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Proof-nets and semantic applications. Alain Lecomte ESSLLI2002. e+. t-. e-. t+. child. Semantic proof nets. child. x:e, child: e t |- child(x) : t hence : child: et |- x.child(x):e t. run :. e+. t-. x. e-. t+. x. child. run :. e+. t-. x. e-. t+. x. child. - PowerPoint PPT Presentation
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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.