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Knowledge Representation and Reasoning University "Politehnica" of Bucharest Department of Computer Science Fall 2011 Adina Magda Florea http://turing.cs.pub.ro/krr_11 curs.cs.pub.ro Master of Science in Artificial Intelligence, 2011-2013

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Lecture 2 Lecture outline Predicate logic Herbrand theorem Powerful inference rules

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1. Predicate logic

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PrL, FOPL Extend PrL, PL Sentential logic of beliefs Uses beliefs atoms B A ( ) Index PL with agents Modal logic Modal operators Logics of knowledge and belief Modal operators B and K Dynamic logic Modal operators for actions Temporal logic Modal operators for time Linear time Branching time CTL logic Branching time and action BDI logic Adds agents, B, D, I Linear model Structured models Situation calculus Adds states, actions Symbol level Knowledge level Description Logics Subsumption relationships

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1. Predicate logic - Syntax variables function symbols terms predicate symbols atoms Boolean connectives quantifiers The function symbols and predicate symbols, each of given arity, comprise a signature . A ground term is a term without any variables 13

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Predicate logic - Semantics Universe (aka Domain) : Set U Variables – values in U Function symbols – (total) functions over U Predicate symbols – relations over U Terms – values in U Formulas – Boolean truth values Interpretation of a formula 14

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Algorithmic problems F – a well formed formula, A – an interpretation Validity(F): |= F ? (is F true in every interpretation?) Satisfiability (F): F satisfiable? Entailment (F, G): F |= G? (does F entail G?) Model(A,F): A |=F ? Solve (A,F): find an assignment such that A, |=F Solve (F): find a substitution such that |=F Abduce(F): find G with certain properties: such that G |= F 15

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How can we do it? Suppose we want to prove H |= G (entailment). Equivalently, we can prove that F := H → G is valid. Equivalently, we can prove that ~F, i.e., H ~G is unsatisfiable. This principle of “refutation theorem proving” is the basis of almost all automated theorem proving methods. 16

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How can we do it? Use normal forms Find a way to prove refutation – Herbrand model and theorems 16

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Normal forms Study of normal forms is motivated by: reduction of logical concepts efficient data structures for theorem proving. The main problem in first-order logic is the treatment of quantifiers. The normal form transformations are intended to eliminate many of them. 17

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Normal forms Prenex normal form CNF Eliminate existential quantifiers and conjunctions => Normal form matrix prefix 18

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Herbrand universe Herbrand universe Herbrand base Ground instances of a clause (set of clauses) S belongs to H = 19

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Examples 20

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Herbrand interpretation H-interpretation 21

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Herbrand interpretation H-interpretation I* correspinding to an interpretation I for a set of clauses S 22

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Herbrand theorem Lemma. If an interpretation I over a domain D satisfies the set of clauses S (i.e., the set S is satisfiable under that interpretation), then any H-interpretation I* corresponding to I also satisfies S Theorem. A set of clauses S is inconsistent iff S is false for any H- interpretation of S 23

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Semantic Trees Semantic trees Complete semantic trees Herbrand base of S is 24

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Semantic Trees Herbrand base of S is Complete semantic tree 25

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Closed semantic trees 26

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Closed semantic trees Failure nodes Inference nodes 27

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Review of definitions Basic instance of a clause C in S = an instance C' of a clause C from S obtained by replacing the variables it contains with elements of the Herbrand universe Complete semantic tree (CST) = for any leaf of the tree, I(N) contains either Ai or ~Ai (i.e., any path from the root to a leaf contains all elements of the Herbrand base either in positive or negative form) Failure node in a ST = I(N) falsifies a basic instance of a clause in S and I(N') does not falsify any instance of a clause in S, with N' the immediate predecesor of N Closed semantic tree = any branch has (ends in) a failure nod Inference node in a ST = all the node's successors are failure nodes 28

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Herbrand's theorem Idea: to test if a set S of clauses is unsatisfiable we have to test if S is unsatisfiable only for H-interpretations (interpretations over the Herbrand universe) First version of HT A set of clauses S is unsatisfiable iff for any semantic tree of S there exists a finite closed semantic tree (any complete semantic tree of S is a closed semantic tree) 29

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Herbrand's theorem Second version of HT A set of clauses S is unsatisfiable iff there exists a finite set S' of ground instances of S which is unsatisfiable (the Herbrand base of S is unsatisfiable) 30

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Powerful inference rules - Resolution Resolution Binary resolution Factorization General resolution Paramodulation 31

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Resolution Theorem. Resolution is sound. That is, all derived formulas are entailed by the given ones Theorem: Resolution is refutationally complete. That is, if a clause set is unsatisfiable, then Resolution will derive the empty clause eventually. If a clause set is unsatisfiable and closed under the application of resolution inference rule then it contains the empty clause. 35

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Completeness of Resolution complete semantic tree closed semantic tree inference node failure node 36

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Completeness of Resolution closed semantic tree inference node failure node inference node 37

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Completeness of Resolution Lifting lemma If C1' si C2' are basic instances of C1 and C2 and then there is a clause C such that C' is an instance of C Theorem: Resolution is refutationally complete. A set of clauses S is unsatisfiable iff there is a deduction of the empty clause from S. 38

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Factorization Resolution si not complete without factorization 32

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Factorization – Russell's antinomy A barber shaves men if and only if they do not shave themselves. Should the barber shave himself or not? (A1) ~Shaves(x,x) Shaves(barber,x) (A2) Shaves(barber,y) ~Shaves (y,y) (C1) Shaves(x,x) Shaves (barber,x) (C2) ~Shaves (barber,y) ~Shaves (y,y) (Res1) ~Shaves (barber,x) Shaves (barber,x) (Res2) Shaves(barber,barber) ~Shaves (barber,barber) (FC1): Shaves (barber,barber) (FC2): ~Shaves (barber,barber) See also http://en.wikipedia.org/wiki/Russell%27s_paradox 32

