11a. Predicate modification and adjectives

11a. Predicate modification and adjectives
CAS LX 502 11a. Predicate modification and adjectives

Is hungry As a starting point, we’ve been considering is hungry to be an intransitive verb. Really, though, is is the verb, hungry is an adjective. An individual can either be hungry or not hungry. That is, hungry is either true or false of an individual. Hungry is a function from individuals to truth values, <e,t>. In is hungry, the verb is is not contributing any meaning, it’s just there to link up the subject and the adjective.

Bond is hungry Let’s tweak our syntax so that is hungry is comprised of is and hungry, and let’s say that is has no semantic value, that it is meaningless. VP  Vbe Adj Vbe  is Adj  hungry, happy, tall [Vbe]M,g = — [hungry]M,g =  x [ x  F(hungry) ]

Bond is hungry To interpret this we want is to be ignored. To be precise, we can slightly modify Pass-Up so that it applies to this case. Pass-Up If a node b has only one daughter with a semantic value, a, then [b]M,g = [a]M,g S NP VP NP Vbe Adj Bond is hungry

 x [ x  F(hungry) ] <e,t>
Bond is hungry Pass-Up If a node b has only one daughter with a semantic value, a, then [b]M,g = [a]M,g S NP VP NP Vbe Adj Bond is hungry  x [ x  F(hungry) ] <e,t>

Bond is hungry  x [ x  F(hungry) ] <e,t>
Pass-Up If a node b has only one daughter with a semantic value, a, then [b]M,g = [a]M,g S NP VP NP Vbe Adj Bond is hungry  x [ x  F(hungry) ] <e,t>

Bond is hungry  x [ x  F(hungry) ] <e,t>
Pass-Up If a node b has only one daughter with a semantic value, a, then [b]M,g = [a]M,g S NP VP NP Vbe Adj Bond is hungry  x [ x  F(hungry) ] <e,t>  x [ x  F(hungry) ] <e,t>

 x [ x  F(hungry) ] <e,t>
Bond is hungry Pass-Up If a node b has only one daughter with a semantic value, a, then [b]M,g = [a]M,g S NP VP NP Vbe Adj Bond is hungry F(Bond) ] <e>

 x [ x  F(hungry) ] <e,t>
Bond is hungry F(Bond) ] <e> Pass-Up If a node b has only one daughter with a semantic value, a, then [b]M,g = [a]M,g Functional application [g a b ]M,g = [b]M,g ( [a]M,g ) or [a]M,g ( [b]M,g ) whichever is defined S NP VP NP Vbe Adj Bond is hungry

 x [ x  F(hungry) ] <e,t>
Bond is hungry F(Bond) ] <e> S Functional application [g a b ]M,g = [b]M,g ( [a]M,g ) or [a]M,g ( [b]M,g ) whichever is defined [S]M,g = [VP]M,g ( [NP]M,g ) =  x [ x  F(hungry) ] ( [NP]M,g ) =  x [ x  F(hungry) ] ( F(Bond) ) = F(Bond)  F(hungry) NP VP NP Vbe Adj Bond is hungry

Every hungry fish is happy
By separating is from hungry, we’ve isolated a category of adjectives, which also appear in noun phrases modifying a common noun, as in every hungry fish. Now that we have adjectives, we can turn a common noun like fish into a more descriptive common noun like hungry fish… inching closer to actual English. NC  Adj NC

Nemo is a fish One more detour before we continue: What is the contribution of a in Nemo is a fish? We have a listed as a quantifier, meaning essentially the same as some, e.g., A fish likes every book. Some fish likes every book. A a b means that there is an x that for which both a and b hold. Every a b means that for every x, being a implies also being b.

Nemo is a fish But does Nemo is a fish really mean ‘There is an x that is a fish, and Nemo is that x’? It doesn’t really feel like that. Also, notice that every cannot be used here: *Nemo is every happy fish.

Nemo is a fish What it seems like intuitively is that a is not adding anything to the meaning either. That, like is, a is just meaningless, passing along the meaning of the common noun. So, let’s allow for that by building in a “dummy determiner” that has no meaning and shows up only when the verb is is. VP  Vbe NPpred NPpred  Detdummy NC Detdummy  a [DETdummy a ]M,g = —

 x [ x  F(fish) ] <e,t>
Nemo is a fish There’s nothing new or fancy going on here, just more use of Pass-Up. [fish]M,g =  x [ x  F(fish) ] [DETdummy a]M,g = — [is]M,g = — S NP VP NP Vbe NPpred Nemo is Detdummy NC a fish  x [ x  F(fish) ] <e,t>

Nemo is a fish F(Nemo) <e>  x [ x  F(fish) ] <e,t>
Then, as before: [S]M,g = [VP]M,g ( [NP]M,g ) =  x [ x  F(fish) ] ( [NP]M,g ) =  x [ x  F(fish) ] ( F(Nemo) ) = F(Nemo)  F(fish) NP VP NP Vbe NPpred Nemo is Detdummy NC a fish  x [ x  F(fish) ] <e,t>

What the meaning of is is
Is is always meaningless? It seems to be in Nemo is a fish. But what about in Nemo is the President? Or A hungry fish is a happy fish? The “meaningless” kind of is we’ll call predicative. The “equals” kind of is we’ll call equative.

Equative be The equative is is kind of like a conjunction that means “equals” and seems to be able to equate any two NPs. We might give the rule as (perhaps limiting a and b to NPs): [is]M,g = b [ a [ [a]M,g = [b]M,g ] ]

Nemo is a happy fish We added a rule to allow for adjectives to attach to common nouns: NC  Adj NC So, we should be able to draw a structure for Nemo is a happy fish.

