logic is everywhere Logik ist überall Hikmat har Jaga Hai Mantık her yerde la logica è dappertutto lógica está em toda parte Non-Monotonic Reasoning

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Logika ada di mana-mana Hikmat har Jaga Hai logika je svuda Mantık her yerde la logica è dappertutto

1 Non-Monotonic Reasoning 1 Introduction logika je všude 2 Closed World Assumption 3 Completion logic is everywhere 4 Circumscription la lógica está por todas partes 5 Default Logic 6 Answer Set Programming Logika ada di mana-mana Hikmat har Jaga Hai logika je svuda Mantık her yerde la logica è dappertutto Logik ist überall Logica este peste tot lógica está em toda parte la logique est partout Non-Monotonic Reasoning 1

2 Introduction The Missionaries and Cannibals Puzzle Three missionaries and three cannibals come to a river. A rowboat that seats two is available. If the cannibals ever outnumber the missionaries on either bank of the river, the missionaries will be eaten. How shall the missionaries and the cannibals cross the river? Representation M CB, where M denotes the number of missionaries on the left bank of the river, C denotes the number of cannibals on the left bank of the river, and B denotes the number of rowboats on the left bank of the river. Solution (331, 220, 321, 300, 311, 110, 221, 020, 031, 010, 021, 000) Can it be derived as a logical consequence of a first order formalization? Non-Monotonic Reasoning 2

3 Problems Unless it can be deduced that an object is present, we conjecture that it is not present. Unless there is something wrong with the boat or something else prevents the boat from using it, it can be used to cross the river. Non-Monotonic Reasoning 3

4 Non-Monotonic Logics A logic A, L, = is said to be non-monotonic iff there exist K, K and G such that K = G and K K = G, where K and K are sets of formulas in L and G is a formula in L. Propositional and first order logic are monotonic Exercise Non-Monotonic Reasoning 4

5 Closed World Assumption Open world assumption (OWS): The only answers given to a query are those that can be obtained from proofs of the query, given the knowledge base. Closed world assumption (CWS): certain additional answers are admitted as a result of a failure to prove a result. Example K = {lectures(steffen, cl001), lectures(steffen, cl005), lectures(michael, cl002), lectures(michael, cl005), lectures(heiko, cl004), lectures(horst, cl003)}. query OWS CWS K = ( X) lectures(steffen, X) yes yes K = lectures(michael, cl006) no yes Non-Monotonic Reasoning 5

6 The Formal Theory Let A, L, = be a first order logic. Let K L be a satisfiable set of formulas. C(K) = {G K = G} is the theory or closure of K. Let K = { A A is a ground atom in L and K = A}. C CWA (K) = C(K K) is the theory of K under the closed world assumption. Non-Monotonic Reasoning 6

7 Satisfiability Is C CWA (K) satisfiable? Consider K = {leakyvalve puncturedtube} K = leakyvalve K = puncturedtube { leakyvalve, puncturedtube} K K K {leakyvalve puncturedtube, leakyvalve, puncturedtube} K K is unsatisfiable! Theorem 11.2 Let K be a satisfiable set of formulas in Skolem normal form. C CWA (K) is satisfiable iff K admits a least Herbrand model. Non-Monotonic Reasoning 7

8 Models and the Closed World Assumption Let M = (D, I) and M = (D, I ) be two models of K. Let R be the set of relation symbols in L and P R. M is a submodel of M wrt P (M P M ) iff D = D and I, I are identical except that for all q P we find q I q I. If P = R then we write M M instead of M P M. A model M of K is minimal iff for all models M of K we find that M M implies M = M. M M iff M M and M M. A model M of K is the least model of K iff for all models M of K we find M M implies M M. The closed world assumption eliminates non-least models! Non-Monotonic Reasoning 8

9 Remarks K = A cannot be decided in first-order logic! There are several extensions of the closed world assumption. Non-Monotonic Reasoning 9

