7 October 2005 Foundations of Logic and Constraint Programming 1 Unification An overview Need for Unification Ranked alfabeths and terms. Substitutions

7 October 2005 Foundations of Logic and Constraint Programming 1

Unification

Anoverview

• NeedforUnification

• Rankedalfabethsandterms.

• Substitutions

• Unifiersandmgus

• MartelliMontanariAlgorithm


Why unification ?

Inpropositionallogictwoclausesmayresolveiftheyhavetwocomplementaryliterals.

In the propositional case, two literals are the same if they have the same proposition(negatedinexactlyoneofthem).

In first order logic, the situation is more complex, due to the existence ofvariables.Wemayassumethatvariablesareuniversallyquantified.

Inthiscase,toapplyresolution,thepropositiondonothavetobethesameinthecomplementaryliteralsbelongingtotwoclauses.

Intuitively,sincethevariablesareuniversallyquantifiedandmaybereplacedbyany“objects”,thepredicatemustrefertothesameobjects.Forexample

A P P BA B

q(X) p(X,b) P(a,Y) r(Y)q(a) r(b)


Term Universe

Informally, the purpose of unification is to findwhat the common objects are.Suchgoalmaybeacomplishedbypurelysyntacticmeans,throughprocessingofthetermsappearinginthecommonpredicates.

Atermiscomposedofoneormoresymbolsfrom• Variablesymbols(convention:startingwithuppercase)• AlphabetisafinitesetΣofsymbols(convention:startingwithlowercase)

• Toeverysymbolanaturalnumber0isassigned(itsarityorrank).• Σ(n)denotesthesubsetofΣwitharityn• Asymbolwitharity0(belongingtoΣ(0))isaconstant.

Given a set of variablesV and a ranked alphabet of function symbolsF, thetermuniverseTUF,V(overFandV)isthesmallestsetToftermssuchthat

a)V T

b)f T,iffF(0)

c)f(t1,t2, ... ,tn) Tiff F(n)andt1,t2, ... , tn T


Ground terms and Herbrand Universe

ThesetofvariablesappearinginatermtisdenotedbyVar(t) Var(t):setofvariablesintermt

Atermisgroundifitcontainsnovariables tisground:Var(t) =

IfFisarankedalphabetoffunctionsymbols,thenHUF,theHerbrandUniverseoverF,isthesetoftermsthatcanbeformedwithoutvariables

HUF:TUF,

Example:GivenF = {a, f/1, g/2}itisHUF= = { a, f(a), g(a,a),

g(a,f(a)), g(f(a),a), g(g(a,a),a), g(a, g(a,a)), f(f(a)), f(g(a,a)), g(f(a),f(a)), g(f(a),g(a,a)),

g(g(a,a),f(a)), g(g(a,a),g(a,a)),... }

Asubtermsoft,isatermthatisincludedint s=sub_term(t):sisatermthatisasub-stringoft


Substitutions

Informally a variable (universally quantified) may be substituted by a term(withsomeconstraints).Moreformally,given

• V,asetofVariables• XV,asubsetofvariablestobesubstituted• F,arankedalphabet

asubtitutionθ:X→TUF,Vwithxθ(x)foranyx X.

Thefollowingnotationiscommonlyusedforasubstitutionθ

θ = {x1/t1, x2/t2, ... xn/tn },where

1. X = {x1, x2, ... xn },

2. θ(xi) = ti for every xi X

Itisalsoconvenienttodefine Emptysubstitutionε:n=0 θisagroundsubstitution:t1, t2, ... , tn aregroundterms.

θisapurevariablesusbtitution:t1, t2, ... , tn arevariables.

θisarenaming:{t1, t2, ... , tn }{x1, x2, ... xn }.


Substitutions

Forasubtitutionθitisalsoconvenienttoconsider

Thedomainofthesubstitutionisthesetofvariablestobesubstituted

Dom(θ ):{x1, x2, ... xn }.

TheRangeofthesubstitutionisthesetoftermsforwhichthevariablearesubstituted

Range(θ ):{t1, t2, ... , tn }

TheRanofthesubstitutionisthesetofvariablesappearingintheseterms

Ran(θ ):Var(Range(θ ) )

Thevariablesofthesubstitutionarethoseinitsdomainorinitsran.

