The Resolution Principle for First Order Logic

Resolution for FOL The Resolution Principle for First Order Logic Completeness of Resolution

Examples of Resolution

Deletion Strategy Summary

Resolution Resolution for FOL [Chang-Lee Ch. 5.5] for FOL

Completeness Completeness of the resolution principle [Chang-Lee Ch. of Resolution 5.6] Examples of Resolution Examples of resolution [Chang-Lee Ch. 5.7] Deletion Strategy Deletion Strategy [Chang-Lee Ch. 5.8] Resolution Principle for FOL

The Resolution Principle for First Order Logic Brief Recap. Resolution for FOL We introduced resolution as a refutation procedure for

Completeness prop. logic of Resolution

Examples of We know how to match literals containing variables using Resolution unication and substitutions Deletion Strategy We will see how to use these concepts to obtain a refutation procedure for FOL Factor

The Resolution Principle for First Order Logic

Resolution for FOL Denition (Factor) Completeness of Resolution If two ore more literals (with the same sign) in a clause C have Examples of a most general unier , then C is called a factor for C. If C Resolution σ σ σ

Deletion is a unit clause then it is called a unit factor. Strategy Example

The Resolution Principle for First Order Logic

Resolution for FOL Example (unit factor)

Completeness of Resolution Consider C = P(x) ∨ P(a). Examples of a x is a MGU for P x and P a . Resolution σ = { / } ( ) ( ) Deletion Cσ = P(a) is a unit factor of C Strategy Example II

The Resolution Principle for First Order Logic

Resolution for FOL Example (factor)

Completeness of Resolution Consider C = P(x) ∨ P(f (y)) ∨ ¬Q(x). Examples of f y x is a MGU for P x and P f y . Resolution σ = { ( )/ } ( ) ( ( )) Deletion Cσ = P(f (y)) ∨ ¬Q(f (y)) is a factor of C Strategy Binary Resolvent

The Resolution Principle for First Order Logic Denition (Binary Resolvent)

Resolution Given two clauses C and C (called parent clauses) with no for FOL 1 2

Completeness variables in common. Let L1 and L2 be two literals in C1 and of Resolution C2 respectively. If L1 and ¬L2 have a MGU σ then the clause Examples of Resolution C L C L Deletion ( 1σ − 1σ) ∪ ( 2σ − 2σ) Strategy is a binary resolvent of C1 and C2. L1 and L2 are the literals solved upon. Example: Binary Resolvent

The Resolution Example (Binary Resolvent) Principle for First Order Consider the two clauses C P x Q x and Logic 1 = ( ) ∨ ( ) C2 = ¬P(a) ∨ R(x).

Resolution Since x appears in both we will rename x with y in for FOL C2 = P(a) ∨ R(y) Completeness of Resolution Choose L1 = P(x) and L2 = ¬P(a). Examples of Resolution L1 and ¬L2 = P(a) have the MGU σ = a/x

Deletion Strategy (C1σ − L1σ) ∪ (C2σ − L2σ) = ({P(a), Q(a)} − {P(a)}) ∪ ((¬P(a), R(y)) − {¬P(a)}) = ({Q(a)} ∪ {R(y)} = {Q(a), R(y)} = Q(a) ∨ R(y)

Q(a) ∨ R(y) is the binary resolvent and P(x), ¬P(a) are the literals resolved upon Resolvent

The Resolution Principle for First Order Logic Denition (Resolvent)

Resolution for FOL Given two clauses C1 and C2 (parent clauses) a resolvent is one Completeness of the following binary resolvents: of Resolution

Examples of a binary resolvent of C1 and C2 Resolution a binary resolvent of C1 and a factor of C2 Deletion Strategy a binary resolvent of a factor of C1 and C2

a binary resolvent of a factor of C1 and a factor of C2 Example: Resolvent

The Resolution Principle for First Order Logic

Example (Resolvent) Resolution for FOL Consider the two clauses C1 = P(x) ∨ P(f (y)) ∨ R(g(y)) and Completeness of Resolution C2 = ¬P(f (g(a))) ∨ Q(b). Examples of C 0 P f y R g y is a factor of C Resolution 1 = ( ( )) ∨ ( ( )) 1 0 Deletion Cr = R(g(g(a))) ∨ Q(b) is a binary resolvent of C1 and C2 Strategy Therefore Cr is a resolvent of C1 and C2 Completeness of Resolution

The Resolution Principle for First Order Logic Completeness of resolution

Resolution Resolution is an inference rule that produce resolvents from for FOL sets of clauses Completeness of Resolution It is more ecient than previous proof procedure (e.g. Examples of Resolution Gilmore + DPLL) Deletion Resolution is complete: if the set S of clauses is Strategy unsatisable using resolution we will always manage to obtain Example

The Resolution Principle for Example (Trapezoid) First Order Logic Show that alternate interior angles formed by a diagonal of a trapezoid are equal.

