α-formulas β-formulas

Size: px

Start display at page:

Download "α-formulas β-formulas"

Tobias Carter
6 years ago
Views:

1 α-formulas Logic: Compendium TDDD88/ Andrzej Szalas IDA, University of Linköping October 25, 2017 Rule α α 1 α 2 ( ) A 1 A 1 ( ) A 1 A 2 A 1 A 2 ( ) (A 1 A 2 ) A 1 A 2 ( ) (A 1 A 2 ) A 1 A 2 ( ) A 1 A 2 A 1 A 2 A 2 A 1 Factorization Rule (fctr): remove redundant duplicates. E.g. p,q,p,r,r is simplified to p,q,r. Andrzej Szalas Slide 1 of 35 Andrzej Szalas Slide 3 of 35 Propositional calculus: tableaux β-formulas Semantic tableau for a formula A A semantic tableau T is a tree with each node labeled with a set of formulas, where T represents A in such a way that A is equivalent to the disjunction of formulas appearing in all leaves, assuming that sets of formulas labeling leaves are interpreted as conjunctions of their members. Rule β β 1 β 2 ( ) (B 1 B 2 ) B 1 B 2 ( ) B 1 B 2 B 1 B 2 ( ) B 1 B 2 B 1 B 2 ( ) (B 1 B 2 ) (B 1 B 2 ) (B 2 B 1 ) Andrzej Szalas Slide 2 of 35 Andrzej Szalas Slide 4 of 35

2 Literals and complementary formulas Construction of a semantic tableau for a formula C A literal is an atom (propositional variable) or the negation of an atom. Atoms are called positive literals and their negations are called negative literals. For any formula A, {A, A} is a complementary pair of formulas. A is the complement of A and A is the complement of A. Remark Observe that the conjunction of a complementary pair of formulas is equivalent to false. Initially, T consists of a single node labeled with {C}. If T is completed then no further construction is possible. Otherwise chose a leaf, say l, labeled with S containing a non-literal and chose from S a formula D which is not a literal and: if D is an α-formula then create a successor of l and label it with ( S {D} ) {α 1,α 2 } if D is a β-formula then create two new successors of l, the first one labeled with ( S {D} ) {β 1 } and the second one labeled with ( S {D} ) {β 2 }. Andrzej Szalas Slide 5 of 35 Andrzej Szalas Slide 7 of 35 Closed and open leaves Proving with semantic tableaux A leaf is called closed if it contains a complementary pair of formulas. If a leaf consists of literals only and contains no complementary pair of literals then we call it open. A tableau is completed if all its leaves are open or closed. To prove that a formula A is a tautology we construct a completed tableau for its negation A. If the tableau is closed then A is a tautology (its negation is not satisfiable). Otherwise A is not a tautology. Andrzej Szalas Slide 6 of 35 Andrzej Szalas Slide 8 of 35

3 NNF: Negation Normal Form CNF: Conjunctive Normal Form Recall that a literal is a formula of the form A or A, where A is an atomic formula. A literal of the form A is called positive and of the form A is called negative. We say that formula A is in negation normal form, abbreviated by Nnf, iff it contains no other connectives than,,, and the negation sign appears in literals only. A clause is any formula of the form A 1 A 2... A k, where k 0 and A 1,A 2,...,A k are literals. The empty clause (when k = 0), denoted by, is equivalent to F. A Horn clause is a clause in which at most one literal is positive. Formula A is in conjunctive normal form, abbreviated by Cnf, if it is a conjunction of clauses. It is in clausal form if it is a set of clauses (considered to be an implicit conjunction of clauses). Andrzej Szalas Slide 9 of 35 Andrzej Szalas Slide 11 of 35 Transforming formulas into NNF Transforming formulas into CNF Any propositional formula can be equivalently transformed into the Nnf by replacing subformulas according to the table below, until Nnf is obtained. Rule Subformula Replaced by 1 A B ( A B) (A B) 2 A B A B 3 A A 4 (A B) A B 5 (A B) A B Any propositional formula can be equivalently transformed into the Cnf: 1 Transform the formula into Nnf 2 Replace subformulas according to the table below, until Cnf is obtained. Rule Subformula Replaced by 6 (A B) C (A C) (B C) 7 C (A B) (C A) (C B) Andrzej Szalas Slide 10 of 35 Andrzej Szalas Slide 12 of 35

