Semantics Boot Camp. Contents. Elizabeth Coppock. Compilation of handouts from NASSLLI 2012, Austin Texas (revised December 15, 2012) 1 Semantics 2

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1 Semantics Boot Camp Elizabeth Coppock Compilation of handouts from NASSLLI 2012, Austin Texas (revised December 15, 2012) Contents 1 Semantics 2 2 Predicate calculus Syntax of Predicate Calculus Atomic symbols Syntactic composition rules Semantics of Predicate Calculus Sets Ordered pairs and relations Functions Set theory as meta-language Models and interpretation functions Interpretation rules English as a formal language A fragment of English with set denotations Models and the Lexicon Composition rules A fragment of English satisfying Frege s conjecture Transitive and ditransitive verbs Fun with Functional Application Rick Perry is conservative Rick Perry is in Texas Rick Perry is proud of Texas Rick Perry is a Republican

2 5 Predicate Modification Rick Perry is a conservative Republican Austin is a city in Texas The definite article The negative square root of Top-down evaluation Bottom-up style Predicate calculus: now with variables! 32 8 Relative clauses 34 9 Quantifiers Typee? Solution: Generalized quantifiers The problem of quantifiers in object position The problem An in situ approach A Quantifier Raising approach Arguments in favor of the movement approach Free and Bound Variable Pronouns Toward a unified theory of anaphora Assignments as part of the context Our fragment of English so far Composition Rules Additional principles Lexical items Semantics Semantics: The study of meaning. What is meaning? How can you tell whether somebody or something understands? Does Google understand language? An argument that it does not: Google can t do inferences: (1) Everyone who smokes gets cancer No heavy smoker avoids cancer 2

3 (cf. No alleged smoker avoids cancer...) A hallmark of a system or agent that understands language / grasps meaning is that it can do these kinds of inferences. Another way to put it: A good theory of meaning should be able to account for the conditions 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 is true, B is true too. E.g. Obama was born in 1961 entails Obama was born in the 1960s. 2. Presupposition (semantics/pragmatics): A presupposes B if and only if an utterance of A takes B for granted. E.g. Sue has finally stopped smoking presupposes that Sue smoked in the recent past. 3. Conversational Implicature (pragmatics): A conversationally implicates B if and only if a hearer can infer B from an utterance of A by making use of the assumption that the speaker is being cooperative. E.g. Some of the students passed conversationally implicates that not all of the students passed. The primary responsibility for a theory of semantics is to account for the conditions under which one sentence entails another sentence. Other nice things a theory of semantics could do: Account for presuppositions, contradictions, equivalences, semantic ill-formedness, distribution patterns. Strategy: Assign truth conditions to sentences. The truth conditions are the conditions under which the sentence is true. Knowing the meaning of a sentence does not require knowing whether the sentence is in fact true; it only requires being able to discriminate between situations in which the sentence is true and situations in which the sentence is false. (Cf. Heim & Kratzer s bold first sentence: To know the meaning of a sentence is to 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 which B is true, then A entails B. 3

4 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 exists between formal and natural languages.... In the present paper I shall accordingly present a precise treatment, culminating in a theory of truth, of a formal language that I believe may reasonably be regarded as a fragment of ordinary English.... The treatment given here will be found to resemble the usual syntax and model theory (or semantics) [due to Tarski] of the predicate calculus, but leans rather heavily on the intuitive aspects of certain recent developments in intensional logic [due to Montague himself]. (Montague 1970b, p.188 in Montague 1974) 2 Predicate calculus Predicate calculus is a logic. Logics are formal languages, and as such they have syntax and semantics. Syntax: specifies which expressions of the logic are well-formed, and what their syntactic categories are. Semantics: specifies which objects the expression correspond to, and what their semantic categories are. 2.1 Syntax of Predicate Calculus Atomic symbols Formulas are built up from atomic symbols drawn from the following syntactic categories: 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,>,

5 function constants: unary function constants: MOTHER, ABSOLUTE VALUE,... binary function constants: DISTANCE,+,, logical connectives:,,,, quantifiers:, I write constants in SMALL CAPS and variables in italics. But I will ignore quantifiers and variables for now (they will be discussed later) Syntactic composition rules How to build complex expressions (Greek letters are metalanguage variables): Ifπ is an-ary predicate andα 1...α n are terms, then π(α 1,...α n ) is an atomic formula. Ifπ is a unary predicate andαis a term, thenπ(α) is an atomic formula. Ifπ is a binary predicate andα 1 andα 2 are terms, thenπ(α 1,α 2 ) is an atomic formula. Ifα 1...α n are terms, andγ is a function constant with arityn, thenγ(α 1,...,α n ) is a term. If φ is a formula, then φ is a formula. If φ is a formula and ψ is a formula, then[φ ψ] is a formula, and so are [φ ψ],[φ ψ], and[φ ψ]. Expression Syntactic category JOHN,MARY (individual) constant x variable HAPPY, EVEN unary predicate constant LOVE, > binary predicate constant LOVE(JOHN, MARY) (atomic) formula HAPPY(x) (atomic) formula x > 1 (atomic) formula MOTHER unary function constant MOTHER(JOHN) term 5

