1. CAS LX 502 10b. Binding

2. Syntactic base rules (F2) S  NP VPVP  Vt NP S  S ConjPVP  Vi ConjP  Conj SNP  Det N C S  Neg SNP  N P Det  the, a, everyN P  Pavarotti, Loren, Bond, Nemo, Dory, Blinky, Semantics, The Last Juror, he n, she n, it n, him n, her n, himself n, herself n, itself n. Conj  and, or Vt  likes, hates Vi  is boring, is hungry Neg  it is not the case thatN C  book, fish, man, woman

3. [Pavarotti] M,g = F(Pavarotti) (any N P ) [is boring] M,g = x [ x  F(is boring) ] (any N C or Vi) [likes] M,g = y [ x [  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)] ] ] Pass-Up [   ] M,g = [  ] M,g Functional application [   ] M,g = [  ] M,g ( [  ] M,g ) or [  ] M,g ( [  ] M,g ) whichever is defined Quantifier Raising [ S X NP Y ]  [ S NP [ S i [ S X t i Y ] ] ]

4. Binding Among our N P s, we have a couple of special types: –self anaphora: himself n, herself n, itself n pronouns: him n, her n, it n, he n, she n. Unlike Bond, Loren, Pavarotti, these N P s don’t denote a specific individual in U, regardless of the model; they “pick up their reference” from elsewhere.

5. Binding We think of the denotation of these elements as dependent—their reference is provided by something else. In the case of pronouns, it is provided by “pointing” (courtesy of the assignment function g). With –self anaphora, the reference of …self is bound to the reference of another (generally preceding) NP. Bond 1 likes himself 1. He 1 is hungry 1 

6. Bond 1 likes himself 1 It is common, when discussing reference of pronouns and –self anaphora to write sentences with a subscript on each NP under consideration. What Bond 1 likes himself 1 really means is: The N P s Bond and himself both point to the same individual (hence the coindexation). Bond intrinsically names a particular individual. Himself has no intrinsic reference, it’s all in the connections (via the index) to other N P s. Therefore, himself also points to the individual Bond points to.

7. Bond 1 likes himself 1 This is a useful notation for describing the intended meanings of sentences, but this is not something F2 can give us. N P  Bond N P  himself n The intended meaning of N P  himself n is that N P can be rewritten as himself n for some index n. Doesn’t matter which, pick one. In fact, pick 1. Notice the lack of any N P  Bond n rule.

8. Every fish likes itself In fact, –self anaphora (and pronouns too) can pick up their reference from quantifiers like every fish too. We might write this informally as: Every fish 1 likes itself 1. …although this is just as un-directly-generable by F2 as Bond 1 likes himself 1 was. An example of a bound pronoun, slightly outside of F2, would be every fish 1 thinks that Loren likes him 1.

9. Pavarotti likes himself Our syntactic rules allow us to build and evaluate this sentence, but with a twist. Let’s see. S NPVP S  NP VP

10. Pavarotti likes himself Our syntactic rules allow us to build and evaluate this sentence, but with a twist. Let’s see. S NPVP VtNP S  NP VP VP  Vt NP

11. Pavarotti likes himself Our syntactic rules allow us to build and evaluate this sentence, but with a twist. Let’s see. S NPVP VtNPNP NP NPNP S  NP VP VP  Vt NP NP  N P

12. Pavarotti likes himself Our syntactic rules allow us to build and evaluate this sentence, but with a twist. Let’s see. S NPVP Vt likes NPNP Pavarotti NP NPNP S  NP VPVt  likes VP  Vt NPN P  Pavarotti NP  N P

13. Pavarotti likes himself N P  himself n The subscript n means: pick an index, any index. Let’s pick 3. S NPVP Vt likes NPNP Pavarotti NP NPNP himself 3 S  NP VPVt  likes VP  Vt NPN P  Pavarotti NP  N P N P  himself n

