1Resolution Theorem ProvingComputational Logic Lecture 10Michael Genesereth Autumn 20052PlanFirst Lecture - Resolution PreliminariesRelational Clausal FormUnificationSecond Lecture - ResolutionResolution Principle and FactoringSatisfiability and Logical EntailmentAnswer ExtractionReductionThird Lecture - Resolution StrategiesElimination Strategies (tautology elimination, subsumption, …)Restriction Strategies (ancestry-filtering, set of support, …)Fourth Lecture- Ordered ResolutionOrdered ResolutionModel Elimination23Propositional Resolution€ {ϕ1,...,ϕ,...,ϕm}{ψ1,...,¬ϕ,...,ψn}{ϕ1,...,ϕm,ψ1,...,ψn}4Relational Resolution I€ {ϕ1,...,ϕ,...,ϕm}{ψ1,...,¬ψ,...,ψn}{ϕ1,...,ϕm,ψ1,...,ψn}σwhere σ= mgu(ϕ,ψ)35Example€ { p(a, y),r(y)}{¬p(x,b)}{r(y)}{x ← a, y ← b}{r(b)}6Example€ { p(a, y), r(y)}{¬p(x, f (x)),q(g(x))}{r(y),q(g(x))}{x ← a, y ← f (a)}{r( f (a)),q(g( a))}47ExampleEverybody loves somebody. Everybody loves alover. Show that everybody loves everybody.€ ∀x.∃y.loves(x , y)∀u.∀v.∀w.(loves(v,w) ⇒ loves(u,v))¬∀x.∀y.loves(x, y)8Example (continued)€ ∀x.∃y.loves(x , y)∀u.∀v.∀w.(loves(v,w) ⇒ loves(u,v))¬∀x.∀y.loves(x, y){loves(x, f (x))}{¬loves(v,w),loves(u,v)}{¬loves( jack, jill)}59Example (concluded)€ 1. {loves(x, f (x))} Premise2. {¬loves(v,w),loves(u,v)} Premise3. {¬loves( jack, jill)} Goal4. {loves(u, x)} 1,25. {} 4,310Harry and RalphEvery horse can outrun every dog. Some greyhoundcan outrun every rabbit. Show that every horse canoutrun every rabbit.€ ∀x.∀y.(h(x)∧ d(y) ⇒ f (x, y))∃y.(g(y)∧∀z.(r(z) ⇒ f (y, z)))∀y.(g(y) ⇒ d(y))∀x.∀y.∀z.( f (x, y)∧ f (y,z) ⇒ f (x, z))¬∀x.∀y.(h(x)∧ r(y) ⇒ f (x, y))611Harry and Ralph (continued)€ ∀x.∀y.(h(x)∧ d(y) ⇒ f (x, y))∃y.(g(y)∧∀z.(r(z) ⇒ f (y, z)))∀y.(g(y) ⇒ d(y))∀x.∀y.∀z.( f (x, y)∧ f (y, z) ⇒ f (x, z)){¬h(x),¬d(y), f (x, y)}{g(greg)}{¬r(z), f (greg,z)}{¬g(y),d(y)}{¬f (x, y),¬f (y, z), f (x, z)}12Harry and Ralph (continued)€ ¬∀x.∀y.(h(x)∧ r(y) ⇒ f (x, y)){h(harry)}{r(ralph)}{¬f (harry,ralph)}713Harry and Ralph (concluded)€ 1. {¬h(x),¬d(y), f (x, y)}2. {g(greg)}3. {¬r(z), f (greg, z)}4. {¬g(y),d(y)}5. {¬f (x , y),¬f (y, z), f (x,z)}6. {h(harry)}7. {r(ralph)}8. {¬f (harry,ralph)}€ 9. {d(greg)}10. {¬d(y), f (harry, y)}11. { f (harry,greg)}12. { f (greg,ralph)}13. {¬f (greg, z), f (harry, z)}14. { f (harry,ralph)}15. {}14ExampleGiven:∃x.∀y.(p(x,y) ⇔ q(x,y))∀x.∃y.(p(x,y) ∨ q(x,y))Prove:∃x.∃y.(p(x,y) ∧ q(x,y))815Example (continued)∃x.∀y.(p(x,y) ⇔ q(x,y)) ∃x.∀y.((¬p(x,y) ∨ q(x,y)) ∧ (p(x,y) ∨ ¬q(x,y))) (¬p(a,y) ∨ q(a,y)) ∧ (p(a,y) ∨ ¬q(a,y)) {¬p(a,y), q(a,y)} {p(a,y), ¬q(a,y)}∀x.∃y.(p(x,y) ∨ q(x,y)) p(x, f(x)) ∨ q(x, f(x)) {p(a,f(x)), q(a,f(x))}16Example (continued)Negate the goal: ∃x.∃y.(p(x,y) ∧ q(x,y)) → ¬∃x.∃y.(p(x,y) ∧ q(x,y))Convert to Clausal Form: ¬∃x.∃y.(p(x,y) ∧ q(x,y)) ∀x.∀y.¬(p(x,y) ∧ q(x,y)) ∀x.∀y.(¬p(x,y) ∨ ¬ q(x,y)) ¬p(x,y) ∨ ¬q(x,y) {¬p(x,y), ¬q(x,y)}917Example (concluded)1. {¬p(a,y), q(a,y)} Premise2. {p(a,y), ¬q(a,y)} Premise3. {p(x, f(x)), q(x, f(x))} Premise4. {¬p(x,y), ¬q(x,y)} Negated Goal5. {q(a, f(a))} 1, 36. {p(a, f(a))} 2, 37. {¬p(a, f(a))} 4, 58. {} 6, 718Example∀x.p(x) ⇒ ∀y.p(y)1019Example (continued)¬(∀x.p(x) ⇒ ∀ y.p(y)) I: ¬(¬∀x.p(x) ∨ ∀ y.p(y)) N: ¬¬∀x.p(x) ∧ ¬∀y.p(y)∀x.p(x) ∧ ∃y.