10 Completion Can we add more complex formulas than negative ground atoms to a knowledge base? Consider F = {tweedy/0, john/0} and R = {penguin/1}. Let K = {penguin(tweedy)}. M 1 = {penguin(tweedy)} and M 2 = {penguin(tweedy), penguin(john)} are models for K. M 1 M 2. How can the least model be computed? Another example: K = {penguin(tweedy), penguin(john)}. And another one: K = {( X) ( fly(x) fly(x))}. Non-Monotonic Reasoning 10

11 Solitary Clauses An occurrence of a predicate symbol p/n in a clause C is said to be positive iff we find terms t i, 1 i n, such that p(t 1,..., t n ) C, negative iff we find terms t i, 1 i n, such that p(t 1,..., t n ) C. A set K of clauses is said to be solitary wrt p/n iff for each clause C K we find that if C contains a positive occurrence of p/n then C does not contain another occurrence of p/n. The clause [ fly(tweedy), fly(john), penguin(tweedy), penguin(john)] is solitary in fly/1 and not solitary in penguin/1. Non-Monotonic Reasoning 11

12 The Completion Algorithm Input A set K of clauses and a predicate symbol p/m. Output The completion formula C K,p of K with respect to p. 1 Replace each clause {L 1,..., L n, p(t 1,..., t m )} K by ( X)(( Y ) (X 1 t 1... X m t m L 1... L n ) p(x)), where X = X 1,..., X m is a sequence of new variables, Y is a sequence of those variables which occur in the clause, and all occurrences of double negation are removed. 2 Let {( X) (C i p(x)) 1 i k} be the set of all formulas where p occurs in the conclusion. Return the completion formula C K,p = ( X) (C 1... C k p(x)). Non-Monotonic Reasoning 12

13 An Example K = { penguin(y ) bird(y ), bird(tweedy), penguin(john) } K 1 = { ( X)(( Y ) (X Y penguin(y )) bird(x)), ( X) (X tweedy bird(x)), penguin(john) } ( X) (( Y ) (X Y penguin(y )) X tweedy bird(x)) C K,bird = ( X) (( Y )(X Y penguin(y )) X tweedy bird(x)) Non-Monotonic Reasoning 13

14 The Equational System E C E C = {( X, Y ) f(x) g(y ) f/n, g/m F and f g} {( X) t[x] X t X} {( X, Y ) ( n i=1 X i Y i f(x) f(y )) f/n F} {( X) X X} {( X, Y ) ( n i=1 X i Y i f(x) f(y )) f/n F} { ( n i=1 X i Y i p(x) p(y )) p/n R} Non-Monotonic Reasoning 14

15 Predicate Completion Let K be a set of formulas, which is solitary in p. The predicate completion C C (K, p) of p is defined as C C (K, p) = {G K {C K,p } E C = G}. Theorem 11.3 Let K be a set of clauses which is solitary in p. If K is satisfiable, then so is C C (K, p). Non-Monotonic Reasoning 15

16 Parallel Completion and Logic Programming K = {bird(tweedy), ( X) (bird(x) abnormal(x) fly(x))} Normal program clauses: p(t) A 1... A m A m+1... A n. A normal logic program is a set of normal program clauses. p/n is defined in the logic program K iff K contains a clause with p/n occurring in its head. Let R D be the set of defined predicate symbols. The completion C C (K) of a normal logic program K with defined predicate symbols R D is defined as C C (K) = {G K {C K,p p R D } E C {( X) p(x) p R \ R D } = G}. Non-Monotonic Reasoning 16

17 Stratified Logic Programs Consider R, F and V. A level mapping is a total mapping l : R N. l(p) is the level of p. A normal logic program K is stratified iff in each clause of the form p(t) p 1 (s 1 )... p m (s m ) p m+1 (s m+1 )... p n (s n ) of K we find l(p) l(p i ), 1 i m, and l(p) > l(p j ), m < j n. Theorem 11.4 Let K be a stratified normal logic programs. Then C C (K) is satisfiable. Non-Monotonic Reasoning 17

