Proof-nets and semantic applications

Proof-nets and semantic applications

Alain Lecomte

ESSLLI2002

Semantic proof nets

• child

child

e+ t- e- t+

x:e, child: et |- child(x) : thence : child: et |- x.child(x):et

• run :

child

e+ t- e- t+

x

x

• run :

child

e+ t- e- t+

x

x

• run :

child

e+ t- e- t+

x

x

• run :

child

e+ t- e- t+

x

x

x.child(x)

x

• run :

child

e+ t- e- t+

x

x

x.child(x)

x

each, every…

• A determiner like every, each… decomposes into :– A quantifier, for instance : type : (et)t– A connective, for instance :type : t(tt)

• needs two predicates (e t) for obtaining one proposition (t)

• A determiner is therefore of type

(et)((et)t)

• A determiner is therefore associated with a sequent:

• Its « semantic » is represented by its proof

))(()()(),( ttetettettt

deduction

))(()()(),(

)()(),(,)(),(,,

)(,,)(,,,)(,,,,)(,,,)(,,

,

ttetettettt

ttettettttetttettttete

tttettttetettttteteettttteteeetttttetetttttttttttttt

ttee

ee

C

remark

• With a very remarkable step : an application of the contraction rule!

necessity of working inside Intuitionistic linear logic with exponentials

• The exact sequent which encodes the determiner is :

))(()()(),( ttttt!ettt --o |-- --o--o --o--o --o--o--o !e!e

Exponentials

CA

AA

!,

!,!,

DA

A

!,

,

WA

!,

Exponentials (one-sided)

CA

AA

?,

?,?,

DA

A

?,

,

WA?,

Representation of the proof

c

(!e –o t) –o ((!e –o t) –o t)

every child

c

child (!e –o t) –o t

every child likes to play

c

child

likes to play

t

Application

A –o B +

A+- +

Application

A - B +

Abstraction

A +B -

Abstraction

A +B -

B –o A +

Syntactic proof-nets

• Proof-nets for Lambek calculus

• Like PN for MILL +– condition on semi-planarity

every child plays1) unfolding

(s/(np\s))/n -every

n -child

np\s -plays

s +

(s/(np\s)) - n + s -np +

s - np\s +

np -s+

every child plays2) links

(s/(np\s))/n -every

n -child

np\s -plays

s +

(s/(np\s)) - n + s -np +

s - np\s +

np -s+

Attention!

2) links

(s/(np\s))/n -every

n -child

np\s -plays

s +

(s/(np\s)) - n + s -np +

s - np\s +

np -s+ WRONG !

Parsing

• through homomorphism– H(s) = t– H(np) = !e– H(n) = !e –o t– H(A/B) = H(B\A) = H(B) –o H(A)

every child plays

(s/(np\s))/n -every

n -child

np\s -plays

s +

(s/(np\s)) - n + s -np +

s - np\s +

np -s+

every child plays3) homomorphism

(!e –o t) –o ((!e –o t) –ot))every

!e –o tchild

!e –o tplays

t+

(!e –o t) –o t !e –o t t!e

t !e –o t +

!et

• semantic recipes– child : x.child(x) – every : P.Q.(x.(P(x)Q(x))– plays : x.play(x)

• represented by proof-nets :

child

e+ t- !e- t+d

• represented by proof-nets :

plays

e+ t- !e- t+d

every

c

(!e –o t) –o ((!e –o t) –o t)

plugging lexical semantic types to the homomorphic PN by cut

(!e –o t) –o ((!e –o t) –ot))every

!e –o tchild

!e –o tplays

t+

(!e –o t) –o t !e –o t t!e

t !e –o t +

!et

child

e+ t- !e- t+d

CUT

(!e –o t) –o ((!e –o t) –ot))every

!e –o tplays

t+

(!e –o t) –o t t!e

t !e –o t +

!et

child

d

!e t

(!e –o t) –o ((!e –o t) –ot))every

!e –o tplays

t+

(!e –o t) –o t t!e

t !e –o t +

!et

child

d

!e t

plays

e+ t- !e- t+d

CUT

(!e –o t) –o ((!e –o t) –ot))every

plays t+

(!e –o t) –o t t!e

t !e –o t +

!et

child

d

!e t

d

(!e –o t) –o ((!e –o t) –ot))every

plays t+

(!e –o t) –o t t!e

t !e –o t +

!et

child

d

!e t

d

PNevery

CUT

plays t+

t!e

child

d

d

c

plays t+

t!e

child

d

d

c

plays t+

t!e

child

d

d

c

plays t+

t!e

child

d

d

c

x

plays t+

t!e

child

d

d

c

x

plays t+

t!e

child

d

d

c

x

plays

plays t+

t!e

child

d

d

c

x

plays

child

plays t+

t!e

child

d

d

c

x

plays

child

child(x)