Resolution T (x, y, z, w) is true i xyzw are the vertices of a trapezoid. for FOL P(x, y, u, v) is true i line segment xy is parallel to line segment uv. Completeness of Resolution E(x, y, z, u, v, w) is true i the angle xyz is equal to uvw. Examples of Resolution Axioms:

Deletion A1 (∀x)(∀y)(∀u)(∀v)(T (x, y, u, v) → P(x, y, u, v)) Strategy , A2 , (∀x)(∀y)(∀u)(∀v)(P(x, y, u, v) → E(x, y, v, u, v, y)). A3 , T (a, b, c, d). We want to proove that G , E(a, b, d, c, d, b) holds, given A1, A2, A3. Show that, by using resolution we can refute A1 ∧ A2 ∧ A3 ∧ ¬G Resolution and Semantic trees

The Resolution Principle for First Order Logic

Resolution and Semantic trees Resolution for FOL Resolution is deeply related to semantic trees Completeness of Resolution Resolution generates clauses that can be used to prune Examples of branches of semantic trees Resolution

Deletion Semantic trees can be used to prove completeness of Strategy resolution Example

The Resolution Principle for Example (resolution and semantic trees) First Order Logic Consider the set of clauses S = {P, ¬P ∨ Q, ¬P ∨ ¬Q}. We can nd a closed semantic tree with 5 nodes. Using resolution Resolution for FOL we can obtain: Completeness P Q P Q of Resolution ¬ ∨ ¬ ∨ ¬

Examples of ¬P Resolution Consider the set S0 S C, we can nd a closed semantic tree Deletion = ∪ Strategy with 3 nodes. Using resolution we can obtain: ¬PP 00 0 Consider the set S = S ∪ we can nd a closed semantic tree with one node. Semantic tree and completeness of Resolution

The Resolution Principle for First Order Logic

Resolution Semantic trees and Resolution for FOL A similar reasoning can be used to prove the completeness Completeness of Resolution of Resoluton Examples of Given a set of unsatisable clauses: Resolution

Deletion 1 Construct a closed semantic tree Strategy 2 Force the tree to collapse while building a resolution proof. Lifting lemma

The Resolution Theorem Principle for 0 0 First Order Lifting Lemma If C1 and C2 are instances of C1 and C2 Logic 0 0 0 0 respectively, and if C is a resolvent of C1 and C2, then C is an instance of C (resolvent of C1 and C2). Resolution for FOL

Completeness Example of Resolution Consider C P x Q x and Examples of 1 = ( ) ∨ ( ) Resolution C2 = ¬P(f (y)) ∨ ¬P(z) ∨ R(y). Deletion 0 Strategy C1 = P(f (a)) ∨ Q(f (a)) is an instance of C1 0 C2 = ¬P(f (a)) ∨ R(a) is an instance of C2 0 0 0 C3 = Q(f (a)) ∨ R(a) is a resolvent for C1 and C2 0 Lifting Lemma ⇒ ∃ C3 such that C3 is an instance of C3. For example, C3 = Q(f (y)) ∨ R(y) is a resolvent for C1 0 and C2 and C3 is an instance of C3 Lifting lemma: proof

The Resolution Principle for First Order Logic Lifting Lemma

If necessary we rename variables in C1 or C2 so that Resolution for FOL variables in C1 are all dierent from variables in C2. Completeness Let L0 and L0 be the literals resolved upon of Resolution 1 2 0 0 0 0 0 0 0 Examples of C = (C1γ − L1γ) ∪ (C2γ − L2γ), γ MGU for L1, L2. Resolution Since C 0 and C 0 are instances of C 0 and C 0 we can write Deletion 1 2 1 2 Strategy 0 0 C1 = C1θ and C2 = C2θ where θ is one substitution. 1 Ri Let Li , ··· , Li denote the literals in Ci corresponding to 0 1 Ri 0 Li (i.e. Li θ, ··· , Li θ = Li ) Lifting lemma: proof II

