Propositional Logic: Contents

Transcription

1 Propositional Logic

2 Propositional Logic: Contents Syntax and Semantics of Propositional Logic Satisfiability (SAT) Tableau Algorithm for SAT Structural induction Semantic consequence and logical equivalence Conjunctive and disjunctive normal form (CNF and DNF) Logic in Computer Science 2

3 Formulas of propositional Logic The alphabet of propositional logic consists of an infinite set p 1, p 2,... of atomic formulas; the logical connectives: ( not ), called negation; ( and ), called conjunction; ( or ), called disjunction; brackets: ( and ). Remarks: atomic formulas are also called propositional variables; we use letters p, q, r and indexed letters q 1, q 2,... to denote atomic formulas. Logic in Computer Science 3

4 Formulas of propositional logic The set P of all formulas of propositional logic is defined inductively: all atomic formulas are formulas; if P is a formula, then P is a formula; if P and Q are formulas, then (P Q) is a formula; if P and Q are formulas, then (P Q) is a formula; Nothing else is a formula. Remarks: So, formulas are just strings over a certain alphabet without truth values or meaning. We use P, Q, R and indexed letters such as P 1, P 2,..., Q 1, Q 2,... to denote formulas of propositional logic. Logic in Computer Science 4

5 Truth Values An interpretation I is a function which assigns to any atomic formula p i a truth value I(p i ) {0, 1}. If I(p i ) = 1, then p i is called true under the interpretation I. If I(p i ) = 0, then p i is called false under the interpretation I. Given an assignment I we can compute the truth value of compound formulas step by step using so-called truth tables. Logic in Computer Science 5

6 Truth tables: negation The negation P of a formula P is true when P is false and false otherwise: Definition Suppose an interpretation I is given and we know the value I(P ). Then the value I( P ) is computed by I( P ) = { 0 if I(P ) = 1 1 if I(P ) = 0 Corresponding truth table: P P Logic in Computer Science 6

7 Truth tables: conjunction The conjunction (P Q) is true if and only if both P and Q are true. Definition Suppose an interpretation I is given and we know I(P ) and I(Q). Then I(P Q) = { 1 if I(P ) = 1 and I(Q) = 1 0 if I(P ) = 0 or I(Q) = 0 Corresponding truth table: P Q (P Q) Logic in Computer Science 7

8 Truth tables: disjunction The disjunction (P Q) is true if and only if P is true or Q is true. Definition Suppose an interpretation I is given and we know I(P ) and I(Q). Then I(P Q) = { 1 if I(P ) = 1 or I(Q) = 1 0 if I(P ) = 0 and I(Q) = 0 Corresponding truth table: P Q (P Q) Logic in Computer Science 8

9 Truth under an interpretation So, given an interpretation I, we can compute the truth value I(P ) of any formula P under I. If I(P ) = 1, then P is called true under the interpretation I. If I(P ) = 0, then P is called false under the interpretation I. Logic in Computer Science 9

10 Example List the Interpretations I such that P = ((p 1 p 2 ) p 3 ) is true under I. P is true under I 1, I 2, and I 3, where I 1 (p 1 ) = I 1 (p 2 ) = I 1 (p 3 ) = 1, I 2 (p 1 ) = I 2 (p 3 ) = 1 and I 2 (p 2 ) = 0, I 3 (p 1 ) = I 3 (p 2 ) = 0 and I 3 (p 3 ) = 1. p 1 p 2 p 3 p 2 (p 1 p 2 ) P Logic in Computer Science 10

11 Truth table for ( P Q) P Q P ( P Q) ( P Q) represents the assertion if P is true, then Q is true. Define a new connective by: (P Q) = ( P Q). In what follows we use (P Q) as an abbreviation for ( P Q). Logic in Computer Science 11