¬p(y) S: ∀x.p(x) ∧ ∃y.¬p(y) E: ∀x.p(x) ∧ ¬p(a) A: p(x) ∧ ¬p(a) D: p(x) ∧ ¬p(a) O: {p(x)} and {¬p(a)}20Example (concluded)Resolution:1. {p(x)} Premise2. {¬p(a)} Premise3. {} 1,2 {x←a}1121Problem€ { p(a, x)}{¬p(x,b)}Failure22Relational Resolution II€ {ϕ1,...,ϕ,...,ϕm}{ψ1,...,¬ψ,...,ψn}{ϕ1τ,...,ϕmτ,ψ1,...,ψn}σwhere σ= mgu(ϕτ,ψ)where τ is a variable renaming on ϕ1223Example€ { p(a, x)}{¬p(x,b)}Failure€ { p(a, y)}{¬p(x,b)}{}{x ← a, y ← b}24Solution With Repeated Renaming1. {r(a,b,u1)} Premise2. {r(b,c,u2)} Premise3. {r(c,d,u3)} Premise4. {r(x,z,f(v)),¬r(x,y,f(f(v))),¬r(y,z,f(f(v)))} Premise5. {¬r(a,d,w)} Goal6. {¬r(a,y6,f(f(v6))),¬r(y6,d,f(f(v6)))} 4,57. {¬r(b,d,f(f(v7)))} 1,68. {¬r(b,y8,f(f(f(v8)))),¬r(y8,d,f(f(f(v8))))} 4,79. {¬r(c,d,f(f(f(v9))))} 2,810. {} 3,91325Problem Without Repeated Renaming1. {r(a,b,u1)} Premise2. {r(b,c,u2)} Premise3. {r(c,d,u3)} Premise4. {r(x,z,f(v)),¬r(x,y,f(f(v))),¬r(y,z,f(f(v)))} Premise5. {¬r(a,d,w)} Goal6. {¬r(a,y,f(f(v))),¬r(y,d,f(f(v)))} 1,47. {¬r(b,d,f(f(v)))} 1,68. Failure 4,726Problem€ {p (x), p(y)}{¬p(u),¬p(v)}{p (y),¬p(v)}{p (x),¬p(v)}{p (y),¬p(u)}{p (x),¬p(u)}1427FactorsIf a subset of the literals in a clause Φ has a mostgeneral unifier γ, then the clause Φ' obtained byapplying γ to Φ is called a factor of Φ.Clause{p(x),p(f(y)),r(x,y)}Factors{p(f(y)),r(f(y),y)}{p(x),p(f(y)),r(x,y)}28Relational Resolution III (Final Version)€ ΦΨ((Φ'−{φ})τ∪ (Ψ'−{¬ψ}))σwhere φ∈ Φ', a factor of Φ where ¬ψ∈ Ψ', a factor of Ψ where σ= mgu(ϕτ,ψ)where τ is a variable renaming on ϕ1529Example€ {p (x), p(y)}{¬p(u),¬p(v)}{p (y),¬p(v)}{p (x),¬p(v)}{p (y),¬p(u)}{p (x),¬p(u)}€ {p (x)}{¬p(u)}{}30Need for Original Clauses1. {p(a,y), p(x,b)} Premise2. {¬p(a,d)} Premise3. {¬p(c,b)} Premise4. {p(x,b)} 1,25. {} 3,41. {p(a,y), p(x,b)} Premise2. {¬p(a,d)} Premise3. {¬p(c,b)} Premise4. {p(a,b)} Factor of 11631ProvabilityA resolution derivation of a clause ϕ from a set Δ ofclauses is a sequence of clauses terminating in ϕ inwhich each item is(1) a member of Δ or(2) the result of applying the resolution to earlier items.A sentence ϕ is provable from a set of sentences Δ byresolution if and only if there is a derivation of theempty clause from the clausal form of Δ∪{¬ϕ}.A resolution proof is a derivation of the empty clausefrom the clausal form of the premises and the negationof the desired conclusion.32Soundness and CompletenessMetatheorem: Provability using the RelationalResolution Principle is sound and complete forRelational Logic (without equality).1733Determining Logical EntailmentTo determine whether a set Δ of sentences logicallyentails a closed sentence ϕ, rewrite Δ∪{¬ϕ} inclausal form and try to derive the empty clause.34ExampleShow that {(p(x) ⇒ q(x)), p(a)} logically entails ∃z.q(z).€
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