18 Negation as Failure We do not want to compute with the only-if parts and E C! { A A C(K)} = { A A C C (K)} Replace /1 by /1 called negation as failure. K = {bird(tweedy), fly(x) bird(x) abnormal(x)} Non-Monotonic Reasoning 18

19 Finitely Failed Search Trees A search tree is finitely failed iff it is finite and each leaf is labelled as a failure. K = { abnormal(x) broken wing(x), abnormal(x) ratite(x), ratite(x) ostrich(x), ratite(x) emu(x), ratite(x) kiwi(x) } abnormal(tweedy) broken wing(tweedy) ratite(tweedy) kiwi(tweedy) emu(tweedy) ostrich(tweedy) Non-Monotonic Reasoning 19

20 SLDNF-Resolution Let G be a goal clause consisting of positive and negative literals, K a normal logic program, L be the selected literal in G and A be a ground atom. If L is positive, then each SLD-resolvent of G using L and some new variant of a clause in K is also an SLDNF-resolvent. If L is a ground negative literal, i.e. L = A, and the query A finitely fails with respect to K and SLDNF-resolution, then the SLDNF-resolvent of G is obtained from G by deleting L. If L is a ground negative literal, i.e. L = A, and the query A suceeds with respect to K and SLDNF-resolution, then the SLDNF-derivation of G fails. If L is negative and non-ground, then without loss of generality we may assume that each literal in G is negative and non-ground. In this case G is said to be blocked. Theorem 11.5 Let K be a normal logic program. SLDNF-resolution is sound with respect to the completion of K. Non-Monotonic Reasoning 20

21 Classical Negation vs. Negation as Failure cross train cross train Non-Monotonic Reasoning 21

22 Circumscription All approaches mentioned so far cannot handle p(a) p(b) or ( X) green(x). How can we compute their minimal models? We want to conjecture that the tuples (X 1,..., X m ) which can be shown to satisfy a relation p/m are all the tuples satisfying p/m. In other words, we want to circumscribe p/m. G{p Q} is the string obtained from the formula G by replacing each occurrence of p/m by the predicate variable Q/m. The circumscription of p in G: Circ(G, p) = (G{p Q} ( X) (Q(X) p(x))) ( X)(p(X) Q(X)) Circ(G, p) is a schema. Non-Monotonic Reasoning 22

23 Example 1 Let G = isblock(a) isblock(b) isblock(c). Then Circ(G, isblock) = ((Q(a) Q(b) Q(c)) ( X) (Q(X) isblock(x))) ( X) (isblock(x) Q(X)). This schema can be instantiated replacing Q(X) by (X a X b X c), where /2 denotes syntactic equality. Let G be the corresponding instance of Circ(G, isblock). Note Q(a) = a a a b a c Q(b) = b a b b b c Q(c) = c a c b c c Non-Monotonic Reasoning 23

24 Example 1 Continued Remember G = isblock(a) isblock(b) isblock(c). Hence G ( X) (X a X b X c isblock(x)) ( X) (isblock(x) X a X b X c) Let I be a Herbrand-interpretation with I = {G, G }. [( X) (X a X b X c isblock(x)] I = iff for all t T (F) we find [t a t b t c isblock(t)] I =. ( ) case t {a, b, c} [t a t b t c] I = and [isblock(t)] I =, thus, ( ) holds. case t {a, b, c} [t a t b t c] I =, thus, ( ) holds. Therefore, we obtain {G, G } = ( X) (isblock(x) (X a X b X c)). Non-Monotonic Reasoning 24

25 Non-monotonicity Circumscription is non-monotonic. Reconsider Example 1. {G, G } = isblock(d). Now let H = G isblock(d). Then {H, Circ(H, isblock)} = isblock(d). Non-Monotonic Reasoning 25

