1. Intro to AI Fall 2002 © L. Joskowicz 1 Introduction to Artificial Intelligence LECTURE 11: Nonmonotonic Reasoning Motivation: beyond FOL + resolution Closed-world assumption Default rules and theories Ref: “Logical Foundations of AI”, Genesereth and Nilsson, Morgan Kauffman, 1987.

2. Intro to AI Fall 2002 © L. Joskowicz 2 Knowledge representation with FOL + resolution FOL + resolution have limitations in the kind of sentences and deductions we can make. –cannot express uncertainty –cannot make unsound but likely deductions –cannot revise conclusions in light of new knowledge –cannot make conclusions based on the entire state of the KB

3. Intro to AI Fall 2002 © L. Joskowicz 3 Motivation: incompleteness If we cannot prove P or ~ P from KB, what should we conclude? KB: neighbor(israel,jordan) neighbor(israel,egypt) neighbor(lebanon,syria) … Query: neighbor(israel,morroco) Cannot conclude anything unless there is an explicit statement (negation) in KB!

4. Intro to AI Fall 2002 © L. Joskowicz 4 Motivation: exceptions All rules have exceptions! For each, we must forsee all of them and explicitly state them: “all birds fly”  X bird(X) => flies(X) “except ostriches”  X bird(X) /\ ~ostrich(X) => flies(X) “except newborns”  X bird(X) /\ ~ostrich(X) /\ ~ newborn(X) => flies(X) We would like to conclude flies(X) from bird(X) unless something is abnormal with X  X bird(X) /\ ~Abnormal(X) => flies(X)

5. Intro to AI Fall 2002 © L. Joskowicz 5 Motivation: changes We assumed that all clauses in KB are true and remain true. What if we later discover that this is not the case? How do we revise conclusions already made?  X citizen(X) /\ income(X,Y) => pay_tax(X,Y) As the rules change, we need to revise all the intermediate conclusions! We would like to identify only those that indeed need revision

6. Intro to AI Fall 2002 © L. Joskowicz 6 Possible extensions Language: make it more expressive without loosing some of its computational properties Semantics: revise the concept of truth value Inferencing: design new inference rules to deal with exceptions, uncertainty, etc Minimal extensions to FOL+new inference rules!

7. Intro to AI Fall 2002 © L. Joskowicz 7 Nonmonotonic logics (NML) FOL + resolution is monotonic: if KB |= P then (KB U {Q}) |= P for all consistent KB and all Q obtained from KB by applying resolution. The number of statements known to be true is strictly increasing over time Non-monotonic: “jump” to conclusions that can later be withdrawn (defeasible conclusions)

8. Intro to AI Fall 2002 © L. Joskowicz 8 Consistency in NML A deduction rule R is consistent iff: KR |= R c KR U {c} |= Ø iff KR |= Ø T R (KB) = the set of all conclusions from KB using inference rule R (transitive closure). Note: R is applied in parallel!

9. Intro to AI Fall 2002 © L. Joskowicz 9 Nonmonotonic frameworks 1. Closed World Assumption if you cannot prove P or ~ P from KB, add ~ P to KB. 2. Default Rules new inference rules on how to augment KB 3. (Predicate completion -- Circumscription) compute a formula which says how KB should be satisfied 4. (Truth Maintenance Systems) methods to maintain consistency in a KB where statements are constantly added and deleted

10. Intro to AI Fall 2002 © L. Joskowicz 10 1. Closed-world assumption (CWA) if you cannot prove P or ~ P from KB, add ~ P to KB KB |= c and KB |= ~c then add ~c Idea: if you cannot prove P, assume it is false. This means you assume you know everything there is to know about the world (e.g. the world is closed). This is the semantics of databases and of PROLOG.

11. Intro to AI Fall 2002 © L. Joskowicz 11 Complete theories A theory T is a set of sentences closed under logical implication (like transitive closure) T is complete if either every ground literal in the language or its negation is in the theory KB: neighbor(israel,jordan) neighbor(israel,egypt)  X  Yneighbor(X,Y) neighbor(Y,X) T(KB) is not complete because neither neighbor(egypt,jordan) or ~neighbor(egypt,jordan) is in T(KB).

12. Intro to AI Fall 2002 © L. Joskowicz 12 Theory completion Given an incomplete KB, include the negation of a ground literal when the ground literal does not follow from KB KB: neighbor(israel,jordan) neighbor(israel,egypt)  X  Y neighbor(X,Y) neighbor(Y,X) The atom ~ neighbor(egypt,jordan) will be added Is this always consistent?NO!