6 2.2 Semantics of Predicate Calculus Each expression belongs to a certain semantic type. The types of our predicate calculus are: individuals, sets, relations, functions, and truth values Sets Set. An abstract collection of distinct objects which are called the members or elements of that set. Elements may be concrete (like the beige 1992 Toyota Corolla I sold in 2008, David Beaver, or your computer) or abstract (like the number 2, the English phoneme /p/, or the set of all Swedish soccer players). The elements of a set are not ordered, and there may be infinitely many of them or none 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. Empty set. We write is a member of with. The empty set, written or{}, is the set containing no elements. Subset. A is a subset of B, written A B, if and only if every member of A is a member of B. A B iff for all x: if x Athen x B. Proper subset. A is a proper subset of B, written A B, if and only if A is a subset of B and A is not equal tob. A B iff (i) for all x: if x Athen x B and (ii) A B. Powerset. The powerset of A, written (A), is the set of all subsets of A. (A)={S S A} Set Union. The union of A and B, written A B, is the set of all entities x such that x is a member of A or x is a member of B. A B={x x Aor x B} 6

7 Set Intersection. The intersection of A and B, written A B, is the set of all entities x such that x is a member of A and x is a member of B. A B={x x Aand x B} Ordered pairs and relations Ordered pair. Sets are not ordered. {Bart, Lisa} ={Lisa, Bart} But the elements of an ordered pair written a,b are ordered. Here, a is the first member and b is the second member. We can also have ordered triples. Bart, Lisa Lisa, Bart {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 set of ordered tuples. For example, the older-than relation among Simpsons kids: { Bart, Lisa, Lisa, Maggie, Bart, Maggie } Note that this is a set. How many elements does it have? Domain. The domain of a relation is the set of entities that are the first member of some ordered pair in the relation. Range. The range of a relation is the set of entities that are the second member of some ordered pair in the relation Functions Function. A function is a special kind of relation. A relation R from A tob is a function 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 of R is equal toa 7

8 A function gives a single output for a given input. Are these relations functions? f 1 ={ Bart, Lisa, Lisa, Maggie, Bart, Maggie } f 2 ={ Bart, Lisa, Lisa, Maggie } Easier to see when you notate them like this: f 1 = Bart Lisa Lisa Maggie Maggie f 2 =[ Bart Lisa Lisa Maggie ] Function notation. Just as sets can be specified either by listing the elements or by description, functions can be described either by listing the ordered pairs that are members of the relation or by description. Example: x is the argument variable x+1 is the value description [x x+1] Function application. F(a) denotes the result of applying function F to argument a or F of a or F applied to a. If F is a function that contains the ordered pair a, b, then: F(a)=b This means that given a as input, F gives b as output. The result of applying a function specified descriptively to its argument can normally be written as the value description part, with the argument substituted for the argument variable. [x x+1](4)=4+1=5 8

9 2.2.4 Set theory as meta-language All these set-theoretic symbols are formal symbols, but they are not part of the language for which we are giving a semantics. They are being used to characterize the values that expressions of predicate calculus will have. In that sense, we are using the language of set theory as a meta-language Models and interpretation functions Interpretation with respect to a model. Expressions of predicate calculus are interpreted in models. Models consist of a domain of individuals D and an interpretation function I which assigns values to all the constants: M= D,I A valuation function[[]] M, built up recursively on the basis of the basic interpreation function I, assigns to every expression α of the language (not just the constants) a semantic value[[α]] M. Here are two models,m r andm f (r for real, andf for fantasy / fiction / fake ): They share the same domain: M r = D,I r M f = D,I f D={Maggie, Bart, Lisa} In M r, Bart is happy, but Maggie and Lisa are not: In M f, everybody is happy: I r (HAPPY)={Bart} I f (HAPPY)={Maggie, Bart, Lisa} Both interpretation functions map the constant MAGGIE to the individual Maggie: I r (MAGGIE)=Maggie I f (MAGGIE)=Maggie 9

10 What is the semantic value of HAPPY(MAGGIE)? It should come out as false in M r, and true inm f. So what we want to get is: [[HAPPY(MAGGIE)]] Mr =0 [[HAPPY(MAGGIE)]] M f =1 What tells us this? We need rules for determining the values of complex expressions from the values of their parts. So far all we have is rules for determining the values of constants Interpretation rules Constants If M= D,I and α is a constant, then[[α]] M = I(α). Complex terms Ifα 1...α n are terms andγ is a function constant with arityn, then[[γ(α 1,...α n )]] M is[[γ]] M ([[α 1 ]] M,...,[[α n ]] M ). Atomic formulae If π is an n-ary predicate and α 1...α n are terms,[[π(α 1,...,α n )]] M =1iff [[α 1 ]] M,...,[[α n ]] M [[π]] M If π is a unary predicate and α is a term, then[[π(α)]] M =1iff [[α]] M [[π]] M Negation [[ φ]] M =1if[[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. Because HAPPY is a unary predicate and MAGGIE is a term, we can use the rule for atomic formulae to figure out the value of HAPPY(MAGGIE). [[HAPPY(MAGGIE)]] M f =1iff[[MAGGIE]] M f [[HAPPY]] M f, i.e. iff Maggie {Maggie, Bart, Lisa}. 10