4 Resolution rule for propositional calculus Resolution method Resolution method has been introduced by Robinson (1965) and is considered the most powerful automated proving technique. Resolution rule, denoted by (res), is formulated as follows: α L, L β α β where L is a literal and α,β are clauses. The position of L and L in clauses does not matter. The method, where formula A is to be proved 1 transform A into the conjunctive normal form 2 try to derive the empty clause by applying resolution (res) and factorization (fctr): if the empty clause is obtained then A is a tautology, if the empty clause cannot be obtained no matter how (res) and (fctr) are applied, then conclude that A is not a tautology. Andrzej Szalas Slide 13 of 35 Andrzej Szalas Slide 15 of 35 Factorization rule for propositional calculus DeMorgan s laws for quantifiers Reminder Factorization rule, denoted by (fctr): Remove from a clause redundant repetitions of literals. Example P P Q P Q P Q DeMorgan laws for quantifiers are: [ x A(x)] [ x A(x)] [ x A(x)] [ x A(x)] 1 It is not the case that all animals are large is equivalent to Some animals are not large. 2 It is not the case that some animals are plants is equivalent to All animals are not plants. 3 x y z[friend(x,y) likes(y,z)] is equivalent to x y z [friend(x,y) likes(y,z)]. Andrzej Szalas Slide 14 of 35 Andrzej Szalas Slide 16 of 35

5 Other laws for quantifiers δ-formulas 1 Null quantification. Assume variable x is not free in formula P. Then: 1 x[p] is equivalent to P; 2 x[p] is equivalent to P; 3 x[p Q(x)] is equivalent to P x[q(x)]; 4 x[p Q(x)] is equivalent to P x[q(x)]. 2 Pushing quantifiers past connectives: 1 x[p(x) Q(x)] is equivalent to x[p(x)] x[q(x)]; 2 x[p(x) Q(x)] is equivalent to x[p(x)] x[q(x)] δ-formulas Rule δ δ(a) ( ) x A(x) A(a) ( ) x A(x) A(a) where a is a fresh constant. Andrzej Szalas Slide 17 of 35 Andrzej Szalas Slide 19 of 35 γ-formulas Tableaux Rule γ γ(a) ( ) x A(x) A(a) ( ) x A(x) A(a) where a is an arbitrary constant. Construction of a semantic tableau for formula A Initially T consists of a single unmarked node labeled with {A}. Until possible choose anon-closed leaf l labeled with U(l) and apply: if U(l) contains a pair of complementary formulas then mark the leaf closed; otherwise: Andrzej Szalas Slide 18 of 35 Andrzej Szalas Slide 20 of 35