The Resolution Principle for First Order Logic Lifting Lemma

1 Ri assume i > 1 obtain a MGU λi for Li , ··· , Li . and let Resolution 1 for FOL Li = Li λi for i = 1, 2. Completeness then L is a literal in factor C of C . of Resolution i i λi i 1 Examples of assume i = 1 then λi = and Li = Li λi . Resolution Let Deletion λ = λ1 ∪ λ2 Strategy 0 Then Li is an instance of Li 0 0 Since L1 and L2 are uniable then L1 and L2 are uniable. Let σ be a MGU of L1 and L2 Lifting lemma: proof III

The Resolution Principle for First Order Logic Proof.

Resolution (Lifting Lemma) for FOL 1 R1 Completeness Let C = (C1(λ ◦ σ) − ({L1, ··· , L1 })(λ ◦ σ)) ∪ ((C2(λ ◦ of Resolution 1 R2 σ) − ({L2, ··· , L2 })(λ ◦ σ))) Examples of Resolution 0 1 R1 Then C = (C1(θ ◦ γ) − ({L1, ··· , L1 })(θ ◦ γ)) ∪ ((C2(θ ◦ Deletion 1 R2 Strategy γ) − ({L2, ··· , L2 })(θ ◦ γ))) is an instance of C as λ ◦ σ is a more general unier than θ ◦ γ Completeness of Resolution

The Resolution Principle for First Order Logic

Resolution for FOL Theorem (Completeness of Resolution) Completeness of Resolution A set S of clauses is unsatisable i there is a resolution Examples of Resolution deduction of the empty clause from S Deletion Strategy Completeness of Resolution: proof ⇐

The Resolution . Principle for ⇐ First Order If there is a resolution deduction of the empty clause from S Logic then S is unsatisable

Resolution Suppose there is a deduction of from S. Let for FOL R1, R2, ··· , Rk be the resolvents in the deduction. Completeness of Resolution Assume S is satisable then there is I |= S. Examples of Resolution Assume Ri is resolvent of Cu and Cv , notice that I |= S Deletion therefore I |= Cu ∧ Cv Strategy Since resolution is an inference rule then if I |= Cu ∧ Cv then I |= Ri for all resolvents However, one of the resolvents is therefore S must be unsatisable. Completeness of Resolution: proof ⇒

The Resolution Principle for First Order ⇒ Logic If S is unsatisable then there is a resolution deduction of the empty clause from S. Resolution for FOL Suppose S is unsatisable, and let A = {A1, A2, A3, ···} Completeness be the atome set for S. of Resolution Examples of Let T be a complete semantic tree for S. Resolution

Deletion By Herbrand's theorem (version I) T has a nite closed Strategy sematic tree T 0 0 If T consists only of one root node then must be in S, because no other clauses can be falsied at the root of a semantic tree, Thus the theorem is true. Completeness of Resolution: proof ⇒

The Resolution Principle for First Order ⇒ Logic Assume T 0 has more than one node. T 0 must have at least one inference node Resolution for FOL This is because, otherwise, every node would have at least Completeness of Resolution one non failure descendent and thus T 0 would have an

Examples of innite branch (and thus not be a closed tree). Resolution 0 Deletion Let N be an inference node in T , and let N1 and N2 be Strategy the failure nodes immediately below N.

Let I (N) = {m1, m2, ··· , mn}, I (N1) = {m1, m2, ··· , mn, mn + 1}, I (N2) = {m1, m2, ··· , mn, ¬mn + 1} Completeness of Resolution: proof ⇒

The Resolution Principle for ⇒ First Order Logic 0 0 Since N not a failure node, there exist C1 and C2, ground instances of C1 and C2 such that: 0 0 Resolution C1 and C2 are both not falsied by I (N) for FOL 0 0 C1 and C2 are falsied by I (N1) and I (N2) respectively. Completeness of Resolution 0 0 Therefore C1 contains ¬mn+1 and C2 contains mn+1 Examples of 0 0 Resolution Let L1 = ¬mn+1 and L2 = mn+1 and 0 0 0 0 0 Deletion C = (C1 − L1) ∪ (C2 − L2) Strategy 0 0 0 C must be false in I (N) because both (C1 − L1) and 0 0 (C2 − L2) are. By the lifting lemma we can then nd a resolvent C of C1 0 and C2 such that C is a (ground) instance of C Completeness of Resolution: proof ⇒

