17 October 2006 Foundations of Logic and Constraint Programming 1 Declarative Semantics An overview...

17 October 2006 Foundations of Logic and Constraint Programming 1

Declarative Semantics

Anoverview

• Algebras(semanticsforterms)

• Interpretations(semanticsforprograms)

• SoundnessofSLD-Resolution

• CompletenessofSLD-Resolution

• LeastHerbrandModels

• ComputingleastHerbrandModels


Algebras

Informally,analgebraassignsvaluesofconstantsandfunctionstoelementsofa

certaindomain(forexample,theusualalgebraforthenaturalnumbersN,assigns

tofunction+(3,1))thevalue4).Moreformally

- GivenasetofVariablesV,andarankedalphabetFoffunctionsymbols,

AnalgebraJforF(orF-algebraorpre-interpretationforF)consistsof:

1. domain:non-emptysetD

2. Assignmentofamapping

fJ : Dn D

toeveryf F (n)(withn0)

- StateσoverD:mappingσ : V D

- Extensionofσ toTUF,V:

σ: TUF,V D suchthatforevery f F (n)

σ(f(t1, ... , tn)) = fJ (σ(t1), ... , σ(tn))


Interpretations

Informally,aninterpretationassignstrueorfalsetopredicatesoveritsarguments,

which are elements of a domain (or terms pre-interpreted by an algebra). For

example,(+(3,1),2)istrueintheusualinterpretationof/2).Moreformally

- GivenarankedalphabetFoffunctionsymbols,andarankedalphabetofpredicatesymbols

- AninterpretationIforFandconsistsof:

1. AlgebraJforF(withdomainD)

2. AssignmentofarelationpI D D ... D toeveryp (n)(withn0)

- Note:GivenapredicatepwithnargumentsinD,thisdefinitionisequivalent

tomappingthepredicateinto

• trueifthen-tupleisamemberofintherelationpI

• falseifthen-tupleisnotamemberoftherelationpI

n


Interpretations

LetPaddbethe”add-program”

add(0,M,M). add(s(M),N,s(K)):- add(M,N,K).

I1, I2, I3, I4, I5and I6areall legitimate (ifnotall “conventional”) interpretations forthesetsF={0,s/1}and={add/3}

- I1 : DI1 = N, 0I1 = 0, sI1(n) = n+1 f.e. n N

addI1 = {(m,n,m+n) | m, n N }

Note:Thisistheusualinterpretationwheretermsareinterpretedtonumbersandaddis“addition”

- I2 : DI2 = N, 0I2 = 0, sI2(n) = n+1 f.e. n N

addI2 = {(m,n,m*n) | m, n N }

Note:Inthisinterpretation,“additionisinfactmultiplication”.

- I3 : DI3 = HU{s,0}, 0I3 = 0, sI3(t) = s(t) f.e. t HU{s,0},

addI3 = {(sm(0), sn(0), sm+n(0),) | m, n N }

Note:ThisistheinterpretationofadditioninthePeanoalgebraforintegers.


Interpretations


add(0,M,M).add(s(M),N,s(K)):- add(M,N,K).


addI4 = { }

Note:Inthisinterpretation,“Peanonumberscannotbeadded”.


addI5 = (HU{s,0} )3

Note:Inthisinterpretation,“anythinggoes”,sinceanytripleoftermsisinthemodel.

- I6 : DI6 = {0,1} 0I6 = 0, sI6(n) = n f.e. n { 0,1}

addI6 = {(m,n,n) | m, n { 0,1}

Note: in this interpretation forabooleandomain,s is the identity function,andadd“selectsofthesecondargument”.


Logical Truth (2)

GivenanexpressionE,andaninterpretationI:

Letx1,...,xkbethevariablesoccurringinE

x1, xkE:UniversalclosureofE(abreviatedE

x1, xkE:ExistentialclosureofE(abreviated E

I |= E :ifI |= σ E foreverystateσ

I |= E :ifI |= σ E forsomestateσ

WhenI |= Ewesaythat

• I isamodelof E;or

• E istruein I.


