CAS LX 502 9 b Formal semantics Pronouns

CAS LX 502 9 b. Formal semantics Pronouns and quantifiers

Bond is hungry S N VP Bond Vi is hungry • [N]M = F(Bond) • [VP]M = [Vi]M = … is hungry … Bond Loren Pavarotti … F x [ x F(is hungry)] • [S]M = [VP]M ( [N]M ) = F(Bond) x [ x F(is hungry)] = F(Bond) F(is hungry) U

Bond is hungry S N VP Bond Vi is hungry … Bond Loren Pavarotti … F 1 • [S]M 1 = F 1(Bond) F 1(is hungry) = Bond {Bond, Loren} • In the specific situation M 1. U 1

He is hungry S N VP He Vi is hungry … Bond Loren Pavarotti … F • We don’t have he in our lexicon yet, but if we did, how should we interpret it? • This sentence could mean different things (have different truth conditions), in the same situation, depending on who we’re pointing at. U

He 1 is hungry S N VP He 1 Vi is hungry … Bond Loren Pavarotti … F g 1 2 3 … • When writing a sentence like he is hungry, the standard practice is to indicate the “pointing” relation by using a subscript on he: • He 1 is hungry. • The idea here is that this is interpreted in conjunction with a “pointing function” that tells us who “ 1” points to. U

He 1 is hungry S N VP He 1 Vi F is hungry … Bond Loren Pavarotti … g 1 1 2 3 … • Where different people are being pointed to, we use different subscripts: • He 1 likes her 2, but he 3 hasn’t noticed. • The “pointing function” goes by the more official name assignment function, and is generally referred to as g. U

He 1 is hungry S N VP He 1 Vi is hungry F … is hungry … Bond Loren Pavarotti … g 1 1 2 3 … • The assignment function g fits into the system much like the valuation function F does. F maps lexical items into the universe of individuals, g maps subscripts into the universe of individuals. • [ n]M, g = g(n) U

He 1 is hungry S N VP He 1 Vi is hungry F 1 … is hungry … Bond Loren Pavarotti … g 1 1 2 3 … • [ n]M, g = g(n) • [S]M 1, g 1 = g 1(he 1) F 1(is hungry) = Bond {Bond, Loren} U 1

Quantifiers • Quantifiers (everyone, someone, noone) allow us to state generalizations. • Someone is boring. • Everyone is hungry. • When we say everyone is hungry, we’re saying that for each individual x, x is hungry. • We can think of this as follows: Run through the universe of individuals, pointing at each one in turn, and evaluate s/he is hungry. If it is true for every pointing, then everyone is hungry is true.

Bond likes everyone • So, what we’re after is something like this: • x U [Bond likes x in M] • That is, we have to convert everyone into a pronoun and interpret the S, with a pronoun in it, with every “pointing” that we can do. • To do this, we will introduce a rule called Quantifier Raising for sentences with quantifiers in them that will accomplish just that.

Syntactic base rules S NP VP VP Vt NP S S Conj. P VP Vi Conj. P Conj S NP Det NC S Neg S NP Det the, a, every Conj and, or Vt likes, hates Vi is boring, is hungry NP Pavarotti, Loren, Bond, Nemo, Dory, Blinky, Semantics, The Last Juror, hen, shen, itn, himn, hern, himselfn, herselfn, itselfn. Neg it is not the case that NC book, fish, man, woman

Bond likes every fish S • Not much new here, except that we’ve added some words we can play with, including some common nouns, and some determiners (Det) to use to build quantifiers. NP VP NP Vt Bond likes NP Det NC every fish

Bond likes every fish • Back to the problem of quantifiers. • Consider the meaning of Bond likes every fish. It should be something like: • For every x in U that is a fish, Bond likes x. Or: x U [x is a fish in M Bond likes x in M]

Bond likes every fish • x U [x is a fish in M Bond likes x in M] • Notice that our sentence is basically here, but with x instead of every fish. The meaning of every fish is kind of “factored out” of the sentence and used to set the value of x. • In order to get this interpretation, we’re going to introduce a transformation. A new kind of rule.

