Inference in First Order Logic

Inference in First Order Logic Introduction to Artificial Intelligence CS440/ECE448 Lecture 12 New homework out today, due March 2

Last lecture • Examples of sentences • Simple inference rules • Inference with quantifiers This lecture • Substitutions and unification • Generalized Modus Ponens • Resolution and refutation Reading • Chapter 9

Substitutions Given a sentence  and binding list , the result of applying the substitution to  is denoted by Subst(, ). Example: s = {x/Sam, y/Pam}  = Likes(x,y) Subst({x/Sam, y/Pam}, Likes(x,y)) = Likes(Sam, Pam)

Inference with Universal Quantifiers Universal Elimination (UE):  x  Subst({x/}, ) [ must be a ground term (i.e., no variables).] Example Imagine Universe of Discourse is {Rodrigo, Sharon, David, Jonathan}  x At (x, UIUC)  OK(x) At(Rodrigo, UIUC)  OK(Rodrigo) substitution list

Inference with Existential Quantifiers Existential Elimination:   Subst({/k},  ) where k is a new (Skolem) constant. Example From x Teaches(x, CS348) we can infer that Teaches(PorkyPig, CS348) if the symbol PorkyPig hasnot appeared before.

A simple proof • Premises: • Bob is a buffalo • Pat is a pig • Buffaloes outrun pigs • Prove that “Bob outruns Pat”. • Buffalo(Bob) • Pig(Pat) • x,y Buffalo(x)Pig(y)Faster(x,y) 4. Buffalo(Bob)  Pig(Pat) AI 1, 2 5. Buffalo(Bob)  Pig(Pat) Faster(Bob, Pat) UE 3, {x/Bob, y/Pat} 6. Faster(Bob, Pat) MP 4, 5

Search with primitive inference rules • Operators are inference rules. • States are sets of sentences. • Goal test checks state to see if it contains query sentence.. • Problem: branching factor huge, esp. for UE. • AI, UE, MP is a common inference pattern. • Idea: find a substitution that makes the rule premise match some known facts.  A single, more powerful inference rule.

Unification A substitution  unifies atomic sentences p and q if Subst(, p)= Subst(, q) (also denoted p=q). p q  Knows(John,x) Knows(John, Jane) { x/Jane } { y/John, x/Amy } Knows(John,x) Knows(y, Amy) { y/John, x/Mother(John) } Knows(John,x) Knows(y, Mother(y)) Idea: Unify rule premises with known facts, apply unifier to conclusion. E.g., if we know both q and Knows(John,x)  Likes(John,x) then we conclude: Likes(John,Jane) Likes(John,Amy) Likes(John,Mother(John))

A bit more on substitutions A substitution  is a set of pairings of variables with terms: • = { v1/term1, v2/term2, … } such that: • each variable is paired at most once; and • a variable’s pairing term may not contain the variable directly or indirectly [ e.g. can’t have substitution { x/f(y), y/f(x) } ]. When unifying expressions  and , the variable names in  and the variable names in  should be disjoint. • UNIFY(Loves (John, x), Loves (x, Jane)) ! No unifier • UNIFY(Loves (John, x), Loves (y, Jane)) ! = {x/Jane,y/John}

Most General Unifier • A substitution  is more general than  iff there is a substitution  such that =. • e.g.  = { z/ F(w) } is more general than  = { z/F(C) } since  = { w/C } • A most general unifier  of  and  is such that, for any unifier  of  and , there exists a substitution  =  = (). • Informally, the most general unifier (MGU) imposes the fewest constraints on the terms (contains the most variables). • MGU’s are unique up to variable reordering and changes of naming of variables in terms. • The unification procedure on p. 278 gives the MGU.

Generalized Modus Ponens (GMP) p1’, p2’,…, pn’, (p1p2… pn  q) wherepi’ = pi q for all i For example, let p1’ = Faster (Bob, Pat) p2’ = Faster (Pat, Steve) Faster (x, y)  Faster (y, z)  Faster (x, z) Unify p1’ and p2’ with the premise  = { x/Bob, y/Pat, z/Steve } Apply substitution to the conclusion q = Faster(Bob, Steve) pi and q atomic sentences Universally quantified variables

A GMP Inference Engine • Perform inference search using only one inference rule, Generalized Modus Ponens. • Every sentence in the database should be put in canonical form • an atomic sentence, or • p1  p2 … pn  q where piand q are atomic sentences. • These are Horn sentences. Can convert other sentences to Horn sentences using Existential Elimination and And-Elimination. • All variables are assumed to be universally quantified. • Previous proof is now one step.

Forward chaining When a new fact P is added to the KB: For each rule such that P unifies with a premise, if the other premises are known then add the conclusion to the KB and continue chaining. Forward chaining is data-driven, e.g., inferring properties and categories from percepts.

Forward Chaining Example • White: Facts added in turn • Yellow: The result of implication of rules. • Buffalo(x) Pig(y)  Faster(x, y) • Pig(y) Slug(z)  Faster(y, z) • Faster(x, y) Faster(y, z)  Faster(x, z) • Buffalo(Bob) [ Unifies with 1-a ] • Pig(Pat) [ Unifies with 1-b, GMP Fires ] [ Unifies with 2-a ] • Faster(Bob,Pat) [ Unifies with 3-a, 3-b ] • Slug(Steve) [ Unifies with 2-b, GMP Fires ] • Faster(Pat, Steve) [ Unifies with 3-b and with 6,GMP Fires ] • Faster (Bob, Steve) • …

Backward chaining p1  p2 … pn  q • When a query q is examined: if a matching fact q' is known, return the unifier; for each rule whose consequent q' matchesq, attempt to prove each premise of the rule by backward chaining. • (Some added complications in keeping track of the unifiers.) • (More complications help to avoid infinite loops.) • Two versions: (1) find any solution, (2) find all solutions. • Backward chaining is the basis for “logic programming,” e.g., Prolog.