26 Example 2 Let G = p(a) p(b). Then Circ(G, p) = ((Q(a) Q(b)) ( X) (Q(X) p(x))) ( X) (p(x) Q(X)). This schema can be instantiated replacing Q(X) by X a, where /2 denotes syntactic equality. We obtain G 1 = ((a a b a) ( X) (X a p(x))) ( X) (p(x) X a) ( X) (X a p(x)) ( X) (p(x) X a) p(a) ( X) (p(x) X a). Non-Monotonic Reasoning 26

27 Example 2 Continued Remember G = p(a) p(b) and G 1 p(a) ( X) (p(x) X a). Circ(G, p) can also be instantiated replacing Q(X) by X b and we obtain G 2 p(b) ( X) (p(x) X b). Note Hence {H 1 H 2, H 1 H 1, H 2 H 2 } = H 1 H 2. {G, G 1, G 2 } = ( X) (p(x) X a) ( X) (p(x) X b). Non-Monotonic Reasoning 27

28 The Main Result Theorem 11.6 Let G be an instance of Circ(G, p). G holds in all models of G which are minimal in {p/m}. Proof Consider (G{p Q} ( X) (Q(X) p(x))) ( X)(p(X) Q(X)). Let I be a model for G which is minimal in {p/m}. Let p /m be a predicate symbol such that [G{p p } ( X) (p (X) p(x))] I =. To show [( X)(p(X) p (X))] I =. Suppose [( X)(p(X) p (X))] I =. Hence p I p I. Because [( X) (p (X) p(x))] I = we find p I p I. We conclude p I p I. Because [G{p p }] I = we can construct a model I for G which is identical to I for all predicate letters different from p and with p I = p I. Because p I = p I p I this contradicts the minimality of I in {p/m}. Non-Monotonic Reasoning 28

29 Remarks G follows minimally from K with respect to p/m, written K = {p} G, iff G holds in all models of K which are minimal in {p/m}. Corollary 11.7 Let G be an instance of Circ(G, p). If {G, G } = H then {G} = {p} H. Circumscribing a predicate may lead to an unsatisfiable theory. Under certain circumstances circumscription can be reduced to first order reasoning. Many extensions are known. Non-Monotonic Reasoning 29

30 Default Logic Most objects of sort s have property p. Object o is of sort s. Does object o have property p? Most birds are flying. Tweedy is a bird. Does Tweedy fly? A first order formalization: ( X) (bird(x) penguin(x) ostrich(x)... fly(x). Problems We do not know all exceptions. We cannot conlude that Tweedy does not belong to one of the exceptions. Idea We would like to conclude the Tweedy flies by default. Non-Monotonic Reasoning 30

31 Default Reasoning Unless any information to the contrary is known we assume that exceptions are not logical consequences (CWA), we finitely failed to prove exceptions (NAF), it is consistent to assume that... (Default Logic). Default rules bird(x) : fly(x) / fly(x). Exceptions { ( X) (penguin(x) fly(x)), ( X) (ostrich(x) fly(x)),... } But how is consistency defined? Few objects of sort s have property p: man(x) : moon(x) / moon(x). Non-Monotonic Reasoning 31

32 Default Rules Let A, L, = be a first order logic. A default rule is any expression of the form G : G 1,..., G n / H or G : G 1,..., G n. H G is called prerequisite, G 1,..., G n are called justifications, H is called consequent. A default rule is said to be closed iff all formulas occurring in it are closed. It is said to be open iff it is not closed. It is a scheme representing the set of its ground instances. Non-Monotonic Reasoning 32

33 Default Rules Special Cases If G is missing, then G. If n = 0, then this is a rule in A, L, =. If n = 1 and G 1 = H, then the default rule is said to be normal. If n = 1 and G 1 = H H, then the default rule is said to be semi-normal. Non-Monotonic Reasoning 33