Transformations • The syntactic base rules that we have allow us to construct trees. • A transformation takes a tree and alters it, resulting in a new tree. • The particular transformation we are going to adopt here (Quantifier Raising) takes an NP like every fish and attaches it to the top of the tree, leaving an abstract pronoun behind. Then we will write our semantic rules to interpret that structure.

Quantifier Raising S S NP NP Vt Bond likes NP Det NC every fish S Det NC 1 every fish S NP VP NP Vt Bond likes t 1

Quantifier Raising • Quantifier Raising [S X NP Y ] [S NP [S i [S X ti Y ] ] ] S • That is: S … NP S i S … ti …

Interpreting quantifiers • Now comes the tricky part: How do we assign a semantic interpretation to the structure? (It is easier—nay, possible—now that we have the QR rule, but let’s see why). • Remember, what we’re after is: x U [x is a fish in M Bond likes x in M]

Interpreting quantifiers • Let’s start with the lower S. We know how to interpret that, it is essentially just Bond likes it 1. S NP S Det NC 1 every fish • < Bond, g(1) > F(likes) S NP VP NP Vt Bond likes t 1

Interpreting quantifiers • The purpose of the 1 S node is to make a predicate out of this sentence. • The predicate will be, in effect, things Bond likes. • Goal: NP S Det NC 1 every fish S NP VP NP Vt Bond likes 1 [< Bond, g(1) > F(likes)] t 1

Interpreting quantifiers • The interpretation of [1]M, g, S then, will be a function that takes a sentence (type <t>) and returns a predicate (type <e, t>). • [1]M, g is type <t, <e, t>>. • [1]M, g = S [ x [ [S]M, g[1/x] ] ] NP S Det NC 1 every fish S NP VP NP Vt Bond likes t 1

[ M, g[1/x] ] • To understand what is going to happen here, we need to introduce one more concept, the modified assignment function g[1/x]. • Remember that the assignment function maps subscripts to individuals, so that we can interpret pronouns like he 2. • So, g 1 (an assignment function for a particular pointing situation) might map 1 to Pavarotti, 2 to Nemo, 3 to Loren, and so forth.

[ M, g[1/x] ] • g 1 maps 1 to Pavarotti, 2 to Nemo, 3 to Loren, … • A modified assignment function g[i/x] is an assignment function that is just like the original assignment function except that instead of whatever g mapped i to, g[i/x] maps i to x instead. That is: • g 1(1) = Pavarotti, g 1(2) = Nemo • g 1[2/Bond](1) = Pavarotti, g 1[2/Bond](2) = Bond

[ M, g[1/x] ] • g 1(1) = Pavarotti, g 1(2) = Nemo • g 1[2/Bond](1) = Pavarotti, g 1[2/Bond](2) = Bond • The reason that this is useful is that to interpret every fish, we want to go through all of the fish, and check whether Bond likes it is true when we point to each fish. • It is a pronoun, whose interpretation is dependent on who we are pointing to, so we need to be able to change who we point to (accomplished by modifying the assignment function).

Interpreting quantifiers • [1]M, g = S [ x [ [S]M, g[1/x] ] S ] NP • [S ]M, g = S Det NC 1 every fish S NP [S]M, g S [ x [ [S]M, g[1/x] ] ] NP x [ [S]M, g[1/x] ] = x [<Bond, g[1/x](1)> F(likes) ] = Bond x [<Bond, x> F(likes)] VP Vt likes t 1

Interpreting quantifiers • Great, now we have S [S ]M, g as a predicate that means things Bond likes. • Now, let’s turn to every fish. • Fish is a predicate, true of fish; that is: x [x F(fish)] NP S Det NC 1 every fish S NP VP NP Vt Bond likes t 1

Interpreting quantifiers • What we’re looking for is a way to verify that every individual that is a fish is also an individual that Bond likes. • That is: x U [x is a fish x is a thing Bond likes] • [S ]M, g is the predicate things Bond likes. [NC]M, g is the predicate fish. S NP S Det NC 1 every fish S NP VP NP Vt Bond likes t 1

Interpreting quantifiers • Informally, every takes two predicates, and yields true if everything that satisfies the first predicate also satisfies the second. • <<e, t>, t>> • [every]M, g = S NP S Det NC 1 every fish P [ Q [ x U [P(x) Q(x)] ] ] S NP VP NP Vt Bond likes t 1