12 Truth table for ((P Q) (Q P )) P Q (P Q) (Q P ) ((P Q) (Q P )) ((P Q) (Q P )) represents the assertion P is true if and only if Q is true. Define a new connective by: (P Q) := ((P Q) (Q P )). In what follows we use (P Q) as an abbreviation for ((P Q) (Q P )). Logic in Computer Science 12

13 Satisfiability Definition A formula P is satisfiable if and only if there exists an interpretation I such that I(P ) = 1. Examples Every atomic formula p is satisfiable: given p, take the interpretation I with I(p) = 1. p is satisfiable: take the interpretation I with I(p) = 0. Then I( p) = 1. (p p) is not satisfiable: for any interpretation I, I(p p) = 0. (p q) is satisfiable: take the interpretation I with I(p) = 1 and I(q) = 0. Logic in Computer Science 13

14 Satisfiability and Puzzles (1) Isaac and Albert were excitedly describing the result of the Third Annual International Science Fair Extravaganza in Sweden. There were three contestants, Louis, Rene, and Johannes. Isaac reported that Louis won the fair, while Rene came in second. Albert, on the other hand, reported that Johannes won the fair, while Louis came in second. In fact, neither Isaac nor Albert had given a correct report of the results of the science fair. Each of them had given one true statement and one false statement. What was the actual placing of the three contestants? (Credits: based on slides by Andrei Voronkov) Logic in Computer Science 14

15 Encoding into SAT We take atomic formulas L1, L2, L3, R1, R2, R3, J1, J2, J3 with the intuitive meaning: L1: Louis came in first, L2: Louis came in second, L3: Louis came in third. R1: Rene came in first, R2: Rene came in second, R3: Rene came in third. J1: Johannes came in first, J2: Johannes came in second, J3: Johannes came in third. We represent the information about Isaac s report using the formula J: J = ((L1 R2) ( L1 R2)) We represent the information about Albert s report using the formula A: A = ((J1 L2) ( J1 L2)) Logic in Computer Science 15

16 Encoding into SAT We have to encode additional information. Namely, everybody comes in at exactly one place: represent this using (P 1 P 2 ), where P 1 = ((L1 L2 L3) (R1 R2 R3) (J1 J2 J3)) and P 2 = ( (L1 L2) (L1 L3) (L2 L3) (R1 R2) ) Only one person can come in first, etc: represent this using Q, where Q = ( (L1 R1) (L2 R2) (L3 R3) (R1 J1) ) Any interpretation I with I(J A P 1 P 2 Q) = 1 corresponds to a possible placing of the three contestants. Logic in Computer Science 16

17 Note on Conjunctions and Disjunctions On the previous slide, we have used formulas (P 1 P n ) and (P 1 P n ) which we have not defined yet. Note that according to the syntax of propositional formulas we use brackets whenever we form the conjunction/disjunction of two formulas. (P 1 P n ) is defined by induction over n as follows: (P 1 ) = P 1 ; (P 1 P n+1 ) = ((P 1 P n ) P n+1 ); (P 1 P n ) is defined by induction over n as follows: (P 1 ) = P 1 ; (P 1 P n+1 ) = ((P 1 P n ) P n+1 ); Logic in Computer Science 17

18 Checking Satisfiability (SAT) We want an algorithm that checks whether a given propositional formula is satisfiable. In other words, for a given P, we search for an interpretation I such that I(P ) = 1. If this search is successful, then the output of the algorithm should be yes, P is satisfiable. If no such interpretation can be found, then the output of the algorithm should be no, P is not satisfiable. Logic in Computer Science 18

19 SAT applications SAT has numerous applications in computer and information science. Here are some: Circuit design: e.g., when are two circuits equivalent? Model checking: does a program represented as a graph structure satisfy its specification? Planning in artificial intelligence; Haplotyping in bioinformatics: derive haplotype data from genotype data. Logic in Computer Science 19

