣x is a human member of the Simpsons family}

Element. We write ‘is a member of’ with∈.

Empty set. Theempty set,written∅ or {}, is the set containing no elements.

Subset. A is asubsetof B, writtenA ⊆ B, if and only if every member ofA is amember ofB.

A ⊆ B iff for all x: if x ∈ A thenx ∈ B.

Proper subset. A is aproper subsetof B, writtenA ⊂ B, if and only ifA is asubset ofB andA is not equal toB.

A ⊂ B iff (i) for all x: if x ∈ A thenx ∈ B and (ii)A ≠ B.

Powerset. Thepowersetof A, written℘(A), is the set of all subsets ofA.

℘(A) = {S∣S ⊆ A}Set Union. Theunionof A andB, writtenA ∪B, is the set of all entitiesx suchthatx is a member ofA or x is a member ofB.

A ∪B = {x∣x ∈ A or x ∈ B}6

Set Intersection. The intersectionof A andB, writtenA ∩ B, is the set of allentitiesx such thatx is a member ofA andx is a member ofB.

A ∩B = {x∣x ∈ A andx ∈ B}2.2.2 Ordered pairs and relations

Ordered pair. Sets are not ordered.

{Bart,Lisa} = {Lisa,Bart}But the elements of anordered pairwritten ⟨a, b⟩ are ordered. Here,a is thefirstmemberandb is thesecond member.

⟨Bart,Lisa⟩ ≠ ⟨Lisa,Bart⟩We can also have ordered triples.

{Bart,Lisa,Maggie} = {Maggie,Lisa,Bart}⟨Bart,Lisa,Maggie⟩ ≠ ⟨Maggie,Lisa,Bart⟩

Relation. A binary relation is a set of ordered pairs. In general, a relation is a setof ordered tuples. For example, the ‘older-than’ relation among Simpsons kids:

{⟨Bart,Lisa⟩, ⟨Lisa,Maggie⟩, ⟨Bart,Maggie⟩}Note that this is aset. How many elements does it have?

Domain. Thedomainof a relation is the set of entities that are the first memberof some ordered pair in the relation.

Range. The rangeof a relation is the set of entities that are the second memberof some ordered pair in the relation.

2.2.3 Functions

Function. A function is a special kind of relation. A relationR fromA toB is afunction if and only if it meets both of the following conditions:

1. Each element in the domain is paired with just one element in the range.

2. The domain ofR is equal toA

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A function gives a singleoutput for a giveninput .

Are these relations functions?

f1 = {⟨Bart,Lisa⟩, ⟨Lisa,Maggie⟩, ⟨Bart,Maggie⟩}f2 = {⟨Bart,Lisa⟩, ⟨Lisa,Maggie⟩}

Easier to see when you notate them like this:

f1 =

⎡⎢⎢⎢⎢⎢⎣Bart → Lisa

→ MaggieLisa → Maggie

⎤⎥⎥⎥⎥⎥⎦f2 = [ Bart → Lisa

Lisa → Maggie]

Function notation. Just as sets can be specified either by listing the elements orby description, functions can be described either by listing the ordered pairs thatare members of the relation or by description. Example:

[x ↦ x + 1]• x is theargument variable

• x + 1 is thevalue description

Function application. F (a) denotes ‘the result of applying functionF to argu-menta’ or F of a’ or ‘F applied toa’. If F is a function that contains the orderedpair ⟨a, b⟩, then:

F (a) = bThis means that givena as input,F givesb as output.

The result of applying a function specified descriptively toits argument can nor-mally be written as the value description part, with the argument substituted for theargument variable.

[x ↦ x + 1](4) = 4 + 1 = 5

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2.2.4 Set theory as meta-language

All these set-theoretic symbols are formal symbols, but they are not part of thelanguage for which we are giving a semantics. They are being used to characterizethevaluesthat expressions of predicate calculus will have.

In that sense, we are using the language of set theory as a meta-language.

2.2.5 Models and interpretation functions

Interpretation with respect to a model. Expressions of predicate calculus areinterpretedin models. Models consist of a domain of individualsD and an inter-pretation functionI which assigns values to all theconstants:

M = ⟨D,I⟩A valuation function[[]]M , built up recursively on the basis of the basic inter-preation functionI, assigns to every expressionα of the language (not just theconstants) asemantic value[[α]]M .

Here are two models,Mr andMf (r for “real”, andf for “fantasy”/“fiction”/“fake”):

Mr = ⟨D,Ir⟩Mf = ⟨D,If ⟩

They share the same domain:

D = {Maggie,Bart,Lisa}In Mr, Bart is happy, but Maggie and Lisa are not:

Ir(HAPPY) = {Bart}In Mf , everybody is happy:

If(HAPPY) = {Maggie,Bart,Lisa}Both interpretation functions map the constant MAGGIE to the individual Maggie:

Ir(MAGGIE) =Maggie

If(MAGGIE) =Maggie

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What is the semantic value ofHAPPY(MAGGIE)? It should come out as false inMr, and true inMf . So what we want to get is:

[[HAPPY(MAGGIE)]]Mr= 0

[[HAPPY(MAGGIE)]]Mf= 1

What tells us this? We need rules for determining the values of complex expres-sions from the values of their parts. So far all we have is rules for determining thevalues of constants.

2.2.6 Interpretation rules

• ConstantsIf M = ⟨D,I⟩ andα is a constant, then[[α]]M = I(α).

• Complex termsIf α1...αn are terms andγ is a function constant with arityn, then[[γ(α1, ...αn)]]Mis [[γ]]M ([[α1]]M , ..., [[αn]]M ).

• Atomic formulaeIf π is ann-ary predicate andα1...αn are terms,[[π(α1, ..., αn)]]M = 1 iff

⟨[[α1]]M , ..., [[αn]]M ⟩ ∈ [[π]]MIf π is a unary predicate andα is a term, then[[π(α)]]M = 1 iff

[[α]]M ∈ [[π]]M• Negation[[¬φ]]M = 1 if [[φ]]M = 0; otherwise[[¬φ]]M = 0.

• Connectives[[φ ∧ ψ]]M = 1 if [[φ]]M = 1 and [[ψ]]M = 1; 0 otherwise. Similarly for[[φ ∨ ψ]]M , [[φ → ψ]]M , and[[φ↔ ψ]]M .

Example. BecauseHAPPY is a unary predicate and MAGGIE is a term, we canuse the rule for atomic formulae to figure out the value ofHAPPY(MAGGIE).[[HAPPY(MAGGIE)]]Mf = 1 iff [[MAGGIE]]Mf ∈ [[HAPPY]]Mf ,i.e. iff Maggie∈ {Maggie,Bart,Lisa}.

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[[HAPPY(MAGGIE)]]Mr= 1 iff [[MAGGIE]]Mr

∈ [[HAPPY]]Mr ,i.e. iff Maggie∈ {Bart}.How about:¬HAPPY(BART) ∨ LOVE(BART,MAGGIE) ?

3 English as a formal language

Montague: “I reject the contention that an important theoretical difference existsbetween formal and natural languages.”

So we will put (parsed) English inside the denotation brackets, instead of logic.

Example:

(2)

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

S

NP VP

N V

Barack smokes

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M

(Wrong: [[Barack smokes]]M )

Following Heim & Kratzer, I use bold face for object languageinside denotationbrackets here, but often I am lazy about this.

3.1 A fragment of English with set denotations

3.1.1 Models and the Lexicon

Again, a modelM = ⟨D,I⟩ is a pair consisting of a domainD and an interpretationfunction I. For simplicity, let us assume that all of our models have thesamedomainD. There are two important subsets ofD:

• De, the set of individuals

• Dt, the set of truth values, 0 and 1

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As in predicate calculus, models come with an interpretation function that specifiesthe values of all the constants.

For concreteness, assume:

(3) De = {Barack Obama, Angela Merkel, Hugh Grant}The constants of natural language are the lexical items: theproper names, the in-transitive verbs, the adjectives, etc. Thus the interpretation function I will maplexical items to their values in a model.

