# Inference in first- order logic. Outline Reducing first-order inference to propositional inference...

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Inference in first- order logic Slide 2 Outline Reducing first-order inference to propositional inference Unification Generalized Modus Ponens Forward chaining Backward chaining Resolution Slide 3 Universal instantiation (UI) Every instantiation of a universally quantified sentence is entailed by it: v Subst({v/g}, ) for any variable v and ground term g(without any variable) E.g., x King(x) Greedy(x) Evil(x) yields: King(John) Greedy(John) Evil(John) King(Richard) Greedy(Richard) Evil(Richard) King(Father(John)) Greedy(Father(John)) Evil(Father(John)). Slide 4 Existential instantiation (EI) For any sentence , variable v, and constant symbol k that does not appear elsewhere in the knowledge base: v Subst({v/k}, ) E.g., x Crown(x) OnHead(x,John) yields: Crown(C 1 ) OnHead(C 1,John) provided C 1 is a new constant symbol, called a Skolem constant Slide 5 Reduction to propositional inference Suppose the KB contains just the following: x King(x) Greedy(x) Evil(x) King(John) Greedy(John) Brother(Richard,John) Instantiating the universal sentence in all possible ways, we have: King(John) Greedy(John) Evil(John) King(Richard) Greedy(Richard) Evil(Richard) King(John) Greedy(John) Brother(Richard,John) The new KB is propositionalized: proposition symbols are King(John), Greedy(John), Evil(John), King(Richard), etc. Slide 6 Reduction contd. Every FOL KB can be propositionalized so as to preserve entailment A ground sentence is entailed by new KB iff entailed by original KB Idea: propositionalize KB and query, apply resolution, return result Problem: with function symbols, there are infinitely many ground terms, e.g., Father(Father(Father(John))) Slide 7 Reduction contd. Theorem: Herbrand (1930). If a sentence is entailed by an FOL KB, it is entailed by a finite subset of the propositionalized KB Idea: For n = 0 to do create a propositional KB by instantiating with depth-$n$ terms see if is entailed by this KB Problem: works if is entailed, loops if is not entailed Theorem: Turing (1936), Church (1936) Entailment for FOL is semidecidable algorithms exist that say yes to every entailed sentence no algorithm exists that also says no to every nonentailed sentence. Slide 8 Problems with propositionalization Propositionalization seems to generate lots of irrelevant sentences. Example from: x King(x) Greedy(x) Evil(x) King(John) y Greedy(y) Brother(Richard,John) it seems obvious that Evil(John), but propositionalization produces lots of facts such as Greedy(Richard) that are irrelevant With p k-ary predicates and n constants, there are pn k instantiations. Slide 9 Unification We can get the inference immediately if we can find a substitution such that King(x) and Greedy(x) match King(John) and Greedy(y) = {x/John,y/John} works Unify(,) = if = p q Knows(John,x) Knows(John,Jane) Knows(John,x)Knows(y,OJ) Knows(John,x) Knows(y,Mother(y)) Knows(John,x)Knows(x,OJ) Standardizing apart eliminates overlap of variables, e.g., Knows(John,z 27 ) Knows(z 17,OJ) {x/Jane} {x/OJ, y/John} {x/Mother(John),y/John} No substitution possible yet. Slide 11 Unification Examples UNIFY(Knows(John,x), Knows(John,Jane)) = {x/Jane} UNIFY(Knows(John,x), Knows(y,Bill)) = {x/Bill, y/John} UNIFY(Knows(John,x), Knows(y,Mother(y))= {y/John, x/Mother(John) UNIFY(Knows(John,x), Knows(x,Elizabeth)) = fail Last example fails because x would have to be both John and Elizabeth We can avoid this problem by standardizing: The two statements now read UNIFY(Knows(John,x), Knows(z,Elizabeth)) This is solvable: UNIFY(Knows(John,x), Knows(z,Elizabeth)) = {x/Elizabeth,z/John} Slide 12 Unification To unify Knows(John,x) and Knows(y,z) Can use = {y/John, x/z } Or = {y/John, x/John, z/John} The first unifier is more general than the second. There is a single most general unifier (MGU) that is unique up to renaming of variables. MGU = { y/John, x/z } Slide 13 Unification Unification algorithm: Recursively explore the two expressions side by side Build up a unifier along the way Fail if two corresponding points do not match Slide 14 The unification algorithm Slide 15 Slide 16 Simple Example Brother(x,John) Father(Henry,y) Mother(z, John) Brother(Richard,x) Father(y,Richard) Moth er(Eleanore,x) Slide 17 Generalized Modus Ponens (GMP) p 1 ', p 2 ', , p n ', ( p 1 p 2 p n q) q p 1 ' is King(John) p 1 is King(x) p 2 ' is Greedy(y) p 2 is Greedy(x) is {x/John,y/John} q is Evil(x) q is Evil(John) GMP used with KB of definite clauses (exactly one positive literal) All variables assumed universally quantified where p i ' = p i for all i Slide 18 Soundness of GMP Need to show that p 1 ', , p n ', (p 1 p n q) q provided that p i ' = p i for all I Lemma: For any sentence p, we have p p by UI 1. (p 1 p n q) (p 1 p n q) = (p 1 p n q) 2. p 1 ', \; , \;p n ' p 1 ' p n ' p 1 ' p n ' 3. From 1 and 2, q follows by ordinary Modus Ponens Slide 19 Storage and Retrieval Use TELL and ASK to