CS 2710 Foundations of AI. Lecture 12. Inference in FOL. CS 2710 Foundations of AI. Optimization competition winners.
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1 Lecture 12 Inference in FOL Milos Hauskrecht 5329 Sennott Square Optimization competition winners CS2710 Homework 3
2 Simulated annealing 1. Valko Michal 2. Zhang Yue 3. Villamarin Ricardo Genetic Algorithm 1. Jung Sung-Young 2. Zhang Yue 3. Mills Bryan
3 Logical inference in FOL Logical inference problem: Given a knowledge base KB (a set of sentences) and a sentence α, does the KB semantically entail α? KB = α? In other words: In all interpretations in which sentences in the KB are true, is also α true? Logical inference problem in the first-order logic is undecidable!!!. No procedure that can decide the entailment for all possible input sentences in a finite number of steps. Logical inference problem in the Propositional logic Computational procedures that answer: = α KB? Three approaches: Truth-table approach Inference rules Conversion to the inverse SAT problem Resolution-refutation
4 Inference in FOL: Truth table approach Is the Truth-table approach a viable approach for the FOL? NO! Why? It would require us to enumerate and list all possible interpretations I I = (assignments of symbols to objects, predicates to relations and functions to relational mappings) Simply there are too many interpretations Inference in FOL: Inference rules Is the Inference rule approach a viable approach for the FOL? Yes. The inference rules represent sound inference patterns one can apply to sentences in the KB What is derived follows from the KB Caveat: we need to add rules for handling quantifiers
5 Inference rules Inference rules from the propositional logic: Modus ponens A B, A B Resolution A B, B C A C and others: And-introduction, And-elimination, Orintroduction, Negation elimination Additional inference rules are needed for sentences with quantifiers and variables Must involve variable substitutions Sentences with variables First-order logic sentences can include variables. Variable is: Bound if it is in the scope of some quantifier x P(x) Free if it is not bound. x P( Q( x) y is free Sentence (formula) is: Closed if it has no free variables y x P( Q( x) Open if it is not closed Ground if it does not have any variables Likes( Jane)
6 Variable substitutions Variables in the sentences can be substituted with terms. (terms = constants, variables, functions) Substitution: Is represented by a mapping from variables to terms { 2 x1 / t1, x2 / t, K} Application of the substitution to sentences SUBST ({ x / Sam, y / Pam}, Likes( x, ) = Likes( Sam, Pam) SUBST ({ x / z, y / fatherof ( John)}, Likes( x, ) = Likes( z, fatherof ( John)) Inference rules for quantifiers Universal elimination x φ ( x) φ ( a) substitutes a variable with a constant symbol Existential elimination. x φ ( x) φ ( a) a - is a constant symbol x Likes( x, IceCream) Likes( Ben, IceCream) Substitutes a variable with a constant symbol that does not appear elsewhere in the KB x Kill( x, Victim) Kill( Murderer, Victim)
7 Inference rules for quantifiers Universal instantiation (introduction) φ x is not free in φ x φ Introduces a universal variable which does not affect φ or its assumptions Sister ( Amy, Jane) x Sister( Amy, Jane) Existential instantiation (introduction) φ ( a) a is a ground term in φ xφ ( x) x is not free in φ Substitutes a ground term in the sentence with a variable and an existential statement Likes( Ben, IceCream) x Likes( x, IceCream) Unification Problem in inference: Universal elimination gives many opportunities for substituting variables with ground terms x φ ( x) a - is a constant symbol φ ( a) Solution: Try substitutions that may help Use substitutions of similar sentences in KB Unification takes two similar sentences and computes the substitution that makes them look the same, if it exists UNIFY ( p, q) = σ s.t. SUBST( σ, p) = SUBST ( σ, q)
8 Unification. Examples. Unification: UNIFY ( p, q) = σ s.t. SUBST( σ, p) = SUBST ( σ, q) Examples: UNIFY ( Knows( Knows( Jane)) = { x / Jane} UNIFY( Knows( Knows( y, Ann)) =? UNIFY ( Knows ( Knows ( y, MotherOf =? ( )) UNIFY ( Knows ( Knows ( x, Elizabeth )) =? Unification. Examples. Unification: UNIFY ( p, q) = σ s.t. SUBST( σ, p) = SUBST ( σ, q) Examples: UNIFY ( Knows( Knows( Jane)) = { x / Jane} UNIFY ( Knows( Knows( y, Ann)) = { x / Ann, y / John} UNIFY ( Knows ( Knows ( y, MotherOf =? ( )) UNIFY ( Knows ( Knows ( x, Elizabeth )) =?