For example, letM1 = ⟨D,I1⟩, and letI1 be defined such that:

(4) a. I1(Barack) = Barack Obama

b. I1(Angela) = Angela Merkel

c. I1(Hugh) = Hugh Grant

d. I1(smokes) = {Barack Obama, Angela Merkel}e. I1(drinks ) = {Barack Obama, Hugh Grant}f. I1(likes) = {⟨Barack Obama, Hugh Grant⟩,⟨Hugh Grant, Barack

Obama⟩,⟨Hugh Grant, Hugh Grant⟩ }For example, letM2 = ⟨D,I2⟩, and letI2 be defined such that:

(5) a. I2(Barack) = Barack Obama

b. I2(Angela) = Angela Merkel

c. I2(Hugh) = Hugh Grant

d. I2(smokes) = {Angela Merkel}e. I2(drinks ) = {}f. I2(likes) = {⟨Hugh Grant, Hugh Grant⟩}

Here, as in predicate calculus, we are defining the semantic values of intransitiveverbs as sets, and we are defining the semantic values of transitive verbs as rela-tions. But stay tuned; later we will interpret intransitiveverbs as thecharacteristicfunctionsof sets instead, in order to explore Frege’s conjecture, that all semanticcomposition can be done with one single rule.

Now we want to define avaluation function that assigns semantic values to allexpressions of the language, not just constants. For constants, it’s easy; just use theinterpretation function. For example:

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(6) [[smokes]]M1 = I1(smokes) = {Barack Obama, Angela Merkel}In general, ifα is a constant andM = ⟨D,I⟩ is a model, then:

(7) Interpretation Rule: Lexical Terminals (LT)[[α]]M = I(α)

But what about complex expressions like the tree in (10)?

3.1.2 Composition rules

For non-branching nodes, we use the semantic value of the daughter:

(8) Interpretation Rule: Non-branching Nodes

If α has the formγ

βthen [[α]]M = [[β]]M .

Sentences are true iff the value of the subject NP is an element of the value of theset denoted by the VP:

(9) Interpretation Rule: Sentences

If α has the formS

β γthen [[α]]M = 1 iff [[ β]]M ∈ [[γ]]M .

So it will turn out that:

(10)

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

S

NP VP

N V

Barack smokes

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M1

= 1

because Barack Obama is an element of [[smokes]]M1 .

But what do we do about transitive verbs?

(11) Interpretation Rule: Transitive VPs

If α has the formVP

β γthen [[α]]M = ???

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We have said that the verb denotes a relation, but what does the verb combined withthe object denote? Intuititively, it is a relation that is incompletely saturated. Wecan solve this problem by treating transitive verbs as functions that, when appliedto an individual (the one denoted by the object NP), yield something correspondingto a set of individuals. Generalizing this idea, we will alsotreat VPs as functions.When applied to their subject NP, they will give us a truth value. So intransitiveverbs will denote functions from individuals to truth values and intransitive verbswill denote functions from individuals to functions from individuals to truth values.

3.2 A fragment of English satisfying Frege’s conjecture

Scientific question: What semantic interpretation rules do we need in order tocalculate the values of complex expressions from the valuesof their parts?

Frege’s conjecture: We only need one composition rule: Functional application.

This will help us with our VP problem, but we have to change howverbs areinterpreted a little bit. Now we will have interpretations like the following. Recallthat before we said thatI1(smokes) = {Barack Obama, Angela Merkel}. Now wewill treat smokesas thecharacteristic functionof that set, with domainD.

(12) f is thecharacteristic function of a setS iff for all x in the relevant domain,f(x) = 1 if x ∈ S, andf(x) = 0 otherwise.

(13) I1(smokes) = {⟨Barack Obama,1⟩, ⟨Angela Merkel,1⟩, ⟨Hugh Grant,0⟩}

=

⎡⎢⎢⎢⎢⎢⎣Barack Obama → 1

Angela Merkel → 1

Hugh Grant → 0

⎤⎥⎥⎥⎥⎥⎦Hence, by the Lexical Terminals rule given in (7):

(14) [[smokes]]M1 =

⎡⎢⎢⎢⎢⎢⎣Barack Obama → 1

Angela Merkel → 1

Hugh Grant → 0

⎤⎥⎥⎥⎥⎥⎦Now we have to change our interpretation rule for sentences:

(15) Interpretation Rule: Sentences

If α has the formS

β γthen [[α]]M = [[γ]]M ([[β]]M ).

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And we can add an interpretation rule for VPs:

(16) Interpretation Rule: VPs

If α has the formVP

β γthen [[α]]M = [[β]]M ([[γ]]M ).

Instead of treating transitive verbs as relations, we will say that the semantic valueof the transitive verblikes is a function from individuals to functions from individ-uals to truth valuesFor example:

(17) I1(likes) =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

Barack Obama →

⎡⎢⎢⎢⎢⎢⎣Barack Obama → 0

Angela Merkel → 0

Hugh Grant → 1

⎤⎥⎥⎥⎥⎥⎦Angela Merkel →

⎡⎢⎢⎢⎢⎢⎣Barack Obama → 0

Angela Merkel → 0

Hugh Grant → 0

⎤⎥⎥⎥⎥⎥⎦Hugh Grant →

⎡⎢⎢⎢⎢⎢⎣Barack Obama → 1

Angela Merkel → 0

Hugh Grant → 1

⎤⎥⎥⎥⎥⎥⎦

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦Notice that (16) and (15) are very similar. Both say that the value of a tree whosetop node is a branching node is the value of one daughter – the one that denotes afunction – applied to the value of the other daughter, which works out if it denotessomething in the domain of the function. So we can combine these two rules intoone single rule.1

(18) Interpretation Rule: Functional Application (FA)If α is a branching node,{β, γ} is the set ofα’s daughters, and[[β]]M is afunction whose domain contains[[γ]]M , then[[α]]M = [[β]]M ([[γ]]M ).

Since FA does not make any reference to the syntactic categories of the nodes inthe phrase to be interpreted, and only makes reference to thesemantic type of thosephrases, it is atype-driven interpretation rule .

1The phrasefunction(al) applicationhas two meanings: The process of applying a function toan argument, and thecomposition rulethat allows us to compute the semantic value of aphrasegiven the semantic values of its parts. I will refer to the composition rule using title capitalization(‘Functional Application’) and the process of applying a function to its argument using lowercaseletters (‘function(al) application’).

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Types. All of our denotations are individuals, truth values, or functions. Individ-uals have typee and they are in the domain of individuals. Put more formally:

(19) Barack Obama∈ De

The truth values 0 and 1 have typet and they are elements ofDt.

A function from individuals to truth values is type⟨e, t⟩. By (14), [[smokes]]M1 isof type⟨e, t⟩ because its domain is the set of individuals, and its range isthe set oftruth values. Hence:

(20) [[smokes]]M1∈ D⟨e,t⟩

[[ likes]] will have type⟨e, ⟨e, t⟩⟩ because its domain is the set of individuals and itsrange is D⟨e,t⟩. So:

(21) [[likes]]M1∈ D⟨e,⟨e,t⟩⟩

Inventory of types. We will only need the following types for the meanings ofexpressions in natural language:

• e, the type of individuals

• t, the type of truth values

• ⟨σ, τ⟩, whereσ andτ are types

Function notation

For the purposes of talking about the composition rules thatare necessary for build-ing up the meanings of sentences from the meanings of their parts, it is not nec-essary to think about which individuals are actually mappedto 1 or 0 in in whichactual models. So we make our lives easier and skip all that bywriting:

(22) [[smokes]]M =[x ↦ 1 iff x smokes inM ]

whereM is an arbitrary model, and the natural language sentence ‘x smokes’means that the individual x is designated as a smoker according to the interpre-tation functionI in M .

In fact, we can just leave offM entirely for the present purposes:

(23) [[smokes]] = [x ↦ 1 iff x smokes]

16

The domain of the function denoted bysmokesis the set of individuals. Generally,we use ‘x’ as a variable over individuals, so the domain of thefunction is clear fromour choice of variable. But sometimes it is useful to be able to state conditions onthe domain. Taking inspiration from Heim and Kratzer, we will specify domainrestrictions with a colon, using expressions of the form:

[α ∶ φ ↦ γ]where

• α is theargument variable, as before (a letter that stands for an arbitraryargument of the function we are defining)

• φ is thedomain condition (places a condition on possible values forα)

• γ is thevalue description(specifies the value that our function assigns toα)

So the more expanded notation would be:2

(24) [[smokes]] = [x : x ∈ De ↦ 1 iff x smokes]

We will sometimes abbreviate this as:

(25) [[smokes]] = [x ∈ De ↦ 1 iff x smokes]

And we will leave off the domain condition entirely when we use the variables x,y, and z.