14. Pavarotti likes himself What should this mean? S NPVP Vt likes NPNP Pavarotti NP NPNP himself 3

15. Pavarotti likes himself What should this mean? So, we need himself 3 to denote Pavarotti. Our interpretation rules say it refers to g(3)—whoever “3” points to. S NPVP Vt likes NPNP Pavarotti NP NPNP himself 3 [  i ] M,g = g(i) [himself 3 ] M,g = g(3)

16. Pavarotti likes himself In less formal circumstances, this interpretation is written like: Pavarotti 3 likes himself 3. Where we give the proper name Pavarotti a matching index. In F2, we can’t give Pavarotti a subscript. First, no rule provides for that. Second, the interpretation of anything 3 is g(3). We lose the connection to Pavarotti. Right? S NPVP Vt likes NPNP Pavarotti NP NPNP himself 3 [  i ] M,g = g(i) [himself 3 ] M,g = g(3)

17. Pavarotti likes himself So that’s the puzzle: How can Pavarotti still refer to this guy and simultaneously bind the anaphor himself 3 by matching its index? It turns out, there’s a straightforward way to do this, and we can do it with the F2 we already have. S NPVP Vt likes NPNP Pavarotti NP NPNP himself 3 [  i ] M,g = g(i) [himself 3 ] M,g = g(3)

18. Pavarotti likes himself The answer? QR. All that QR says is: find an NP inside an S, pick an index, attach the index and the NP to the top, and replace the original NP with a t that shares the index. Pavarotti is an NP. So, hey, why not? Quantifier Raising [ S X NP Y ]  [ S NP [ S i [ S X t i Y ] ] ] S t3t3 VP Vt likes NP NPNP himself 3 S S 3 NP NPNP Pavarotti

19. Pavarotti likes himself Using a little foresight, we’ll pick 3 for our index when we do QR as well. So, what does that leave us with? [  i ] M,g = g(i) So, [himself 3 ] M,g and [t 3 ] M,g both = g(3). Using the rules as usual, the lower [S] M,g is true if g(3) likes g(3). S t3t3 VP Vt likes NP NPNP himself 3 [S] M,g =  F(likes) S S 3 NP NPNP Pavarotti

20. Pavarotti likes himself We have a node that is true if g(3) likes g(3). We want to combine it with the index node 3. We want the result to be the property of liking oneself. We can then attribute to Pavarotti the property of liking oneself, and the sentence will be true if Pavarotti likes himself. S t3t3 VP Vt likes NP NPNP himself 3 [S] M,g =  F(likes) S S 3 NP NPNP Pavarotti

21. Pavarotti likes himself [i] M,g = S [ x [ [S] M,g[i/x] ] ] [3] M,g = S [ x [ [S] M,g[3/x] ] ] We know that [S] M,g =  F(likes) What if we evaluate that not with the assignment function g but with g[3/x] instead? That is, with an assignment function where 3 points to x? g[3/x] (1) = g(1); g[3/x] (2) = g(2); g[3/x] (3) = x; g[3/x] (4) = g(4); … S t3t3 VP Vt likes NP NPNP himself 3 [S] M,g =  F(likes) S S 3 NP NPNP Pavarotti We know this for sure; that’s what g[3/x] means.

22. Pavarotti likes himself [i] M,g = S [ x [ [S] M,g[i/x] ] ] [3] M,g = S [ x [ [S] M,g[3/x] ] ] We know that [S] M,g =  F(likes) [S] M,g[3/x] =  F(likes) =  F(likes) S t3t3 VP Vt likes NP NPNP himself 3 [S] M,g =  F(likes) S S 3 NP NPNP Pavarotti

23. Pavarotti likes himself [i] M,g = S [ x [ [S] M,g[i/x] ] ] [3] M,g = S [ x [ [S] M,g[3/x] ] ] [S] M,g[3/x] =  F(likes) So, now, we can figure out what [S] M,g is: [3] M,g is a function, it takes an S as its argument. [3] M,g ( [S] M,g ) = x [ [S] M,g[3/x] ] = x [  F(likes) ] S t3t3 VP Vt likes NP NPNP himself 3 [S] M,g = x [  F(likes) ] S S 3 NP NPNP Pavarotti