Logical Truth (3)

GivensetsofexpressionsSandT,andaninterpretationI:

I isamodelof S (written I |= S )

:I |= E foreveryE S

T is a semantic (orlogical)consequenceof S, (written S |= T )

:everymodelofSisamodelofT

GivenaprogramP,aqueryQ0andasubstitution

| var(Q0) isacorrect answer substitution ofQ0

:P |= Q0

Q0 is acorrect instanceofQ0

:P |= Q0


Models (Example)



- I1 : DI1 = N, 0I1 = 0, sI1(n) = n+1 ;addI1 = {(m,n,m+n) | m, n N }

- I1 |= PaddasI1 |= c foreveryclause c of Paddandstateσ : V N.Infact

(i)(σ(0), σ(x), σ(x)) add I1;and

(ii)if(σ(M), σ(N), σ(K)) add I1 then(σ(M)+1, σ(N), σ(K)+1) add I1

- I2 : DI2 = N, 0I2 = 0, sI2(n) = n+1 ; addI2 = {(m,n,m*n) | m, n N }

- I2 | Padd.Tocheckthisletσ(M) = 1.ThenI2 |σ add(0,M,M)since(σ(0), σ(M), σ(M))=(0,1,1)addI2 (inotherwords,0*11)

- I3 : DI3 = HU{s,0}, 0I3 = 0, sI3(t)=s(t) ; addI3={(sm(0),sn(0),sm+n(0),)|m,n N }

- I3 |= Padd like for I1. SincethedomainD is theHerbrandUniverseof terms,

thisinterpretationisa(least)HerbrandModel.


Models (Example)


- I4 : DI4 = HU{s,0}, 0I4 = 0, sI4(t) = s(t) f.e. t HU{s,0}, addI4 = { }

- I4 | Padd.Tocheckthisletσ(M) = 1.ThenI4 |σ add(0,M,M)since(σ(0),

σ(M), σ(M))=(0,1,1)addI4 = { }.

- I5 : DI5 = HU{s,0}, 0I5 = 0, sI5(t) = s(t) f.e. t HU{s,0}, addI5 = (HU{s,0} )3

- I5 | Padd as any triple, (σ(0), ỤỴ(M), σ(M)) or (s(σ(M)), σ(N)), s(σ(K)),

belongs to addI5. SinceDI5 is the Herbrand Universe, I5 is a Herbrand

Model.However,itisnotaleastmodel,sinceaddI5includesmoretriples

thanstrictlynecessary(addI2 addI5)

- I6 : DI6 = {0,1} 0I6 = 0, sI6(n) = n;addI6 = {(m,n,n) | m, n { 0,1}

- I6 |= Padd.LikeforI1.

(i)(σ(0), σ(x), σ(x)) add I6;and

(ii)if(σ(M), σ(N), σ(K)) add I6 then(σ(M), σ(N), σ(K)) add I6


Towards Soundness of SLD-Resolution (1)

- Lemma 4.3 (i)

- LetQ c Q’beanSLD-derivationstepandQ← Q’theresultantassociated

withit.Then

c |= Q← Q’

Proof.

- LetQ = A, B, C withselectedatom B. Let H ← B betheinputclause c , andQ’ = (A,B,C) .Then

c |= H ← B (variantofc)

implies c |= H ← B (instance)

implies c |= B ← B (unifiesBandH)

implies c |= (A,B,C) ← (A,B,C) (unchanged“context”:A,C)


Towards Soundness of SLD-Resolution (2)

Lemma 4.3 (ii)

LetbeanSLD-derivationofP {Q0}.ForiletRi betheresultantofleveliof.Then

P |= Ri

Proof. Let=Q0 1 Q1 ... Qn n+1Qn+1. Theproofisbyinductiononn.

- i=0: R0 =Q0 ← Q0 =“true”. Hence,P |= R0.

- i=1: R1 =Q01← Q1 . ByLemma4.3(i)itisP |= R1.

- i~>i+1:AssumethatP |= Ri andQii+1← Qi+1

- LetQi i+1← Qi+1 andRi = Q0 12i← Qi;Then

- Ri+1 = Ri i+1= Q0 12i+1 ← Qi+1

- Hence,byLemma4.3(i) andtheinductionhipothesis,itis P |= Ri+1.