The lexical entry for a transitive verb looks like this:

(26) [[likes]] = [x ↦ [ y ↦ 1 iff y likes x]]

(Notice that it’s ‘y likes x’, not ‘x likes y’, because the verb combinesfirst with itsobject,andthenwith its subject.)

To express the result of applying the function to an argument, we write just thevalue description, with the argument substituting for all instances of the argu-ment variable.

(27) [x↦ [ y ↦ 1 iff y likes x]](Barack Obama)= [y ↦ 1 iff y likes Barack Obama]

2In Heim and Kratzer’s notation, the same function would be written: ‘λx : x ∈ De . x smokes’.We avoidλ notation in the meta-language here, because it is more customary in mathematics todescribe functions by description using the↦ symbol, and it avoids confusion over whether ourmetalanguage expressions are supposed to be treated as object language expressions of some variantof Church’s lambda calculus.

17

Note: Two kinds of formal stuff. Notice that (lambdified) English, along withset-theory talk, is themeta-language,the language with which we describe the se-mantic values that object language expressions may have.

So there’s two kinds of formal stuff:

• meta-language (↦, symbols of set theory)

• object language: when we were giving a syntax and semanticsfor predicatecalculus, symbols of predicate calculus were part of the object language.Now that we are giving a semantics for natural language directly, logic is nolonger part of the picture.

Example. Overview of how the compositional derivation of the truth conditionsfor Barack smokeswill go:

(28) S:t (FA)

NP:e (NN) VP: ⟨e, t⟩ (NN)

N: e (NN) V: ⟨e, t⟩ (NN)

Barack: e (TN) smokes: ⟨e, t⟩ (TN)

Compositional derivation of the truth conditions:

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

S

NP VP

N V

Barack smokes

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

VP

V

smokes

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎛⎜⎜⎜⎜⎜⎝

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

NP

N

Barack

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

M0⎞⎟⎟⎟⎟⎟⎠by Functional Application

18

=

⎡⎢⎢⎢⎢⎣⎡⎢⎢⎢⎢⎣

V

smokes

⎤⎥⎥⎥⎥⎦⎤⎥⎥⎥⎥⎦

⎛⎜⎜⎜⎜⎜⎝

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

NP

N

Barack

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

M0⎞⎟⎟⎟⎟⎟⎠by Non-branching Nodes

= [[ smokes ]] ([[ Barack ]]) by Non-branching Nodes (× 3)

= [x ↦ 1 iff x smokes]([[Barack]]) by Terminal Nodes

= [x ↦ 1 iff x smokes](Barack) by Terminal Nodes

= 1 iff Barack Obama smokes. by function application

Note: The last step does not involve any composition rules; we’re done breakingdown the tree and it’s just a matter of simplifying the expression at this point.

3.3 Transitive and ditransitive verbs

Semantics for transitive verbs. Example:

S

NP VP

N V NP

Angela likes N

Barack

Exercise: Derive the truth conditions for this using FA for all the branching nodes.Start by decorating the tree with the semantic the VP should be a function fromindividuals to truth values (type⟨e, t⟩).

Ditransitive verbs. Suppose we have a phrase structure rule for ditransitive verbslike tell, that generates sentences likeX told Y about Z. It could generate trees likethis for example:

19

(29)

S

NP VP

N V NP PP

Barack Obama told N P NP

Angela Merkel about N

Barack Obama

What type shouldtell be? What order should the arguments come in? This is wherelinking theorycomes in.

4 Fun with Functional Application

Composition rules

For branching nodes:

(30) Functional Application (FA)If α is a branching node and{β, γ} the set of its daughters, then, for anyassignmenta, if [[ β]] is a function whose domain contains [[γ]], then [[α]] =[[β]]([[ γ]]).

For non-branching nodes:

(31) Non-branching Nodes (NN)If α is a non-branching node andβ its daughter, then, for any assignmenta,[[α]]=[[ β]].

(32) Terminal Nodes (TN)If α is a terminal node occupied by a lexical item, then [[α]] is specified inthe lexicon.

4.1 Rick Perry is conservative

(33) [[Rick Perry]] = Rick Perry

(34) [[conservative]] = [x↦ 1 iff x is conservative]

20

What to do withis? How about just an identity function:

(35) [[is]] = [f ∈ D⟨e,t⟩ ↦ f]

(36) S:t

NP:e

N: e

Rick Perry

VP: ⟨e, t⟩V: ⟨⟨e, t⟩, ⟨e, t⟩⟩

is

A: ⟨e, t⟩conservative

[[[ S [NP [N Rick Perry ] ] [VP [V is ] [A conservative ] ] ]]]

= [[ [ VP [V is ] [A conservative ] ]]]([[ [NP [N Rick Perry ] ]]]) FA

= [[[ V is ]]]([[[ A conservative ]]])([[ [ NP [N Rick Perry ] ]]]) FA

= [[is]]([[conservative]]) ([[Rick Perry]]) NN

= [f ∈ D⟨e,t⟩ ↦ f]([ x ↦ 1 iff x is conservative])(Rick Perry) TN

= [x ↦ 1 iff x is conservative](Rick Perry) function application

= 1 iff Rick Perry is conservative (function application)

4.2 Rick Perry is in Texas

(37) [[in]] = [y ↦ [x ↦ 1 iff x is in y]]

(38) [[Texas]] = Texas

21

S:e

NP:e

N: e

Rick Perry

VP: ⟨e, t⟩

V: ⟨⟨e, t⟩, ⟨e, t⟩⟩is

PP:⟨e, t⟩P: ⟨e, ⟨e, t⟩⟩

in

NP:e

N: e

Texas

4.3 Rick Perry is proud of Texas

(39) [[proud]] = [y↦ [x ↦ 1 iff x is proud of y]]

(40) [[of]] = [x ↦ x]

S: t

NP:e

N: e

Rick Perry

VP

V: ⟨⟨e, t⟩, ⟨e, t⟩⟩is

AP: ⟨e, ⟨e, t⟩⟩A: ⟨e, ⟨e, t⟩⟩

proud

PP:e

P: ⟨e, e⟩of

NP:e

N: e

Texas

4.4 Rick Perry is a Republican

(41) [[Republican]] = [x↦ 1 iff x is a Republican]

Hey, let’s make the indefinite article vacuous too:

22

(42) [[a]] = [f ∈ D⟨e,t⟩ ↦ f]

(43) S:t

NP:e

N: e

Rick Perry

VP: ⟨e, t⟩

V: ⟨⟨e, t⟩, ⟨e, t⟩⟩is

NP: ⟨e, t⟩D: ⟨⟨e, t⟩, ⟨e, t⟩⟩

a

N′: ⟨e, t⟩Republican

5 Predicate Modification

5.1 Rick Perry is a conservative Republican

(44) S:t

NP:e

N: e

Rick Perry

VP: ⟨e, t⟩

V: ⟨⟨e, t⟩, ⟨e, t⟩⟩is

NP: ⟨e, t⟩

D: ⟨⟨e, t⟩, ⟨e, t⟩⟩a

N′: ⟨e, t⟩

A: { ⟨e, t⟩⟨⟨e, t⟩, ⟨e, t⟩⟩ }conservative

N′: ⟨e, t⟩Republican

Our lexical entry forconservativefrom above is type⟨e, t⟩:(45) [[conservative1 ]] = [x ↦ 1 iff x is conservative]

23

If we want to use Functional Application here, we needconservativeto be a func-tion of type⟨⟨e, t⟩, ⟨e, t⟩⟩.(46) [[conservative2 ]] = [f ∈ D⟨e,t⟩ ↦ [x ↦ 1 iff f(x) = 1 and x is conservative]]

Now, [[[N′ [A conservative2 ] [ N Republican ] ]]]

= [[[ A conservative2 ]]]([[[ N Republican ]]]) FA

= [[conservative2 ]]([[Republican]]) NN

= f ∈ D⟨e,t⟩ ↦ x↦ 1 iff f(x) = 1 and x is conservative( x ↦ 1 iff x is a Republican) TN

= x ↦ 1 iff x is a Republican and x is conservative function application

This is Montague’s strategy. All adjectives are⟨⟨e, t⟩, ⟨e, t⟩⟩ for him, and in pred-icate position, they combine with a silent noun. Despite theungrammaticality of*Rick Perry is conservative Republican.