Interpreting quantifiers • [every]M, g = P [ Q [ x U [P(x) Q(x)] ] ] NP • [NC]M, g = [fish]M, g = y [y F(fish)] Det • [NP]M, g = S S NC 1 every fish [every]M, g ( [fish]M, g ) = y [y F(fish)] P [ Q [ x U [P(x) Q(x)] ] ] = Q [ x U [x y [y F(fish)] Q(x)] ] = Q [ x U [x F(fish) Q(x)] ] S NP VP NP Vt Bond likes t 1

Interpreting quantifiers • [NP]M, g = Q [ x U [x F(fish) Q(x)] ] S • [S ]M, g = NP y [<Bond, y> F(likes)] • [S]M, g = S Det NC 1 every fish y [<Bond, y> F(likes)] Q [ x U [x F(fish) Q(x)] ] = x U [x F(fish) x y [<Bond, y> F(likes)]] = x U [x F(fish) <Bond, x> F(likes) ] S NP VP NP Vt Bond likes t 1

Phew • And, we’ve done it. We’ve derived the truth conditions for Bond likes every fish: • x U [x F(fish) <Bond, x> F(likes) ] • For every individual x in U, if x is a fish, then Bond likes x.

New semantic rules • The new semantic rules we needed (lexical entries, we’re still using the same Functional Application and Pass-up rules) were: • [every]M, g = P [ Q [ x U [P(x) Q(x)] ] ] • [i]M, g = S [ x [ [S]M, g[i/x] ] ]

Loren hates a book • We’ve worked out an interpretation for a single quantificational determiner, every, but we can in a parallel way give a meaning to a (as in a book). • The meaning we want for Loren hates a book is that there is some individual x such that x is a book and Loren hates x: • x U [x F(book) <Loren, x> F(hates) ]

Loren hates a book • Without running through all of the steps again, what we want here is for [a]M, g to take two predicates (here, book, and things Loren hates), and be true if there is some individual that satisfies both: • [a]M, g = P [ Q [ x U [P(x) Q(x)] ] ]

Recap—step one • When faced with a sentence like A fish likes Nemo, the first thing to do is use the syntactic base rules to construct a tree. S • S NP VP • NP Det NC NP VP • Det a • NC fish NC Vt NP Det • VP Vt NP • Vt likes NP a fish likes • NP Nemo

Recap—step two S • Then, for the quantificational NPs (those with every or a NP as a determiner), use the QR transformation to attach the quantifier Det to the top of the tree. a • Quantifier Raising [S X NP Y ] [S NP [S i [S X ti Y ] ] ] S NC fish 1 S t 1 VP Vt NP likes NP Nemo

Recap—step three S • And then work your way up the tree, evaluating first the lexical entries at the bottom of the tree: NP S Det NC a fish • [Nemo]M, g = Nemo • • • [fish]M, g = x[x F(fish)] [t 1]M, g = g(1) [likes]M, g = y[ x[<x, y> F(likes)]] [a]M, g = P [ Q [ x U [P(x) Q(x)] ] ] [1]M, g = S [ x [ [S]M, g[1/x] ] ] 1 S t 1 VP Vt NP likes NP Nemo

Recap—step three S • Work out parent nodes by applying functions to arguments NP Det • [VP]M, g = a [Vt]M, g ( [NP]M, g ) = Nemo y[ x[<x, y> F(likes)]] = x[<x, Nemo> F(likes)] • [S]M, g = [VP]M, g ( [t 1]M, g ) = g(1) x[<x, Nemo> F(likes)] = <g(1), Nemo> F(likes) S NC fish 1 S t 1 VP Vt NP likes NP Nemo

Recap—step three S • Work out parent nodes by applying functions to arguments • [NP]M, g = [Det]M, g ( [NC]M, g ) = NP S Det NC a fish 1 S t 1 z[z F(fish)] P [ Q [ x U [P(x) Q(x)] ] ] = Q [ x U [x z[z F(fish)] Q(x)] ] ] = Q [ x U [x F(fish) Q(x)] ] ] • [S ]M, g = [1]M, g ( [S]M, g ) = <g(1), Nemo> F(likes) S [ x [ [S]M, g[1/x] ] ] = x [<g[1/x](1), Nemo> F(likes) ] = x [<x, Nemo> F(likes) ] VP Vt NP likes NP Nemo

