UIUC CS 497: Section EA Lecture #3 Reasoning in Artificial Intelligence Professor: Eyal Amir Spring Semester 2004

Last Time SAT checking using DPLL (instantiate, propagate, backtrack) Entailment/SAT checking using Resolution (create more and more clauses until KB is saturated) Formal verification uses mainly SAT checking such as DPLL, but also sometimes resolution

From Homework You Should Know Deduction theorem for FOL Language of FOL Soundness, completeness, and incompleteness theorems Models of FOL

Today Reasoning procedure for FOL –Proving entailment using Resolution Application du jour: Temporal Reasoning Applications we will not touch –Spatial reasoning, formal verification, mathematics, planning, NLP, …

First-Order Theories Signature L: –Function symbols (f(x,y)) –Predicate (relation) symbols (P(x,A)) –Constant symbols (A,B,C,…) FOL language: quantification over objects

Model Theory: Reminder Structure/Interpretation: –U = Universe of elements –I = Mapping of Constant symbols to elements in U Predicate symbols to relations over U Function symbols to functions over U M T - M satisfies T –T is a theory, i.e., a set of FOL sentences in language L for which M is an interpretation ╨

Logical Entailment ? ╨     ╨  L={man, woman, loves} M 1 = U 1 ={Sue,Kim,Pat} I 1 [man]={Pat} I 1 [woman]={Sue,Kim} I 1 [loves]={, } ? M1  ╨ ╨ 

Clausal Form Every FOL formula is consistency- equivalent to conjunction of F.O. clauses. First-order clause –Only universal quantifiers (which are implicit) –Disjunction of literals (atoms or their negation)

Conversion to Clausal Form 1.“  ” replaced by “  ”, ”  ” 2.Negations in front of atoms 3.Standardize variables (unique vars.) 4.Eliminate existentials (Skolemization) 5.Drop all universal quantifiers 6.Move disjunctions in (put into CNF) 7.Rename vars. (standardize vars. apart)

Take a Breath Until now: –First-order logic basics –How to convert a general FOL sentence to clausal form From now: Resolution theorem proving –Search in the space of proofs Later: Temporal reasoning

Resolution Theorem Proving Given: –KB – a set of first-order sentences –Query Q – a logical sentence Calling procedure: 1.Add  Q to KB 2.Convert KB into clausal form 3.Run theorem prover. If we prove contradiction, return T.

Resolution Theorem Proving 1.Add  Q to KB 2.Convert KB into clausal form 3.Run theorem prover. If we prove contradiction, return T. Deduction theorem: KB Q iff KB   Q FALSE ╨╨

Resolution Theorem Proving 1.Add  Q to KB 2.Convert KB into clausal form 3.Run theorem prover. If we prove contradiction, return T. Deduction theorem: KB Q iff KB   Q FALSE ╨╨

First-Order Resolution Resolution inference rule: C1: P(t1,…,tk)  C1’(t1,…,tk) C2:  P(s1,…,sk)  C2’(s1,…,sk) mgu(, ) = {r1,…,rn} C3: (C1’  C2’) {r1,…,rn}

First-Order Resolution Resolution algorithm (saturation): 1.While there are unresolved C1,C2: (1)Select C1, C2 in KB (2)If C1, C2 are resolvable, resolve them into a new clause C3 (3)Add C3 to KB (4)If C3={ } return T. 2.STOP C1: P(t1,…,tk)  C1’(t1,…,tk) C2:  P(s1,…,sk)  C2’(s1,…,sk) mgu(, ) = {r1,…,rn} C3: (C1’  C2’) {r1,…,rn}

Resolution in Action On board Negated Query KB C1: P(t1,…,tk)  C1’(t1,…,tk) C2:  P(s1,…,sk)  C2’(s1,…,sk) mgu(, ) = {r1,…,rn} C3: (C1’  C2’) {r1,…,rn}

Resolution in Action On board Negated Query KB C1: P(t1,…,tk)  C1’(t1,…,tk) C2:  P(s1,…,sk)  C2’(s1,…,sk) mgu(, ) = {r1,…,rn} C3: (C1’  C2’) {r1,…,rn}