Alternative strategy: Use another composition rule.

(47) Predicate Modification (PM)If α is a branching node,{β, γ} is the set ofα’s daughters, and[[β]] and[[γ]]are both inD⟨e,t⟩, then[[α]] = [x ↦ 1 iff [[β]](x) = 1 and[[γ]](x) = 1

Now, [[[N′ [A conservative1 ] [ N Republican ] ]]]

= [x ↦ 1 iff [[[ A conservative1 ]]](x) = [[[ N Republican ]]](x) = 1] PM

= [x ↦ 1 iff [[conservative1 ]](x) = [[Republican]](x) = 1] NN

= [x ↦ 1 iff x is conservative and x is a Republican] function application

5.2 Austin is a city in Texas

We can also use Predicate Modification with PP modifiers:

24

(48) S:t

NP:e

N: e

Austin

VP: ⟨e, t⟩

V: ⟨⟨e, t⟩, ⟨e, t⟩⟩is

NP: ⟨e, t⟩

D: ⟨⟨e, t⟩, ⟨e, t⟩⟩a

N′: ⟨e, t⟩

N′: ⟨e, t⟩city

PP:⟨e, t⟩P: ⟨e, ⟨e, t⟩⟩

in

NP:e

Texas

6 The definite article

What if we havethe governor of Texasinstead ofRick Perry?

6.1 The negative square root of 4

Regardingthe negative square root of 4,Frege says, “We have here a case in whichout of a concept-expression [i.e., an expression whose meaning is of type⟨e, t⟩] acompound proper name is formed [that is to say, the entire expression is of typee]with the help of the definite article in the singular, which isat any rate permissiblewhen one and only one object falls under the concept.”

“Permissible”: the denotes a function of type⟨⟨e, t⟩, e⟩ that is onlydefinedforinput predicates that characterize one single entity. In other words,thepresupposesexistence and uniqueness. This can be implemented as a restriction on the domain(highlighted with boldface type).

(49) [[the]] = [f ∈ D⟨e,t⟩ : there is exactly one x such that f(x) = 1↦ the uniquey such that f(y) = 1]

So the is not a function fromD⟨e,t⟩ to De; it is a partial function from D⟨e,t⟩ toDe. But we can still give it the type⟨⟨e, t⟩, e⟩ if we interpret this to allow partial

25

functions.

To flesh out Frege’s analysis of this example further, Heim and Kratzer suggest thatsquare rootis a “transitive noun”, with a meaning of type⟨e, ⟨e, t⟩⟩, and that “of isvacuous,[[square root]] applies to 4 via Functional Application, and the result ofthat composes with[[negative]] under predicate modification.”

(50) [[negative]] = [x↦ 1 iff x is negative]

(51) [[square root]] = [y↦ [x ↦ 1 iff x is the square root of y]]

(52) [[of]] = [x ∈ De ↦ x]

(53) [[four]] = 4

So the constituents will have denotations of the following types:

NP: e

D: ⟨⟨e, t⟩, e⟩ N: ⟨e, t⟩

the: ⟨⟨e, t⟩, e⟩ A: ⟨e, t⟩ N: ⟨e, t⟩

negative:⟨e, t⟩ N: ⟨e, ⟨e, t⟩⟩ PP:e

square root:⟨e, ⟨e, t⟩⟩ P: ⟨e, e⟩ NP: e

of: ⟨e, e⟩ N: e

four: e

6.2 Top-down evaluation

To compute the value “top-down”, we put the whole tree in one big old pair ofdenotation brackets, and use composition rules to break down the tree. Here I amputting the name of the rule I used as a subscript on the equalssign, because I don’thave enough room to put them off to the right.

26

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

NP

D N

the A N

negative N PP

square root P NP

of N

four

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=FA

⎡⎢⎢⎢⎢⎢⎢⎣

⎡⎢⎢⎢⎢⎢⎢⎣

D

the

⎤⎥⎥⎥⎥⎥⎥⎦

⎤⎥⎥⎥⎥⎥⎥⎦

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

N

A N

negative N PP

square root P NP

of N

four

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

27

=NN [[ the ]]

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

N

A N

negative N PP

square root P NP

of N

four

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=PM [[ the ]]

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

x ↦

⎡⎢⎢⎢⎢⎢⎢⎣

⎡⎢⎢⎢⎢⎢⎢⎣

A

negative

⎤⎥⎥⎥⎥⎥⎥⎦

⎤⎥⎥⎥⎥⎥⎥⎦(x) =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

N

N PP

square root P NP

of N

four

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(x) = 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=NN,FA [[ the ]]

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

x ↦ [[ negative]] (x) = [[square root]]

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

PP

P NP

of N

four

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(x) = 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

28

=FA [[ the ]]⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝x ↦ [[ negative]] (x) = [[square root]]

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎡⎢⎢⎢⎢⎢⎢⎣

⎡⎢⎢⎢⎢⎢⎢⎣

P

of

⎤⎥⎥⎥⎥⎥⎥⎦

⎤⎥⎥⎥⎥⎥⎥⎦

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

NP

N

four

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠(x) = 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠=NN [[ the ]] (x ↦ [[ negative]] (x) = [[square root]] ([[ of ]] ([[ four ]])) (x) = 1)Now we are done breaking down the tree. No more composition rules.

[[the]]([x ↦ 1 iff [[negative]](x) = [[square root ]]([[of]]([[four]]))(x) = 1])

= [[the]]([x ↦ 1 iff [[negative]](x) = [[square root ]]([[of]](4))(x) = 1])

= [[the]]([x ↦ 1 iff [[negative]](x) = [[square root ]]([x↦ x](4))(x) = 1])

= [[the]]([x ↦ 1 iff [[negative]](x) = [[square root ]](4)(x) = 1])

= [[the]]([x ↦ 1 iff [[negative]](x) = [y↦ [z↦ 1 iff z is a square root of y]](4)(x) = 1])

= [[the]]([x ↦ 1 iff [[negative]](x) = [z↦ 1 iff z is a square root of 4](x) = 1])

= [[the]]([x ↦ 1 iff [[negative]](x) = 1 and x is a square root of 4])

= [[the]]([x ↦ 1 iff [[z↦ 1 iff z is negative]](x) = 1 and x is a square root of 4])

= [[the]]([x ↦ 1 iff x is negative and x is a square root of 4])

= [ f ∈ D⟨e,t⟩ : there is exactly one x such that f(x) = 1↦ the unique y such that f(y)= 1]([x ↦ 1 iff x is negative and x is a square root of 4])

= the unique y such that y is negative and x is a square root of 4

= -2

29

6.3 Bottom-up style

NP: e

D: ⟨⟨e, t⟩, e⟩ N: ⟨e, t⟩

the: ⟨⟨e, t⟩, e⟩ A: ⟨e, t⟩ N: ⟨e, t⟩

negative:⟨e, t⟩ N: ⟨e, ⟨e, t⟩⟩ PP:e

square root:⟨e, ⟨e, t⟩⟩ P: ⟨e, e⟩ NP: e

of: ⟨e, e⟩ N: e

four: e

We need to compute a semantic value for every subtree/node. For convenience, wecan group the nodes according to the string of words that theydominate, and startfrom the most deeply embedded part of the tree.

Note: I reserve the right to abbreviate e.g. [PP [P of ] [ NP [N four ] ] ] as [PP of four],omitting all but the outermost brackets.