Var(θ ):Dom(θ) Ran(θ )

TheprojectionofasubstitutiontoasubsetofitsdomainYDom(θ)

θ | Y :{y/t | y/t θ and y Y)


Application of Substitutions

A few notational conventions and definitions are commonly used when asubstitutionθisapplied

IfxisavariableandxDom(θ), thenxθ :θ(x)

IfxisavariableandxDom(θ), thenxθ :x

f(t1, t2, ... , tn )θ :f(t1θ, t2θ, ... , tn θ )

t is an instance of s :thereisasubstitutionθ with sθ = t

s is more general than t :t isaninstanceof s

t is a variant of s :thereisarenaming θ with sθ = t

Lemma 2.5

t is a variant of s ifft is an instance of sands is an instance of t.


Composition of Substitutions

Duringaresolutionproof,severalsubstitutionsareperformedinsequence,inwhichvariablesaresuccessivelyreplacedbytermswithothervariables.Thenotion of composition captures the intuition of end result of thesesubstitutions.

Definition(2.6) Thecompositionofsubstitutionsθandσ isdefinedas

(θ σ) x :x (θ)σ foreveryvariable x

Lemma (2.3)

Let θ = {x1/t1, x2/t2, ... xn/tn },σ = {y1/s1, y2/s2, ... ym/sm }. Thenθ σcanbeconstructedfromthesequence

x1/t1 σ, x2/t2 σ, ... xn/tn σ, , y1/s1, y2/s2, ... ym/sm

1. Byremovingallbindings x2/t2 σ wherex2 = t2 σ

2. Byremovingallbindings yj/sj whereyj {x1, x2, ... xn}

3. Byformingasubstitutionfromtheresultingsequence.


An Ordering of Substitutions

Asubstitutionrestrictsthevaluesavariablemaytake,andsomoreso,ifmoresubstitutionsareappliedinsequence.Thisaugmentedrestrictioniscapturedbytheorderingofsubstitutions.

Definition(2.6)

Letθandσbesubstitutions.Then

θismoregeneralthanσ:σ = θ forsomesubstitution.

(equivalently,σ isaspecialisationofθ,orθcanbespecialisedtoσ).

Examples:

θ = {x/y} ismoregeneralthan σ = {x/a, y/a} sincefor = {y/a} itis σ = θ. Since, y = a,then θ = {x/yσ , y/a} = {x/a, y/a} = σ.

θ = {x/y} isnotmoregeneralthan σ = {x/a}, sinceeverysuchthatθ = σ shouldcontainapairy/a, andthatpairshouldappearinθ .

x/a {x/y} y/a y Dom(σ ) = Dom (θ)

Intuitively, σ = {x/a} canbespecialisedto{x/a, y/b}but θ = {x/y} cannot,soθ cannotbemoregeneralthan σ.


Unifiers

The resolution of two clauses does not require the occorrence ofcomplementary literals thatareequal (apart from thenegation)but rather thatthecomplementaryliteralsareapplicabletothesameobjects.Hence,thetermsintheliteralsmustbeunifiable,i.e.equalafterperformingsomesubstitution.

Definition(2.9)

Givensubstitution θandtermssandt

θisaunifierofsandt:sθ = tθ.

s and t are unifiable :sθ = tθ forsome θ.

Manyunifiersmayexistbetweentwoterms,butitisspeciallyinteresting(e.g.Inresolution)toconsiderthemostgeneralunifier(mgu).

θisamguofsandt:sθ = tθ s σ = t σ σ = θ forsome .

Thedefinitionofunifiersforpairsofterms,maybeextendedforsetsofpairsofterms.


Most General Unifiers

• The literals to be unified may have more than one argument. The followingdefinitionsarerelevantinthiscase.

- Lets1, s2, ... , sn, t1, t2, ... , tn,beterms;si - tidenotethe(ordered)pair(si, ti); and E = {s1 ≈ t1, s2 ≈ t2, ... Sn ≈ tn }. Then

θisaunifierofE:siθ = tiθ for all i 1..n.

θisanmguofE:σ isaunifierofE σ = θ forsome .