9 Unification. Examples. Unification: UNIFY ( p, q) = σ s.t. SUBST( σ, p) = SUBST ( σ, q) Examples: UNIFY ( Knows( Knows( Jane)) = { x / Jane} UNIFY ( Knows( Knows( y, Ann)) = { x / Ann, y / John} UNIFY ( Knows ( Knows ( y, MotherOf ( )) = { x / MotherOf ( John), y / John} UNIFY ( Knows ( Knows ( x, Elizabeth )) =? Unification. Examples. Unification: UNIFY ( p, q) = σ s.t. SUBST( σ, p) = SUBST ( σ, q) Examples: UNIFY ( Knows( Knows( Jane)) = { x / Jane} UNIFY ( Knows( Knows( y, Ann)) = { x / Ann, y / John} UNIFY ( Knows ( Knows ( y, MotherOf ( )) = { x / MotherOf ( John), y / John} UNIFY ( Knows ( Knows ( x, Elizabeth )) = fail
10 Generalized inference rules. Use substitutions that let us make inferences Example: Modus Ponens If there exists a substitution σ such that SUBST A 1 ( σ, Ai ) = SUBST ( σ, Ai ') A 2 K An B A1 ', A SUBST ( σ, B), 2 for all i=1,2, n ', K A Substitution that satisfies the generalized inference rule can be build via unification process Advantage of the generalized rules: they are focused only substitutions that allow the inferences to proceed n ' Resolution inference rule Recall: Resolution inference rule is sound and complete (refutation-complete) for the propositional logic and CNF A B, A C B C Generalized resolution rule is sound and refutation complete for the first-order logic and CNF w/o equalities (if unsatisfiable the resolution will find the contradiction) φ1 φ2 K φk, ψ1 ψ 2 Kψ n SUBST( σ, φ K φ φ K φ ψ K ψ Example: 1 σ = UNIFY ( φ, ψ ) i 1 i+ 1 i k j 1 fail P( x) Q( Q( John) S( P( John) S( j 1 ψ j+ 1 Kψ ) n
11 Inference with the resolution rule Proof by refutation: Prove that KB, α is unsatisfiable resolution is refutation-complete Main procedure (steps): 1. Convert KB, α to CNF with ground terms and universal variables only 2. Apply repeatedly the resolution rule while keeping track and consistency of substitutions 3. Stop when empty set (contradiction) is derived or no more new resolvents (conclusions) follow Conversion to CNF 1. Eliminate implications, equivalences ( p q) ( p q) 2. Move negations inside (DeMorgan s Laws, double negation) ( p q) p q ( p q) p q x p x p x p x p p p 3. Standardize variables (rename duplicate variables) ( x P( x)) ( x Q ( x)) ( x P( x)) ( y Q ( )
12 Conversion to CNF 4. Move all quantifiers left (no invalid capture possible ) ( x P( x)) ( y Q ( ) x y P( x) Q ( 5. Skolemization (removal of existential quantifiers through elimination) If no universal quantifier occurs before the existential quantifier, replace the variable with a new constant symbol y P( A) Q ( P( A) Q ( B) If a universal quantifier precede the existential quantifier replace the variable with a function of the universal variable x y P( x) Q ( x P( x) Q ( F ( x)) F ( x) - a Skolem function Conversion to CNF 6. Drop universal quantifiers (all variables are universally quantified) x P( x) Q ( F ( x)) P( x) Q ( F ( x)) 7. Convert to CNF using the distributive laws p ( q r) ( p q) ( p r) The result is a CNF with variables, constants, functions