The Resolution Principle for First Order Logic ⇒ Let T 00 be the closed semantic tree associated to S ∪ C. Resolution for FOL T 00 is obtained by T 0 removing all noded which are below Completeness the rst node where C 0 is falsied of Resolution 00 0 Examples of T has fewer nodes than T Resolution We can apply this process until the closed semantic tree Deletion Strategy consists only of the root node. This is possible only when is derived, therefore there is deduction of from S. Example I

The Resolution Principle for First Order Logic Example Resolution for FOL A1 P → S Completeness , of Resolution A2 , S → U Examples of Resolution A2 , P Deletion G U Strategy , Show that (A1 ∧ A2 ∧ A3) |= G. Example II

The Resolution Principle for First Order Logic

Resolution Example for FOL

Completeness of Resolution F , (∀x)(∀y)(P(x, f (y)) ∨ P(y, f (x))) Examples of G u v P u f v P v f u Resolution , (∃ )(∃ )( ( , ( )) ∧ ( , ( ))) Deletion Strategy Show that F |= G. Example III

The Resolution Principle for First Order Logic Example (quack and doctors)

Resolution Show that F1 ∧ F2 |= G, where for FOL Some patients like all doctors Completeness of Resolution F1 ∃x(P(x) ∧ ∀y(D(y) → L(x, y)) Examples of , Resolution No patient likes any quack Deletion F x P x y Q y L x y Strategy 2 , ∀ ( ( ) → ∀ ( ( ) → ¬ ( , ))) No doctor is a quack F3 , ∀xD(x) → ¬Q(x) Exercise I

The Resolution Principle for First Order Logic

Resolution Exercise for FOL Show that F F G, where Completeness 1 ∧ 2 |= of Resolution F1 (∀x)(C(x) → (W (x) ∧ R(x))) Examples of , Resolution F2 , (∃x)(C(x) ∧ O(x)) Deletion G x O x R x Strategy , (∃ )( ( ) ∧ ( )) Exercise II

The Resolution Principle for First Order Logic Exercise

Resolution Students and votes for FOL Premise: Students are citizens. Completeness of Resolution Conclusion: Students' votes are citizens votes. Examples of Students are citizens Resolution Deletion F1 (∀y)(S(y) → C(y)) Strategy , Students' votes are citizens votes F2 , (∀x)((∃y)(S(y) ∧ V (x, y)) → (∃z)(C(z) ∧ V (x, z))) Deleting Clauses

The Resolution Principle for First Order Logic

Need for deleting clauses Resolution for FOL Resolution is complete (Binary resolution + factorisation) Completeness of Resolution Resolution is more ecient than earlier methods (e.g., Examples of Gilmore + DPLL) Resolution

Deletion computational issue: Repeated application of resolution Strategy generates irrelevant and redundant clauses Applying Resolution

The Resolution Principle for First Order Logic Computing resolvents

Resolution Need a deterministic method to apply resolution for FOL

Completeness Deterministic strategy to compute resolvents of Resolution Straightforward strategy: Examples of Resolution compute resolvents for all possible pairs Deletion add resolvents to S Strategy repeat until appears Called Level Saturation Level Saturation

The Resolution Principle for First Order Logic

Resolution Level Saturation Denition for FOL 0 1 2 Completeness Generate the sequence S , S , S , ··· of Resolution S0 = S Examples of Resolution i S = {Resolvents of C1 and C2|C1 ∈ Deletion 0 i−1 i−1 Strategy (S ∪ · · · ∪ S ) and C2 ∈ S }, i = 1, 2, 3, ··· Level Saturation: Procedure

The Resolution Principle for First Order Logic Level Saturation Procedure Resolution for FOL At every step i > 0 0 i 1 Completeness List all clauses in {S ∪ · · · ∪ S − } in order of Resolution