SetsEandE’areequivalent

:θisaunifierofEθisaunifierofE’

E = {x1 ≈ t1, x2 ≈ t2, ... xn ≈ tn } is solved :

x1, x1arepairwisedistinctvariables(1ij,n)

noxI, occurs in tj (1i,j,n)


Most General Unifiers

Lemma (2.15)

IfE = {x1 ≈ t1, x2 ≈ t2, ... xn ≈ tn } issolved,then θ = {x1/t1, x2/t2, ... xn/tn },isanmguofE.

Proof:

1)xi θ = ti= ti θ

2)Foreveryunifier σ of E, it is xiσ = ti σ. But ti= ti θ andso xiσ = xi θ σ; forevery i1..n.Additionally,x σ = x θ σ =x for everyx {x1, x2, ... , xn}. Hence, σ = θ σandθ isthusmoregeneralthan σ.

• Oncetheunificationofpredicatesandtermswithmorethanoneargumentisdefined, it is important toobtainanalgorithmtoperformtheunificationof twoterms.

• Infact,itiseasiertodefineanalgorithmthatunifiesasetofpairsofterms–theMartelli–Montanarialgorithm.


Martelli – Montanari algorithm

Martelli-Montaneri algorithm (MM)

Let E be a set of pairs of terms. As long as it is possible choosenondeterministicallyapairinEandperformtheappropriateaction:

a) f(s1,s2, .. , sn) ≈ f(t1,t2, .. , tn)

→replace,inE,bys1 ≈ t1,s1 ≈ t1,...,sn ≈ tn• f(s1,s2, .. , sn) ≈ g(t1,t2, .. , tn)wheref g

→haltwithfailure(clash)• x ≈ x

→deletethepair• t ≈ x wheretisnotavariable

→replacebyx ≈ t • x ≈ t wherex Var(t) andxoccursinsomeotherpair

→performsubstitution{x / t } inthosepairs • x ≈ t wherex Var(t)andx t

→haltwithfailure(occurs check)

Thealgorithmterminateswhennoactioncanbeperformed.


Martelli – Montanari algorithm

Theorem 2.16

Iftheoriginalsethasaunifier,thealgorithmMMsuccessfullyterminatesandproduces a solved set E´ that is equivalent to E; otherwise the algorithmterminateswithfailure.

Proof Steps:

1. Provethatthealgorithmterminates.

2. Provethateachactionreplacesthesetofpairsbyanequivalentone.

3. Prove that the algorithm terminateswith failure, then the set of pairs at themomentoffailurehasnounifiers*.• Given2,thesetE’isequivalenttoE,soEalsohasnounifiers

4. Prove that if the algorithm terminates successfully, the final set of pairs issolved**.• Bylemma2.15,ifthefinalsetofpairsissolved,thenitisanmgu.


Substitution Ordering

Beforegoing througheachof thesteps, thenotionofwell-foundedorderings mustbedefinedforrelationsonsets.

Intuitively,evenifthesetisinfinite,anwellfoundedrelationguaranteesthatthesetscanbeorderedsothattheyhavean“initial”element.

Firstsomedefinitionsonrelations

Definitions

RrelationonsetA :R A A

R reflexive :(a,a) R forall a A

R irreflexive :(a,a) R forall a A

R antisymmetric :(a,b) R and(b,a) R impliesthat a = b.

R transitive :(a,b) R and(b,c) R impliesthat(a,c) R .


Lexicographic Ordering

The lexicographic ordering<n (n>1) is defined inductively on the set Nn of n-tuplesofnaturalnumbers.

BaseClause

(a1) <1 (b1) :a1 < b1

InductiveClause

(a1, a2, ... , an+1) <n+1 (b1, b2, ... , bn+1) :

(a1, a2, ... , an) <n+1 (b1, b2, ... , bn) ;or

(a1, a2, ... , an) = (b1, b2, ... , bn) and an+1 <n bn+1

Examples:(5,3,9)<3(5,3,11),(2,4,8)<3(1,3,5).

Thelexicographicordering(Nn,<n)iswell-founded

Irreflexivepartialordering(irreflexibleandtransitive)with Lowestelement:(0,0,...,0)


Well founded orderings

Oncedefinedthebasicpropertiesofrelationssomepropertiesmaybedefinedontheorderingstheyinduceinparticularsets..

Definitions

(A, R)is a (reflexive) partial ordering:R isareflexive,antisymmetric,andtransitiverelationon A

(A, R)is a irreflexive partial ordering:R isairreflexiveandtransitiverelationon A

(A, R)is well-founded ordering:R isairreflexivepartialordering;and

thereisnoinfinitedescendingchain ... a2 R a1 R a0.