Soundness of SLD-Resolution

Theorem 4.4

If there exists a successful SLD-derivation ofP {Q0 } with CAS (computed answer substitution) ,then

P |= Q0

Proof.

Let=Q0 1 Q1 ... Qn-1 n□ beasuccessfulSLD-derivation. Lemma4.3(ii)

appliedtotheresultantoflevelnofimplies

P |= Q01n

andQ01n = Q01n | Var(Q0)) = Q0

- This theorem thus guarantees that the answers computed by a successful derivation are logical consequences of the program.

- Theconverseresult,istheobjectofthecompletenesstheorem.


Intuitive Meaning of Queries

- To prepare the addressing of completeness, the next corollary simply states

thatifanSLD-derivationexists,thentheprogramimpliestheexistentialclosure

ofthequery(someinstanceofthequery).

Corollary 4.5

IfthereexistsasuccessfulSLD-derivationofP {Q0} thenP |= Q0.

Proof.

Theorem4.4.impliesP |= Q0 forsomeCAS.Now,

P |= Q0 .

- impliesforeveryinterpretationI: I |= P,thenI |= Q0

- impliesforeveryinterpretationI: I |= P,thenI |=(Q0 )

- impliesforeveryinterpretationI: I |= P,thenI |=Q0

- impliesP |= Q0


Towards Completeness of SLD-Resolution

- ToassesscompletenessofSLD-Resolutionitisconvenienttocharacterisethe

set of sintactically derivable queries. For this purpose the concepts of term

modelsandimplication treesaredefined.

- Themain idea is to interpret the set of functional symbols into “themselves”,

rather,avoindingtheirmappingintoadifferentdomain.Forexample,insteadof

amappingsuchas

- “0”0,”s(0)”1,“s(s(0))”2,...aterminterpretation/modelwilladopt- “0”0,“s(0)”s(0),“s(s(0))”s(s(0)),...

- Formally,foranexpressionE,andasetofexpressionsS

- inst(E) :setofallinstancesofE

- inst(S) :setofallinstancesofallelementsE S

- ground(E) :setofallgroundinstancesofE

- ground(S) :setofallgroundinstancesofallelementsE S


Term Models

ForasetofvariablesV,functionsymbolsFandpredicatesymbols

- Theterm algebraJforFisdefinedasfollows:

1. DomainD = TUF,V

2. Mapping fJ: (TUF,V)n TUF,V assignedtoevery f F(n) with

fJ(t1, ... , tn) : f(t1, ... , tn)

Aterm interpretationIforFand consistsof:

1. AtermalgebraforF

2. I TB,F,V (set of atoms that are true; equivalently it can be seen as an

assignmentofarelation pI (TUF,V)ntoeveryp(n)).

Aterm modelIofasetSofexpressions

: aterminterpretationIthatisamodelof S


Herbrand Models

TheHerbrand algebraJforFisdefinedforgroundtermsasfollows:

1. DomainD = HUF

2. Mapping fJ: (HUF)n HUF assignedtoevery f F(n) with

fJ(t1, ... , tn) : f(t1, ... , tn)

AHerbrand interpretationIforFand consistsof:

1. HerbrandalgebraforF

2. I HB,F (setofgroundatomsthataretrue;equivalentlyitcanbeseenasan

assignmentofarelation pI (HUF)ntoeveryp(n)).

AHerbrand modelIofasetSofexpressions

: aHerbrandinterpretationIthatisamodelof S

Aleast Herbrand modelIofasetSofexpressions

: I isaHerbrandmodelof S and I I’ forallHerbrandModels I’ ofS


Implication Trees

Animplication treeT withrespecttoprogramP

:

- Tisafinitetree

- ThenodesofTareatoms

- IfAisanodewithdirectdescendantsB1, ... , BnthentheremustbesomeA ← B1, ... , Bn inst(P).

- IfAisaleaf,thenA ← inst(P).

- Bisaground implication treewrt.ProgramP

:

BisanimplicationtreewrtPandallnodesaregroundatoms.


