1Herbrand MethodComputational Logic Lecture 7Michael Genesereth Autumn 20052Propositional Interpretationsp q r0 0 00 0 10 1 00 0 11 0 01 0 11 1 01 1 1For a language with n constants, there are 2n interpretations.23Relational Interpretations |i| a b r{,} {}{,} {}{,} {}{,} {, }{,} {}{,} {}{,} {}{,} {, }{,} {}{,} {}{,} {}{,} {, } . . .Infinitely many interpretations.4Logical EntailmentA set of premises logically entails a conclusion if andonly if every interpretation that satisfies the premisesalso satisfies the conclusion.In the case of Propositional Logic, the number ofinterpretations is finite, and so it is possible to checklogical entailment directly in finite time.In the case of Relational Logic, the number ofinterpretations is infinite, and so a direct check oflogical entailment is not feasible.35Good NewsGiven any set of sentences, there is a special subsetof interpretations called Herbrand interpretations.Under certain conditions, checking just theHerbrand interpretations suffices to determinelogical entailment.Since there are fewer Herbrand interpretations thaninterpretations in general, checking just theHerbrand interpretations is less work than checkingall interpretations.6HHHHerbrandThe Herbrand universe for a set of sentences inRelational Logic (with at least one object constant) isthe set of all ground terms that can be formed fromjust the constants used in those sentences. If there areno object constants, then we add an arbitrary objectconstant, say a.The Herbrand base for a set of sentences is the set ofall ground atomic sentences that can be formed usingjust the constants in the Herbrand universe.47Herbrand InterpretationA Herbrand interpretation for a function-freelanguage is an interpretation in which (1) the universeof discourse is the Herbrand universe for the languageand (2) each object constant maps to itself. |i|={a,b} i(a) = a i(b) = b i(r) = {〈a,a〉, 〈a,b〉}8Herbrand Interpretations |i| a b r{a,b} a b {}{a,b} a b {〈a,a〉}{a,b} a b {〈a,b〉}{a,b} a b {〈b,a〉}{a,b} a b {〈b,b〉}{a,b} a b {〈a,a〉, 〈a,b〉}{a,b} a b {〈a,a〉, 〈b,a〉}{a,b} a b {〈a,a〉, 〈b,b〉}{a,b} a b {〈a,b〉, 〈b,a〉}{a,b} a b {〈a,b〉, 〈b,b〉}{a,b} a b {〈b,a〉, 〈b,b〉}{a,b} a b {〈a,a〉, 〈a,b〉, 〈b,a〉}{a,b} a b {〈a,a〉, 〈a,b〉, 〈b,b〉}{a,b} a b {〈a,a〉, 〈b,a〉, 〈b,b〉}{a,b} a b {〈a,b〉, 〈b,a〉, 〈b,b〉}{a,b} a b {〈a,a〉, 〈a,b〉, 〈b,a〉, 〈b,b〉}59Herbrand TheoremHerbrand Theorem: A set of quantifier-free sentencesin Relational Logic is satisfiable if and only if it has aHerbrand model.Construction of Herbrand model h given i.The model assigns every object constant to itself.The interpretation for relation constant ρ is the set ofall tuples of object constants τ1 ,…, τn such that isatisfies the sentence ρ(τ1 ,…, τn).10ExampleInterpretation |i|={, } i(a) = i(b) = i(r) = {〈,〉, 〈,〉}Herbrand Base {r(a,a), r(a,b), r(b,a), r(b,b)}Herbrand Interpretation |i|={a,b} i(a) = a i(b) = b i(r) = {〈a,b〉, 〈b,b〉}611ExampleInterpretation |i|={, , } i(a) = i(b) = i(r) = {〈,〉, 〈,〉, 〈,〉}Herbrand Base{r(a,a), r(a,b), r(b,a), r(b,b)}Herbrand Interpretation |i|={a,b} i(a) = a i(b) = b i(r) = {〈a,b〉, 〈b,b〉}12Herbrand TheoremHerbrand Theorem: A set of quantifier-free sentencesis satisfiable if and only if it has a Herbrand model thatsatisfies it.Proof. Assume the set of sentences contains at least one object constant. If aset of quantifier-free sentences is satisfiable, then there is a model thatsatisfies it. Take the intersection of this model with the Herbrand base. Bydefinition, this is a Herbrand model. Moreover, it is easy to see that itsatisfies the sentences. If the sentences are ground, it must agree with theoriginal interpretation on all of the sentences, since they are all ground andmention only the constants common to both interpretations. If the sentencescontain variables, the instances must all be true, including those in which thevariables are replaced only by elements in the Herbrand universe.If there is no object constant, then create a tautology involving a newconstant (say a) and add to the set. This does not change the satisfiability ofthe sentences but satisfies proof above. QED713Basic Herbrand MethodDefinition: Add negation of conclusion to thepremises to form the satisfaction set. Loop overHerbrand interpretations. Cross out each interpretationthat does not satisfy the sentences in the satisfactionset. If all Herbrand interpretations are crossed out, bythe Herbrand Theorem, the set is unsatisfiable.Sound and Complete: Negating the conclusion leads toa contradiction; therefore, the premises logically entailthe conclusion.Termination: If there are only finitely many Herbrandinterpretations, the process halts.14Special Cases Ground Relational Logicno variables, no functions, no quantifiers Universal Relational Logicno functions, no quantifiersfree variables implicitly universally quantified Existential Relational Logicno functions Functional Relational Logicno quantifiers815ExamplePremises:Conclusion:Satisfaction Set:p(a) ⇒ q(a)p(b) ⇒ q(b)p(a) ∨ p(b)q(a) ∨ q(b)p(a) ⇒ q(a)p(b) ⇒ q(b)p(a) ∨ p(b)¬(q(a) ∨ q(b)) p q{} {}{} {a}{} {b}{} {a,b}{a} {}{a} {a}{a} {b}{a} {a,b}{b} {}{b} {a}{b} {b}{b} {a,b}{a,b} {}{a,b} {a}{a,b} {b}{a,b} {a,b}16Special Cases Ground Relational Logicno variables, no functions, no quantifiers Universal Relational Logicno functions, no quantifiersfree variables implicitly universally quantified Existential Relational Logicno functions Functional Relational Logicno quantifiers917ExamplePremises:p(x) ⇒ q(x)p(a) ∨ q(a)Conclusion:q(a)Satisfaction Set:p(x) ⇒ q(x)p(a) ∨ q(a)¬q(a) p q{} {}{} {a}{} {b}{} {a,b}{a} {}{a} {a}{a} {b}{a} {a,b}{b} {}{b} {a}{b} {b}{b} {a,b}{a,b} {}{a,b} {a}{a,b} {b}{a,b} {a,b}18ProblemPremisesp(x) ⇒ q(x)p(x) ∨ q(x)Conclusionq(x)What is the Satisfaction set? ¬q(x) No. This says q is false for all
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