Semantics 1

(1)

Semantics 1

May 24, 2012

Gerhard J¨ager

(2)

The copula verb be

Different uses of be

(1) Tully is Cicero. ❀ predicative is proper noun (2) Cicero is a politician. ❀ predicative is indefinite NP (3) Cicero is in Rome. ❀predicative is PP

(4) Cicero is old. ❀ predicative is AP

(May 24, 2012) Semantics 1 Gerhard J¨ager 2 / 16

(3)

Equative be

(1) Tully is Cicero.

is❀λyλxλs.x=y S

λs.t’=c’

NP t’

VP λxλs.x=c’

N t’

Tully

V λyλxλs.x=y

is

NP c’

N c’

(4)

Equative be

Equative be also accounts for quantifiers in predicative position.

(1) Tully is a philosopher.

S

λs.∃x(philosopher’(s, x)∧t’=x)

NP

λQλs.∃x(philosopher’(s, x)∧Q(s, x))

S λs.t’=x₁

NP t’

VP λxλs.x=x₁ N

t’

Tully

V λyλxλs.x=y

is

NP x₁

λs.∃x(philosopher’(s, x)∧t’=x)≡λs.philosopher’(s,t’)

(5)

Predicative be

(1) Tully is old.

is❀λP λxλs.P(s, x) S

λs.old’(s,t’)

NP t’

VP λxλs.old’(s, x)

N t’

Tully

V λP λxλs.P(s, x)

is

AP old’

A old’

(6)

Predicative and attributive use of adjectives

predicative use:

(1) Tully is old. ❀λs.old’(s,t’) attributive use:

(2) old man❀λxλs.man’(s, x)∧old’(s, x)

attributive use involves logical conjunction∧ that is missing in predicative use

Where does this semantic content come from?

(7)

The syntactic solution

Syntax: NP₁→AP,NP₂

Semantics: kNP₁k=λxλs.kNP₂k(s, x)∧ kAPk(s, x) Disadvantage:

does not work for all attributive adjectives:

(1) fake doctor (2) alleged winner (3) imaginary singers

(8)

The lexical solution

Lexical rule

If the lexicon contains an adjective A with the meaning λ~yλxλs.α(s, x)

for some predicate α, then the lexicon also contains an adjectiveA with the meaning

λ~yλP λxλs.P(s, x)∧α(s, x, ~y)

There is no consensus which solution is correct. In this course we will work with the lexical solution.

0(The notation~yrepresents a (possibly empty) sequence of additional arguments.)

(9)

Prepositions

Just like APs, PPs have a predicative and a attributive use (plus an adverbial use, that will not be covered here)

same systematic relationship between predicative and attributive use as above:

inpred ❀λyλxλs.in’(s, x, y)

inattr ❀λyλP λxλs.P(s, x)∧in’(s, x, y)

(10)

Predicative use

S λs.in’(s,p’,s’)

NP p’

Peter

VP λxλs.in’(s, x,s’)

V λP λxλs.P(s, x)

is

PP λxλs.in’(s, x,s’)

P

λyλxλs.in’(s, x, y) in

NP s’

Stuttgart

(11)

Attributive use

N

λxλs.pub’(s, x)∧in’(s, x,s’)

N λxλspub’(s, x)

pub

PP

λP λxλs.P(s, x)∧in’(s, x,s’)

P

λyλP λxλs.P(s, x)∧in’(s, x, y) in

NP s’

Stuttgart

(12)

Inverse linking

(1) A pub in every city opened. ❀

λs.∀y(city’(s, y)→ ∃x(pub’(s, x)∧in’(s, x, y)∧open’(s, x)))

S

NP₁ every city

S

NP₂ S

D a

N NP₂ VP

opened N

pub

PP

P in

NP₁

(13)

Inverse linking

if we do QR in the reverse order...

S

NP2 S

D a

N NP₁

every city

S

N pub

PP NP2 VP

opened P

in

NP1

NP₁ (every city) ends up not c-commanding its trace⇒ illicit movement!

semantics would come out as

λs.∃x(pub’(s, x)∧in’(s, x,x₁)∧ ∀y(city’(s, y)→open’(s, x))) unbound variable (corresponds to non-c-commanded trace)

(14)

Inverse Linking

(1) Some pub in every town offers every beer.

S-Structure

S

NP VP

D some

N V

offers

NP

N pub

PP D

every

N beer

P in

NP

D every

N town

(15)

Inverse linking

LF 1/2/3 S

NP₁ every city

S

NP₂ S

D a

N NP₂ VP

N pub

PP V

offers

NP every beer P

in

NP₁

(16)

Inverse Linking

(1) Some pub in every town offers every beer.

with our current tools, we can derive three readings:

λs.∀z(beer’(s, z)→ ∀y(town’(s, y)→

∃x(pub’(s, x)∧in’(s, x, y)∧offer’(s, x, z)))) λs.∀y(town’(s, y)→ ∀z(beer’(s, z)→

∃x(pub’(s, x)∧in’(s, x, y)∧offer’(s, x, z))))

λs.∀y(town’(s, y)→ ∃x(pub’(s, x)∧in’(s, x, y)∧ ∀z(beer’(s, z)→ offer’(s, x, z))))

two more readings are possible but cannot be derived so far:

λs.∀z(beer’(s, z)→ ∃x(pub’(s, x)∧ ∀y(town’(s, y)→ in’(s, x, y))∧offer’(s, x, z)))

λs.∃x(pub’(s, x)∧ ∀y(town’(s, y)→in’(s, x, y))∧ ∀z(beer’(s, z)→ offer’(s, x, z)))