20 Satisfiability checking based on Truth Tables Here is an algorithm checking satisfiability that is directly based on truth tables: 1. Let P be the input formula; 2. Using truth tables, compute the value I(P ) for all interpretations I; 3. if an I is found such that I(P ) = 1, then output P is satisfiable ; 4. If no such I is found, output P is not satisfiable. If P is not satisfiable, then this algorithms requires the computation of I(P ) for 2 n many interpretations I, where n is the number of atomic formulas in P. Thus the running time of this algorithm is exponential. Major open problem in computer science: does there exist an algorithm checking satisfiability that runs in polynomial time? This problem is also known as the P=NP problem. Logic in Computer Science 20

21 Tableau Method Intuition: to check satisfiability of P, we apply tableau rules to P that make explicit the constraints that P imposes on formulas occuring in P (subformulas). If all sequences of rule applications lead to an obviously unsatisfiable constraint, then P is unsatisfiable. If at least one sequence of rule applications leads to a constraint that cannot be decomposed further and does not contain an obviously unsatisfiable set of constraints, then P is satisfiable. A constraint S is a finite set of propositional formulas. S is satisfiable if there exists an interpretation I such that I(P ) = 1 for all P S. Logic in Computer Science 21

22 Tableau method (Intuition) To check satisfiable of P, one starts with constraint {P }. Then, one applies rules that reflect the following facts: if (P Q) is satisfiable, then {P, Q, (P Q)} is satisfiable; if P is satisfiable, then {P, P } is satisfiable; if (P Q) is satisfiable, then { P, Q, (P Q)} is satisfiable. if (P Q) is satisfiable, then {P, (P Q)} is satisfiable or {Q, (P Q)} is satisfable; if (P Q) is satisfiable, then { P, (P Q)} is satisfiable or { Q, (P Q)} is satisfiable; To avoid branching, we first consider satisfiability of formulas not containing any (P Q) and (P Q). Logic in Computer Science 22

23 Ingredients of the algorithm (partial) A constraint S is a finite set of propositional formulas; A constraint S contains a clash if there exists a formula P such that P S and P S. A non-branching completion rule is of the form S = S, where S, S are constraints. A constraint S is complete if no completion rule is applicable to S. Logic in Computer Science 23

24 Completion Rules (partial) Assume that S does not contain a clash (if it does, no rule is applicable). ( -rule) ( -rule) ( -rule) S = S {P, Q} if (a) (P Q) S and (b) {P, Q} S. S = S {P } if (a) P S and (b) P S. S = S { P, Q} if (a) (P Q) S and (b) { P, Q} S. Logic in Computer Science 24

25 Example 1 We check satisfiability of P = (( p q) r). Set S 0 = {(( p q) r)}. An application of = gives S 1 = S 0 {( p q), r}. An application of = gives S 2 = S 1 { p, q} An application of = gives S 3 = S 2 {r} Note that S 3 = {P, ( p q), r, p, q, r}. Logic in Computer Science 25

26 Example 1 (continued) S 3 = {P, ( p q), r, p, q, r}. No completion rule is applicable to S 3 ; Thus, S 3 is complete. S 3 does not contain any clash. Thus, the output is P is satisfiable. S 3 describes an interpretation I under which P is true. Namely, we set for any atomic formula x from P : I(x) = 1 if, and only if, x S 3. Thus, I(q) = I(r) = 1 and I(p) = 0. Then I(P ) = 1. Logic in Computer Science 26

27 Example 2 We check satisfiability of P = ((p q) p). Set S 0 = {((p q) p)}. An application of = gives S 1 = S 0 {(p q), p}. An application of = gives S 2 = S 1 {p, q} An application of = gives S 3 = S 2 { p} Note that S 3 = {P, (p q), p, p, q, p}. Thus S 3 contains a clash: p S 3 and p S 3 and we output P is not satisfiable. Logic in Computer Science 27

28 Tableau Path (partial) A sequence S 0, S 1,..., S n of constraints is a tableau path if for any i < n at least one of the following conditions is satisfied: S i = S i+1 S i = S i+1 S i = S i+1 Logic in Computer Science 28

