1	Inference in first-order logic Chapter 9
2	Outline Reducing first-order inference to propositional inference Unification Generalized Modus Ponens Forward chaining Backward chaining Resolution
3	Universal instantiation (UI) Every instantiation of a universally quantified sentence is entailed by it: "v α Subst({v/g}, α) for any variable v and ground term g E.g., "x King(x) Ù Greedy(x) Þ Evil(x) yields: King(John) Ù Greedy(John) Þ Evil(John) King(Richard) Ù Greedy(Richard) Þ Evil(Richard) King(Father(John)) Ù Greedy(Father(John)) Þ Evil(Father(John)) . . .
4	Existential instantiation (EI) For any sentence α, variable v, and constant symbol k that does not appear elsewhere in the knowledge base: $v α Subst({v/k}, α) E.g., $x Crown(x) Ù OnHead(x,John) yields: Crown(C₁) Ù OnHead(C₁,John) provided C₁ is a new constant symbol, called a Skolem constant
5	Reduction to propositional inference Suppose the KB contains just the following: "x King(x) Ù Greedy(x) Þ Evil(x) King(John) Greedy(John) Brother(Richard,John) Instantiating the universal sentence in all possible ways, we have: King(John) Ù Greedy(John) Þ Evil(John) King(Richard) Ù Greedy(Richard) Þ 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.
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)))
7	Reduction contd. Theorem: Herbrand (1930). If a sentence α 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 α is entailed by this KB Problem: works if α is entailed, loops if α is not entailed Theorem: Turing (1936), Church (1936) Entailment for FOL is semidecidable (algorithms exist that say yes to every entailed sentence, but no algorithm exists that also says no to every non-entailed sentence.)
8	Problems with propositionalization Propositionalization seems to generate lots of irrelevant sentences. E.g., from: "x King(x) Ù Greedy(x) Þ 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 p·n^k instantiations.
9	Unification We can get the inference immediately if we can find a substitution θ such that King(x) and Greedy(x) match King(John) and Greedy(y) θ = {x/John,y/John} works Unify(α,β) = θ if αθ = βθ p q θ 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(z₁₇,OJ)
10	Unification We can get the inference immediately if we can find a substitution θ such that King(x) and Greedy(x) match King(John) and Greedy(y) θ = {x/John,y/John} works Unify(α,β) = θ if αθ = βθ p q θ Knows(John,x) Knows(John,Jane) {x/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(z₁₇,OJ)
11	Unification We can get the inference immediately if we can find a substitution θ such that King(x) and Greedy(x) match King(John) and Greedy(y) θ = {x/John,y/John} works Unify(α,β) = θ if αθ = βθ p q θ Knows(John,x) Knows(John,Jane) {x/Jane}} Knows(John,x) Knows(y,OJ) {x/OJ,y/John}} Knows(John,x) Knows(y,Mother(y)) Knows(John,x) Knows(x,OJ) Standardizing apart eliminates overlap of variables, e.g., Knows(z₁₇,OJ)
12	Unification We can get the inference immediately if we can find a substitution θ such that King(x) and Greedy(x) match King(John) and Greedy(y) θ = {x/John,y/John} works Unify(α,β) = θ if αθ = βθ p q θ Knows(John,x) Knows(John,Jane) {x/Jane}} Knows(John,x) Knows(y,OJ) {x/OJ,y/John}} Knows(John,x) Knows(y,Mother(y)) {y/John,x/Mother(John)}} Knows(John,x) Knows(x,OJ) Standardizing apart eliminates overlap of variables, e.g., Knows(z₁₇,OJ)
13	Unification We can get the inference immediately if we can find a substitution θ such that King(x) and Greedy(x) match King(John) and Greedy(y) θ = {x/John,y/John} works Unify(α,β) = θ if αθ = βθ p q θ Knows(John,x) Knows(John,Jane) {x/Jane}} Knows(John,x) Knows(y,OJ) {x/OJ,y/John}} Knows(John,x) Knows(y,Mother(y)) {y/John,x/Mother(John)}} Knows(John,x) Knows(x,OJ) {fail} Standardizing apart eliminates overlap of variables, e.g., Knows(z₁₇,OJ)
14	Unification To unify Knows(John,x) and Knows(y,z), θ = {y/John, x/z } or θ = {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 }
15	The unification algorithm
16	The unification algorithm
17	Generalized Modus Ponens (GMP) p₁', p₂', … , p_n', ( p₁ Ù p₂ Ù … Ù p_n Þq) qθ p₁' is King(John) p₁ is King(x) p₂' is Greedy(y) p₂is Greedy(x) θ is {x/John,y/John} q is Evil(x) q θ is Evil(John) GMP used with KB of definite clauses (exactly one positive literal) All variables assumed universally quantified