13 Resolution example KB P( w) Q ( w), Q( S (, P( x) R( x), R( S (, α S(A) Resolution example KB P( w) Q ( w), Q( S (, P( x) R( x), R( S (, α S(A) { y / w} P( w) S ( w)
14 Resolution example KB P( w) Q ( w), Q( S (, P( x) R( x), R( S (, α S(A) { y / w} P( w) S ( w) { x / w} S ( w) R( w) Resolution example KB P( w) Q ( w), Q( S (, P( x) R( x), R( S (, α S(A) { y / w} P( w) S ( w) { x / w} S ( w) R( w) { z / w} S (w)
15 Resolution example KB P( w) Q ( w), Q( S (, P( x) R( x), R( S (, α S(A) { y / w} P( w) S ( w) { x / w} S ( w) R( w) { z / w} KB = α S (w) { w / A} Contradiction Dealing with equality Resolution works for first-order logic without equalities To incorporate equalities we need an additional inference rule Demodulation rule σ = UNIFY φ i, t ) ( 1 φ1 φ2k φk, SUBST({ SUBST( σ, t )/ SUBST( σ, t 1 fail Example: P ( f ( a)), f ( x) = x P( a) Paramodulation rule: more powerful Resolution+paramodulation give a refutation-complete proof theory for FOL 2 t1 = t2 )}, φ K φ 1 i 1 φ K φ i+ 1 k
16 Sentences in Horn normal form Horn normal form (HNF) in the propositional logic a special type of clause with at most one positive literal ( A B) ( A C D ) Typically written as: A clause with one literal, e.g. A, is also called a fact A clause representing an implication (with a conjunction of positive literals in antecedent and one positive literal in consequent), is also called a rule Generalized Modus ponens: ( B A) (( A C ) D ) is the complete inference rule for KBs in the Horn normal form. Not all KBs are convertible to HNF!!! Horn normal form in FOL First-order logic (FOL) adds variables and quantifiers, works with terms Generalized modus ponens rule: σ = a substitution s.t. i SUBST( σ, φi ') = SUBST( σ, φi ) φ1', φ 2 ' K, φ n ', φ1 φ 2 Kφ n τ SUBST ( σ, τ ) Generalized modus ponens: is complete for the KBs with sentences in Horn form; Not all first-order logic sentences can be expressed in this form
17 Forward and backward chaining Two inference procedures based on modus ponens for Horn KBs: Forward chaining Idea: Whenever the premises of a rule are satisfied, infer the conclusion. Continue with rules that became satisfied. Typical usage: If we want to infer all sentences entailed by the existing KB. Backward chaining (goal reduction) Idea: To prove the fact that appears in the conclusion of a rule prove the premises of the rule. Continue recursively. Typical usage: If we want to prove that the target (goal) sentence α is entailed by the existing KB. Both procedures are complete for KBs in Horn form!!! Forward chaining example Forward chaining Idea: Whenever the premises of a rule are satisfied, infer the conclusion. Continue with rules that became satisfied Assume the KB with the following rules: KB: R1: Steamboat ( x) Sailboat ( R2: R3: F1: F2: F3: Sailboat ( RowBoat ( Faster ( y, Faster ( x, Faster ( y, Steamboat (Titanic ) Sailboat (Mistral ) RowBoat(PondArrow) Theorem: Faster ( Titanic, PondArrow )?