Examples of compute all resolvents by comparing every clause Resolution 0 i−1 i−1 C1 ∈ {S ∪ · · · ∪ S } with a clause C2 ∈ S that is Deletion Strategy listed after C1. append computed resolvents to the end of the list Example: Level Saturation

The Resolution Principle for First Order Logic

Resolution for FOL Example (Level Saturation) Completeness of Resolution Consider the set of clauses Examples of Resolution S = {P ∨ Q, ¬P ∨ Q, P ∨ ¬Q, ¬P ∨ ¬Q} Deletion Strategy Level Saturation: Problems

The Resolution Principle for First Order Logic Problems with Level Saturation Generation of many irrelevant and redundant clauses Resolution for FOL Tautologies

Completeness Clauses repeatadly generated of Resolution Tautologies have no impact on satisability Examples of Resolution Tautologies are true in every interpretations Deletion If S is unsatisable, S 0 obtained from S removing Strategy tautologies is unsatisable Tautologies can create other irrelevant clauses We need a deletion strategy that maintains completeness Subsumption

The Resolution Denition (Subsumption) Principle for First Order Logic A clause C subsumes a clause D i there is a substitution σ such that Cσ ⊆ D. D is called a subsumed clause. Resolution for FOL Example (Subsumption) Completeness of Resolution Consider the two clauses C = P(x) and D = P(a) ∨ Q(a). Examples of Resolution Consider the substitution σ = {a/x}. Deletion Strategy Cσ = P(a) therefore Cσ ⊆ D C subsumes D.

Note If C is identical to D or if C is an instance of D then C subsumes D. Deletion Strategy

The Resolution A deletion strategy Principle for First Order Logic Delete any tautology and any subsumed clause whenever possible A Complete deletion strategy Resolution for FOL The above deletion strategy is complete if it is used with the level

Completeness saturation method of Resolution For each step i > 0: Examples of Resolution 1 List clauses in S0 ··· Si−1 in order Deletion Strategy 2 Compute resolvents by comparing any clause in 0 i−1 i−1 C1 ∈ S ··· S with a clause C2 ∈ S which il listed after C1 3 When a resolvent C is computed, append C to the list only if C is not a tautology and C is not subsumed by any claus in the list. Otherwise delete C. Example: Level Saturation deleting clauses

The Resolution Principle for First Order Logic

Resolution for FOL Example (Level Saturation Deleting clauses) Completeness of Resolution Consider the set of clauses Examples of Resolution S = {P ∨ Q, ¬P ∨ Q, P ∨ ¬Q, ¬P ∨ ¬Q} Deletion Strategy Checking redundant clauses

The Resolution Principle for First Order Logic Redundant clauses Need to check: Resolution for FOL 1 whether a clause is a tautology Easy Completeness of Resolution 2 whether a clause is subsumed by another clause need an

Examples of algorithm Resolution

Deletion Checking tautology Strategy Directly check whether there is a complementary pair in the clause. No substitutions involved. Checking Subsumption

The Resolution Principle for First Order Logic Preliminaries Consider two clauses C and D. Resolution for FOL Let θ = {a1/x1, ··· an/xn} where: {x1, ··· , xn} are all Completeness variables occurring in D and a a are new distinct of Resolution { 1, ··· , n} constants not occurring in C or D. Examples of Resolution Suppose D = L1 ∨ L2 ∨ · · · Lm then Deletion Strategy Dθ = L1θ ∨ L2θ ∨ · · · ∨ Lmθ Note that Dθ is a ground clause. ¬Dθ = ¬L1θ ∧ · · · ∧ ¬Lmθ (using de morgan's law) Subsumption Algorithm

The Resolution Principle for First Order Logic Algorithm

1 Let W = {¬L1θ · · · ¬Lmθ} Resolution 0 for FOL 2 Set k = 0 and U = {C} Completeness 3 If Uk contains of Resolution Examples of Yes: terminate; C subsumes D Resolution Otw: let Deletion Uk+1 Resolvents of C and C C Uk and C W Strategy = { 1 2| 1 ∈ 2 ∈ } 4 If Uk+1 is empty Yes: terminate; C does not subsume D Otw: k = k + 1 go to step 3. Example: Subsumption algorithm

The Resolution Principle for First Order Logic

Resolution for FOL Example (Subsumption algorithm)