Pig(y) Slug(z)  Faster (y, z) Slimy(a) Creeps(a)  Slug(a) Pig(Pat) Slimy(Steve) Creeps(Steve) Backward Chaining Example

Soundness & Completeness • An inference procedure is sound if the conclusion is true in all cases where the premises are true. • Easy to show the Generalized Modus Ponens (GMP) is sound. • Completeness: If KB entails , can GMP be used to infer ? PhD(x)  HighlyQualified(x)  PhD(x)  EarlyEarnings(x) HighlyQualified(x)  Rich(x) EarlyEarnings(x)  Rich(x) Should be able to infer Rich(Me) [or Rich(Kiki) for that matter], but can’t since  PhD(x)  EarlyEarnings(x)is not a Horn sentence. • Therefore, GMP is NOT complete!

Refutation Proofs • Given • a knowledge base KB (collection of true sentences), • a proposition P, we wish to prove that P is true. • Proof by contradiction (refutation): • Assume that P is FALSE (i.e., that P is TRUE). • Show that a contradiction arises. • A complete approach to refutation can be obtained using a single inference rule: resolution.

Resolution Inference Rule • Idea: If  is true or  is true and  is false or  is true then  or  must be true • Basic resolution rule from propositional logic:          • Can be expressed in terms of implications   ,       • Note that Resolution rule is a generalization of Modus Ponens ,    is equivalent to TRUE  ,     TRUE  

Generalized Resolution Generalized resolution rule for first order logic (with variables) If pj can be unified with qk, then we can apply the resolution rule: p1  …  pj …  pm q1  …  qk …  qn Subst(, (p1  …  pj-1  pj+1 …  pm q1  …  qk-1  qk+1 …  qn)) where  = Unify (pj, qk) • Example: KB:  Rich(x)  Unhappy(x) Rich(Me) Substitution:  = { x/Me } Conclusion: Unhappy(Me)

Canonical Form • For generalized Modus Ponens, entire knowledge base is represented as Horn Sentences. • For resolution, entire database will be represented using Conjunctive Normal Form (CNF) • Any first order logic sentence can be converted to a Canonical CNF form. • Note: Can also do resolution with implicative form, but let’s stick to CNF.

Converting any FOL to CNF • Literal = (possibly negated) atomic sentence, e.g.,  Rich(Me) • Clause = disjunction of literals, e.g.,  Rich(Me)  Unhappy(Me) • The KB is a conjunction of clauses • Any FOL sentence can be converted to CNF as follows: • Replace P  QbyP  Q • Move inwards to literals, e.g., x P becomes x P • Standardize variables, e.g., (x P)  (x Q) becomes (x P)  (y Q) • Move quantifiers left in order, e.g., x P  y Q becomes x y P  Q • Eliminate  by Skolemization (next slide) • Drop universal quantifiers • Distribute  over  , e.g., (P  Q)  R becomes (P  R)  (Q R) • Flatten nested conjunctions & disjunctions, e.g. (P  Q)  R  P  Q  R

Skolemization (Thoralf Skolem 1920) • The process of removing existential quantifiers by elimination. • Simple case: No universal quantifiers.  Existential Elimination Rule • For example: x Rich(x) becomes Rich(G1) where G1 is a new ``Skolem constant'‘. • More tricky when  is inside .

Skolemization – continued • More tricky when  is inside  E.g., ``Everyone has a heart'' x Person(x)   y Heart(y)  Has(x,y) • Incorrect: x Person(x)  Heart(H1)  Has(x,H1) This means everyone has the same heart calledH1. • Problem is that for each person, we need another “heart” – i.e., consider the “heart” to be a function of the person. • Correct:  Person(x)  Heart(H(x))  Has(x,H(x)) where H is a new symbol (``Skolem function'') • Skolem function arguments: all enclosing universally quantified variables.

Resolution proof p1  …  pj …  pm q1  …  qk …  qn Subst(, (p1  …  pj-1  pj+1 …  pm q1  …  qk-1  qk+1 …  qn)) • To prove : • Negate . • Convert to CNF. • Add to CNF KB. • Infer contradiction using the resolution rule (a contradiction is detected when resolution derives the empty clause). • E.g., to prove Rich(Me),add Rich(Me) to the CNF KB, then: PhD(x)  HighlyQualified(x) PhD(x)  EarlySalary(x) HighlyQualified(x)  Rich(x) EarlySalary(x)  Rich(x)

Resolution Proof  Rich(Me) PhD(x)  HighlyQualified(x) PhD(x)  EarlySalary(x) HighlyQualified(x)  Rich(x) EarlySalary(x)  Rich(x)

All birds fly. No bird swims Pete is a bird Does Pete Fly? The Knowledge base 1. Bird(x) Flies(x) 2. Bird(y)Swims(y) Bird(Pete) The query  Flies ( Pete) Applying Resolution 5. Swims(Pete) 2,3  Bird(Pete) 1,4 { } 3,6 True-or-False Question

Decidability How hard is it to determine if KB entails ? • Propositional logic (zeroth order) is decidable: • Can determine whether or not KB entails  in finite time. • Second order logic is undecidable: • Cannot determine whether KB entails  in finite time. • First order logic is semi-decidable: • If KB entails  or , then a proof will be found in finite time. • But, if KB neither entails  or , then proof process may never terminate.

Resolution • Resolution is sound. • Resolution refutation is complete. (See Section 9.5 for proof.) • Resolution (and any proof procedure) is at best semi-decidable. • Note: checking consistency of a database is also semidecidable.

Inference in First Order Logic