18 Forward chaining example KB: R1: R2: R3: Steamboat ( x) Sailboat ( Sailboat ( RowBoat ( Faster ( y, Faster ( x, Faster ( y, F1: Steamboat (Titanic ) F2: Sailboat (Mistral ) F3: RowBoat(PondArrow)? Forward chaining example KB: R1: R2: R3: Steamboat ( x) Sailboat ( Sailboat ( RowBoat ( Faster ( y, Faster ( x, Faster ( y, F1: Steamboat (Titanic ) F2: Sailboat (Mistral ) F3: RowBoat(PondArrow) Rule R1 is satisfied: F4: Faster( Titanic, Mistral)
19 Forward chaining example KB: R1: R2: R3: Steamboat ( x) Sailboat ( Sailboat ( RowBoat ( Faster ( y, Faster ( x, Faster ( y, F1: Steamboat (Titanic ) F2: Sailboat (Mistral ) F3: RowBoat(PondArrow) Rule R1 is satisfied: F4: Faster( Titanic, Mistral) Rule R2 is satisfied: F5: Faster( Mistral, PondArrow) Forward chaining example KB: R1: R2: R3: Steamboat ( x) Sailboat ( Sailboat ( RowBoat ( Faster ( y, Faster ( x, Faster ( y, F1: Steamboat (Titanic ) F2: Sailboat (Mistral ) F3: RowBoat(PondArrow) Rule R1 is satisfied: F4: Faster( Titanic, Mistral) Rule R2 is satisfied: F5: Faster( Mistral, PondArrow) Rule R3 is satisfied: F6: Faster( Titanic, PondArrow)
20 Backward chaining example Backward chaining (goal reduction) Idea: To prove the fact that appears in the conclusion of a rule prove the antecedents (if part) of the rule & repeat recursively. KB: R1: R2: R3: F1: F2: F3: Steamboat ( x) Sailboat ( Sailboat ( RowBoat ( Faster ( y, Faster ( x, Faster ( y, Steamboat (Titanic ) Sailboat (Mistral ) RowBoat(PondArrow) Theorem: Faster ( Titanic, PondArrow ) Backward chaining example Steamboat(Titanic) Faster ( Titanic, PondArrow ) R1 F1: Steamboat (Titanic ) F2: Sailboat (Mistral ) F3: RowBoat(PondArrow) Sailboat(PondArrow) Steamboat ( x) Sailboat ( Faster ( Titanic, PondArrow ) { x / Titanic, y / PondArrow}
21 Backward chaining example Steamboat(Titanic) Faster( Titanic, PondArrow) R1 R2 Sailboat(Titanic) F1: Steamboat (Titanic ) F2: Sailboat (Mistral ) F3: RowBoat(PondArrow) Sailboat (PondArrow) RowBoat(PondArrow) Sailboat ( RowBoat ( Faster ( y, Faster ( Titanic, PondArrow ) { y / Titanic, z / PondArrow} Backward chaining example Faster( Titanic, PondArrow) R1 R2 R3 F1: Steamboat (Titanic ) F2: Sailboat (Mistral ) F3: RowBoat(PondArrow) Steamboat(Titanic) Sailboat(Titanic) Faster( y, PondArrow) Faster( Titanic, Sailboat (PondArrow) RowBoat(PondArrow) Steamboat(Titanic) R1 R2 Sailboat(Mistral) Sailboat(Mistral) RowBoat(PondArrow)
22 Backward chaining Faster( Titanic, PondArrow) y must be bound to R1 the same term R2 R3 Steamboat(Titanic) Sailboat(Titanic) Faster( y, PondArrow) Faster( Titanic, Sailboat(PondArrow) RowBoat(PondArrow) Steamboat(Titanic) R1 R2 Sailboat(Mistral) Sailboat(Mistral) RowBoat(PondArrow) Backward chaining The search tree: AND/OR tree Special search algorithms exits (including heuristics): AO, AO* Faster ( Titanic, PondArrow ) R1 R3 R2 Steamboat(Titanic) Sailboat(Titanic) Faster( y, PondArrow) Faster( Titanic, Sailboat(PondArrow) RowBoat(PondArrow) Steamboat(Titanic) R1 R2 Sailboat(Mistral) Sailboat(Mistral) RowBoat (PondArrow)
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