6 Tableaux PNF: Prenex Normal Form Construction of a semantic tableau for formula A if U(l) does not contain a pair of complementary literals and contains non-literals, choose a non-literal A U(l) or a γ-formula from any node on the path from l to the root, and: if A is an α- or β-formula, apply rules provided in slide 7; if A is a γ-formula, create a child node l for l and label l with U(l ) = (U(l) {A}) {γ(a)}, where a preferably is a constant appearing in U(l); if A is a δ-formula, create a child node l for l and label l with U(l ) = (U(l) {A}) {δ(a)}, where a is a constant not appearing in U(l). We say A is in prenex normal form, abbreviated by Pnf, if all its quantifiers (if any) are in its prefix, i.e., it has the form: Q 1 x 1 Q 2 x 2... Q n x n [A(x 1,x 2,...,x n )], where n 0, Q 1,Q 2,...,Q n are quantifiers (, ) and A is quantifier-free. 1 A(x) B(x,y) C(y) is in Pnf 2 x y [A(x) B(x,y) C(y)] is in Pnf 3 A(x) x [B(x,y) C(y)] as well as x [A(x) B(x,y) yc(y)] are not in Pnf. Andrzej Szalas Slide 21 of 35 Andrzej Szalas Slide 23 of 35 Transforming formulas into NNF Transforming formulas into PNF NNF: Negation Normal Form Recall (see Slide 9) that in negation normal form, Nnf, negation ca appear only before atoms. Any first-order formula can be equivalently transformed into the Nnf by replacing subformulas according to the table given in Slide 10 and rules given below. Rule Subformula Replaced by 6 x A(x) x[ A(x)] 7 x A(x) x[ A(x)] Any predicate formula can be equivalently transformed into the Pnf 1 Transform the formula into Nnf; 2 Replace subformulas according to the table below, until Pnf is obtained, where Q denotes any quantifier, or. Rule Subformula Replaced by 12 Qx[A(x)], A(x) without variable z Qz[A(z)] 13 x A(x) x B(x) x [A(x) B(x)] 14 x A(x) x B(x) x [A(x) B(x)] 15 A Qx B, where A contains no x Qx (A B) 16 A Qx B, where A contains no x Qx (A B) Andrzej Szalas Slide 22 of 35 Andrzej Szalas Slide 24 of 35

7 Skolem normal form The unification algorithm A formula is in the Skolem normal form iff it is in the Pnf and contains no existential quantifiers. Transforming formulas into the Skolem normal form Eliminate existential quantifiers from left to right: 1 when we have xa(x) then remove x and replace x in A by a new constant symbol 2 when we have x 1... x k xa(x 1,...,x k,x) then remove x and replace x in A by a term f(x 1,...,x k ), where f is a new function symbol. Andrzej Szalas Slide 25 of 35 Input: expressions e,e Output: substitution of variables which makes e and e identical or inform that such a substitution does not exist. 1 traverse trees corresponding to expressions e,e ; 2 if the trees are identical then stop; 3 let t and t be subtrees that have to be identical, but are not: if t and t are function symbols/constants then conclude that the substitutions do not exist and stop; otherwise t or t is a variable; let t be a variable, then substitute all occurrences of t by the expression represented by t assuming that t does not occur in t (if it occurs then conclude that the substitutions do not exist and stop); 4 change the trees, according to the substitution determined in the previous step and repeat from step 2. Andrzej Szalas Slide 27 of 35 Unification Resolution rule for the predicate calculus Given two expressions, unification depends on substituting variables by expressions so that both input expressions become identical. If this is possible, the given expressions are said to be unifiable. 1 To unify expressions father(x) and father(mother(john)), it suffices to substitute x by expression mother(john). 2 To unify expressions (x +f(y)) and ( 2 z +f(3)), it suffices to substitute x by 2 z and y by 3. 3 Expressions father(x) and mother(father(john)) cannot be unified. Resolution rule, denoted by (res), is formulated for first-order clauses as follows: L 1 ( t 1 )... L k 1 ( t k 1 ) L k ( t k ) L k ( t k ) M 1( s 1 )... M l ( s l ) L 1 ( t 1 )... L k 1 ( t k 1 ) M 1( s 1)... M l ( s l ), where: L 1,...,L k,m 1,...,M l are literals; t k and t k are unifiable; primed expressions are obtained from non-primed expressions by applying substitutions unifying t k and t k. Andrzej Szalas Slide 26 of 35 Andrzej Szalas Slide 28 of 35

