Denotational semantics Object language Operational semantics 1 - - PowerPoint PPT Presentation

▶

Jun 11, 2023 17 likes •375 views

Denotational semantics Object language Operational semantics 1 Cheating? State passing save , load Add register 1 Cheating? State passing save , load set! Add register 1 Cheating? Continuation passing reset catch , throw , both ,

SLIDE 1

1

Denotational semantics Object language Operational semantics

SLIDE 2

1

Cheating?

State passing save, load Add register

SLIDE 3

1

Cheating?

State passing save, load set! Add register

SLIDE 4

1

Cheating?

Continuation passing catch, throw, both, someone, everyone reset shift abort Add metacontext

SLIDE 5

1

The rest of this course

Multi-level continuations Donkey anaphora: scope and binding reset shift Quotation

Day 3 Day 5 Day 4

SLIDE 6

2

Donkey anaphora is in-scope binding

Chris Barker and Chung-chieh Shan ESSLLI, 6 August 2008 Semantics and Pragmatics 1(1):1–42, 2008

SLIDE 7

3

Donkey anaphora

If a donkey eats, it sleeps. Every farmer who owns a donkey beats it.

SLIDE 8

3

Donkey anaphora

If a donkey eats, it sleeps. Every farmer who owns a donkey beats it. A donkey pronoun is a pronoun that lies outside the antecedent of a conditional (or the restrictor of a quantifier) yet covaries with an indefinite (or some other quantifier) inside it.

SLIDE 9

3

Donkey anaphora

If a donkey eats, it sleeps. Every farmer who owns a donkey beats it. A donkey pronoun is a pronoun that lies outside the antecedent of a conditional (or the restrictor of a quantifier) yet covaries with an indefinite (or some other quantifier) inside it.

SLIDE 10

3

Donkey anaphora is in-scope binding

If a donkey eats, it sleeps. Every farmer who owns a donkey beats it. A donkey pronoun is a pronoun that lies outside the antecedent of a conditional (or the restrictor of a quantifier) yet covaries with an indefinite (or some other quantifier) inside it. Our claim: the indefinite takes scope over and binds the donkey pronoun as usual. Every boy loves his mother.

SLIDE 11

4

Why not?

Quantifier scope is clause-bound? But not indefinites. A donkey eats. It sleeps.

SLIDE 12

4

Why not?

Quantifier scope is clause-bound? But not indefinites. A donkey eats. It sleeps. Binding requires c-command? Just evaluation order. Every boy’s mother loves him.

SLIDE 13

4

Why not?

Quantifier scope is clause-bound? But not indefinites. A donkey eats. It sleeps. Binding requires c-command? Just evaluation order. Every boy’s mother loves him. How to get the right truth conditions? not ∃d. (donkey d)∧

(eats d) → (sleeps d)

SLIDE 14

4

Why not?

Quantifier scope is clause-bound? But not indefinites. A donkey eats. It sleeps. Binding requires c-command? Just evaluation order. Every boy’s mother loves him. How to get the right truth conditions? not ∃d. (donkey d)∧

(eats d) → (sleeps d)
but

¬∃d. (donkey d)∧(eats d)∧¬(sleeps d) A donkey takes scope over the entire conditional but under if. A donkey sleeps if it eats.

SLIDE 15

5

Our account

Compositional truth conditions: if, every, most, usually, strong/weak. Key: multiple levels of continuations Plan: Everyone loves someone. (surface scope) Everyone loves his mother. If a donkey eats, it sleeps.

SLIDE 16

5

Our account

Compositional truth conditions: if, every, most, usually, strong/weak. Key: multiple levels of continuations Plan: Everyone loves someone. (surface scope) Everyone loves his mother. If a donkey eats, it sleeps. Everyone loves someone. (inverse scope) If a farmer owns a donkey, he beats it. Every farmer who owns a donkey beats it. Most farmers who own a donkey beat it.

SLIDE 17

6

A expression x Lift = ⇒ B (AB) expression λc. c(x)

SLIDE 18

6

A expression x Lift = ⇒ B (AB) expression λc. c(x) C ((A/B)D) left L D (BE) right R = ⇒ C (AE) left right λc. L(λf. R(λx. c(fx)))

SLIDE 19

6

A expression x Lift = ⇒ B (AB) expression λc. c(x) C ((A/B)D) left L D (BE) right R = ⇒ C (AE) left right λc. L(λf. R(λx. c(fx))) C (BD) left L D ((B\A)E) right R = ⇒ C (AE) left right λc. L(λx. R(λf. c(fx)))

SLIDE 20

6

A expression x Lift = ⇒ B (AB) expression λc. c(x) A (SS) expression F Lower = ⇒ A expression F(λx. x) C ((A/B)D) left L D (BE) right R = ⇒ C (AE) left right λc. L(λf. R(λx. c(fx))) C (BD) left L D ((B\A)E) right R = ⇒ C (AE) left right λc. L(λx. R(λf. c(fx)))