ExamplesPartialOrderings: (N, ), (Z, ), (({1,2,3,4}), )IrreflexivepartialOrderings: (N, <), (Z, <), (({1,2,3,4}), )WellFounded: (N, <), (({1,2,3,4}), )NotWellFounded: (Z, <),


Algorithm MM terminates

1. MM terminates (1)

Letthefollowingdenotationsapply

VariablexsolvedinE:

x ≈ t E,andthisistheonlyoccurrenceof x in E

Uns(E) :

numberofvariablesinEthatareunsolved

lfun(E):

number of occurrences of function symbols in the first components ofpairsinE

card(E):

numberofpairsinE

Letusconsidertuple(Uns(E) , lfun(E), card(E)).ToproveterminationofMM,itissufficient toprovethateachaction inMMreducesthistuple.Fromthewell-foundnessof3tuples,therecursionmustterminate(intheworstcase,itcannotgobelowthelowestelement).


Algorithm MM terminates

1. MM terminates (2) Letusconsidertuple(u,l,c)=(Uns(E) , lfun(E), card(E)) toshowthattheMM

operationsexceptb)andf),leadingtofailure,alwaysleadtoalowertuple.

a) f(s1,s2, .. , sn) ≈ f(t1,t2, .. , tn)→replace,inE,bys1 ≈ t1,s2 ≈ t2,...,sn ≈ tn

(u,l,c) → (u-k, l-1, c+n-1)forsomek 1..n

a) x ≈ x→deletethepair

(u,l,c) → (u-k, l, c-1)forsomek 0..1

• t ≈ x wheretisnotavariable→replacebyx ≈ t

(u,l,c)→(u-k1, l-k2, c)forsomek1 0..1 and k2 1

a) x ≈ t wherex Var(t) andxoccursinsomeotherpair→performsubstitution{x / t } inthosepairs

(u,l,c)→(u-1, l+k, c)forsomek 1


Algorithm MM is correct

2. MM replaces the set of pairs by an equivalents set

a) f(s1,s2, .. , sn) ≈ f(t1,t2, .. , tn)→replace,inE,bys1 ≈ t1,s1≈ t1,...,sn ≈ tn

c) x ≈ x→deletethepair

d) t ≈ x wheretisnotavariable→replacebyx ≈ t

• Thisisclearlytrueforactionsa),c)andd)above.Foractione)

e)whenx ≈ t wherex Var(t) andxoccursinsomeotherpair→performsubstitution{x/t } inthosepairs

consider E {x ≈ t } and E{ x/t } {x ≈ t } then

θ is a unifier of E {x ≈ t } iff (θ is a unifier of E) and xθ = tθ iff (θ is a unifier of E {x/t} ) and xθ = tθ iff θ is a unifier of E {x/t} {x ≈ t }


Algorithm MM Result

3. If MM terminates with failure, then the set of pairs at the moment of failure has no unifiers

• InMMterminateswithfailure,thelastactiontakeninE’waseitherb)ord).

b) f(s1,s2, .. , sn) ≈ g(t1,t2, .. , tn)wheref g

→haltwithfailure(clash)

f) x ≈ t wherex Var(t)andx t→haltwithfailure(occurs check)

• Giventhedefinitionofunification,theremustbenounifierofE’.

Any substitution θ applied to f(s1,s2, .. , sn) θ would result in a term withfunctorf,notg,sonounifierθexists.

Nosubstitutionθmaycontainapair x/t wherex Var(t)andx t,aswouldbeneededtounifyxandt.


Algorithm MM Result

4. If the algorithm terminates successfully, the final set of pairs is solved

• WhenalgorithmMMterminates,withsuccess,noleftcomponentsofthepairsinE’arefunctions(otherwise,actionsa)orb)ord)wouldapply).

a) f(s1,s2, .. , sn) ≈ f(t1,t2, .. , tn)→replace,inE,bys1 ≈ t1,s1≈ t1,...,sn ≈ tnd) t ≈ x wheretisnotavariable→replacebyx ≈ t

• AllpairsinE’areintheformx ≈ t,wherexisavariable.Furthermoret x(otherwiseactionc)wouldapply)

c) x ≈ x→deletethepair

• Nopairotherthanx ≈ tcontainsvariablex,otherwiseactione)wouldapply

e) x ≈ t wherex Var(t) andxoccursinsomeotherpair→performsubstitution{x / t } inthosepairs

• Hence,E’hastheformE’ = {x1 ≈ t1, x2 ≈ t2, ... xn ≈ tn },wherexi Var(tj), for 1 i j n, andisthusinsolvedform.