13. Intro to AI Fall 2002 © L. Joskowicz 13 Completion inconsistency Completion can lead to inconsistent theories: KB: p(a) \/ p(b) neither KB |= ~ p(a), p(a) nor KB |= ~p(b), p(b) follow So we add ~ p(a) and ~ p(b) to KB: KB’: p(a) \/ p(b) ~ p(a) ~ p(b) KB’ is inconsistent! Modify the completion rule for consistency

14. Intro to AI Fall 2002 © L. Joskowicz 14 CWA theorem (1) Augment a consistent KB with a new set of sentences (beliefs), to obtain a new consistent set CWA(KB). Theorem: CWA(KB) is consistent iff for every disjunction p 1 \/ p 2 \/ …. \/ p n, that follows from KB, where p i is a positive-ground literal, there is at least one p i such that KB |= p i Eq: CWA(KB) is inconsistent iff there are positive ground literals p 1, … p n such that KB |= p 1 \/ p 2 \/ …. \/ p n but for all i, KB |  p i.

15. Intro to AI Fall 2002 © L. Joskowicz 15 CWA theorem (2) Intuition: add all of ~ p i except one, so no contradiction occurs! Proof: Let KB assumed be the set of all conclusions ~p derived with CWA rule: ~p is in KB assumed iff KB |= p and KB |= ~p CWA(KB) is inconsistent only if KB U KB assumed is. Then, there is a finite subset of KB assumed that contradicts KB. Let this subset be L = {~p 1,…,~p n }. Then KB |= p 1 \/ p 2 \/ …. \/ p n, the negation of L. Since each ~p i is in KB assumed by the definition of KB assumed KB |= p i contradiction!

16. Intro to AI Fall 2002 © L. Joskowicz 16 Consistent CWA rule Complete a KB by including all ground literals that do not contradict the theorem. Important: define the constant atoms first Ex1: KB: p(a) \/ p(b) is not consistent Ex2: KB:  X p(X) \/ q(X) p(a) q(b) for atoms a,b, the augmentation is ~q(a) and ~p(b) OK for atom c, the augmentation is inconsistent: NOT OK (p(X) \/ q(X)) |  p(c) or (p(X) \/ q(X)) |  q(c)

17. Intro to AI Fall 2002 © L. Joskowicz 17 CWA consistency for Horn clauses In general, testing for consistency to see what negated ground literals to add to KB can be expensive! Not so for Horn clauses: Theorem: CWA(KB) is always consistent when KB is a consistent set of Horn clauses. Follows from the fact that Horn clauses have a single positive literal. Variant: define CWA for a subset of clauses only.

18. Intro to AI Fall 2002 © L. Joskowicz 18 Restricted CWA Define the predicates on which CWA is applied KB:  X q(X) => p(X) q(a) p(b) \/ r(b) If we apply CWA to p(X), we will conclude only ~p(b), which keeps consistency (~r(b) cannot be inferred)

19. Intro to AI Fall 2002 © L. Joskowicz 19 Other assumptions Domain closure assumption: Limit the constant terms in the language to be those that can be named using constant and function symbols in KB. Strong assumption : allows replacing universal quantifiers by finite conjunctions and disjunctions.  X p(X) is ( X=a1 \/ X = a2…) /\ p(X) Unique names assumption: if ground terms cannot be proved equal, assume they can be assumed unequal. p(f1(a)) = p(f2(a)) where f1 and f2 are Skolem functions

20. Intro to AI Fall 2002 © L. Joskowicz 20 2. Default rules and theories Define a nonstandard, nonmonotonic set D of inference rules to augment the basic KB. The augmentation of KB with D, denoted E (KB,D) is the theory that contains the usual conclusions + those obtained by applying D on KB. The default rules in D are of the form: bird(X) : flies(X) flies(X) “if X is a bird, and it is consistent to believe that X can fly, then it can be believed that X can fly”.

21. Intro to AI Fall 2002 © L. Joskowicz 21 Default rules semantics  (X):  (X)  (X) If there is an instance x of X for which the ground instance  (x) follows from E (KB,D) and for which  (x) is consistent with E (KB,D), then include  (x) in E (KB,D). Default rules are useful to express beliefs that are usually but not necessarily true In general, E (KB,D) is not unique!

22. Intro to AI Fall 2002 © L. Joskowicz 22 Example 1 KB: bird(tweety)  X ostrich(X) => ~flies(X) D: bird(X): flies(X) flies(X) then flies(tweety) is in E (KB,D). If we add ostrich(tweety), then we cannot deduce flies(tweety) because it is not consistent with KB.

23. Intro to AI Fall 2002 © L. Joskowicz 23 Example 1 (continued) KB: feathers(tweety) D:bird(X) : flies(X) feathers(X): bird(X) flies(X)bird(X) Default proof that flies(tweety). If we add ostrich(tweety) ostrich(X) => ~flies(X) ostrich(X) => feathers(X) we cannot (as expected) prove that flies(tweety).