Bottom-up derivation

four

• [[four]] = 4 TN

• [[[ NP [N four] ]]] = [[[ N four]]] = [[four]] = 4 NN

of

• [[[ P of ]]] = [[of]] = x ↦ x TN, NN

of four

• [[[ PP of four]]] = [[[ P of ]]]([[[ NP 4]]]) FA= [x ↦ x](4) = 4

30

square root

• [[[ N square root]]] = [[square root]] NN= [y ↦ [x ↦ 1 iff x is a square root of y]] TN

square root of four

• [[[ N square root of four]] = [[[N square root]]]([[[PP of four]]]) FA= [y ↦ [x ↦ 1 iff x is a square root of y]](4)= [x ↦ 1 iff x is a square root of 4]= [x ↦ 1 iff x ∈ {2,-2}]

negative

• [[[ A negative]]] = [[negative]] = x↦ 1 iff x is negative TN, NN

negative square root of four

• [[[ N negative square root of four]]]= x ↦ 1 iff [[[ A negative]]](x) = [[[N square root of four]]](x)=1 PM= x↦ 1 iff x is a square root of 4 and x is negative= x↦ 1 iff x ∈ {-2}

the

• [[[ D the ]]] = [[the]] TN, NN= f ∈ D⟨e,t⟩ : there is exactly one x s t. f(x) = 1↦ the y such that f(y) = 1

the negative square root of four

• [[[ NP the negative square root of four]]]= [[[ D the ]]]([[[ N negative square root of four]]]) FA= [f ∈ D⟨e,t⟩ : there is exactly one x such that f(x) = 1↦ the unique y suchthat f(y) = 1](x↦ 1 iff x ∈ {-2})= the unique y such that y∈ {-2}= -2

Note that we used the same composition rules as we did using the top-down style!

This style is a little bit simpler and cleaner. But it is not asconvenient when youhave to manipulate variable assignments... which we will cover next!

31

7 Predicate calculus: now with variables!

Previously, I ignored formulas of predicate calculus with quantifiers and variableslike these:

∀x HAPPY(x) ‘for all x, x is happy’∃x ¬HAPPY(x) ‘there exists anx such that it is not the case thatx is happy’

In order to interpret formulas with variables, we need to make interpretation rela-tive to a modeland an assignment function:

[[φ]]M,g

An assignment function assigns individuals to variables. Examples:

g1 =

⎡⎢⎢⎢⎢⎢⎣x → Maggiey → Bartz → Bart

⎤⎥⎥⎥⎥⎥⎦

g2 =

⎡⎢⎢⎢⎢⎢⎣x → Barty → Bartz → Bart

⎤⎥⎥⎥⎥⎥⎦Informally, ∀x HAPPY(x) is true iff: no matter which individual we assign tox,HAPPY(x) is true. In other words, for all elements in the domaind, d ∈ [[HAPPY]].Informally, ∃x HAPPY(x) is true iff: we can find some individual to assign toxsuch thatHAPPY(x) is true. In other words, there is some element in the domaind

such thatd ∈ [[HAPPY]].The assignment function determines whatx is assigned to. Formally:

[[x]]M,g= g(x)

This in turn influences the value of a formula containingx in whichx is not boundby a quantifier (a formula in whichx is free).

Let’s interpretHAPPY(x) using the reality modelMr and the assignment functionsg1 andg2.

[[HAPPY(x)]]Mr ,g1 = 1

iff [[x]]Mr,g1 ∈ [[HAPPY]]Mr ,g1

32

iff g1(x) ∈ Ir(HAPPY)iff Maggie ∈ {Bart}.

Maggie/∈ {Bart} so[[HAPPY(x)]]Mr ,g1 = 0.

[[HAPPY(x)]]Mr ,g2 = 1

iff [[x]]Mr,g2 ∈ [[HAPPY]]Mr ,g2

iff g2(x) ∈ Ir(HAPPY)iff Bart ∈ Ir(HAPPY).Bart ∈ {Bart} so[[HAPPY(x)]]Mr ,g2 = 1.

Intuitively, this means that∃x HAPPY(x) is true, but∀x HAPPY(x) is false inMr.

New interpretation rules:

• ConstantsIf α is a constant, then[[α]]M,g = I(α).

• Variables – all new!If α is a variable, then[[α]]M,g = g(α).

• Atomic formulaeIf π is ann-ary predicate andα1, ...αn are terms, then[[π(α1, ..., αn)]]M,g

=

1 iff ⟨[[α1]]M,g, ..., [[αn]]M,g⟩ ∈ [[π]]M,g

If π is a unary predicate andα is a term, then[[π(α)]]M,g= 1 iff [[α]]M,g

∈[[π]]M,g.

• Negation[[¬φ]]M,g= 1 if [[φ]]M,g

= 0; otherwise[[¬φ]]M,g= 0.

• Connectives[[φ ∧ ψ]]M,g= 1 if [[φ]]M,g

= 1 and[[ψ]]M,g= 1; 0 otherwise. Similarly for[[φ ∨ ψ]]M,g, [[φ → ψ]]M,g, and[[φ↔ ψ]]M,g.

• Universal quantification – all new![[∀vφ]]M,g= 1 iff for all d ∈ D, [[φ]]M,g′

= 1, whereg′ is an assignmentfunction exactly likeg except thatg′(v) = d.

33

• Existential quantification – all new![[∃vφ]]M,g= 1 iff there is ad ∈ D such that[[φ]]M,g′ , whereg′ is an assign-

ment function exactly likeg except thatg′(v) = d.

[[∀x HAPPY(x)]]Mr ,g1 = 1

iff for all d ∈D, [[HAPPY(x)]]Mr ,g′= 1, whereg′ is an assignment function exactly

like g1 except thatg′(x) = dThis can be falsified by settingd equal to Maggie, sog′(x) =Maggie.[[HAPPY(x)]]Mr ,g

′= 0 in this case.

But there is ad ∈ D such that[[HAPPY(x)]]Mr ,g′, whereg′ is an assignment func-

tion exactly likeg1 except thatg′(x) = d.

As shown above, there is such ad: Bart.

8 Relative clauses

Heim and Kratzer use assignment functions for the interpretation of relative clausessuch as the following:

(54) The carthat Joe bought is very fancy.

(55) The womanwho admires Joeis very lovely.

Semantically, relative clauses are just like adjectives:

(56) Thered car is very fancy.

(57) TheSwedishwoman is very lovely.

They are type⟨e, t⟩ and combine via Predicate Modification.

(58) NP:⟨e, t⟩NP: ⟨e, t⟩

car

CP:⟨e, t⟩that Joe bought

CP stands for “Complementizer Phrase” and Heim and Kratzer assume the follow-ing syntax for relative clause CPs:

34

(59) CP

whichi C′

C

that

S

DP

Joe

VP

V

bought

DP

ti

(60) CP

whoi C′

C

that

S

DP

ti

VP

V

likes

DP

Joe

The text that is struck out likeso isdeleted. Heim and Kratzer assume that eitherthe relative pronounwhichor whoor the complementizerthat is deleted.

Interpretation of variables

(61) Traces Rule (TR)If αi is a trace andg is an assignment, [[αi]]g = g(i)

g1 =

⎡⎢⎢⎢⎢⎢⎣1 → Maggie2 → Bart3 → Maggie

⎤⎥⎥⎥⎥⎥⎦g2 =

⎡⎢⎢⎢⎢⎢⎣1 → Lisa2 → Bart3 → Maggie

⎤⎥⎥⎥⎥⎥⎦[[ t1]]g1 = Maggie

[[ t1]]g2 = Lisa

So now we interpret everything with respect to an assignment.

35

(62) [[ [VP [V abandoned ] [DP t1 ] ]]] g = x ↦ 1 iff x abandonedg(1)But there areassignment-independentdenotations too.

(63) Bridge to assignment-independence (BI)For any treeα, α is in the domain of [[]] iff for all assignmentsg andg′,[[α]]g= [[α]]g

′.

If α is in the domain of [[]], then for all assignmentsg, [[α]]= [[ α]]g.

So we can still have assignment-independent lexical entries like:

(64) [[laugh]] = x ↦ 1 iff x laughs

and then by (63), we have:

(65) [[laugh]]g1 = x ↦ 1 iff x laughs

(66) [[laugh]]g2 = x ↦ 1 iff x laughs

We need to redo the composition rules now too:

(67) Lexical Terminals (LT)If α is a terminal node occupied by a lexical item, then [[α]] is specified inthe lexicon.

(68) Non-branching Nodes (NN)If α is a non-branching node andβ its daughter, then, for any assignmentg,[[α]]g=[[β]]g.

(69) Functional Application (FA)If α is a branching node and{β, γ} the set of its daughters, then, for anyassignmentg, if [[ β]]g is a function whose domain contains [[γ]]g, then [[α]]g

= [[β]]g([[γ]]g).

(70) Predicate Modification (PM)If α is a branching node and{β, γ} the set of its daughters, then, for anyassignmentg, if [[ β]]g and [[γ]]g are both functions of type⟨e, t⟩, then [[α]]g

= x ↦ 1 iff [[ β]]g(x) = [[γ]]g(x) = 1.