Implication Trees (Example)

LetPaddbethe”add-program”,


and n N,Vbeasetofvariablesandt TU{s,0},V.

LetT =

Then

- Tisanimplicationtreewrt.Padd.

- Ifadditionallyt HU{s,0}, then T isagroundimplicationtree.

add(sn(0), t, sn(t))

add(sn-1(0), t, sn-1(t))

add(s(0), t, s(t))

add(0, t, t)



LetPmultbethe”multiply-program”,

mult(0,X,0).mult(s(M),N, K):- mult(M,N,Z), add(Z,N,K).

and n N,Vbeasetofvariablesandt TU{s,0},V. . LetT =

Then

- Tisanimplicationtreewrt.Padd.

- T isalsoagroundimplicationtree.

mult(s(s(0)),s(0),s(s(0)))

mult(s(0),s(0),s(0)) add(s(0),s(0),s(s(0)))

mult(0,s(0),0) add(0,s(0),s(0)) add(0,s(0),s(0))


Implication Trees and Term Models

Lemma 4.7

ConsiderterminterpretationI,atomAandprogramP

• I |= Aiffinst (A) I

• I |= AiffforeveryA ← B1, ... , Bn inst(P), {B1, ... , Bn} I impliesA I.

Lemma 4.12

Theterminterpretation

C (P) : {A | A istherootofsomeimplicationtree wrt.P}

isamodelofP.


Ground Implication Trees and Herbrand Models

Lemma 4.26

ConsiderHerbrandinterpretationI,atomAandprogramP

• I |= Aiffground (A) I

• I |= AiffforeveryA ← B1, ... , Bn ground(P), {B1, ... , Bn} I impliesA I.

Lemma 4.28

TheHerbrandinterpretation

M(P) : {A | A istherootofsomegroundimplicationtree wrt.P}

isamodelofP.



LetPaddbethe”add-program”belowandVasetofvariables.


Theterminterpretation

C (P) = {add(sn(0), t, sn(t)) | n N, t TU{s,0},V }

= {add(sn(0), sm(v), sm+n(v)) | n,m N, v V {0} }

isamodelof Padd.

TheHerbrandinterpretation

M(P) = {add(sn(0), t, sn(t)) | n N, t HU{s,0} }

= {add(sn(0), sm(0), sm+n(0)) | n,m N}

isamodelof Padd.


Answer Substitutions

LetPaddbethe”add-program”belowandthequeryQ=add(s(0), u, s(u)) add(0,M1,M1).

add(s(M2),N,s(K)):- add(M2,N,K).

•{u / s(s(v))}isacorrectanswersubstitutionofQ,since

Padd |=Qadd(s(0), s(s(V)), s(s(s(V)))) .

- However, there isnoSLD-derivation leading toQ. In fact,queryQ leads tothefollowingSLD-derivationofPadd {Q}

add(s(0), u, s(u)) 1 add(0, u, u ) 2 □

where1{M2/0, N/u, K/u} and2{u/M1}. Hence,

=12| {u} ={u/M1}istheComputedAnswerSubstitution(CAS)ofQ.

- Hence,QismoregeneralthanQsince ={M1/s(s(v) ),andCASaremore general than specific correct answer substitutions. Nevertheless, are

therealwaysCAS,possiblymoregeneral,foreverycorrectanswer?

- Thecompleteness result for SLD-resolutionensuresthis.


Completeness of SLD-Resolution (Implication Trees)

First, a convenient lemma is obtained that relates SLD-resolution withimplication trees.

QueryQisn-deep

:everyatominQistherootofanimplicationtree,andnisthetotalnumberofnodesinthesetrees.

Lemma 4.15

SupposeQ isn-deepforsomen 0.ThenforeveryselectionruleR thereexistsasuccessfulSLD-derivationofP {Q} withCASsuchthatQismoregeneralthan Q.

Proof. Byinductiononn(see[Apt97],pg84)


Completeness of SLD-Resolution (1)

Theorem 4.13

SupposeisacorrectanswersubstitutionofQ.ThenforeveryselectionruleR thereexistsasuccessfulSLD-derivationofP {Q} withCASsuchthatQ ismoregeneralthan Q.