34 Default Knowledge Bases A default knowledge base is a pair K D, K W, where K D is a set of at most countably many default rules and K W is a set of at most countably many closed first order formulas over A. A default knowledge base is said to be closed iff all default rules occurring in it are closed. It is said to be open iff it is not closed. Example K D : spouse(x, Y ) htown(y ) Z : htown(x) Z / htown(x) Z, employer(x, Y ) location(y ) Z : htown(x) Z / htown(x) Z. K W : spouse(jane, john), htown(john) munich, employer(jane, tud), location(tud) dresden, ( X, Y, Z) (htown(x) Y htown(x) Z Y Z). Non-Monotonic Reasoning 34

35 Extensions Let K be a set of closed first-order formulas and K D, K W a closed default knowledge base. Intuitively, an extension K of K D, K w should have the properties: K W K, C(K) = K, K should be closed under the application of default rules. Let Γ(K) be the smallest set satisfying the following properties: 1. K W Γ(K). 2. C(Γ(K)) = Γ(K). 3. If G : G 1,..., G n / H K D, G Γ(K) and for all 1 j n we find that G j K then H Γ(K). K is said to be an extension of K D, K W iff Γ(K) = K. The set of extensions of K D, K W is a subset of the set of models for K W. Non-Monotonic Reasoning 35

36 Another Characterization of Extensions Theorem 11.7 Let K D, K W be a closed default knowledge base and K be a set of sentences. Define and for i 1: K 0 = K W K i+1 = C(K i ) {H G : G 1,..., G n / H K D, G K i and for all 1 j n G j K}. Then, K is an extension of K D, K W iff K = i=0 K i. We have to guess extensions! K D = { bird(x) : fly(x) / fly(x) } K W = { bird(tweedy) } K = C({bird(tweedy), fly(tweedy)}) is an extension. Non-Monotonic Reasoning 36

37 Another Example K D : spouse(x, Y ) htown(y ) Z : htown(x) Z / htown(x) Z, employer(x, Y ) location(y ) Z : htown(x) Z / htown(x) Z. K W : spouse(jane, john), htown(john) munich, employer(jane, tud), location(tud) dresden, ( X, Y, Z) (htown(x) Y htown(x) Z Y Z). Its extensions are: and C({spouse(jane, john), htown(john) munich, employer(jane, tud), location(tud) dresden, htown(jane) munich}) C({spouse(jane, john), htown(john) munich, employer(jane, tud), location(tud) dresden, htown(jane) dresden}). Non-Monotonic Reasoning 37

38 Credolous vs. Sceptical Reasoning G follows credolously from K D, K W, in symbols K D, K W = c G, iff there exists an extension K of K D, K W such that G K. G follows sceptically from K D, K W, in symbols K D, K W = s G, iff for all extensions K of K D, K W we find G K. Non-Monotonic Reasoning 38

39 Remarks Default logic is non-monotonic. Extensions are always satisfiable. Extensions may contain counter-intuitive facts. K W = {broken(left arm) broken(right arm)} K D = { : usable(x) broken(x) / usable(x)} There are many approaches extending default logic. Non-Monotonic Reasoning 39

40 Answer Set Programming Example Every student with a GPA of at least 3.8 is eligible. Every minority student with a GPA of at least 3.6 is eligible. No student with a GPA under 3.6 is eligible. The students whose eligibility is not determined by these rules are interviewed by the scholarship committee. K 1 = { eligible(x) highgpa(x), eligible(x) minority(x) fairgpa(x), eligible(x) fairgpa(x), interview(x) eligible(x) eligible(x) } K 2 = { fairgpa(john), highgpa(john) }. What happens with John? Non-Monotonic Reasoning 40

41 Rules and Programs Rules L 1... L k L k+1... L l L l+1... L m L m+1... L n L i are propositional literals. 0 k l m n. If k = l = 0 then rules are called constraints. A program is a set of rules. K 1 K 2. Non-Monotonic Reasoning 41