36

Predicate abstraction. The S in a relative clause is typet. How do we get theCP to have type⟨e, t⟩?(71) CP :⟨e, t⟩

which1 C′:

C:

that

S: t

DP:e

Joe

VP: ⟨e, t⟩V: ⟨e, ⟨e, t⟩⟩

bought

DP:e

t1

Heim and Kratzer:

• The complementizerthat is vacuous;that S= Sor [[that]] = p ∈ Dt ↦ p

• The relative pronoun is vacuous too, but it triggers a special rule called Pred-icate Abstraction

(72) Predicate Abstraction (PA)If α is a branching node whose daughters are a relative pronoun indexediandβ, then [[α]]g = x ↦ [[β]]g

x/i

gx/i is an assignment that is just likeg except thatx is assigned toi.

Note thatx is a variable that is part of themeta-language, bound by themeta-languageoperatorλ, ranging over objects in the domain.

So [[(71)]] = x↦ 1 iff Joe bought x.

In case you don’t believe me:

[[[ CP which1 [C′ [C that ] [S [DP Joe ] [VP [V bought ] [DP t1 ] ] ] ] ]]]

= [[[ CP which1 [C′ [C that ] [S [DP Joe ] [VP [V bought ] [DP t1 ] ] ] ] ]]] g (anyg) BI

= [x ↦ [[[ C′ [C that ] [S [DP Joe ] [VP [V bought ] [DP t1 ] ] ] ]]] gx/1 ] PA

37

= [x ↦ [[[ C that ]]]gx/1

([[[ S [DP Joe ] [VP [V bought ] [DP t1 ] ] ]]] gx/1)] FA

= [x ↦ [[[ C that ]]]([[[ S [DP Joe ] [VP [V bought ] [DP t1 ] ] ]]] gx/1)] BI

= [x ↦ [p ∈ Dt ↦ p]([[[ S [DP Joe ] [VP [V bought ] [DP t1 ] ] ]]] gx/1 ) ] LT

= [x ↦ [[[ S [DP Joe ] [VP [V bought ] [DP t1 ] ] ]]] gx/1 ] function application

= [x ↦ [[[ VP [V bought ] [DP t1 ] ]]] gx/1

([[[ DP Joe ]]]gx/1

)] FA

= [x ↦ [[[ VP [V bought ] [DP t1 ] ]]] gx/1

([[[ DP Joe ]]])] BI

= [x ↦ [[[ VP [V bought ] [DP t1 ] ]]] gx/1

(Joe)] NN, LT

= [x ↦ [[[ V bought ]]]gx/1

([[[ DP t1 ]gx/1

)]](Joe)] FA

= [x ↦ [[[ V bought ]]]([[[DP t1 ]gx/1

)]](Joe)] BI

= [x ↦ [z ↦ [y ↦ 1 iff y bought z]]([[[DP t1 ]]] gx/1

)(Joe)] LT, NN

= [x ↦ [z ↦ [y ↦ 1 iff y bought z]]([[t1]]gx/1

)(Joe)] NN

= [x ↦ [z ↦ [y ↦ 1 iff y bought z]](x)(Joe)] TR

= [x ↦ [y ↦ 1 iff y bought x](Joe)] function application

= [x ↦ 1 iff Joe bought x] function application

9 Quantifiers

How do we analyze sentences like the following:

(73) Somebody is happy.

(74) Everybody is happy.

(75) Nobody is happy.

(76) {Some, every, at least one, at most one, no} linguist is happy.

(77) {Few, some, several, many, most, more than two} linguists are happy.

38

S: t

NP: ?

...

VP: ⟨e, t⟩is happy

9.1 Typee?

Most of the DPs we have seen so far have been of typee:

• Proper names: Mary, John, Rick Perry, 4, Texas

• Definite descriptions: the governor of Texas, the square root of 4

• Pronouns and traces: it,t

Exception: indefinites likea Republicanafter is.

Should words and phrases likeNobodyandAt least one personbe treated as typee? How can we tell?

Predictions of the typee analysis:

• They should validate subset-to-superset inferences

• They should validate the law of contradiction

• They should validate the law of the excluded middle

Subset-to-superset inferences

(78) John came yesterday morning.Therefore, John came yesterday.

This is a valid inference if John is typee. Proof: [[came yesterday morning]]⊆[[came yesterday]] (everything that came yesterday morning came yesterday), andif the subject denotes an individual, then the sentence means that the subject is anelement of the set denoted by the VP. If the first sentence is true, then the subjectis an element of the set denoted by the VP, which means that thesecond sentencemust be true. QED.

(79) At most one letter came yesterday morning.Therefore, at most one letter came yesterday.

This inference is notvalid, soat most one lettermust not be typee.

39

The law of contradiction (¬[P ∧ ¬P ])This sentence is contradictory:

(80) Mount Rainier is on this side of the border, and Mount Rainier is on theother side of the border.

The fact that it is contradictory follows from these assumptions:

• [[Mount Rainier]] ∈ De

• [[is on this side of the border]]∩ [[is on the other side of the border]]= ∅(Nothing is both on this side of the border and on the other side of the border)

• When the subject is typee, the sentence means that it is in the set denotedby the VP

• standard analysis ofand

This sentence is not contradictory:

(81) More than two mountains are on this side of the border, and more than twomountains are on the other side of the border.

Somore than two mountainsmust not be typee.

The law of the excluded middle (P ∨ ¬P )

(82) I am over 30 years old, or I am under 40 years old.

This is a tautology. That follows from the following assumptions:

• [[I]] ∈ De

• [[over 30 years old]]∪ [[under 40 years old]]= D (everything is either over 30years old or under 40 years old)

• When the subject is typee, the sentence means that it is in the set denotedby the VP

• standard analysis ofor

This sentence is not a tautology:

(83) Every woman in this room is over 30 years old, or every woman in this roomis under 40 years old.

Soevery womanmust not be of typee

40

9.2 Solution: Generalized quantifiers

(84) [[nothing]] = f ∈ D⟨e,t⟩ ↦ 1 iff there is no x∈ De such that f(x) = 1

(85) [[everything]] = f∈ D⟨e,t⟩ ↦ 1 iff for all x ∈ De, f(x) = 1

(86) [[something]] = f∈ D⟨e,t⟩ ↦ 1 iff there is some x∈ De such that f(x) = 1

(87) S:t

DP: ⟨⟨e, t⟩, t⟩everything

VP: ⟨e, t⟩V: ⟨e, t⟩vanished

vs. S:t

DP:e

Mary

VP: ⟨e, t⟩V: ⟨e, t⟩vanished

(88) [[every]] = f ∈ D⟨e,t⟩ ↦ [g ∈ D⟨e,t⟩ ↦ 1 iff for all x ∈ De such that f(x) = 1,g(x)=1 ]

(89) [[no]] = f ∈ D⟨e,t⟩ ↦ [g ∈ D⟨e,t⟩ ↦ 1 iff there is no x∈ De such that f(x) = 1and g(x)=1 ]

(90) [[some]] = f∈ D⟨e,t⟩ ↦ [g ∈ D⟨e,t⟩ ↦ 1 iff there is some x∈ De such that f(x)= 1 and g(x)=1 ]

(91) S: t

DP: ⟨⟨e, t⟩, t⟩D: ⟨⟨e, t⟩, ⟨⟨e, t⟩, t⟩⟩

every

NP: ⟨e, t⟩thing

VP: ⟨e, t⟩V: ⟨e, t⟩vanished

41

10 The problem of quantifiers in object position

10.1 The problem

(92) S: t

DP: ⟨⟨e, t⟩, t⟩D: ⟨⟨e, t⟩, ⟨⟨e, t⟩, t⟩⟩

every

NP: ⟨e, t⟩linguist

VP: ⟨e, t⟩V: ⟨e, ⟨e, t⟩⟩

offended

DP:e

John

(93) S: ?????????

DP:e

John

VP: ???????????

V: ⟨e, ⟨e, t⟩⟩offended

DP: ⟨⟨e, t⟩, t⟩D: ⟨⟨e, t⟩, ⟨⟨e, t⟩, t⟩⟩

every

NP: ⟨e, t⟩linguist

Two types of approaches to the problem:

1. Move the quantifier phrase to a higher position in the tree (via QuantifierRaising), leaving a DP trace of typee in object position. (Or simulate move-ment via Cooper Storage, as in Head-Driven Phrase StructureGrammar.)