29 The tableau algorithm (partial) A tableau path S 0,..., S n is complete if S n is complete. A tableau path S 0,..., S n contains a clash if S n contains a clash. To check satisfiability of a formula P, do the following: Generate a tableau path starting with the constraint {P }; If the tableau path is complete and does not contain a clash, then output P is satisfiable. If the tableau path contains a clash, then output P is not satisfiable. Logic in Computer Science 29

30 Example 3 We check satisfiability of P = ((p q) (p q)). Set S 0 = {((p q) (p q))}. An application of = gives S 1 = S 0 {(p q), (p q)}. An application of = gives S 2 = S 1 {p, q} An application of = gives S 3 = S 2 { p, q} S 3 contains a clash: p S 3 and p S 3 and we output P is not satisfiable. Logic in Computer Science 30

31 Analysing the Tableau Algorithm (partial) To show that the tableau algorithm does what it is supposed to do, one has to show the following. Let P be a propositional formula. Termination: The algorithm terminates: there is no infinite tableau path S 0, S 1,... starting with {P }. Soundness: If there exists a complete tableau path S 0, S 1,..., S n with {P } = S 0 and without clash, then P is satisfiable. Completeness: If P is satisfiable, then no tableau path (generated by the three rules above) S 0, S 1,..., S n with {P } = S 0 contains a clash. For the proof, we require definitions and proofs by structural induction. This will be done later. Logic in Computer Science 31

32 Branching How to deal with formulas containing (P Q) or (P Q)? Recall that if (P Q) is satisfiable, then {P, (P Q)} is satisfiable or {Q, (P Q)} is satisfable; if (P Q) is satisfiable, then { P, (P Q)} is satisfiable or { Q, (P Q)} is satisfiable; Thus, we have to explore different ways to satisfy a formula. Logic in Computer Science 32

33 Example 4 We check satisfiability of P = ((p p) (q q)). Set S 0 = {P }. An application of = gives S 1 = S 0 {p p} or S 2 = S 0 {(q q)}. We first try to satisfy S 1. An application of = to S 1 gives S 3 = S 1 {p, p} which contains a clash. We have been unsuccessful. We now try to satisfy S 2. An application of = to S 2 gives S 4 = S 2 {q, q} S 4 does not contain a clash and is complete. Logic in Computer Science 33

34 Example 4 (continued) We have two tableau paths S 0, S 1, S 3 S 0, S 2, S 4 The second path is complete and does not contain a clash. Thus, the output is P is satisfiable. Recall that S 4 = {P, (q q), q}. S 4 also describes an interpretation I under which P is true, namely I(q) = 1. Logic in Computer Science 34

35 Ingredients of the algorithm (complete) A constraint S is a finite set of propositional formulas; A constraint S contains a clash if there exists a formula P such that P S and P S. A non-branching completion rule is of the form S = S, where S, S are constraints. A branching completion rule is of the form S = S 1 or S 2, where S, S 1, S 2 are constraints. A constraint S is complete if no completion rule is applicable to S. Logic in Computer Science 35

36 Completion Rules Assume that S does not contain a clash (if it does, no rule is applicable). ( -rule) S = S {P, Q} if (a) (P Q) S and (b) {P, Q} S. ( -rule) ( -rule) ( -rule) ( -rule) S = S {P } if (a) P S and (b) P S. S = S { P, Q} if (a) (P Q) S and (b) { P, Q} S. S = S {P } or S {Q} if (a) (P Q) S and (b) P S and Q S. S = S { P } or S { Q} if (a) (P Q) S and (b) P S and Q S. Logic in Computer Science 36

37 Tableau Path A sequence S 0, S 1,..., S n of constraints is a tableau path if for any i < n at least one of the following conditions is satisfied: S i = S i+1 S i = S i+1 S i = S i+1 for some S: S i = S i+1 or S or S i = S or S i+1 for some S: S i = S i+1 or S or S i = S or S i+1 Logic in Computer Science 37