42 Answer Sets Remember rules: L 1... L k L k+1... L l L l+1... L m L m+1... L n. Let M be a satisfiable set of literals and K be a program where k = l and n = m, i.e., rules are of the form L 1... L k L l+1... L m. M is said to be closed under K if for every rule of K we find that {L 1,..., L k } M = whenever {L l+1,..., L m } M. M is said to be an answer set for K if M is minimal among the sets closed under K. Example K 3 = { s r, b r }. {s} and {r, b} are answer sets. What happens if we add the constraint s? Non-Monotonic Reasoning 42

43 Reducts and Answer Sets Let K be a program and M a satisfiable set of literals. The reduct K M of K relative to M is the set of rules such that L 1... L k L l+1... L m L 1... L k L k+1... L l L l+1... L m L m+1... L n occurs in K, {L k+1,..., L l } M and {L m+1,..., L n } M =. K M is a program without /1. M is said to be an answer set for K iff M is an answer set for K M. Examples {p q}, { p p}, {p p} {q p q, p, q } What happens if we delete q from the last example? Non-Monotonic Reasoning 43

44 Predicate Symbols, Constants and Variables We allow n-ary predicate symbols ranging over constants and variables. We view rules containing variable occurrences as schemas. K 1 = { eligible(x) highgpa(x), eligible(x) minority(x) fairgpa(x), eligible(x) fairgpa(x), interview(x) eligible(x) eligible(x) } K 2 = { fairgpa(john), highgpa(john) }. Its only answer set is: {fairgpa(john), highgpa(john), interview(john)}. What happens if we add minority(john)? Answer set programming is non-monotonic! Non-Monotonic Reasoning 44

45 Programming with Answer Sets A Hamiltonian cycle is a cyclic tour through a graph visiting each vertex exactly once. The problem of finding a Hamiltonian cycle is known to be NP-complete. Let G be a graph with vertices 0,..., n. Consider an alphabet with F = {0/0,..., n/0} and R = {reachable/1, /2}. Idea WLOG let 0 be the starting vertex of the tour. reachable(i) represents the fact that vertex i is reachable from 0. (i, j) represents the fact that the edge from i to j is in the cycle. Specify a program such that for each answer set M we find: { u, v (u, v) M} is the set of edges in the Hamiltonian cycle. Non-Monotonic Reasoning 45

46 Computing Hamiltonian Cycles Program {(u, v) (u, v) u, v G} { (u, v) (u, w) u, v, u, w G and v w} { (v, u) (w, u) v, u, w, u G and v w} {reachable(u) (0, u) 0, u G} {reachable(v) reachable(u) (u, v) u, v G} { reachable(u) 0 u n} You have to show that the answer sets of this program correspond to Hamiltonian cycles! Non-Monotonic Reasoning 46

47 Answer Set Planning Program: on(b, L, 0) on(b, L, 0) move (B, L, T ) move (B, L, T ) on(b, L, T + 1) move (B, L, T ) on(b, L, T + 1) on(b, L, T ) on(b, L, T + 1) on(b, L, T ) on(b, L, T ) (L L ) move (B, L, T ) on(b, B, T ) (B B ) move (B, B, T ) move (B, L, T ) (B B ) move (B, L, T ) move (B, L, T ) move (B, L, T ) B, L = B, L, B, L = B, L, B, L = B, L on(b, B, T ) on(b, B, T ) (B B ) supported(b, T ) on(b, tab, T) supported(b, T ) on(b, B, T ) supported(b, T ) supported(b, T ) Non-Monotonic Reasoning 47

48 Computing Answer Sets Paradigm shift Logic and constraint programming answer substitution. Answer set programming model, i.e., answer set. Quite successful in recent years. Systems Smodels Dlv DeReS Non-Monotonic Reasoning 48

logic is everywhere Logik ist überall Hikmat har Jaga Hai Mantık her yerde la logica è dappertutto lógica está em toda parte

logic is everywhere Logik ist überall Hikmat har Jaga Hai Mantık her yerde la logica è dappertutto lógica está em toda parte Simple Connectionist Systems Steffen Hölldobler logika je všude International Center for Computational Logic Technische Universität Dresden Germany logic is everywhere Connectionist Networks la lógica