First-Order Resolution Resolution algorithm (saturation): 1.While there are unresolved C1,C2: (1)Select C1, C2 in KB (2)If C1, C2 are resolvable, resolve them into a new clause C3 (3)Add C3 to KB (4)If C3={ } return T. 2.STOP C1: P(t1,…,tk)  C1’(t1,…,tk) C2:  P(s1,…,sk)  C2’(s1,…,sk) mgu(, ) = {r1,…,rn} C3: (C1’  C2’) {r1,…,rn}

First-Order Resolution Rule (2) If C1, C2 are resolvable, resolve them into a new clause C3 If C1,C2 have two literals l1,l2 with same predicates (P) and opposite polarity, and If l1= P(t1,…,tk), l2=  P(s1,…,sk), unifiable with mgu (most general unifier) {r1,…,rn}, then… C1: P(t1,…,tk)  C1’(t1,…,tk) C2:  P(s1,…,sk)  C2’(s1,…,sk) mgu(, ) = {r1,…,rn} C3: (C1’  C2’) {r1,…,rn}

Unification …P(t1,…,tk),  P(s1,…,sk), unifiable with mgu (most general unifier) σ={r1,…rk} Substitution: replace vars. by terms –Term: constant, variable, or a function of terms Composition of substitutions {x/g(w,v)} {w/A,v/f(B,z)} = {x/g(A,f(B,z),w/A,v/f(B,z)} (P(x) v Q(f(x)) v P(g(B,x)) v P(f(y))) {x/B,y/z} (P(B) v Q(f(B)) v P(g(B,B)) v P(f(z))) {x/B,y/z}{x/B,y/z,x/w}{x/B,y/z,z/w}

Unification Unification: find a substitution σ for C1: P(t1,…,tk)  C1’(t1,…,tk) C2:  P(s1,…,sk)  C2’(s1,…,sk) such that P(t1,…,tk)σ = P(s1,…,sk)σ P(A,y,g(x,y)){y/f(A)} = P(z,f(z),g(x,f(w))){z/A,w/A} σ={y/f(A),z/A,w/A} P(A,y,g(x,y)){y/f(w)} = P(z,f(w),g(x,f(w))){z/A} σ={y/f(w),z/A} Most general unifier

Unification Substitution σ1 more general than σ2 if there is substitution γ such that σ1 γ = σ2 P(A,y,g(x,y)){y/f(A)} = P(z,f(z),g(x,f(w))){z/A,w/A} σ={y/f(A),z/A,w/A} P(A,y,g(x,y)){y/f(w)} = P(z,f(w),g(x,f(w))){z/A} σ={y/f(w),z/A} Most general unifier

Unification Substitution σ1 more general than σ2 if there is substitution γ such that σ1 γ = σ2 σ2={y/f(A),z/A,w/A} σ1={y/f(w),z/A} Most general unifier γ={w/A}

Finding the MGU Procedure MGU(x,y) 1.If x=y, return { } 2.If x or y are vars., return MGUvar(x,y) 3.If x or y are const.,or Len(x)=/=Len(y), return F. 4.σ ← { }; For i ← 1 to Len(x) 1.s ← MGU(part(i,x),part(i,y)); if s=F, return F. 2.σ ← compose(σ,s); x ← subst(x,σ); y ← subst(y,σ); 5.Return σ x = P(A,y,g(x,y)) y = P(z,f(w),g(v,w))

Finding the MGU Procedure MGU(x,y) 1.If x=y, return { } 2.If x or y are vars., return MGUvar(x,y) 3.If x or y are const.,or Len(x)=/=Len(y), return F. 4.σ ← { }; For i ← 1 to Len(x) 1.s ← MGU(part(i,x),part(i,y)); if s=F, return F. 2.σ ← compose(σ,s); x ← subst(x,σ); y ← subst(y,σ); 5.Return σ x = P(A,y,g(x,y)) y = P(z,f(w),g(v,w))