11 [[HAPPY(MAGGIE)]] Mr =1iff[[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 exists between formal and natural languages. So we will put (parsed) English inside the denotation brackets, instead of logic. Example: (2) NP N Barack S VP V smokes M (Wrong: [[Barack smokes]] M ) Following Heim & Kratzer, I use bold face for object language inside denotation brackets here, but often I am lazy about this. 3.1 A fragment of English with set denotations Models and the Lexicon Again, a modelm= D,I is a pair consisting of a domaindand an interpretation function I. For simplicity, let us assume that all of our models have the same domain D. There are two important subsets ofd: D e, the set of individuals D t, the set of truth values, 0 and 1 11

12 As in predicate calculus, models come with an interpretation function that specifies the values of all the constants. For concreteness, assume: (3) D e ={Barack Obama, Angela Merkel, Hugh Grant} The constants of natural language are the lexical items: the proper names, the intransitive verbs, the adjectives, etc. Thus the interpretation function I will map lexical items to their values in a model. For example, let M 1 = D,I 1, and let I 1 be defined such that: (4) a. I 1 (Barack)=Barack Obama b. I 1 (Angela)=Angela Merkel c. I 1 (Hugh)=Hugh Grant d. I 1 (smokes)={barack Obama, Angela Merkel} e. I 1 (drinks)={barack Obama, Hugh Grant} f. I 1 (likes)={ Barack Obama, Hugh Grant, Hugh Grant, Barack Obama, Hugh Grant, Hugh Grant } For example, let M 2 = D,I 2, and let I 2 be defined such that: (5) a. I 2 (Barack)=Barack Obama b. I 2 (Angela)=Angela Merkel c. I 2 (Hugh)=Hugh Grant d. I 2 (smokes)={angela Merkel} e. I 2 (drinks)={} f. I 2 (likes)={ Hugh Grant, Hugh Grant } Here, as in predicate calculus, we are defining the semantic values of intransitive verbs as sets, and we are defining the semantic values of transitive verbs as relations. But stay tuned; later we will interpret intransitive verbs as the characteristic functions of sets instead, in order to explore Frege s conjecture, that all semantic composition can be done with one single rule. Now we want to define a valuation function that assigns semantic values to all expressions of the language, not just constants. For constants, it s easy; just use the interpretation function. For example: 12

13 (6) [[smokes]] M 1 =I 1 (smokes)={barack Obama, Angela Merkel} In general, if α is a constant and M = 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)? Composition rules For non-branching nodes, we use the semantic value of the daughter: (8) Interpretation Rule: Non-branching Nodes γ If α has the form then [[α]] M = [[β]] M. β Sentences are true iff the value of the subject NP is an element of the value of the set denoted by the VP: (9) Interpretation Rule: Sentences S If α has the form then [[α]] M =1iff [[β]] M [[γ]] M. β γ So it will turn out that: (10) NP N Barack S VP V smokes M 1 =1 because Barack Obama is an element of [[smokes]] M 1. But what do we do about transitive verbs? (11) Interpretation Rule: Transitive VPs VP If α has the form then [[α]] M =??? β γ 13

14 We have said that the verb denotes a relation, but what does the verb combined with the object denote? Intuititively, it is a relation that is incompletely saturated. We can solve this problem by treating transitive verbs as functions that, when applied to an individual (the one denoted by the object NP), yield something corresponding to a set of individuals. Generalizing this idea, we will also treat VPs as functions. When applied to their subject NP, they will give us a truth value. So intransitive verbs will denote functions from individuals to truth values and intransitive verbs will 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 to calculate the values of complex expressions from the values of 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 how verbs are interpreted a little bit. Now we will have interpretations like the following. Recall that before we said that I 1 (smokes)={barack Obama, Angela Merkel}. Now we will treat smokes as the characteristic function of that set, with domain D. (12) f is the characteristic function of a sets iff for allxin the relevant domain, f(x)=1if x S, and f(x)=0otherwise. (13) I 1 (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]] M 1 = Barack Obama 1 Angela Merkel 1 Hugh Grant 0 Now we have to change our interpretation rule for sentences: (15) Interpretation Rule: Sentences S If α has the form then [[α]] M = [[γ]] M ([[β]] M ). β γ 14