18	Soundness of GMP Need to show that p₁', …, p_n', (p₁ Ù … Ù p_n Þ q) ╞ qθ provided that p_i'θ = p_iθ for all I Lemma: For any sentence p, we have p ╞ pθ by UI (p₁ Ù … Ù p_n Þ q) ╞ (p₁ Ù … Ù p_n Þ q)θ = (p₁θ Ù … Ù p_nθ Þ qθ) p₁', …, p_n' ╞ p₁' Ù … Ù p_n' ╞ p₁'θ Ù … Ù p_n'θ From 1 and 2, qθ follows by ordinary Modus Ponens
19	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
20	Example knowledge base contd. ... it is a crime for an American to sell weapons to hostile nations: American(x) Ù Weapon(y) Ù Sells(x,y,z) Ù Hostile(z) Þ Criminal(x) Nono … has some missiles, i.e., $x Owns(Nono,x) Ù Missile(x): Owns(Nono,M₁) and Missile(M₁) … all of its missiles were sold to it by Colonel West Missile(x) Ù Owns(Nono,x) Þ Sells(West,x,Nono) Missiles are weapons: Missile(x) Þ Weapon(x) An enemy of America counts as "hostile“: Enemy(x,America) Þ Hostile(x) West, who is American … American(West) The country Nono, an enemy of America … Enemy(Nono,America)
21	Forward chaining algorithm
22	Forward chaining proof
23	Forward chaining proof
24	Forward chaining proof
25	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 α is not entailed This is unavoidable: entailment with definite clauses is semidecidable
26	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 Þ 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(M₁) Forward chaining is widely used in deductive databases
27	Hard matching example Colorable() is inferred iff the CSP has a solution CSPs include 3SAT as a special case, hence matching is NP-hard
28	Backward chaining algorithm SUBST(COMPOSE(θ₁, θ₂), p) = SUBST(θ₂, SUBST(θ₁, p))
29	Backward chaining example
30	Backward chaining example
31	Backward chaining example
32	Backward chaining example
33	Backward chaining example
34	Backward chaining example
35	Backward chaining example
36	Properties of backward chaining Depth-first recursive proof search: space is linear in size of proof Incomplete due to infinite loops Þ fix by checking current goal against every goal on stack Inefficient due to repeated subgoals (both success and failure) Þ fix using caching of previous results (extra space) Widely used for logic programming
37	Logic programming: Prolog Algorithm = Logic + Control Basis: backward chaining with Horn clauses + bells & whistles Widely used in Europe, Japan (basis of 5th Generation project) Compilation techniques Þ 60 million LIPS Program = set of clauses = head :- literal₁, … literal_n. criminal(X) :- american(X), weapon(Y), sells(X,Y,Z), hostile(Z). Depth-first, left-to-right backward chaining Built-in predicates for arithmetic etc., e.g., X is Y*Z+3 Built-in predicates that have side effects (e.g., input and output predicates, assert/retract predicates) Closed-world assumption ("negation as failure") e.g., given alive(X) :- not dead(X). alive(joe) succeeds if dead(joe) fails
38	Prolog Appending two lists to produce a third: append([],Y,Y). append([X\|L],Y,[X\|Z]) :- append(L,Y,Z). query: append(A,B,[1,2]) ? answers: A=[] B=[1,2] A=[1] B=[2] A=[1,2] B=[]
39	Resolution: brief summary Full first-order version: l₁ Ú ··· Ú l_k, m₁ Ú ··· Ú m_n (l₁ Ú ··· Ú l_i-1 Ú l_i+1Ú ··· Ú l_k Ú m₁ Ú ··· Ú m_j-1 Ú m_j+1 Ú ··· Ú m_n)θ where Unify(l_i, Øm_j) = θ. The two clauses are assumed to be standardized apart so that they share no variables. For example, ØRich(x) Ú Unhappy(x) Rich(Ken) Unhappy(Ken) with θ = {x/Ken} Apply resolution steps to CNF(KB Ù Øα); complete for FOL
40	Conversion to CNF Everyone who loves all animals is loved by someone: "x ["y Animal(y) Þ Loves(x,y)] Þ [$y Loves(y,x)] 1. Eliminate biconditionals and implications "x [Ø"y ØAnimal(y) Ú Loves(x,y)] Ú [$y Loves(y,x)] 2. Move Ø inwards: Ø"x p ≡ $x Øp, Ø $x p ≡ "x Øp "x [$y Ø(ØAnimal(y) Ú Loves(x,y))] Ú [$y Loves(y,x)] "x [$y ØØAnimal(y) Ù ØLoves(x,y)] Ú [$y Loves(y,x)] "x [$y Animal(y) Ù ØLoves(x,y)] Ú [$y Loves(y,x)]
41	Conversion to CNF contd. Standardize variables: each quantifier should use a different one "x [$y Animal(y) Ù ØLoves(x,y)] Ú [$z Loves(z,x)] Skolemize: a more general form of existential instantiation. Each existential variable is replaced by a Skolem function of the enclosing universally quantified variables: "x [Animal(F(x)) Ù ØLoves(x,F(x))] Ú Loves(G(x),x) Drop universal quantifiers: [Animal(F(x)) Ù ØLoves(x,F(x))] Ú Loves(G(x),x) Distribute Ú over Ù : [Animal(F(x)) Ú Loves(G(x),x)] Ù [ØLoves(x,F(x)) Ú Loves(G(x),x)]
42	Resolution proof: definite clauses