Finding the MGU Procedure MGU(x,y) 1.If x=y, return { } 2.If x or y are vars., return MGUvar(x,y) 3.If x or y are const.,or Len(x)=/=Len(y), return F. 4.σ ← { }; For i ← 1 to Len(x) 1.s ← MGU(part(i,x),part(i,y)); if s=F, return F. 2.σ ← compose(σ,s); x ← subst(x,σ); y ← subst(y,σ); 5.Return σ x = P(A,y,g(x,y)) y = P(z,f(w),g(v,w))

Finding the MGU Procedure MGU(x,y) 1.If x=y, return { } 2.If x or y are vars., return MGUvar(x,y) 3.If x or y are const.,or Len(x)=/=Len(y), return F. 4.σ ← { }; For i ← 1 to Len(x) 1.s ← MGU(part(i,x),part(i,y)); if s=F, return F. 2.σ ← compose(σ,s); x ← subst(x,σ); y ← subst(y,σ); 5.Return σ x = P(A,y,g(x,y)) y = P(z,f(w),g(v,w))

Finding the MGU Procedure MGU(x,y) 1.If x=y, return { } 2.If x or y are vars., return MGUvar(x,y) 3.If x or y are const.,or Len(x)=/=Len(y), return F. 4.σ ← { }; For i ← 1 to Len(x) 1.s ← MGU(part(i,x),part(i,y)); if s=F, return F. 2.σ ← compose(σ,s); x ← subst(x,σ); y ← subst(y,σ); 5.Return σ x = P(A,y,g(x,y)) y = P(z,f(w),g(v,w))

part(3,x) = y part(3,y) = f(w)part(1,x) = P part(1,y) = Ppart(2,x) = A part(2,y) = z Finding the MGU Procedure MGU(x,y) 1.If x=y, return { } 2.If x or y are vars., return MGUvar(x,y) 3.If x or y are const.,or Len(x)=/=Len(y), return F. 4.σ ← { }; For i ← 1 to Len(x) 1.s ← MGU(part(i,x),part(i,y)); if s=F, return F. 2.σ ← compose(σ,s); x ← subst(x,σ); y ← subst(y,σ); 5.Return σ x = P(A,y,g(x,y)) y = P(z,f(w),g(v,w)) part(4,x) = g(x,f(w)) part(4,y) = g(v,w) F Occurs check

Finding the MGU: another example Procedure MGU(x,y) 1.If x=y, return { } 2.If x or y are vars., return MGUvar(x,y) 3.If x or y are const.,or Len(x)=/=Len(y), return F. 4.σ ← { }; For i ← 1 to Len(x) 1.s ← MGU(part(i,x),part(i,y)); if s=F, return F. 2.σ ← compose(σ,s); x ← subst(x,σ); y ← subst(y,σ); 5.Return σ x = P(A,y,g(x,y)) y = P(z,f(w),g(v,f(w)))

Finding the MGU: another example Procedure MGU(x,y) 1.If x=y, return { } 2.If x or y are vars., return MGUvar(x,y) 3.If x or y are const.,or Len(x)=/=Len(y), return F. 4.σ ← { }; For i ← 1 to Len(x) 1.s ← MGU(part(i,x),part(i,y)); if s=F, return F. 2.σ ← compose(σ,s); x ← subst(x,σ); y ← subst(y,σ); 5.Return σ x = P(A,y,g(x,y)) y = P(z,f(w),g(v,f(w)))

Finding the MGU: another example Procedure MGU(x,y) 1.If x=y, return { } 2.If x or y are vars., return MGUvar(x,y) 3.If x or y are const.,or Len(x)=/=Len(y), return F. 4.σ ← { }; For i ← 1 to Len(x) 1.s ← MGU(part(i,x),part(i,y)); if s=F, return F. 2.σ ← compose(σ,s); x ← subst(x,σ); y ← subst(y,σ); 5.Return σ x = P(A,y,g(x,y)) y = P(z,f(w),g(v,f(w)))

Finding the MGU: another example Procedure MGU(x,y) 1.If x=y, return { } 2.If x or y are vars., return MGUvar(x,y) 3.If x or y are const.,or Len(x)=/=Len(y), return F. 4.σ ← { }; For i ← 1 to Len(x) 1.s ← MGU(part(i,x),part(i,y)); if s=F, return F. 2.σ ← compose(σ,s); x ← subst(x,σ); y ← subst(y,σ); 5.Return σ x = P(A,y,g(x,y)) y = P(z,f(w),g(v,f(w)))