plays t+

t!e

child

d

d

c

x

plays

child

child(x)

plays(x)

(x.((child(x),plays(x))))

Logical synthesis:from a formula to a sentence

• the reverse story:

• Start : – a semantic formula – + semantic recipes for lexical entries 1, 2, …

n

• Goal:– A sentence using all these recipes the

meaning of which is

Usual solutions:-term unification

Peter : np : p

kisses : (np\s)/np: x.y.kiss(y,x)

Mary :np : m

?s:kiss(p,m)

GOAL

Peter : np- : p

kisses : (np\s)/np: x.y.kiss(y,x)

Mary :np- : m

?s:kiss(p,m)

np+

s-np+

y.kiss(y, )

kiss(,)

Peter : np- : p

kisses : (np\s)/np: x.y.kiss(y,x)

Mary :np- : m

?s:kiss(p,m)

np+

s-np+

kiss(,)

y.kiss(y, )

Peter : np- : p

kisses : (np\s)/np: x.y.kiss(y,x)

Mary :np- : m

?s:kiss(p,m)

np+

s-np+

kiss(,) = kiss(p,m) = m; = p

kiss(,)

y.kiss(y, )

Peter : np- : p

kisses : (np\s)/np: x.y.kiss(y,x)

Mary :np- : m

?s:kiss(p,m)

np+

s-np+

kiss(,) = kiss(p,m) = m; = p

y.kiss(y, )

kiss(,)

= m

Peter : np- : p

kisses : (np\s)/np: x.y.kiss(y,x)

Mary :np- : m

?s:kiss(p,m)

np+

s-np+

kiss(,) = kiss(p,m) = m; = p

y.kiss(y, )

kiss(,)

= m

= p

Peter : np- : p

kisses : (np\s)/np: x.y.kiss(y,x)

Mary :np- : m

?s:kiss(p,m)

np+

s-np+

kiss(,) = kiss(p,m) = m; = p

y.kiss(y, )

kiss(,)

= m

= p

The net is a proof-net for L, therefore the correct sentence is Peter kisses Mary

But…

• Non decidability for second order

• xf = f(fa) …? – no mgu– two incomparable solutions:

z.z(fa)z.f(za)

the use of PN

• Generation (or « synthesis ») =

research of a proof in L (which is decidable!), helped by a semantic form

The problem

• Find PNL such that:

– plug by cuts : lexical PNs PN1, PN2, …PNn

– cut-elimination

– Given PN

APNL

?

T1 T2 Tn

AH(PNL)

?

HT1 HT2 HTnPN1

PN2

PNn

CUT

CUT

CUT

PN

the execution formula

• Let P be a PN, U the set of its axiom links, the set of its cut links

• u : incidence matrix of U : incidence matrix of

Cut elimination between axioms

e1 e2 e3 e4 e1 e2 e3 e4

|- X, X |- X, X CUT

|- X, X

e1 e2 e3 e4

• after a first step of cut-elimination on axioms : u replaced by uu

• links coming from cut-elimination of level 1:

• To suppress all the links the premisses of which are premisses of a new cut and all the links which had no incident cut:

• uu - 2uu - uu2 + 2u2

= (1- 2)uu(1- 2)

• Idem for links coming from elimination of level 2, 3, …, n, … cuts

• Resulting graph after cut-elimination:

1

22 )1)()(1()1(),(Rek

kuuus

U

?

HT1 HT2 HTnPN1

PN2

PNn

CUT

CUT

CUT

Res(U,)

• U can be calculated from Res(U,) and • Cf. PhD thesis by Sylvain POGODALLA

• http://www.xrce.xerox.com/~pogodalla

• Condition : that each lexical semantics contain at least one constant which intervenes in the global semantic representation.