Proof. LetQ = A1, ... , Am.Then

correctanswersubstitutionofA1, ... , Am

P |= A1, ... , Am

foreveryinterpretation I: if I |= P then I |= A1, ... , Am

C(P) |= A1, ... , Am(sinceC(P) |= P, byLemma4.12)

inst(AiC(P) foreveryi 1, ..., m

AiC(P) foreveryi 1, ..., m

A1, ... , Amis n-deep forsome n0(bydefinitionofC(P) ).

claim (byLemma4.15)


Completeness of SLD-Resolution (2)

The completeness of SLD-resolution is often regarded as the guarantee of

obtainingasuccessfulSLD-derivationforanyquerywhoseexistentialclosure

isimpliedbytheprogram.

Corollary 4.16

SupposeP |= Q. ThenthereexistsasuccessfulSLD-derivationof P {Q}.

Proof. P |= Q

P |= Q forsomesubstitution

isacorrectanswersubstitutionof Q

Claim (byTheorem4.13)

- However,itisinterestingtocharacterise,beforehand,thesetofquerieswhose

existentialclosureisimpliedbytheprogram.Thisisthegoalofthedeclarative

semanticsforlogicprograms.


Least Herbrand Models

Thissetofqueriesisinfacttheleast Herbrand modelofP. Firstwedefineit.

Theorem 4.29

M(P)†istheleastHerbrandmodelofP.

Proof. LetIHbesomeHerbrandmodelofPandlet A M(P).WeproveA IH byinductiononthenumberiofnodesinthegroundimplicationtreewrt. PwithrootA.ThenM(P) IH.

i = 1: A isaleaf A ← ground(P) IH |= A (sinceIH |= P)

A IH

i ~> i+1: A hasdirectdescendants B1, ... , Bn (rootsofsubtrees)

A ← B1, ... , Bn ground(P) and B1, ... , Bn I (HerbrandInterpretation)

A ← B1, ... , Bn ground(P) and IH |= B1, ... , Bn

IH |= A (since IH |= P)

A IH

†(M(P)asdefinedinslide 23)


Ground Equivalence

Nowwecharacterisethesetofgroundatomsimpliedbyaprogram.

Theorem 4.30

ForeverygroundatomA:P |= AiffM(P)|= A

Proof.

:P |= Aand,fromLemma4.28,M(P)|= P. ThenitmustbeM(P) |= A.

: ShowthatforallA M(P),everyinterpretationI: I |= PimpliesI |= A.LetusconsiderHerbrandinterpretationIH = {A | AgroundatomandI |= A}. Then,

I |= P

I |= A ← B1, ... , Bn forall A ← B1, ... , Bn ground(P)

if I |= B1, ... , Bn then I |= A forall A ← B1, ... , Bn ground(P)

if B1 IH, ... , Bn IH, then A IH, forall A ← B1, ... , Bn ground(P)

IH |= P (byLemma4.26;thusIH isaHerbrandmodel)

Forall AM(P),itisA IH , (IH |= A) as M(P) istheleastHerbrandModel.

Hence,I |= A (bydefinitionof IH, A IH implies A IH).


Complete Partial Orderings

Once proven that the least Herbrand models are the set of ground atomsimplied by logic programs, it is interesting to specify alternative forms ofcharacterisingtheseleastHerbrandmodels.

Let(A, )beapartialordering(i.e.reflexive,antisimmetricandtransitive)

• aistheleast elementofX A:

a X, a x forall x X

• aistheleast upper boundofX A (notation a = X ):

a A , x a forall x X and a istheleast suchelementof A

• (A, )isacomplete partial ordering(cpo)containsaleastelement:• Acontainsaleastelement

• Forevery increasingsequenceao a1 a2 ... ofelementsof A, thesetX = { a0, a1, a2, ...}hasaleastupperbound.


Some Properties of Operators

Let(A, )beacompletepartialordering

• Operator T: A A is monotonic:

I J implies T(I) T(J)

• Operator T: A A is finitary:

foreveryinfinitesequenceI0 I1 ...