15 And we can add an interpretation rule for VPs: (16) Interpretation Rule: VPs VP If α has the form β γ then [[α]] M = [[β]] M ([[γ]] M ). Instead of treating transitive verbs as relations, we will say that the semantic value of the transitive verb likes is a function from individuals to functions from individuals to truth values For example: Barack Obama (17) I 1 (likes) = Angela Merkel Hugh Grant Barack Obama 0 Angela Merkel 0 Hugh Grant 1 Barack Obama 0 Angela Merkel 0 Hugh Grant 0 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 whose top node is a branching node is the value of one daughter the one that denotes a function applied to the value of the other daughter, which works out if it denotes something in the domain of the function. So we can combine these two rules into one single rule. 1 (18) Interpretation Rule: Functional Application (FA) If α is a branching node,{β,γ} is the set of α s daughters, and[[β]] M is a function 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 in the phrase to be interpreted, and only makes reference to the semantic type of those phrases, it is a type-driven interpretation rule. 1 The phrase function(al) application has two meanings: The process of applying a function to an argument, and the composition rule that allows us to compute the semantic value of a phrase given 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 lowercase letters ( function(al) application ). 15

16 Types. All of our denotations are individuals, truth values, or functions. Individuals have type e and they are in the domain of individuals. Put more formally: (19) Barack Obama D e The truth values 0 and 1 have type t and they are elements of D t. A function from individuals to truth values is type e,t. By (14), [[smokes]] M 1 is of type e,t because its domain is the set of individuals, and its range is the set of truth values. Hence: (20) [[smokes]] M 1 D e,t [[likes]] will have type e, e,t because its domain is the set of individuals and its range is D e,t. So: (21) [[likes]] M 1 D e, e,t Inventory of types. We will only need the following types for the meanings of expressions in natural language: e, the type of individuals t, the type of truth values σ,τ, whereσ and τ are types Function notation For the purposes of talking about the composition rules that are necessary for building up the meanings of sentences from the meanings of their parts, it is not necessary to think about which individuals are actually mapped to 1 or 0 in in which actual models. So we make our lives easier and skip all that by writing: (22) [[smokes]] M =[x 1iff x smokes inm] where M is an arbitrary model, and the natural language sentence x smokes means that the individual x is designated as a smoker according to the interpretation function I inm. In fact, we can just leave off M entirely for the present purposes: (23) [[smokes]] = [x 1iff x smokes] 16

17 The domain of the function denoted by smokes is the set of individuals. Generally, we use x as a variable over individuals, so the domain of the function is clear from our choice of variable. But sometimes it is useful to be able to state conditions on the domain. Taking inspiration from Heim and Kratzer, we will specify domain restrictions with a colon, using expressions of the form: where [α φ γ] α is the argument variable, as before (a letter that stands for an arbitrary argument of the function we are defining) φ is the domain condition (places a condition on possible values for α) γ is the value description (specifies the value that our function assigns toα) So the more expanded notation would be: 2 (24) [[smokes]] = [x : x D e 1 iff x smokes] We will sometimes abbreviate this as: (25) [[smokes]] = [x D e 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 1iff y likes x]] (Notice that it s y likes x, not x likes y, because the verb combines first with its object, and then with its subject.) To express the result of applying the function to an argument, we write just the value description, with the argument substituting for all instances of the argument variable. (27) [x [y 1iff y likes x]](barack Obama) = [y 1iff y likes Barack Obama] 2 In Heim and Kratzer s notation, the same function would be written: λx : x D e. x smokes. We avoid λ notation in the meta-language here, because it is more customary in mathematics to describe functions by description using the symbol, and it avoids confusion over whether our metalanguage expressions are supposed to be treated as object language expressions of some variant of Church s lambda calculus. 17

18 Note: Two kinds of formal stuff. Notice that (lambdified) English, along with set-theory talk, is the meta-language, the language with which we describe the semantic 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 semantics for predicate calculus, symbols of predicate calculus were part of the object language. Now that we are giving a semantics for natural language directly, logic is no longer part of the picture. Example. Overview of how the compositional derivation of the truth conditions for Barack smokes will 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 M 0 by Functional Application 18

19 NP V = smokes N Barack M 0 by Non-branching Nodes =[[ smokes]]([[ Barack]]) by Non-branching Nodes ( 3) =[x 1 iff x smokes]([[barack]]) =[x 1 iff x smokes](barack) =1iff Barack Obama smokes. by Terminal Nodes by Terminal Nodes by function application Note: The last step does not involve any composition rules; we re done breaking down 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 from individuals to truth values (type e, t ). Ditransitive verbs. Suppose we have a phrase structure rule for ditransitive verbs like tell, that generates sentences like X told Y about Z. It could generate trees like this for example: 19

20 S NP VP N V NP PP (29) Barack Obama told N P NP Angela Merkel about N Barack Obama What type should tell be? What order should the arguments come in? This is where linking theory comes 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 any assignment a, 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 and β its daughter, then, for any assignment a, [[α]]=[[β]]. (32) Terminal Nodes (TN) If α is a terminal node occupied by a lexical item, then [[α]] is specified in the lexicon. 4.1 Rick Perry is conservative (33) [[Rick Perry]] = Rick Perry (34) [[conservative]] = [x 1iff x is conservative] 20

21 What to do with is? How about just an identity function: (35) [[is]] = [f D e,t f] (36) S:t NP:e VP: e,t N:e Rick Perry 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 1iff x is conservative])(rick Perry) TN = [x 1iff 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 1iff x is in y]] (38) [[Texas]] = Texas 21