8 Factorization rule for the predicate calculus Atomic queries about facts The rule Factorization rule, denoted by (fctr): Unify some terms in a clause and remove from the clause all repetitions of literals. 1 (fctr) with x = Jack, y = mother(eve): parent(x, y) parent(jack, mother(eve)) parent(jack, mother(eve)) 2 (fctr) with z = x, u = y: P(x,y) S(y,z,u) P(z,u) P(x,y) S(y,x,y) Atomic queries are of the form name(arg 1,...,arg k ), where arg 1,...,arg k are constants or variables, where is also a variable (without a name, do-not-care variable). 1 likes(john, Marc) does John like Marc? 2 likes(x,marc) who likes Marc? (compute X s satisfying likes(x, Marc)) 3 likes(john,x) whom likes John? 4 likes(x,y) compute all pairs X,Y such that likes(x,y) holds. Andrzej Szalas Slide 29 of 35 Andrzej Szalas Slide 31 of 35 Datalog Rules in Datalog Facts in Datalog Facts in Datalog are represented in the form of relations name(arg 1,...,arg k ), where name is a name of a relation and arg 1,...,arg k are constants. 1 address(john, Kungsgatan 12 ) 2 likes(john, Marc). Rules in Datalog are expressed in the form of (a syntactic variant of) Horn clauses: R( Z): R 1 ( Z 1 ),...,R k ( Z k ) where Z, Z 1,..., Z k are vectors of variable or constant symbols such that any variable appearing on the lefthand side of : (called the head of the rule) appears also at the righthand side of the rule (called the body of the rule). Andrzej Szalas Slide 30 of 35 Andrzej Szalas Slide 32 of 35

9 Rules in Datalog Designing Datalog databases Semantics The intended meaning of the rule is that [R 1 ( Z 1 )... R k ( Z k )] R( Z), where all variables that appear both in the rule s head and body are universally quantified, while those appearing only in the rule s body, are existentially quantified. Example Rule: R(X,c): Q(X,Z),S(Z,X) denotes implication: X { Z [Q(X,Z) S(Z,X)] R(X,c)}. Remark One also often uses the following tautology: [(b h1) (b h2)] [b (h1 h2)]. It allows to substitute: h1 h2: b by two rules: h1: b. h2: b. Andrzej Szalas Slide 33 of 35 Andrzej Szalas Slide 35 of 35 Designing Datalog databases Index Remark To define disjunction in rule s body, one uses the following rules: h: b1. h: b2. These rules are equivalent to (b1 h) (b2 h) which, in turn, is equivalent to: (b1 b2) h. Andrzej Szalas Slide 34 of 35 α-formula, 3 β-formula, 4 δ-formula, 19 γ-formula, 18 atomic query, 31 body of a rule, 32 clausal form, 11 clause, 11 empty, 11 Horn, 11 closed leaf, 6 CNF, 11 complement of a formula, 5 complementary formulas, 5 completed tableau, 6 conjunctive normal form, 11 Datalog, 30 DeMorgan laws for quantifiers, 16 empty clause, 11 36

10 fact in Datalog, 30 factorization, 3, 14 rule, 14, 29 head of a rule, 32 Horn, 11 clause, 11 leaf closed, 6 open, 6 literal, 5, 9 negative, 5, 9 positive, 5, 9 negation normal form, 9, 22 negative literal, 5 negative literal, 9 NNF, 9, 22 open leaf, 6 PNF, 23 positive literal, 5 positive literal, 9 prenex normal form, 23 resolution, 13 method, 15 rule, 13, Robinson, 13 rule in Datalog, 32 semantic tableau, 2 Skolem normal form, 25 tableau completed, 6 unifiable expressions, 26 unification, 26 algorithm, 27 38

Predicate Calculus - Semantic Tableau (2/2) Moonzoo Kim CS Division of EECS Dept. KAIST

Predicate Calculus - Semantic Tableau (2/2) Moonzoo Kim CS Division of EECS Dept. KAIST moonzoo@cs.kaist.ac.kr http://pswlab.kaist.ac.kr/courses/cs402-07 1 Formal construction is explained in two steps