∞n=0 T (In) exists and T (∞

n=0 In) ∞n=0 T(In)

• Operator T: A A is continuous:

Tismonotonicandfinitary

• I isa pre-fixpoint of T :T(I) I

• I isa fixpoint of T :T(I) = I


Iterating of Operators

Let(A, )beaCPO,T: A A and I A

• T↑0(I) :I

• T↑(n+1)(I) :T(T↑n(I) )

• T↑(I) :∞n=0 (T↑n(I) )

T↑ :T↑() (for = 0, 1, 2, ... ,

Bythedefinitionofa CPO:

• Ifthesequence T↑0(I) , T↑1(I) , T↑2(I) , ... isincreasing,then T↑(I) exists

Theorem 4.22

IfTisacontinuousoperatoronaCPO,thenT↑existsandistheleastpre-fixpointofTandtheleastfixpointofT.


Consequence Operator TP

ConsiderCPO({I | I Herbrand Interpretation},

Let PbeaprogramandIaHerbrandinterpretation.Then

• TP(I) :{A | A ← B1, ... , Bn ground(P) , B1, ... , Bn I}

Lemma 4.33

(i) TP is finitary

(i) TP is monotonic

• In“practice”,theTpoperatorisusefultoobtain(least)HerbrandModelswithabottom-upprocedure.


Consequence Operator TP: Example (1)

TP↑0 =

TP↑1 = {sunny, summer}

TP↑2 = {sunny, summer, warm}

TP↑3 = {sunny, summer, warm, happy}

...

TP↑ = {sunny, summer, warm, happy} isthefixpoint

happy ← summer, warm

warm ← sunny

sunny ←

summer ←



Notice that if I is some “incomplete model” of P, then the consequenceoperatoraddsnewatomstoI.Forexample,let

I = {summer, sunny}

then

J = TP(I) = {summer, sunny, warm}

and

TP(I)

happy ← summer, warm

warm ← sunny

sunny ←

summer ←



TP↑0 =

TP↑1 = {num(0)} %0isthe1stnumber

TP↑2 = {num(sk(0))} 0k<2 %0,1arethefirst2numbers

... wheres0(0)=0

TP↑i = {num(sk(0))} 0k<i %0,...,i-1arethefirstinumbers

...

TP↑ (I)= {num(sk(0))} 0k %fixpoint

num(0) ← num(s(x)) ← num(x).



TP↑0=

TP↑1={num(0}

TP↑2={num(0}, num(s(0)),sum(0,0,0)}

TP↑3={num(sk(0)),sum(si(0),sj(0),si+j(0))} 0k<2,0i<1,0j<i

...

TP↑i={num(sk(0)),sum(si(0),sj(0),si+j(0))} 0k<i,0i<i-1,0j<i

...

TP↑w={num(sk(0)),sum(si(0),sj(0),si+j(0))} 0k,0i,0jfixpoint

num(0) ←

num(s(x)) ← num(x).

sum(0,z,z) ← num(z)

sum(s(x),y,s(z)) ← sum(x,y,z).


TP characterization

Lemma 4.32

AHerbrandInterpretationIisamodelofPiff

TP (I) I

Proof.

I |= P

iffforevery A ← B1, ... , Bn ground(P) :

{B1, ... , Bn} I implies A I (by Lemma 4.26)

iff foreverygroundatom A :

A TP (I) implies A I

iff

TP (I) I


Characterization Theorem

Theorem 4.31

Thefollowingsetsareequal

• M(P) (i)

• = least Herbrand model of P (ii)

• = least pre-fixpoint of P (iii)

• = least fixpoint of P (iv)

• = TP↑(v)

• = {A | A is a ground atom, P |= A} (vi)


Success Sets

Thesuccess setofaprogram P

:

{A | A isagroundatom, sucessfulSLD-derivationof P {A}}

Theorem 4.37

Foragroundatom A,thefollowingareequivalent

M(P) |= A (i)

• P |= A (ii)

EverySLD-treeforP {A}issuccessful (iii)

A isinthesuccesssetof P (iv)

17 October 2006 Foundations of Logic and Constraint Programming 1 Declarative Semantics An overview...

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