interact with Inference Engine Implemented with STORE and FETCH STORE(s) stores sentence s FETCH(q) returns all unifiers that the query q unifies with Example: q = Knows(John,x) KB is: Knows(John,Jane), Knows(y,Bill), Knows(y,Mother(y)) Result is 1 ={x/Jane}, 2 =, 3 = {John/y,x/Mother(y)} Slide 20 Storage and Retrieval First approach: Create a long list of all propositions in Knowledge Base Attempt unification with all propositions in KB Works, but is inefficient Need to restrict unification attempts to sentences that have some chance of unifying Index facts in KB Predicate Indexing Index predicates: All Knows sentences in one bucket All Loves sentences in another Use Subsumption Lattice (see below) Slide 21 Storing and Retrieval Subsumption Lattice Child is obtained from parent through a single substitution Lattice contains all possible queries that can be unified with it. Works well for small lattices Predicate with n arguments has a 2 n lattice Structure of lattice depends on whether the base contains repeated variables Knows(John,John) Knows(x,John)Knows(x,x)Knows(John,x) Knows(x,y) Slide 22 Forward Chaining Idea: Start with atomic sentences in the KB Apply Modus Ponens Add new atomic sentences until no further inferences can be made Works well for a KB consisting of Situation Response clauses when processing newly arrived data Slide 23 Forward Chaining First Order Definite Clauses Disjunctions of literals of which exactly one is positive: Example: King(x) Greedy(x) Evil(x) King(John) Greedy(y) First Order Definite Clauses can include variables Variables are assumed to be universally quantified Greedy(y) means y Greedy(y) Not every KB can be converted into first definite clauses Slide 24 Example knowledge base The law says that it is a crime for an American to sell weapons to hostile nations. The country Nono, an enemy of America, has some missiles, and all of its missiles were sold to it by Colonel West, who is American. Prove that Col. West is a criminal Slide 25 Example knowledge base contd.... it is a crime for an American to sell weapons to hostile nations: American(x) Weapon(y) Sells(x,y,z) Hostile(z) Criminal(x) Nono has some missiles, i.e., x Owns(Nono,x) Missile(x): Owns(Nono,M 1 ) and Missile(M 1 ) all of its missiles were sold to it by Colonel West Missile(x) Owns(Nono,x) Sells(West,x,Nono) Missiles are weapons: Missile(x) Weapon(x) An enemy of America counts as "hostile: Enemy(x,America) Hostile(x) West, who is American American(West) The country Nono, an enemy of America Enemy(Nono,America) Slide 26 Forward chaining algorithm Slide 27 Forward chaining proof Slide 28 Slide 29 Slide 30 Properties of forward chaining Sound and complete for first-order definite clauses Datalog = first-order definite clauses + no functions FC terminates for Datalog in finite number of iterations May not terminate in general if is not entailed This is unavoidable: entailment with definite clauses is semidecidable Slide 31 Efficiency of forward chaining Incremental forward chaining: no need to match a rule on iteration k if a premise wasn't added on iteration k-1 match each rule whose premise contains a newly added positive literal Matching itself can be expensive: Database indexing allows O(1) retrieval of known facts e.g., query Missile(x) retrieves Missile(M 1 ) Forward chaining is widely used in deductive databases Slide 32 Hard matching example Colorable() is inferred iff the CSP has a solution CSPs include 3SAT as a special case, hence matching is NP-hard Diff(wa,nt) Diff(wa,sa) Diff(nt,q) Diff(nt,sa) Diff(q,nsw) Diff(q,sa) Diff(nsw,v) Diff(nsw,sa) Diff(v,sa) Colorable() Diff(Red,Blue) Diff (Red,Green) Diff(Green,Red) Diff(Green,Blue) Diff(Blue,Red) Diff(Blue,Green) Slide 33 Backward Chaining Improves on Forward Chaining by not making irrelevant conclusions Alternatives to backward chaining: restrict forward chaining to a relevant set of forward rules Rewrite rules so that only relevant variable bindings are made: Use a magic set Example: Rewrite rule: Magic(x) American(x) Weapon(x) Sells(x,y,z) Hostile(z) Criminal(x) Add fact: Magic(West) Slide 34 Backward Chaining Idea: Given a query, find all substitutions that satisfy the query. Algorithm: Work on lists of goals, starting with original query Algo finds every clause in the KB that unifies with the positive literal (head) and adds remainder (body) to list of goals Slide 35 Backward chaining algorithm SUBST(COMPOSE( 1, 2 ), p) = SUBST( 2, SUBST( 1, p)) Slide 36 Backward chaining example Slide 37 Slide 38 Slide 39 Slide 40 Slide 41 Slide 42 Slide 43 Slide 44 Properties of backward chaining Depth-first recursive proof search: space is linear in size of proof Incomplete due to infinite loops fix by checking current goal against every goal on stack Inefficient due to repeated subgoals (both success and failure) fix using caching of previous results (extra space) Widely used for logic programming Slide 45 Logic programming: Prolog Algorithm = Logic + Control

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