24. Pavarotti likes himself A point of pedantic clarification: Strictly speaking, [S] M,g is the argument, so in… [3] M,g = S [ x [ [S] M,g[3/x] ] ] …S is really going to be [S] M,g. So, what we’re evaluating is really [ [S] M,g ] M,g[3/x]. We take [S] M,g to mean the value of [S] under the current model and assignment function. If we are evaluating the whole sentence under M,g, that’s the current model and assignment function. But by evaluating [S] M,g under M,g[3/x], we’ve changed the “current assignment function” inside the brackets to be g[3/x]. So, this is effectively the same as [ [S] M,g [3/x] ] M,g[3/x]. You can safely ignore this technicality, and just do things as they were done on the previous slide. This is here only for completeness. S t3t3 VP Vt likes NP NPNP himself 3 [S] M,g = x [  F(likes) ] S S 3 NP NPNP Pavarotti

25. Pavarotti likes himself [S] M,g = x [  F(likes) ] Perfect, now we’re set. Combining the function represented by [S] M,g and the individual represented by the top NP, we have: [S] M,g = [S] M,g ( [NP] M,g ) = x [  F(likes) ] ( F(Pavarotti) ) =  F(likes) S t3t3 VP Vt likes NP NPNP himself 3 [S] M,g = x [  F(likes) ] S S 3 NP NPNP Pavarotti

26. Every man likes himself What’s more, we can easily now interpret Every man likes himself. Same structure, only now with every man instead of Pavarotti. We perform QR just as we did, we interpret every node up to the S just as we did—so [S] M,g is the property of liking oneself. [every] M,g takes first the predicate man, then the predicate likes oneself, and the sentence is true if being a man implies liking oneself. S t3t3 VP Vt likes NP NPNP himself 3 [S] M,g = x [  F(likes) ] S S 3 NP NCNC man Det every

27. Every man likes himself [every] M,g = P [ Q [  x  U [P(x)  Q(x)] ] ] [N C ] M,g = x [ x  F(man) ] [NP] M,g = [Det] M,g ( [N C ] M,g ) In terms of the definition of every, we will replace P with [N C ] M,g. We also need P(x), which will be [N C ] M,g (x), or x [ x  F(man) ] (x), so (replacing xes with xes), P(x) is just x  F(man). [NP] M,g = Q [  x  U [x  F(man)  Q(x)] ] S t3t3 VP Vt likes NP NPNP himself 3 [S] M,g = x [  F(likes) ] S S 3 NP NCNC man Det every

28. Every man likes himself [NP] M,g = Q [  x  U [x  F(man)  Q(x)] ],t> Now, the last step, combining the NP and the S. [S] M,g is the predicate likes oneself, type. [NP] M,g needs a predicate to call Q. So, we replace Q with [S] M,g. We also need Q(x), which is x [  F(likes) ] (x), or (replacing xes with xes),  F(likes). [S] M,g = [NP] M,g ( [S] M,g ) =  x  U [x  F(man)   F(likes)] S t3t3 VP Vt likes NP NPNP himself 3 [S] M,g = x [  F(likes) ] S S 3 NP NCNC man Det every

29. Himself n, herself n, itself n Additional restrictions on –self anaphors: Himself must denote a man, herself must denotes a woman, itself must denote something that is neither a man nor a woman. *Pavarotti 3 likes herself 3. There is also a restriction that the thing that has a matching index must be “nearby” and “higher in the tree”: *Pavarotti 3 is hungry and Loren likes himself 3. *Himself 3 likes Pavarotti 3.

30. Binding theory The “nearby” condition, approximately: A –self anaphor must have an index that matches the index of something within the smallest S that contains it. LX522: Binding theory, Principle A. The “higher in the tree” condition, approximately: The node with an index matching a –self anaphor must be the sister of a node that contains the –self anaphor. LX522: The binder must c-command the –self anaphor. You can verify for yourself that these conditions hold, and you’ll hear more about them in LX522. These are not semantic conditions—our system can interpret sentences violating these conditions just fine. These are syntactic.