Finding the MGU: another example Procedure MGU(x,y) 1.If x=y, return { } 2.If x or y are vars., return MGUvar(x,y) 3.If x or y are const.,or Len(x)=/=Len(y), return F. 4.σ ← { }; For i ← 1 to Len(x) 1.s ← MGU(part(i,x),part(i,y)); if s=F, return F. 2.σ ← compose(σ,s); x ← subst(x,σ); y ← subst(y,σ); 5.Return σ x = P(A,y,g(x,y)) y = P(z,f(w),g(v,f(w)))

part(3,x) = y part(3,y) = f(w)part(1,x) = P part(1,y) = Ppart(2,x) = A part(2,y) = z Finding the MGU: another example Procedure MGU(x,y) 1.If x=y, return { } 2.If x or y are vars., return MGUvar(x,y) 3.If x or y are const.,or Len(x)=/=Len(y), return F. 4.σ ← { }; For i ← 1 to Len(x) 1.s ← MGU(part(i,x),part(i,y)); if s=F, return F. 2.σ ← compose(σ,s); x ← subst(x,σ); y ← subst(y,σ); 5.Return σ part(4,x) = g(x,f(w)) part(4,y) = g(v,f(w)) σ={y/f(w),z/A,v/x} x = P(A,y,g(x,y)) y = P(z,f(w),g(v,f(w)))

Correctness of FOL Resolution Soundness: Resolution is sound for first- order inference Refutation Completeness: Resolution will find the empty clause, if FALSE is entailed No guarantee of termination (saturation), if FALSE not entailed: FOL semi-decidable

Simple Efficiency Improvements Subsumption between clauses: –If clause C subsumes clause D (C entails D), then we can remove D from the KB. {P(A)}{P(A),P(B)} {P(A)}{~P(A)} {P(f(x),A),Q(g(x),B)}{P(f(v),y),Q(g(v),y)} Algorithm for checking subsumption? ╨ ╨ ╨ {P(A)}{P(t)} ╨

Simple Efficiency Improvements Subsumption within the clause: –If literal a subsumes literal b (a entails b), then we can remove a from the clause… –But, notice the variables’ scope {P(A),P(t)} {P(A),~P(B),P(t)} {P(A,x),P(y,B)} {P(x),Q(x),P(A)} {P(A)} {P(A),~P(B)} {P(A,x),P(y,B)} {P(x),Q(x),P(A)}

Properties of Resolution Unifying two literals of length n O(n 2 ) – because of occurs check Finding two resolvable clauses from m clauses of length n: O(m 2 n 2 ) – the simple bound Overall algorithm: –Semi-decidable –Unbounded length of proof as function of n,m

Related to FOL Resolution Clause selection and restriction strategies for resolution (lecture #5, paper #19) Consequence finding (paper #3) Constraint Satisfaction Problem (paper #5) Reasoning with equality (paper #6) DPLL in FOL (paper #7) Decidable fragments of FOL (paper #8)

Summary So Far Resolution theorem proving allows us to find contradictions and explanation. –The deduction theorem tells us how to ask queries from Resolution Next: Temporal Reasoning

Situation Calculus A first-order language for describing the effects of actions and events over time –Constants: S0 – initial state; action constants –Functions: result(, ) –Predicates: “fluents” – properties that change over time at(1, S0) at(x,s)  at(x+1,result(move_fwd,s)) Query: at(1+1,s’)   ans(s’) ?  at(1+1,s’)  ans(s’)

Situation Calculus Requires axioms describing effects and non-effects of actions/events Can be used for planning, projection, diagnosis, filtering (tracking) Frame Problem: the compact and natural- language-like specification of effects of actions Qualification Problem: the preconditions of actions

Notations Substitutions –φσ σ = {x1/t1,…,xk/tk} –φ σ σ = [x1/t1,…,xk/tk]

Next Time Description Logics Requires prior knowledge of: –FOL theorem proving –Tableau theorem proving – will be useful –Frame systems – will be useful