Recap—step three S • Work out parent nodes by applying functions to arguments • Notice that you can change the names of variables freely—to ensure that they don’t conflict. • [S ]M, g = [NP]M, g ( [S ]M, g ) = NP S Det NC a fish 1 S t 1 z [<z, Nemo> F(likes) ] Q [ x U [x F(fish) Q(x)] ] = x U [x F(fish) x z [<z, Nemo> F(likes) ] ] = x U [x F(fish) <x, Nemo> F(likes) ] VP Vt NP likes NP Nemo

Comments about QR • Quantifier Raising [S X NP Y ] [S NP [S i [S X ti Y ] ] ] • As it is stated, QR applies to any NP, whether it is a quantifier or not. So, when do you use QR? • There are certain situations (e. g. , Nemo hates every book) where the structure cannot be interpreted without QR. When the quantifier is the object of a transitive verb, for example. • A transitive verb needs something of type <e>. A quantifier needs something of type <e, t> (and is itself of type <<e, t>, t>>). Neither one can take the other as an argument.

Comments about QR • Quantifier Raising [S X NP Y ] [S NP [S i [S X ti Y ] ] ] • As it is stated, QR applies to any NP, whether it is a quantifier or not. So, when do you use QR? • On the other hand, if you apply QR to a name like Loren, the semantic interpretation becomes more complicated to work out, but the end result (the truth conditions) are the same as if you hadn’t done QR. So, you don’t really need QR to interpret the structure.

Comments about QR • Quantifier Raising [S X NP Y ] [S NP [S i [S X ti Y ] ] ] • As it is stated, QR applies to any NP, whether it is a quantifier or not. So, when do you use QR? • In fact, it turns out that when you have a quantificational NP as a subject (as in Every fish hates The Last Juror), you don’t actually need QR in order to interpret the structure either. • The VP hates The Last Juror is a predicate. Every applies to fish and then can apply to the VP, resulting in the same truth conditions you would have if you applied QR.

Comments about QR • Quantifier Raising [S X NP Y ] [S NP [S i [S X ti Y ] ] ] • As it is stated, QR applies to any NP, whether it is a quantifier or not. So, when do you use QR? • Moreover, if you apply QR to an NP, you still have an NP —you could apply QR again to that same NP if you wanted. Again, like with names, this won’t affect the ultimate interpretation, it will just increase the amount of effort necessary to work out the truth conditions.

Comments about QR • Quantifier Raising [S X NP Y ] [S NP [S i [S X ti Y ] ] ] • As it is stated, QR applies to any NP, whether it is a quantifier or not. So, when do you use QR? • So, the answer is: Use QR as necessary, where it will result in a different interpretation from not using QR (or when not using QR prevents the structure from being interpreted at all).

Every fish likes a book • Some sentences have more than one quantifier. We know that, because a book is the object of a transitive verb, we need to apply QR to a book. We could refrain from applying QR to every fish (the subject) because subjects don’t require QR in order to be interpretable. • The result will be, paraphrasing: There is a book x such that for every y, if y is a fish, then y likes x. This is certainly something the sentence can mean. • However, the sentence can also mean: For every y, if y is a fish, then there is a book x such that y likes x. To get this meaning, you must also apply QR to the subject (after applying QR to the object).

Quantifier Raising [S X NP Y ] [S NP [S i [S X ti Y ] ] ] Pass-Up [b ]M, g = [ ]M, g Functional application [g b ]M, g = [b]M, g ( [ ]M, g ) or [ ]M, g ( [b]M, g ) whichever is defined [Pavarotti]M, g = F(Pavarotti) (any NP) [is boring]M, g = x [ x F(is boring) ] (any NC or Vi) [likes]M, g = y [ x [ <x, y> F(likes) ] ] (any Vt) [and]M, g = y [ x [ x y ] ] (analogous for or) [it is not the case that]M, g = x [ x ] [ i]M, g = g(i) [i]M, g = S [ x [ [S]M, g[i/x] ] ] [every]M, g = P [ Q [ x U [P(x) Q(x)] ] ] [a]M, g = P [ Q [ x U [P(x) Q(x)] ] ]