38 The tableau algorithm A tableau path S 0,..., S n is complete if S n is complete. A tableau path S 0,..., S n contains a clash if S n contains a clash. To check satisfiability of a formula P, do the following: Generate tableau paths starting with the constraint {P }; If there is a tableau path that is complete and does not contain a clash, then output P is satisfiable. If no such tableau path can be found (i.e., all complete tableau paths starting with P contain a clash), then output P is not satisfiable. Logic in Computer Science 38

39 Example 5 We check satisfiability of P = ((( p q) p) q). Set S 0 = {P }. An application of = gives S 1 = S 0 {(( p q) p), q}. An application of = gives S 2 = S 1 {( p q), p}. An application of = gives S 3 = S 2 { p} S 3 contains a clash: p S 3 and p S 3. The other possible application of = to S 2 gives S 4 = S 2 {q} S 4 is complete and does not contain a clash. Logic in Computer Science 39

40 Example 5 (continued) We have two tableau paths S 0, S 1, S 2, S 3 S 0, S 1, S 2, S 4 The second path is complete and does not contain a clash. Thus, the output is P is satisfiable. Recall that S 4 = {P, (( p q) p), p, q, ( p q), q}. S 4 also describes an interpretation I under which P is true. Namely, we set for any atomic formula x from P : I(x) = 1 if, and only if, x S 4. Thus, I(p) = I(q) = 1. Then I(P ) = 1. Logic in Computer Science 40

41 Example 6 We check satisfiability of P = ((( p q) p) q). Set S 0 = {P }. An application of = gives S 1 = S 0 {(( p q) p), q}. An application of = gives S 2 = S 1 {( p q), p}. An application of = gives S 3 = S 2 { p} S 3 contains a clash: p S 3 and p S 3. The other possible application of = gives S 4 = S 2 {q} S 4 contains a clash: q S 4 and q S 4. Logic in Computer Science 41

42 Example 6 (continued) We have two tableau paths S 0, S 1, S 2, S 3 S 0, S 1, S 2, S 4 Both contain a clash. Thus, the output is P is not satisfiable. Logic in Computer Science 42

43 Analysing the Tableau Algorithm To show that the tableau algorithm does what it is supposed to do, one has to show the following. Let P be a propositional formula. Termination: The algorithm terminates: there are only finitely many tableau paths starting with {P }. Soundness: If there exists a complete tableau path S 0, S 1,..., S n with S 0 = {P } without clash, then P is satisfiable. Completeness: If P is satisfiable, then there exists a complete tableau path S 0, S 1,..., S n with S 0 = {P } without clash. For the proof, we require definitions and proofs by structural induction. We first introduce this important concept. Logic in Computer Science 43

44 Definitions by structural induction Many important functions F which have as domain the set of all propositional formulas are defined by specifying the values F (p i ), for all propositional variables p i, F (P Q), given the values F (P ) and F (Q), F (P Q), given the values F (P ) and F (Q), F ( P ), given the value F (P ). Such a definition is called a definition by structural induction. (The idea should be familar from proofs by induction for natural numbers.) Logic in Computer Science 44

45 Example 1: Interpretations The definition of interpretations I was given by structural induction. To define an interpretation I it is sufficient to define I(p i ) for all atomic formulas p i. The values I(P ), P an arbitrary propositional formula, are then given by means of truth tables. In other words, truth tables define the values I(P Q), given the values I(P ) and I(Q), I(P Q), given the values I(P ) and I(Q), I( P ), given the value I(P ). Logic in Computer Science 45

46 Example 2: Subformulas The function sub(p ) giving the subformulas of a formula P is defined by structural induction as follows: sub(p i ) = {p i }, for all atomic formulas p i, sub(p Q) = {(P Q)} sub(p ) sub(q) sub(p Q) = {(P Q)} sub(p ) sub(q) sub( P ) = { P } sub(p ). The set sub(p ) is called the set of subformulas of P. Logic in Computer Science 46