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Prover9 and Mace4 Prover9 is an automated theorem prover for first-order and equational logic Mace4 searches for finite models and counterexamples Prover9 is the successor of the Otter prover. http://www.cs.unm.edu/~mccune/prover9/ http://www.cs.unm.edu/~mccune/prover9/ 33

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Factorization – Russell's antinomy Prover 9 -Shaves(x,x) -> Shaves(barber,x). Shaves(barber,y) -> - Shaves (y,y). 1 -Shaves(x,x) -> Shaves(barber,x) # label(non_clause). [assumption]. 2 Shaves(barber,x) -> -Shaves(x,x) # label(non_clause). [assumption]. 3 Shaves(x,x) | Shaves(barber,x). [clausify(1)]. 4 -Shaves(barber,x) | -Shaves(x,x). [clausify(2)]. 5 Shaves(barber,barber). [factor(3,a,b)]. 6 $F. [factor(4,a,b),unit_del(a,5)]. 34

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Resolution strategies Level resolution Unit resolution Set of support Linear resolution Linear input resolution Hiper-resolution Ordered resolution Framed literal resolution 39

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The Boolean Satisfiability Problem Given: A Boolean formula Can F evaluate to 1 (true)? – Is F satisfiable? – If yes, return values to xi’s (satisfying assignment) that make F true 3-SAT – S.A. Cook 1971 NP-complete problem 2-SAT – polynomial time Horn-SAT 39

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The Boolean Satisfiability Problem Why is SAT important? Theoretical importance: – First NP-complete problem (Cook, 1971) Many practical applications: – Model Checking – Automatic Test Pattern Generation – Combinational Equivalence Checking – Planning in AI – Automated Theorem Proving – Software Verification – … 39

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Powerful inference rule: Paramodulation C1: P(a) C2: a=b If C1 contains a term t and there is a unity clause C2: t=s then we can infer a new clause from C1 by the substitution of a single occurrence of t in C1 with s. Paramodulation is a generalisation of that rule 39

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Paramodulation Be C1 and C2 two clauses, which have no variables in common. If C1: L[t] C1' C2: r = s C2' where L[t] is a literal containing t, C1' and C2' are clauses, and = mgu(t,r), then we can infer by paramodulation L [s ] C1' C2' where L [s ] is obtained by replacing only one single occurrence of t in L with s . Binary paramodulant 40

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Paramodulation Paramodulation with factorization – general paramodulation Paramodulation with resolution is sound and refutationally complete 41

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Example Group axioms % Associativity (x * (y * z)) = ((x * y) * z). % Identity element ((x * e) = x) & ((e * x) = x). % Inverse element ((x * i(x)) = e) & ((i(x) * x)=e). Prove % Right regular element ((f1 * f2) = (f0 * f2)) -> (f1 = f0). 42

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1 x * e = x & e * x = x [assumption]. 2 x * i(x) = e & i(x) * x = e [assumption]. 3 f1 * f2 = f0 * f2 -> f1 = f0 [goal]. 4 x * (y * z) = (x * y) * z. [assumption]. 5 (x * y) * z = x * (y * z). [copy(4),flip(a)]. 6 x * e = x. [clausify(1)]. 7 e * x = x. [clausify(1)]. 8 x * i(x) = e. [clausify(2)]. 9 i(x) * x = e. [clausify(2)]. 10 f0 * f2 = f1 * f2. [deny(3)]. 11 f1 * f2 = f0 * f2. [copy(10),flip(a)]. 12 f0 != f1. [deny(3)]. 13 f1 != f0. [copy(12),flip(a)]. 14 x * (i(x) * y) = y. [para(8(a,1),5(a,1,1)),rewrite([7(2)]),flip(a)]. e * z = x * (i(x) * z) [8,5] z = x * (i(x) * z) [7]y = x * (i(x) * y) x * (i(x) * y) = y [flip]

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15 x * (y * i(x * y)) = e. [para(8(a,1),5(a,1)),flip(a)]. 17 i(x) * (x * y) = y. [para(9(a,1),5(a,1,1)),rewrite([7(2)]),flip(a)]. 22 i(f1) * (f0 * f2) = f2. [para(11(a,1),17(a,1,2))]. 27 i(f1) * f0 = e. [para(22(a,1),15(a,1,2,2,1)),rewrite([5(8),8(7),6(5)])]. 29 f1 = f0. [para(27(a,1),14(a,1,2)),rewrite([6(3)])]. 30 $F. [resolve(29,a,13,a)].

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SAT Solvers Davis-Putnam-Logemann-Loveland (DPLL) algorithm was one of the first SAT search algorithms (Davis and Putnam 1960; Davis, Logemman and Loveland 1962) P & ~Q & (~P Q R) & (P ~S) ~Q & (Q R) R (P ← true, Q ← false, R ← true, S ← false) proving that the original formula is satisfiable. 39

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SAT Solvers A formula may cause the algorithm to branch the search through a branch reaches a dead end the moment a clause is deemed false—a conflicting clause—and all variations of the assignment that has been partially constructed up to this point can be discarded. 39

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SAT Solvers 39 1 R & (P Q) & (~P Q) & (~P ~Q) Given 2 (P Q) & (~P Q) & (~P ~Q) By letting R ← true 3Q & ~QBy letting P ← true 4?Conflict: Q and ~Q cannot both be true 5 (P Q) & (~P Q) & (~P ~Q) Backtrack to (2): R ← true still holds 6~P~PBy letting Q ← true 7trueBy letting ~P be true, i.e., P ← false The formula is satisfiable by the existence of (P ← false, Q ← true, R ← true).

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