SLIDE 21

7

Linear notation Tower notation B (AC) B C A S (DPS) S S DP

SLIDE 22

7

Linear notation Tower notation B (AC) B C A S (DPS) S S DP λc. f[c(x)] f[ ] x λc. ¬∃x. c(mother x) ¬∃x. [ ] mother x

SLIDE 23

8

A expression x Lift = ⇒ B (AB) expression λc. c(x) A (SS) expression F Lower = ⇒ A expression F(λx. x) C ((A/B)D) left L D (BE) right R = ⇒ C (AE) left right λc. L(λf. R(λx. c(fx))) C (BD) left L D ((B\A)E) right R = ⇒ C (AE) left right λc. L(λx. R(λf. c(fx)))

SLIDE 24

8

A expression x Lift = ⇒ B B A expression [ ] x A S S expression f[ ] x Lower = ⇒ A expression f[x] C D A/B left g[ ] f D E B right h[ ] x = ⇒ C E A left right g[h[ ]] f(x) C D B left g[ ] x D E B\A right h[ ] f = ⇒ C E A left right g[h[ ]] f(x)

SLIDE 25

8

A expression x Lift = ⇒ B B A expression [ ] x A S S expression f[ ] x Lower = ⇒ A expression f[x] C D A/B left g[ ] f D E B right h[ ] x = ⇒ C E A left right g[h[ ]] f(x) C D B left g[ ] x D E B\A right h[ ] f = ⇒ C E A left right g[h[ ]] f(x) DP⊲B B DP he λy. [ ] y

SLIDE 26

8

A expression x Lift = ⇒ B B A expression [ ] x A S S expression f[ ] x Lower = ⇒ A expression f[x] C D A/B left g[ ] f D E B right h[ ] x = ⇒ C E A left right g[h[ ]] f(x) A B DP expression f[ ] x Bind = ⇒ A DP⊲B DP expression f([ ]x) x C D B left g[ ] x D E B\A right h[ ] f = ⇒ C E A left right g[h[ ]] f(x) DP⊲B B DP he λy. [ ] y

SLIDE 27

8

A expression x Lift = ⇒ B B A expression [ ] x A S S expression f[ ] x Lower = ⇒ A expression f[x] C D A/B left g[ ] f D E B right h[ ] x = ⇒ C E A left right g[h[ ]] f(x) A B DP expression f[ ] x Bind = ⇒ A DP⊲B DP expression f([ ]x) x C D B left g[ ] x D E B\A right h[ ] f = ⇒ C E A left right g[h[ ]] f(x) DP⊲B B DP he λy. [ ] y S S (S/S)/S if ¬[ ] λpλq. p∧¬q

SLIDE 28

9

Every farmer who owns a donkey beats it

S S N farmer who owns a donkey ∃y. (donkey y)∧ [ ] λz. (farmer z)∧(owns y z)

SLIDE 29

9

Every farmer who owns a donkey beats it

S DP⊲S N farmer who owns a donkey ∃y. (donkey y)∧([ ] y) λz. (farmer z)∧(owns y z)

SLIDE 30

9

Every farmer who owns a donkey beats it

            S S S S DP

every ¬∃x. [ ] λP. Px∧¬[ ] x S DP⊲S N farmer who owns a donkey ∃y. (donkey y)∧([ ] y) λz. (farmer z)∧(owns y z)            

SLIDE 31

9

Every farmer who owns a donkey beats it

            S S S S DP

every ¬∃x. [ ] λP. Px∧¬[ ] x S DP⊲S N farmer who owns a donkey ∃y. (donkey y)∧([ ] y) λz. (farmer z)∧(owns y z)             DP⊲S S S S DP\S beats it λw. [ ] [ ] beats w

SLIDE 32

9

Every farmer who owns a donkey beats it

            S S S S DP

every ¬∃x. [ ] λP. Px∧¬[ ] x S DP⊲S N farmer who owns a donkey ∃y. (donkey y)∧([ ] y) λz. (farmer z)∧(owns y z)             DP⊲S S S S DP\S beats it λw. [ ] [ ] beats w ¬∃x∃y. donkey y∧((farmer x∧owns y x)∧¬(beats y x))

SLIDE 33

10

Most farmers who own a donkey beat it

S S S S DP

most

MOST(λxλp. [ ])

λP. Px∧(p∨[ ]) x

MOST(F)

= #{x | F(x)(FALSE)} #{x | F(x)(TRUE)} > 1 2

SLIDE 34

10

Most farmers who own a donkey beat it (weak)

S S S S DP

most

MOST(λxλp. [ ])

λP. Px∧(p∨[ ]) x

MOST(F)

= #{x | F(x)(FALSE)} #{x | F(x)(TRUE)} > 1 2