22 S: e NP:e VP: e,t N: e Rick Perry V: e,t, e,t is PP: e,t P: e, e,t NP:e in N: e Texas 4.3 Rick Perry is proud of Texas (39) [[proud]] = [y [x 1iff x is proud of y]] (40) [[of]] = [x x] S: t NP:e N:e Rick Perry V: e,t, e,t is VP AP: e, e,t A: e, e,t PP:e proud P: e,e of NP: e N:e Texas 4.4 Rick Perry is a Republican (41) [[Republican]] = [x 1iff x is a Republican] Hey, let s make the indefinite article vacuous too: 22

23 (42) [[a]] = [f D e,t f] (43) S:t NP:e N:e Rick Perry V: e,t, e,t is VP: e,t NP: e,t D: e,t, e,t N : e,t a Republican 5 Predicate Modification 5.1 Rick Perry is a conservative Republican (44) S:t NP:e VP: e,t N: e Rick Perry V: e,t, e,t is NP: e,t D: e,t, e,t N : e,t a A:{ e,t e,t, e,t } conservative N : e,t Republican Our lexical entry for conservative from above is type e,t : (45) [[conservative 1 ]] = [x 1 iff x is conservative] 23

24 If we want to use Functional Application here, we need conservative to be a function of type e,t, e,t. (46) [[conservative 2 ]] = [f D e,t [x 1iff f(x) = 1 and x is conservative]] Now, [[[ N [ A conservative 2 ] [ N Republican ] ]]] = [[[ A conservative 2 ]]]([[[ N Republican ]]]) FA = [[conservative 2 ]]([[Republican]]) NN = f D e,t x 1iff f(x) = 1 and x is conservative ( x 1 iff x is a Republican ) TN = x 1iff 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 predicate position, they combine with a silent noun. Despite the ungrammaticality of *Rick Perry is conservative Republican. Alternative strategy: Use another composition rule. (47) Predicate Modification (PM) Ifαis 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)=1and[[γ]](x)=1 Now, [[[ N [ A conservative 1 ] [ N Republican ] ]]] = [x 1iff [[[ A conservative 1 ]]](x) = [[[ N Republican ]]](x) = 1] PM = [x 1iff [[conservative 1 ]](x) = [[Republican]](x) = 1] NN = [x 1iff 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

25 (48) S: t NP:e VP: e,t N:e Austin V: e,t, e,t NP: e,t is D: e,t, e,t N : e,t a N : e,t PP: e,t city P: e, e,t in NP:e Texas 6 The definite article What if we have the governor of Texas instead of Rick Perry? 6.1 The negative square root of 4 Regarding the negative square root of 4, Frege says, We have here a case in which out of a concept-expression [i.e., an expression whose meaning is of type e,t ] a compound proper name is formed [that is to say, the entire expression is of type e] with the help of the definite article in the singular, which is at any rate permissible when one and only one object falls under the concept. Permissible : the denotes a function of type e, t, e that is only defined for input predicates that characterize one single entity. In other words, the presupposes existence 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 unique y such that f(y) = 1] So the is not a function from D e,t to D e ; it is a partial function from D e,t to D e. But we can still give it the type e,t,e if we interpret this to allow partial 25

26 functions. To flesh out Frege s analysis of this example further, Heim and Kratzer suggest that square root is a transitive noun, with a meaning of type e, e,t, and that of is vacuous,[[square root]] applies to 4 via Functional Application, and the result of that composes with[[negative]] under predicate modification. (50) [[negative]] = [x 1iff x is negative] (51) [[square root]] = [y [x 1iff x is the square root of y]] (52) [[of]] = [x D e 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 of denotation brackets, and use composition rules to break down the tree. Here I am putting the name of the rule I used as a subscript on the equals sign, because I don t have enough room to put them off to the right. 26

27 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

28 = 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

29 = 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 1iff z is a square root of y]](4)(x) = 1]) = [[the]]([x 1 iff [[negative]](x) = [z 1iff 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

30 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 We need to compute a semantic value for every subtree/node. For convenience, we can group the nodes according to the string of words that they dominate, and start from 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: e [[four]] = 4 [[[ NP [ N four] ]]] = [[[ N four]]] = [[four]] = 4 TN NN of [[[ P of ]]] = [[of]] = x x TN, NN of four [[[ PP of four]]] = [[[ P of ]]]([[[ NP 4]]]) = [x x](4) = 4 FA 30

31 square root [[[ N square root]]] = [[square root]] NN = [y [x 1iff x is a square root of y]] TN square root of four [[[ N square root of four]] = [[[ N square root]]]([[[ PP of four]]]) = [y [x 1iff x is a square root of y]](4) = [x 1iff x is a square root of 4] = [x 1iff x {2,-2}] FA negative [[[ A negative]]] = [[negative]] = x 1iff x is negative TN, NN negative square root of four the [[[ N negative square root of four]]] = x 1iff [[[ 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} [[[ 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 such that 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 as convenient when you have to manipulate variable assignments... which we will cover next! 31