Nemo is a happy fish S ? NP VP However, when we try to work out the truth conditions, we run into a problem. NP Vbe NPpred Nemo is Detdummy NC a Adj NC happy fish  x [ x  F(happy) ] <e,t>  x [ x  F(fish) ] <e,t>

Nemo is a happy fish ? <e,t> What type should happy fish be?
Seems like it should be the same as fish. A property (a predicate), true of individuals (<e,t>), that are happy and fish. Nemo is happy and Nemo is a fish. NP VP NP Vbe NPpred Nemo is Detdummy NC a Adj NC happy fish  x [ x  F(happy) ] <e,t>  x [ x  F(fish) ] <e,t>

Nemo is a happy fish ? <e,t>
We want something that, given an individual z, is true if happy is true of z and fish is true of z.  z [ z  F(happy)  z  F(fish) ] NP VP NP Vbe NPpred Nemo is Detdummy NC a Adj NC happy fish  x [ x  F(happy) ] <e,t>  x [ x  F(fish) ] <e,t>

Predicate modification
To make the structure interpretable and to accomplish the desired meaning, we add a third interpretation rule: Predicate modification [a b]M,g =  z [ [a]M,g(z)  [b]M,g(z) ] where a and b are predicates (type <e,t>).

Predicate modification
Predicate modification [a b]M,g =  z [ [a]M,g(z)  [b]M,g(z) ] where a and b are predicates (type <e,t>). For [happy fish]M,g, a will be happy, b will be fish. [happy]M,g =  x [ x  F(happy) ] [fish]M,g =  x [ x  F(fish) ] [happy fish]M,g =  z [ [happy]M,g(z)  [fish]M,g(z) ] =  z [  x [ x  F(happy) ](z)  [fish]M,g(z) ] =  z [ z  F(happy)  [fish]M,g(z) ] =  z [ z  F(happy)   x [ x  F(fish) ](z) ] =  z [ z  F(happy)  z  F(fish) ]

Nemo is a happy fish  z [ z  F(happy)  z  F(fish) ] <e,t>
Now that we have a semantic value for the whole NC, the rest proceeds as in Nemo is a fish from before. Is and a have no semantic value, so [NC]M,g is passed up all the way to [VP]M,g. NP VP NP Vbe NPpred Nemo is Detdummy NC a Adj NC happy fish  x [ x  F(happy) ] <e,t>  x [ x  F(fish) ] <e,t>

 z [ z  F(happy)  z  F(fish) ] <e,t>
Nemo is a happy fish  z [ z  F(happy)  z  F(fish) ] <e,t> F(Nemo) <e> S NP VP NP Vbe NPpred Nemo is Detdummy NC [S]M,g = [VP]M,g ( [NP]M,g ) =  z [ z  F(happy)  z  F(fish) ] ( [NP]M,g ) =  z [ z  F(happy)  z  F(fish) ] ( F(Nemo) ) = F(Nemo)  F(happy)  F(Nemo)  F(fish) Nemo is happy and Nemo is a fish. a Adj NC happy fish

(F3) S  NP VP VP  Vt NP S  S ConjP VP  Vi ConjP  Conj S
NP  Det NC S  Neg S NP  NP VP  Vbe NPpred NC  Adj NC NPpred  Detdummy NC Det  the, a, every NP  Pavarotti, Loren, Bond, Nemo, Dory, Blinky, Semantics, The Last Juror, hen, shen, itn, himn, hern, himselfn, herselfn, itselfn. Conj  and, or Vt  likes, hates Adj  boring, hungry Neg  it is not the case that NC  book, fish, man, woman Detdummy  a Vbe  is

Quantifier Raising [S X NP Y ][S NP [S i [S X ti Y ]]]
Predicate modification [a b]M,g = z [ [a]M,g(z)  [b]M,g(z) ] where a and b are predicates Pass-Up If a node b has only one daughter with a semantic value, a, then [b]M,g = [a]M,g Functional application [g a b ]M,g = [b]M,g ( [a]M,g ) or [a]M,g ( [b]M,g ) [Pavarotti]M,g = F(Pavarotti) (any NP) [boring]M,g =  x [ x  F(boring) ] (any NC or Adj 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 ] [every]M,g = P [ Q [ xU [P(x)  Q(x)] ] ] [a]M,g = P [ Q [ xU [P(x)  Q(x)] ] ] [i]M,g = g(i) [is]M,g = — [i]M,g = S [x [ [S]M,g[i/x] ] ] [DETdummy a]M,g = —

The boring fish There are two more things to add to our system before we call it complete enough for this semester. One is to add an interpretation for the (which our syntax can generate), as in the boring fish. The is a Det but it is different from every: It doesn’t seem to rely on the value of the sentence: Every a b means for each x, if a is true of x, b is also true of x. The a is just an individual, one of which a is true, with the presupposition that there is only one individual of which a is true.

A unique fish However, rather than try to incorporate presuppositions into F3, we’ll instead define the to be a quantifier like every or a except meaning a unique. (This means not presupposing existence and uniqueness, but rather asserting it) [the]M,g = P [ Q [ xU [P(x)  y[P(y)x=y]  Q(x)] ] ] [a]M,g = P [ Q [ xU [P(x)  Q(x)] ] ]

The fish that Bond likes
The last thing to incorporate is the relative clause. Idea: suppose we start with Bond likes the fish and we transform this S into an NP (the fish that Bond likes) by doing something similar to QR. Relative clause transformation: [S X Det NC Y ] [NP Det [Nc NC [S that [S i [S X ti Y ] ] ] ] ]

Relative clause transformation
NP Det NC S NC S  … NP … that S Det NC i S … ti … [that]M,g = —

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