Examples

Somenon-trivialexamples(fromSICStusProlog,wherevariablesarewitteninuppercase)

| ?- f(X,b) = f(a,Y).

X = a,Y = b ? ; no

|?- f(X,f(b)) = f(g(a,Y),Y).

X = g(a,f(b)), Y = f(b) ? ; no

| ?- f(X,f(b,Z)) = f(g(a,Y),Y).

X = g(a,f(b,Z)), Y = f(b,Z) ? ; no

| ?- X = f(Y), Y = f(a).

X = f(f(a)), Y = f(a) ? ; no

| ?- X = f(Y,Z), g(a,Y) = g(Z,b).

X = f(b,a),Y = b,Z = a ? ; no


Unifiers Size Can Be Exponential

Example:

| ?- f(X1) = f(g(X0,X0)).X1 = g(X0,X0) ? ; no

| ?- f(X1,X2) = f(g(X0,X0),g(X1,X1)).X1 = g(X0,X0),X2 = g(g(X0,X0),g(X0,X0)) ? ; no

| ?- f(X1,X2,X3) = f(g(X0,X0),g(X1,X1),g(X2,X2)).X1 = g(X0,X0),X2 = g(g(X0,X0),g(X0,X0)),X3 = g(g(g(X0,X0),g(X0,X0)),g(g(X0,X0),g(X0,X0))) ?

; no. ...

Inthiscase,the“size”ofvariableXi,intermsofoccurrencesofX0,is:

size(X1)=2size(X2)=22=4size(X3)=23=8...size(Xn)=2n


Strong MGUs

Definition

SubstitutionθisIdempotent

:θθ=θ :Mguθoftermssandtisstrong

:σismguofsandtσ=θσ

Theorem 2.16

Anmguisstrongiffitisidempotent.

Proof :

Letθbeastrongmgu.Sinceθisastrongmgu,foranyunifierσitisσ=θσ.Inparticular,forunifierθ,wehaveθ=θθ.Henceθisidempotent.

Letθbeanidempotentmgu.Then,foranyunifierσthereissomeλsuchthatσ=θλ.Butsinceθisidempotentitisσ=θλ=(θθ)λ=θ(θλ)=θσ.Hence,θisstrong.


Strong MGUs

Examples

Themguθ={x/y,y/x}oftwoidenticaltermsisnotstrong.Indeed,θθ={x/y,y/x}°{x/y,y/x}={x/x,y/y,x/y,y/x}=ε

Substitutionθ1={x/u,z/f(u,v),y/v}isIdempotent{x/u,z/f(u,v),y/v}°{x/u,z/f(u,v),y/v}=

{x/u,z/f(u,v),y/v,x/u,z/f(u,v),y/v}={x/u,z/f(u,v),y/v}=θ1

Substitutionθ2={x/u,z/f(u,v),v/y}isnotIdempotent{x/u,z/f(u,v),v/y}°{x/u,z/f(u,v),v/y}=

{x/u,z/f(u,y),v/y,x/u,z/f(u,v),y/v}={x/u,z/f(u,y),y/v}θ2.

Thefollowinglemmajustifiesasimpletesttocheckwhetherasubstitutionmguisidempotent.

Lemma:AsubstitutionθisidempotentiffDom(θ)Ran(θ)1=


Relevant MGUs

Definition Mguθoftermssandtisrelevant

:Var(θ)Var(s)Var(t)

Informaly,arelevantmgudoesnotintroducenewvariables.AllmgusproducedbytheMartelli-Montanarialgorithmarerelevant.Moreover

Theorem 2.22 (Relevance)

Everyidempotentmguisrelevant.

Proof : in [Apt, 1997]

Notethattheconverseisnottrue,i.e.Notallrelevantmgusareidempotent.Forexampleθ={x/y,y/x},whichisarelevantmguoftermss=f(x,y)andt=f(x,y),isnotidempotent.