2. Interpret the quantifier phrasein situ. In this case one can apply a type-shifting operation to change its type.

10.2 An in situ approach

Multiple versions of lexical items:[[everybody1 ]] = f ∈ D⟨e,t⟩ ↦ 1 iff for all persons x∈ D, f(x) = 1[[everybody2 ]] = f ∈ D⟨e,⟨e,t⟩⟩ ↦ [x ∈ De ↦ 1 iff for all persons y∈ De, f(y)(x) = 1 ][[somebody1 ]] = f ∈ D⟨e,t⟩ ↦ 1 iff there is some person x∈ De such that f(x) = 1

42

[[somebody2 ]] = f ∈ D⟨e,⟨e,t⟩⟩ ↦ [x ∈ De ↦ 1 iff there is some person y∈ De suchthat f(y)(x) = 1 ]

(94) S: t

DP:e

John

VP: ⟨e, t⟩

V: ⟨e, ⟨e, t⟩⟩offended

DP: ⟨⟨e, ⟨e, t⟩⟩, ⟨e, t⟩⟩everybody2

(95) S: t

DP: ⟨⟨e, t⟩, t⟩Everybody

VP: ⟨e, t⟩

V: ⟨e, ⟨e, t⟩⟩offended

DP: ⟨⟨e, ⟨e, t⟩⟩, ⟨e, t⟩⟩somebody2

Note: This only gets one of the readings.

We need a neweverybodyfor ternary relations:

(96) S

DP

Ann

VP

V′

V

introduced

DP

everybody

PP

P

to

DP

Maria

What type are the determiners (note:et = ⟨e, t⟩)?

43

(97) S: t

DP:e

John

VP: et

V: ⟨e, et⟩offended

DP: ⟨⟨e, et⟩, et⟩D:⟨et, ⟨⟨e, et⟩, et⟩⟩

every

NP:et

linguist

How do we get thiseveryfrom our normal⟨et, ⟨et, t⟩⟩ every? A lexical rule.

(98) For every lexical itemδ1 with a meaning of type⟨et, ⟨et, t⟩⟩, there is a (ho-mophonous and syntactically identical) itemδ2 with the following meaningof type⟨et, ⟨⟨e, et⟩, et⟩⟩:[[δ2]] = f ∈ D⟨e,t⟩ ↦ [g ∈ D⟨e,et⟩ ↦ [x ∈ De ↦ [[δ1]](f)(z ∈ De ↦ g(z)(x)) ] ]

10.3 A Quantifier Raising approach

Several levels of representation:

• Deep Structure (DS): Where the derivation begins

• Surface Structure (SS): Where the order of the words is whatwe see

• Phonological Form (PF): Where the words are realized as sounds

• Logical Form (LF): The input to semantic interpretation

Transformations map from DS to SS, and from SS to PF and LF. (Since the trans-formations from SS to LF happen “after” the order of the wordsis determined, wedo not see the output of these transformations. These movement operations are inthis sensecovert.)

A transformation called QR (Quantifier Raising) maps the SS structure in (99a) tosomething like the LF structure in (99b)

44

(99) a. S

DP

John

VP

V

offended

DP

D

every

NP

linguist

b. S

DPi

D

every

NP

linguist

S

DP

John

VP

V

offended

DP

ti

Actually, Heim and Kratzer propose the following, so that they can make it workwith Predicate Abstraction:

(100) S

DP

D

every

NP

linguist

1 S

DP

John

VP

V

offended

DP

t1

(101) Predicate Abstraction (PA) (revised)Letα be a branching node with daughtersβ andγ, whereβ dominates onlya numerical index i. Then for any variable assignmentg, [[α]]g =x ∈ De ↦

[[γ]]gx/i

.

Example. Let’s give every node of the tree a unique category label so wecanrefer to the denotation of the tree rooted at that node using the category label.

45

(102) S1

DP1

D

every

NP

linguist

?

1 S2

DP2

John

VP

V

offended

DP3

t1

The task is to analyze the truth conditions of S1 (or, to be more precise, the treerooted at the node labelled S1). The basic idea is straightforward – Predicate Ab-straction at the mystery-category node (labelled ‘?’ here), Pronouns and Traces ruleat the trace, and Functional Application everywhere else – but it is a bit tricky togo between assignment-dependent and assignment-independent denotations. Thetrick is to start with the Bridge to Independence.

[[Sg1]]

=[[S1]]g, for all g BI=[[DP1]]g([[?]]g) FA

=[[DP1]]g(x ↦ [[S2]]gx/1

) PA

=[[DP1]]g(x ↦ [[VP]] gx/1

([[DP2]]gx/1

)) FA

=[[DP1]]g(x ↦ [[V]] gx/1([[DP3]]gx/1

)([[DP2]]gx/1

)) FA

=[[DP1]]g(x ↦ [[offended]]gx/1

([[t1]]gx/1

)([[John]]gx/1

)) NN, TN

=[[DP1]]g(x ↦ [[offended]]gx/1

(x)([[John]]gx/1

)) TR=[[DP1]]g(x ↦ [[offended]](x)([[John]])) BI=[[DP1]]g(x ↦ [y ↦ [z ↦ 1 iff z offended y ]](x)(John)) TN=[[DP1]]g(x ↦ John offended x) β-R=[[D]] g([[NP]]g)(x ↦ 1 iff John offended x) TN=[[every]]g([[linguist]]g )(x ↦ 1 iff John offended x) FA=[[every]]([[linguist]])(x ↦ 1 iff John offended x) BI= [f ∈ D⟨e,t⟩ ↦ g ∈ D⟨e,t⟩ ↦ 1 iff for all y, if f(y) then g(y)]([[linguist]])(x ↦ 1 iff John offended x) TN= [g ∈ D⟨e,t⟩ ↦ 1 iff for all y, if y is a linguist then Q(y)](x↦ 1 iff John offended x) TN,β-R=1 iff for all y, if y is a linguist then John offended x β-R

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10.4 Arguments in favor of the movement approach

Argument #1: Scope ambiguities. In order to get both readings ofEverybodyloves somebody,we have to introduce yet even more complicated types. Scopeambiguities are trivially derived under the movement approach.

Argument #2: Inverse linking. There is one class of examples that cannot begenerated under anin situ approach:

(103) One apple in every basket is rotten.

This does not mean: ‘One apple that is in every basket is rotten’. That is the onlyreading that anin situ analysis can give us.

QR analysis:

(104) S

DP

every basket 1 S

DP

D

one

NP

N

apple

PP

P

in

t1

VP

is rotten

Argument #3: Antecedent-contained deletion

(105) I read every novel that you did.

Like regular VP ellipsis:

(106) I readWar and Peacebefore you did.

except that the antecedent VP is contained in the elided VP!

To create an appropriate antecedent, you have to QR the object.

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Argument #4: Quantifiers that bind pronouns

(107) a. Mary blamed herself.

b. Mary blamed Mary.

(108) a. Every woman blamed herself.

b. Every woman blamed every woman.

(109) No man noticed the snake next to him.

Treat pronouns as variables and use QR⇒ no problem.

(110) Traces and Pronouns Rule (TP)(p. 116)If α is a pronoun or trace andg is an assignment andi is in the domain ofg,[[αi]]g = g(i)

(111) S

DP

D

every

NP

woman

1 S

DP

t1

VP

V

blamed

DP

herself1

But how do we get the truth conditions on the in-situ approach?

(112) [[ [VP [V blamed ] [DP herself1] ] ]] g = x ↦ 1 iff x blamedg(1)How do we combine this withevery woman? We cannot get an assignment-independent denotation.

11 Free and Bound Variable Pronouns

11.1 Toward a unified theory of anaphora

A deictic use of a pronoun:

(113) [after a certain man has left the room:]I am glad he is gone.

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An anaphoric use of a pronoun:

(114) I don’t think anybody here is interested in Smith’s work. He should not beinvited.

“Anaphoric and deictic uses seem to be special cases of the same phenomenon: thepronoun refers to an individual which, for whatever reason,is highly salient at themoment when the pronoun is processed.” (Heim and Kratzer 1998, p. 240)

Hypothesis 1: All pronouns refer to whichever individual is most salient at themoment when the pronoun is processed.