32 7 Predicate calculus: now with variables! Previously, I ignored formulas of predicate calculus with quantifiers and variables like these: x HAPPY(x) for all x, x is happy x HAPPY(x) there exists an x such that it is not the case that x is happy In order to interpret formulas with variables, we need to make interpretation relative to a model and an assignment function: [[φ]] M,g An assignment function assigns individuals to variables. Examples: g 1 = g 2 = x Maggie y Bart z Bart x Bart y Bart z Bart Informally, x HAPPY(x) is true iff: no matter which individual we assign to x, HAPPY(x) is true. In other words, for all elements in the domain d,d [[HAPPY]]. Informally, x HAPPY(x) is true iff: we can find some individual to assign to x such that HAPPY(x) is true. In other words, there is some element in the domain d such that d [[HAPPY]]. The assignment function determines what x is assigned to. Formally: [[x]] M,g =g(x) This in turn influences the value of a formula containing x in whichxis not bound by a quantifier (a formula in whichxis free). Let s interpret HAPPY(x) using the reality modelm r and the assignment functions g 1 and g 2. [[HAPPY(x)]] Mr,g 1 =1 iff[[x]] Mr,g 1 [[HAPPY]] Mr,g 1 32

33 iff g 1 (x) I r (HAPPY) iff Maggie {Bart}. Maggie/ {Bart} so[[happy(x)]] Mr,g 1 =0. [[HAPPY(x)]] Mr,g 2 =1 iff[[x]] Mr,g 2 [[HAPPY]] Mr,g 2 iff g 2 (x) I r (HAPPY) iff Bart I r (HAPPY). Bart {Bart} so[[happy(x)]] Mr,g 2 =1. Intuitively, this means that x HAPPY(x) is true, but x HAPPY(x) is false inm r. New interpretation rules: Constants If 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 formulae Ifπ is ann-ary predicate andα 1,...α n are terms, then[[π(α 1,...,α n )]] M,g = 1 iff [[α 1 ]] M,g,...,[[α n ]] M,g [[π]] M,g If π is a unary predicate and α is a term, then[[π(α)]] M,g = 1 iff[[α]] M,g [[π]] M,g. Negation [[ φ]] M,g =1if[[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, where g is an assignment function exactly like g except that g (v)=d. 33

34 Existential quantification all new! [[ vφ]] M,g =1iff there is a d Dsuch that[[φ]] M,g, where g is an assignment function exactly like g except that g (v)=d. [[ x HAPPY(x)]] Mr,g 1 =1 iff for alld D,[[HAPPY(x)]] Mr,g =1, whereg is an assignment function exactly like g 1 except that g (x)=d This can be falsified by setting d equal to Maggie, so g (x)=maggie. [[HAPPY(x)]] Mr,g =0in this case. But there is a d D such that[[happy(x)]] Mr,g, where g is an assignment function exactly like g 1 except that g (x)=d. As shown above, there is such ad: Bart. 8 Relative clauses Heim and Kratzer use assignment functions for the interpretation of relative clauses such as the following: (54) The car that Joe bought is very fancy. (55) The woman who admires Joe is very lovely. Semantically, relative clauses are just like adjectives: (56) The red car is very fancy. (57) The Swedish woman 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 following syntax for relative clause CPs: 34

35 (59) CP which i C C S that VP Joe V bought t i (60) CP who i C C S that VP t i V likes Joe The text that is struck out like so is deleted. Heim and Kratzer assume that either the relative pronoun which or who or the complementizer that is deleted. Interpretation of variables (61) Traces Rule (TR) If α i is a trace and g is an assignment, [[α i ]] g = g(i) g 1 = 1 Maggie 2 Bart 3 Maggie g 2 = 1 Lisa 2 Bart 3 Maggie [[t 1 ]] g 1 = Maggie [[t 1 ]] g 2 = Lisa So now we interpret everything with respect to an assignment. 35

36 (62) [[ [ VP [ V abandoned ] [ t 1 ] ]]] g = x 1iff x abandoned g(1) But there are assignment-independent denotations too. (63) Bridge to assignment-independence (BI) For any tree α, α is in the domain of [[]] iff for all assignments g and g, [[α]] g = [[α]] g. If α is in the domain of [[]], then for all assignments g, [[α]]= [[α]] g. So we can still have assignment-independent lexical entries like: (64) [[laugh]] = x 1iff x laughs and then by (63), we have: (65) [[laugh]] g 1 = x 1iff x laughs (66) [[laugh]] g 2 = x 1iff 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 in the lexicon. (68) Non-branching Nodes (NN) If α is a non-branching node and β its daughter, then, for any assignment g, [[α]] g =[[β]] g. (69) Functional Application (FA) If α is a branching node and{β,γ} the set of its daughters, then, for any assignment g, 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 any assignment g, if [[β]] g and [[γ]] g are both functions of type e,t, then [[α]] g = x 1iff [[β]] g (x) = [[γ]] g (x) = 1. 36