Completeness of Resolution Consider the two clauses: Examples of C P x Q f x a Resolution = ¬ ( ) ∨ ( ( ), ) Deletion D = ¬P(h(y)) ∨ Q(f (h(y)), a) ∨ ¬P(c) Strategy Subsumption Algorithm: termination

The Resolution Principle for First Order Logic Termination Subsumption algorithm always terminates. Resolution for FOL Each clause CUk+1 is always one litteral smaller than Completeness k of Resolution clauses in U for k = 0, 1, ··· k 1 Examples of This is because U + is obtained by computing the Resolution resolvents of clauses in Uk and W , therefore, if a resolvent Deletion Strategy exists it will always be one literal smaller than the parent clauses. Otw Uk+1 is empty. k k Therefore for some k we will have ∈ U or U is empty. Example II: Subsumption algorithm

The Resolution Principle for First Order Logic

Resolution for FOL Example (Subsumption algorithm)

Completeness of Resolution Consider the two clauses: Examples of C P x x Resolution = ( , ) Deletion D = P(f (x), y) ∨ P(y, f (x)) Strategy Subsumption Algorithm: correctness

The Resolution Principle for First Order Logic

Resolution for FOL

Completeness Theorem (Correctness) of Resolution

Examples of C subsumes D i subsumption algorithm terminates in step 3. Resolution

Deletion Strategy Proof ⇒

The Resolution ⇒. Principle for First Order Logic If C subsumes D then subsumption algorithm terminates in step 3

Resolution If C subsumes D then there is σ such that Cσ ⊆ D for FOL Hence C D Completeness (σ ◦ θ) ⊆ θ of Resolution Therefore literals in C(σ ◦ θ) can be resolved by using unit Examples of Resolution gound clauses in W Deletion But C is an instance of C Strategy (σ ◦ θ) Therefore literals in C can be resolved away by using unit clauses in W k k Therefore we will eventually nd a U such that ∈ U and the algorithm will terminate at step 3. Proof ⇐

The ⇐. Resolution Principle for First Order If the subsumption algorithm terminates in step 3 then C Logic subsumes D If algorithm terminates at step 3 then we have a refutation Resolution of . for FOL Completeness Indicates with R , B the parent clauses, where B W , of Resolution i i i ∈ and with R C; Indicates with R the resolvent Examples of 0 = i+1 Resolution obtained at each step for i = 0, 1, ··· , r Deletion Strategy Let σi be the most general unier for each resolution step. Then C(σ0 ◦ σ1 ◦ · · · ◦ σr ) = {¬B0, ¬B1, · · · ¬Br } ⊆ Dθ Let λ = σ0 ◦ · · · ◦ σr then Cλ ⊆ Dθ. Let σ be the substitution obtained by replacing ai with xi in each component of λ for i = 1, ··· , n Then Cσ ⊆ D therefore C subsumes D. Sol C Subsumes D

Example III: Subsumption algorithm

The Resolution Principle for First Order Logic Example (Subsumption algorithm) Resolution for FOL Consider the two clauses: Completeness of Resolution C = P(x, y) ∨ Q(z) Examples of D = Q(a) ∨ P(b, b) ∨ R(u) Resolution

Deletion Check whether C subsumes D Strategy Example III: Subsumption algorithm

Deletion Check whether C subsumes D Strategy Sol C Subsumes D Sol C Does not subsumes D

Example IV: Subsumption algorithm

The Resolution Principle for First Order Logic Example (Subsumption algorithm) Resolution for FOL Consider the two clauses: Completeness of Resolution C = P(x, y) ∨ R(y, x) Examples of D = P(a, y) ∨ R(z, b) Resolution

Deletion Check whether C subsumes D Strategy Example IV: Subsumption algorithm

Deletion Check whether C subsumes D Strategy Sol C Does not subsumes D Sol C Does not subsumes D but C |= D

Example V: Subsumption algorithm

The Resolution Principle for First Order Logic Example (Subsumption algorithm)

Resolution for FOL Consider the two clauses: Completeness C P x P f x of Resolution = ¬ ( ) ∨ ( ( )) Examples of D = ¬P(x) ∨ P(f (f (x))) Resolution Check whether C subsumes D and whether C D Deletion |= Strategy Example V: Subsumption algorithm

The Resolution Principle for First Order Logic Example (Subsumption algorithm)