It can’t be that simple for all pronouns:

(115) the book such1 that Mary reviewed it1

(116) No1 woman blamed herself1.

So not all pronouns are referential.3

Hypothesis 2: All pronouns are bound variables.

Then in (114) we would have to QRSmithto a position where it QR’sHe in thesecond sentence somehow.

Plus, the strict/sloppy ambiguity exemplified in (117) can be explained by sayingthat on one reading, we have a bound pronoun, and on another reading, we have areferential pronoun.

In the movieGhostbusters,there is a scene in which the three Ghostbusters Dr.Peter Venkman, Dr. Raymond Stanz, and Dr. Egon Spengler (played by Bill Mur-ray, Dan Akroyd, and Harold Ramis, respectively), are in an elevator. They havejust started up their Ghostbusters business and received their very first call, from afancy hotel in which a ghost has been making disturbances. They have their protonpacks on their back and they realize that they have never beentested.

(117) Dr Ray Stantz: You know, it just occurred to me that we really haven’t hada successful test of this equipment.

3Sometimes it is said thatNo womanandherselfare “coreferential” in (116) but this is strictlyspeaking a misuse of the term “coreferential”, because, as Heim and Kratzer point out, “coreferenceimplies reference.”

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Dr. Egon Spengler: I blame myself.Dr. Peter Venkman: So do I.

Strict reading: Peter blames himself.Sloppy reading: Peter blames Egon.

LF of antecedent for sloppy reading:

(118) S

DP

I1 S

DP

t1

VP

V

blame

DP

myself1

LF of antecedent for strict reading:

(119) S

DP

I1

VP

V

blame

DP

myself1

Heim and Kratzer’s hypothesis: All pronouns are variables, and bound pro-nouns are interpreted as bound variables, and referential pronouns are interpretedas free variables.

What does it mean for a variable to be bound or free?

• The formal definition (p. 118): Letαn be an occurrence of a variableα ina treeβ. Thenαn is free inβ if no subtreeγ of β meets the following twoconditions: (i)γ containsαn, and (ii) there are assignmentsg such thatα isnot in the domain of [[]]g, butγ is.

• More intuitively: A variable is free in a treeβ if the value of [[β]]g dependson whatg assigns to the variable’s index.

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• With the Predicate Abstraction rule, we make semantic values independentof assignments, so we can use the following shortcut to determine whether avariable is bound or free: A variable is bound if there is a node that meets thestructural description for Predicate Abstraction dominating it and its index;otherwise it is free.

Examples:

(120) S

DP

[feminine] DP

She1

VP

V

is

A

nice

(121) S

DP

Every boy1 S

DP

t1

VP

V

loves

D

DP

[masculine] D

his1

NP

father

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(122) S

DP

John

VP

V

hates

DP

D

[masculine] D

his1

NP

father

(123) S

DP

John1 S

DP

t1

VP

V

hates

D

DP

[masculine] D

his1

NP

father

11.2 Assignments as part of the context

A consequence: “Treating referring pronouns as free variables implies a new wayof looking at the role of variable assignments. Until now we have asssumed that anLF whose truth-value varied from one assignment to the next could ipso factonotrepresent a felicitous, complete utterance. We will no longer make this assumption.Instead, let us think of assignments as representing the contribution of the utterancesituation.”

(124) Appropriateness ConditionA context c is appropriate for an LFφ only if c determines a variable assign-

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mentgc whose domain includes every index which has a free occurrence inφ.

Now truth and falsity will be relative to contexts:

(125) Truth and Falsity Conditions for UtterancesIf φ is uttered in c and c is appropriate forφ, then the utterance ofφ in c istrue if [[ φ]]gc = 1 andfalseif [[ φ]]gc = 0.

Suppose the contextc1 “determines” the assignmentg1 (i.e., the context containsa bunch of information, among which is the assignmentg1), andg1 is defined asfollows:

g1 =

⎡⎢⎢⎢⎢⎢⎣1 → Kim2 → Kim3 → Sandy

⎤⎥⎥⎥⎥⎥⎦Assume:

• Kim is male

• Sandy is female

• Kim is nice

• Sandy is not nice

[[feminine]] = x : x is female↦ x

Questions:

• Is (120) appropriate forc1?

• Is [[(120)]]g1 defined?

• If so, what is [[(120)]]g1? (I.e. is it true or false?)

12 Our fragment of English so far

12.1 Composition Rules

For branching nodes:

(126) Functional Application (FA)If α is a branching node and{β, γ} the set of its daughters, then, for anyassignmentg, if [[ β]]g is a function whose domain contains [[γ]]g, then [[α]]g

= [[β]]g([[γ]]g).

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(127) Predicate Modification (PM)If α is a branching node and{β, γ} the set of its daughters, then, for anyassignmentg, if [[ β]]g and [[γ]]g are both functions of type⟨e, t⟩, then [[α]]g

= x ∈ De ↦ 1 iff [[ β]]g(x) = [[γ]]g(x) = 1.

(128) Predicate Abstraction (PA)Letα be a branching node with daughtersβ andγ, whereβ dominates onlya numerical index i. Then for any variable assignmentg, [[α]]g =x ∈ De ↦

[[γ]]gx/i

For non-branching and terminal nodes:

(129) Non-branching Nodes (NN)If α is a non-branching node andβ its daughter, then, for any assignmentg,[[α]]g=[[β]]g.

(130) Lexical Terminals (LT)If α is a terminal node occupied by a lexical item, then [[α]] is specified inthe lexicon.

(131) Traces and Pronouns Rule (TP)If α is a pronoun or trace andg is an assignment andi is in the domain ofg,[[αi]]g = a(i)

12.2 Additional principles

(132) Bridge to assignment-independenceFor any treeα, α is in the domain of [[]] iff for all assignmentsg andg′,[[α]]g= [[α]]g

′.

If α is in the domain of [[]], then for all assignmentsg, [[α]]= [[ α]]g.

(133) Quantifier Raising (QR)Surface structures containing quantificational NP like (133a) undergo a coverttransformation to LFs like (133b)

a. S

DP

John

VP

V

offended

DP

D

every

NP

linguist

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b. S

DP

D

every

NP

linguist

1 S

DP

John

VP

V

offended

DP

t1

(134) Appropriateness ConditionA context c is appropriate for an LFφ only if c determines a variable assign-mentgc whose domain includes every index which has a free occurrence inφ.

(135) Truth and Falsity Conditions for UtterancesIf φ is uttered in c and c is appropriate forφ, then the utterance ofφ in c istrue if [[ φ]]gc = 1 andfalseif [[ φ]]gc = 0.

(136) LF Identity Condition on EllipsisA constituent may be deleted at PF only if it is a copy of another constituentat LF.

(137) Indexing ConditionNo LF representation (for a sentence or multisentential text) must containboth bound occurrences and free occurrences of the same index.

12.3 Lexical items

Proper names:

(138) [[Rick Perry]] = Rick Perry

(139) [[Texas]] = Texas

(140) [[four]] = 4

Nouns:

(141) [[Republican]] = x↦ 1 iff x is a Republican

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(142) [[square root]] = y↦ x ↦ 1 iff x is the square root of y

Adjectives:

(143) [[conservative⟨e,t⟩]] = x ↦ 1 iff x is conservative

(144) [[conservative⟨⟨e,t⟩,⟨e,t⟩⟩ ]] = f ∈D⟨e,t⟩↦ x↦ 1 iff f(x) = 1 and x is conservative

(145) [[negative]] = x↦ 1 iff x is negative

(146) [[proud]] = y↦ x ↦ 1 iff x is proud of y

Verbs:

(147) [[is]] = f ∈ D⟨e,t⟩ ↦ f

Prepositions:

(148) [[in]] = y ∈ De ↦ x ↦ 1 iff x is in y

(149) [[of]] = x ∈ De ↦ x

Determiners:

(150) [[a]] = f ∈ D⟨e,t⟩ ↦ f

(151) [[the]] = f ∈ D⟨e,t⟩ : there is exactly one x such that f(x) = 1↦ the unique ysuch that f(y) = 1

Complementizers: [[that]] = p∈ Dt ↦ p

Conclusion

Tiny bit more to do before we’re done with English.

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