24. Intro to AI Fall 2002 © L. Joskowicz 24 Is the default rule: ~ p(X) ~ p(X) the same as CWA? No! KB: p \/ q D: : ~ p and : ~ q ~p ~q CWA(KB) is inconsistent E (KB,D) can be either {p \/ q, ~ p } or {p \/ q, ~ q } However, the union of both is inconsistent! Example 2

25. Intro to AI Fall 2002 © L. Joskowicz 25 Properties of default theories Default theories might have more than one augmentation (see previous example) There are default theories with no augmentations KB = { p(X )}, D is :p(X)/ ~ p(X) Every normal default theory (only D statements of the form  (X):  (X)/  (X) ) has an augmentation If D’  D are sets of normal rules then for any E (KB,D’) there is a E (KB,D) such that E (KB,D’)  E (KB,D). Normal default rules are semi-monotonic.

26. Intro to AI Fall 2002 © L. Joskowicz 26 Example 3: anomalities (1) Typically, drug dealers are adults Typically, adults are employed dealer(X): adult(X)adult(X):employed(X) adult(X)employed(X) dealer(joe) adult(joe)(from default rule 1) employed(joe)(from default rule 2) Question: how to fix this anomality?

27. Intro to AI Fall 2002 © L. Joskowicz 27 Example 3: anomalities (2) Exchange the second rule by: adult(X) : ~dealer(X) /\ employed(X) employed(X) but it is not in normal form! Consider instead the new rules: dealer(X): adult(X) adult(X) /\ ~dealer(X) :employed(X) adult(X) employed(X) adult(X): ~dealer(X) ~dealer(X)

28. Intro to AI Fall 2002 © L. Joskowicz 28 Default rules: observations The difference between CWA(KB) and E(KB,D) CWA: add ~p if consistent with KB Default: add ~p if consistent with E(KB,D) => the order matters!! Inference with normal defaults:  (X):  (X)  (X) Forward: check  (X) at the time of the application Backward: two passes. First, ignore consistency checks and them verify them on the resulting chain

29. Intro to AI Fall 2002 © L. Joskowicz 29 3. Predicate completion The only objects that satisfy a predicate are those who must do so KB: p(a) (a single predicate and atom)  X X= a => p(X)if Assuming there are no other atoms satisfying P:  X p(X) => X = a only if This is the completion of the if formula Def: The completion of P in KB is COMP(KB;p)  KB /\ (  X p(X) => X = a)   X p(X) X = a

30. Intro to AI Fall 2002 © L. Joskowicz 30 Predicate completion: example What about more than one atom? KB: p(a) p(b) COMP(KB;p)   X ((X=a) \/ (X=b)) p(X) In these examples, predicate completion works like a CWA with respect to a predicate Predicate completion is not defined for arbitrary formulas -- only for solitary clauses

31. Intro to AI Fall 2002 © L. Joskowicz 31 Solitary clauses Def: A set of clauses is solitary in p if each clause with a positive occurrence of p has at most one occurrence of p. KB: p(a) \/ q(a) p and q are solitary KB: p(a) \/ q(a) \/ ~p(b)p is not solitary

32. Intro to AI Fall 2002 © L. Joskowicz 32 Predicate completion (1) Rewrite all clauses solitary in p in the form  Y q 1 /\ q 2 /\... /\ q k => p(T) where q i are literals not containing p T is a tuple of terms T = (T 1,...., T n ) p(a,b)T = (a,b) p(c,d)T= (c,d) The q i and t may contain variables, i.e the tuple of variables Y.

33. Intro to AI Fall 2002 © L. Joskowicz 33 Predicate completion (2) The expression is equivalent to:  Y  X (X=T) /\ q 1 /\ q 2 /\... /\ q k ) => p(X) where X is a tuple of variables not occurring in T and (X = T) stands for (X 1 = t 1 /\.... /\ X n = t n ) Since Y appears only in the antecedent of the implication, this is equivalent to:  X (  Y (X=T) /\ q 1 /\ q 2 /\... /\ q k ) => p(X)) We re-write in normal form all such clauses  X E 1 => p(X)....  X E k => p(X)

34. Intro to AI Fall 2002 © L. Joskowicz 34 Predicate completion rule (3) where E i are existentially quantified conjunction of literals Grouping them together, we obtain:  X (E 1 \/… \/ E k ) => p(X) The completion rule is then  X p(X) => (E 1 \/…\/ E k ) COMP(KB;p)  KB /\  X p(X) (E 1 \/…\/ E k ) where E i are the antecedents of the normal forms of the clauses in KB

35. Intro to AI Fall 2002 © L. Joskowicz 35 Predicate completion example Given the KB: 1.  X ostrich(X) => bird(X) all ostriches are birds 2.bird(tweety) tweety is a bird 3. ~ostrich(sam) sam is not an ostrich Question: is sam a bird? bird is solitary in 1. Its completion is: 4.  X (X = tweety) \/ ostrich(X) bird(X) We can now prove ~bird(sam)

36. Intro to AI Fall 2002 © L. Joskowicz 36 More on nonmonotonic logics Many different formalisms to deal with inferences of this type –circumscription: extension of predicate completion –Truth Maintenance Systems To learn more, see advanced courses –deduction systems –advanced logics and AI courses