37 Predicate abstraction. CP to have type e,t? The S in a relative clause is type t. How do we get the (71) CP : e,t which 1 C : C: S:t that :e VP: e,t Joe V: e, e,t :e Heim and Kratzer: bought t 1 The complementizer that is vacuous; that S = S or [[that]] = p D t p The relative pronoun is vacuous too, but it triggers a special rule called Predicate Abstraction (72) Predicate Abstraction (PA) If α is a branching node whose daughters are a relative pronoun indexed i and β, then [[α]] g = x [[β]] gx/i g x/i is an assignment that is just like g except that x is assigned to i. Note that x is a variable that is part of the meta-language, bound by the metalanguage operator λ, ranging over objects in the domain. So [[(71)]] = x 1iff Joe bought x. In case you don t believe me: [[[ CP which 1 [ C [ C that ] [ S [ Joe ] [ VP [ V bought ] [ t 1 ] ] ] ] ]]] = [[[ CP which 1 [ C [ C that ] [ S [ Joe ] [ VP [ V bought ] [ t 1 ] ] ] ] ]]] g (any g) BI = [x [[[ C [ C that ] [ S [ Joe ] [ VP [ V bought ] [ t 1 ] ] ] ]]] gx/1 ] PA 37

38 = [x [[[ C that ]]] gx/1 ([[[ S [ Joe ] [ VP [ V bought ] [ t 1 ] ] ]]] gx/1 )] FA = [x [[[ C that ]]]([[[ S [ Joe ] [ VP [ V bought ] [ t 1 ] ] ]]] gx/1 )] BI = [x [p D t p]([[[ S [ Joe ] [ VP [ V bought ] [ t 1 ] ] ]]] gx/1 ) ] LT = [x [[[ S [ Joe ] [ VP [ V bought ] [ t 1 ] ] ]]] gx/1 ] function application = [x [[[ VP [ V bought ] [ t 1 ] ]]] gx/1 ([[[ Joe ]]] gx/1 )] FA = [x [[[ VP [ V bought ] [ t 1 ] ]]] gx/1 ([[[ Joe ]]])] BI = [x [[[ VP [ V bought ] [ t 1 ] ]]] gx/1 (Joe)] NN, LT = [x [[[ V bought ]]] gx/1 ([[[ t 1 ] gx/1 )]](Joe)] FA = [x [[[ V bought ]]]([[[ t 1 ] gx/1 )]](Joe)] BI = [x [z [y 1iff y bought z]]([[[ t 1 ]]] gx/1 )(Joe)] LT, NN = [x [z [y 1iff y bought z]]([[t 1 ]] gx/1 )(Joe)] NN = [x [z [y 1iff y bought z]](x)(joe)] TR = [x [y 1iff y bought x](joe)] function application = [x 1iff 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

39 S:t NP:?... VP: e,t is happy 9.1 Type e? Most of the s we have seen so far have been of type e: 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 like a Republican after is. Should words and phrases like Nobody and At least one person be treated as type e? How can we tell? Predictions of the type e 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 type e. Proof: [[came yesterday morning]] [[came yesterday]] (everything that came yesterday morning came yesterday), and if the subject denotes an individual, then the sentence means that the subject is an element of the set denoted by the VP. If the first sentence is true, then the subject is an element of the set denoted by the VP, which means that the second sentence must be true. QED. (79) At most one letter came yesterday morning. Therefore, at most one letter came yesterday. This inference is not valid, so at most one letter must not be type e. 39

40 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 the other side of the border. The fact that it is contradictory follows from these assumptions: [[Mount Rainier]] D e [[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 type e, the sentence means that it is in the set denoted by the VP standard analysis of and This sentence is not contradictory: (81) More than two mountains are on this side of the border, and more than two mountains are on the other side of the border. So more than two mountains must not be type e. 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]] D e [[over 30 years old]] [[under 40 years old]]= D (everything is either over 30 years old or under 40 years old) When the subject is type e, the sentence means that it is in the set denoted by the VP standard analysis of or This sentence is not a tautology: (83) Every woman in this room is over 30 years old, or every woman in this room is under 40 years old. So every woman must not be of type e 40

41 9.2 Solution: Generalized quantifiers (84) [[nothing]] = f D e,t 1 iff there is no x D e such that f(x) = 1 (85) [[everything]] = f D e,t 1 iff for all x D e, f(x) = 1 (86) [[something]] = f D e,t 1 iff there is some x D e such that f(x) = 1 (87) S:t vs. S: t : e,t,t VP: e,t :e VP: e,t everything V: e,t Mary V: e,t vanished vanished (88) [[every]] = f D e,t [g D e,t 1 iff for all x D e such that f(x) = 1, g(x)=1 ] (89) [[no]] = f D e,t [g D e,t 1 iff there is no x D e such that f(x) = 1 and g(x)=1 ] (90) [[some]] = f D e,t [g D e,t 1 iff there is some x D e such that f(x) = 1 and g(x)=1 ] (91) S:t : e,t,t VP: e,t D: e,t, e,t,t every NP: e,t thing V: e,t vanished 41