Asobservedbefore,θθ=ε,andε is themgubtainedfromtheMMalgorithmwhenpresentedwithE={f(x,y)≈f(x,y)}.


Properties of MGUs

Thefollowingtheoremscanalsobeprovedandareusefullater(see[Apt97])

Lemma 2.23 (Equivalence):

Letθ1beanmguofasetofequationsE.Thenforeverysubstitutionθ2,θ2isanmguofEiffθ2=θ1ρ,whereρisarenamingsuchthat

Var(ρ)Var(θ2)Var(θ2).

Lemma 2.24 (Iteration) :

LetE1andE2betwosetsofequations.Supposethatθ1isanmguofE1,andθ2is anmguofE2 θ1. Thenθ1θ2 is anmguofE1E2.Moreover, ifE1E2 isunifiable,thenanmguθ1ofE1 existsandforanyforanymguθ1ofE1anmguθ2ofE2 θ1existsaswell.

Corollary 2.25 (Switching) :

LetE1andE2betwosetsofequations.Supposethatθ1isanmguofE1,andθ2isanmguofE2θ1.ThenE2isunifiable,andforeverymguθ1’ofE2 thereexistsanmguθ2’ofE2 θ1’such thatθ1’θ2 ’=θ1θ2 .Moreover,θ2’canbesochosensuchthatVar(θ2’)Var(E1)Var(θ1’)Var(θ1θ2).


Occurs Check in Prolog

Theoccurs checkisusuallynotimplemented(bydefault)inProlog,since Itmakesunificationsignificantlyharder Ithappensquiterarely

Ofcourse,unificationbehavesstrangelywhentheoccurscheckisneeded.Forexample,inSICStus,wemayobserve

| ?- X = f(X).X = f(f(f(f(f(f(f(f(f(f(...)))))))))) ? ; no

Evenworse

| ?- X = f(X).X = f(f(f(f(f(f(f(f(f(f(f(... % LOOP <CTRL-C>

Toavoid this incorrectbeheviour,mostPrologsystemsprovidethepossibilityofusingcorrect,ifinnefficientunification.



InSICStus,theexplicitunificationbetweentwotermsSandTcanbeexpressedeitheras

S = T,thedefaultmode,incorrectimplementationofunification;or unify_with_occurs_check(S,T),correctimplementation,inneficient.

Now,itis

| ?- unify_with_occurs_check(X,f(X)).no

| ?- unify_with_occurs_check(X,f(Y)).X = f(Y) ? ; no

It isup to theuser toselect theappropriate form.Forexample, the testoranemptydifference listLcouldbesimplyexpressedasT-T.But if the list isnotempty,andcontaintsnelements,thenittakestheform

[a1,a2,...,an|T]-T

in which case the default unification (without occurs check) would not workproperly,sinceitwouldtrytounifyTwith[a1,a2,...,an|T].



This behaviour can be observed with the rev_diff/2 predicate to reverse adifferencelist

rev_diff(L-L,L-L). % incorrect versionrev_diff([H|T]-Z,X-W):- % according to diff_cat Z \== [H|T], % C=X, D=W if Y = [H|D] rev_diff(T-Z,X-[H|W]).

leadingtothefollowinginteraction

| ?- rev_diff0([1,2|T]-T,L).L = [1,2,1,2,1,2,1,2,1|...]-[1,2,1,2,1,2,1,2,1|...],T = [1,2,1,2,1,2,1,2,1,2|...] ? ;L = [2,1|_A]-_A,T = [2,1|_A] ? ;no

Ifthefirstclauseiswritteninsteadas

rev_diff(L-T,L-L):- unify_with_occurs_check(L,T).

thenthecorrectbehaviourisobserved

| ?- rev_diff([1,2|T]-T,L). L = [2,1|_A]-_A,T = [2,1|_A] ? ;no


Exercises for the Week

Solveexercisesbelowfrom[Apt97],chapter2

2.1(pag.21);

2.2(pag.21);

2.3(pag.22);

2.4(pag.24);

2.5(pag.26);

2.6(pag.33);

2.8(pag.36);

Optionally

2.9(pag.37);

2.10(pag.39);

Documents

7 October 2005 Foundations of Logic and Constraint Programming 1 Unification An overview Need for Unification Ranked alfabeths and terms. Substitutions