31. Pronouns vs. –self anaphors Often, when you want co-reference but can’t use a –self anaphor, you can use a pronoun instead: Pavarotti 3 is hungry and Loren likes him 3. *Pavarotti 3 likes him 3. F2 does not distinguish between pronouns and –self anaphors, the interpretation is the same: [him 3 ] M,g = [himself 3 ] M,g = g(3). So, where a pronoun is appropriate, where a –self anaphor is appropriate, these are governed by separate constraints on what structures are valid as structures—syntactic constraints.

32. VP anaphora In a manner somewhat similar to the way herself picks up the referent of a higher NP, English has a phrase do too that seems to pick up the denotation of a preceding VP. Loren likes Bond and Pavarotti does too. The VP in the first sentence is a predicate, meaning is a Bond-liker: x [  F(like) ] The VP in the second sentence, does too, is interpreted as if it were just the same as the VP in the first sentence: Loren is a Bond-liker and Pavarotti is a Bond-liker.

33. VP anaphora Every man likes Loren and Nemo does too. Being a man implies being a Loren-liker, and Nemo is a Loren-liker. A fish likes every book. There is a fish z such that for every book x, z likes x For every book x, there is a fish z such that z likes x

34. A fish likes every book Every book is a quantifier in object position (type,,t>>) and must undergo QR in order to be interpretable. A fish can either undergo QR or not. If it does not, we have the one-uncritical-fish interpretation. If it does, we have the to- each-its-own interpretation.

35. and Loren does too A fish likes every book. [A fish [ 1 [every book [2 [t 1 likes t 2 ]]]] [every book [ 2 [a fish likes t 2 ]]] Loren likes every book. [every book [ 2 [Loren likes t 2 ]]] A fish likes every book and Loren does too. Parallelism: Where a VP is anaphoric to another, the two sentences must have a parallel structure. Scope economy: Only do QR if it makes a difference in the meaning.

36. And we’re done! We’ve described English! Well, not exactly. Actually, there are plenty of English sentences we don’t have a formal interpretation procedure for yet. Here’s one: Most fish are boring.

37. Most fish are boring The first steps are simple enough. Most looks like it is a determiner like every, so we can add Det  most to our syntactic base rules. And it seems to mean something kind of similar. It should take two predicates and return a truth value; that is, most takes fish then are boring and says something about individuals for which fish is true being individuals for which is boring is true too.

38. Most fish are boring Every says: given predicates P and Q, for each individual x in U, P(x)  Q(x). Most is less than every, so maybe: Given predicates P and Q, for over half of the individuals x in U, P(x)  Q(x). That was easy. But wait. What if there are 2 fish in U and 47 books? Certainly if x is a book, fish(x)  Q(x) is true for any Q, given that false  anything is true. If that’s the case, then every fish is happy can be true, even while most fish are unhappy is true (after all, 47 of 49 xes are such that being a fish implies being unhappy). Hmm.

39. Most fish are boring What we need is to be checking only the fish, not all of the individuals in the universe. We don’t care about non-fish when we are evaluating most. Of the fish, are most such that they are also boring?

40. Thinking in terms of sets A predicate like is boring or fish defines a set, the set of boring individuals, or the set of fish. In fact, the F function already gave us that set, and our interpretation rule turned it into a function. [is boring] M,g = x [ x  F(is boring) ]

41. Most fish are boring When we think in terms of sets, what most fish are boring is saying is that the boring fish outnumber the non-boring fish. The boring fish are individuals in both F(fish) and F(is boring). The nonboring fish are individuals in F(fish) but not in F(is boring). Most(A,B) = |A  B| > |A-B| U F(fish) F(is boring)

42. Most fish are boring For predicates P and Q, P = {x: P(x)} Q = {x: Q(x)} P  Q = {x: P(x)  Q(x)} P - Q = {x: P(x)  Q(x)} Most(A,B) = |A  B| > |A-B| So, we can write most as: [most] M,g = P [ Q [ | {x: P(x)  Q(x) } | > | {x: P(x)   Q(x) } | ] ] U F(fish) F(is boring)

43.             