42 10 The problem of quantifiers in object position 10.1 The problem (92) S: t : e,t,t VP: e,t D: e,t, e,t,t NP: e,t V: e, e,t :e every linguist offended John (93) S:????????? :e VP:??????????? John V: e, e,t : e,t,t offended 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 Quantifier Raising), leaving a trace of typeein object position. (Or simulate movement via Cooper Storage, as in Head-Driven Phrase Structure Grammar.) 2. Interpret the quantifier phrase in situ. In this case one can apply a typeshifting operation to change its type An in situ approach Multiple versions of lexical items: [[everybody 1 ]] = f D e,t 1 iff for all persons x D, f(x) = 1 [[everybody 2 ]] = f D e, e,t [x D e 1 iff for all persons y D e, f(y)(x) = 1 ] [[somebody 1 ]] = f D e,t 1 iff there is some person x D e such that f(x) = 1 42

43 [[somebody 2 ]] = f D e, e,t [x D e 1 iff there is some person y D e such that f(y)(x) = 1 ] (94) S: t :e VP: e,t John V: e, e,t offended : e, e,t, e,t everybody 2 (95) S: t : e,t,t VP: e,t Everybody V: e, e,t offended : e, e,t, e,t somebody 2 Note: This only gets one of the readings. We need a new everybody for ternary relations: (96) S VP Ann V PP V P introduced everybody to Maria What type are the determiners (note: et = e, t )? 43

44 (97) S: t :e VP:et John V: e,et : e,et,et offended D: et, e,et,et every NP:et linguist How do we get this every from our normal et, et,t every? A lexical rule. (98) For every lexical item δ 1 with a meaning of type et, et,t, there is a (homophonous and syntactically identical) item δ 2 with the following meaning of type et, e,et,et : [[δ 2 ]] = f D e,t [g D e,et [x D e [[δ 1 ]](f)(z D e 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 what we 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 transformations from SS to LF happen after the order of the words is determined, we do not see the output of these transformations. These movement operations are in this sense covert.) A transformation called QR (Quantifier Raising) maps the SS structure in (99a) to something like the LF structure in (99b) 44

45 (99) a. S VP John V offended b. S D every NP linguist i S D NP VP every linguist John V offended Actually, Heim and Kratzer propose the following, so that they can make it work with Predicate Abstraction: (100) S t i D NP 1 S every linguist VP John V offended (101) Predicate Abstraction (PA) (revised) Letαbe a branching node with daughters β andγ, whereβ dominates only a numerical index i. Then for any variable assignment g, [[α]] g =x D e [[γ]] gx/i. Example. Let s give every node of the tree a unique category label so we can refer to the denotation of the tree rooted at that node using the category label. 45 t 1

46 (102) S 1 1? D NP 1 S 2 every linguist 2 VP John V offended 3 t 1 The task is to analyze the truth conditions of S 1 (or, to be more precise, the tree rooted at the node labelled S 1 ). The basic idea is straightforward Predicate Abstraction at the mystery-category node (labelled? here), Pronouns and Traces rule at the trace, and Functional Application everywhere else but it is a bit tricky to go between assignment-dependent and assignment-independent denotations. The trick is to start with the Bridge to Independence. [[S g 1 ]] =[[S 1 ]] g, for all g BI =[[ 1 ]] g ([[?]] g ) FA =[[ 1 ]] g (x [[S 2 ]] gx/1 ) PA =[[ 1 ]] g (x [[VP]] gx/1 ([[ 2 ]] gx/1 )) FA =[[ 1 ]] g (x [[V]] gx/1 ([[ 3 ]] gx/1 )([[ 2 ]] gx/1 )) FA =[[ 1 ]] g (x [[offended]] gx/1 ([[t 1 ]] gx/1 )([[John]] gx/1 )) NN, TN =[[ 1 ]] g (x [[offended]] gx/1 (x)([[john]] gx/1 )) TR =[[ 1 ]] g (x [[offended]](x)([[john]])) BI =[[ 1 ]] g (x [y [z 1iff z offended y ]](x)(john)) TN =[[ 1 ]] g (x John offended x) β-r =[[D]] g ([[NP]] g )(x 1iff John offended x) TN =[[every]] g ([[linguist]] g )(x 1iff 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 46

47 10.4 Arguments in favor of the movement approach Argument #1: Scope ambiguities. In order to get both readings of Everybody loves somebody, we have to introduce yet even more complicated types. Scope ambiguities are trivially derived under the movement approach. Argument #2: Inverse linking. There is one class of examples that cannot be generated under an in 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 only reading that an in situ analysis can give us. QR analysis: (104) S every basket 1 S VP D NP is rotten one N PP apple P in t 1 Argument #3: Antecedent-contained deletion (105) I read every novel that you did. Like regular VP ellipsis: (106) I read War and Peace before you did. except that the antecedent VP is contained in the elided VP! To create an appropriate antecedent, you have to QR the object. 47