47 Subformulas Compute sub(p ) for P = ((p 1 p 2 ) p 3 ). sub(p ) = {P } sub(p 1 p 2 ) sub( p 3 ) = {P } {(p 1 p 2 )} sub(p 1 ) sub( p 2 ) sub( p 3 ) = {P, (p 1 p 2 )} {p 1 } { p 2 } sub(p 2 ) { p 3 } sub(p 3 ) = {P, (p 1 p 2 ), p 1, p 2, p 2, p 3, p 3 } Logic in Computer Science 47

48 Example 3: Length of a formula The function L(P ) giving the length of a formula is defined by structural induction as follows: L(p i ) = 1, for all atomic formulas p i, L(P Q) = 1 + L(P ) + L(Q), L(P Q) = 1 + L(P ) + L(Q), L( P ) = 1 + L(P ). L(P ) is called the length of formula P. Logic in Computer Science 48

49 Length of a formula Compute L(P ) for P = (p 0 p 1 ). L(P ) = 1 + L(p 0 p 1 ) = L(p 0 ) + L( p 1 ) = L(p 1 ) = 5. Logic in Computer Science 49

50 Proofs by structural induction Statements about objects defined by structural induction can often be proved by structural induction. We illustrate this proof method by means of the following example. ( sub(p ) denotes the number of subformulas of P.) Theorem For every formula P : sub(p ) L(P ). Proof The proof is by structural induction. In other words, we show: 1. for all atomic formulas p i : sub(p i ) L(p i ); 2. for all formulas P and Q: if sub(p ) L(P ) and sub(q) L(Q), then sub(p Q) L(P Q); 3. for all formulas P and Q: if sub(p ) L(P ) and sub(q) L(Q), then sub(p Q) L(P Q); 4. for every formula P : if sub(p ) L(P ), then sub( P ) L( P ). Logic in Computer Science 50

51 We now check (1.)-(4.): 1. sub(p i ) = 1 1 = L(p i ). 2. Suppose sub(p ) L(P ) and sub(q) L(Q). Then sub(p Q) = {P Q} sub(p ) sub(q) 1 + sub(p ) + sub(q) 1 + L(P ) + L(Q) = L(P Q). 3. Exercise. 4. Exercise. Logic in Computer Science 51

52 Termination of the tableau algorithm Assume P is given. We have to show that there are only finitely many tableau paths {P } = S 0, S 1,..., S n. Let sub (P ) = sub(p ) { Q Q sub(p )}. Now observe for any tableau path {P } = S 0, S 1,..., S n : S 0 S 1 S n sub (P ). Hence the length of any tableau path {P } = S 0, S 1,..., S n is not greater than sub (P ) and the number of such tableau paths is not greater that sub (P ) sub (P ). (Here, by X Y we denote that X is a proper subset of Y.) Logic in Computer Science 52

53 Soundness of the tableau algorithm Let {P } = S 0, S 1,..., S n be a complete tableau path such that S n does not contain a clash. We define an interpretation I by { 1 if pi S I(p i ) = n 0 if p i S n We show the following claim by structural induction: Claim 1 I(Q) = 1 for all Q S n ; I(Q) = 0 for all Q S n. Since P S n, we obtain I(P ) = 1. Thus P is satisfiable. Logic in Computer Science 53

54 The steps of the structural induction We have to show: 1. Claim 1 holds for all atomic formulas p i ; 2. if Claim 1 holds for P 1 and P 2, then Claim 1 holds for (P 1 P 2 ); 3. if Claim 1 holds for P 1 and P 2, then Claim 1 holds for (P 1 P 2 ); 4. if Claim 1 holds for Q, then it holds for Q. Logic in Computer Science 54

55 Proof of Point 1 Let p i be an atomic formula. We have to show (a) I(p i ) = 1 if p i S n ; (b) I(p i ) = 0 if p i S n. Point (a) follows by definition of I. For Point (b), assume that p i S n. Since S n does not contain a clash, p i S n. Hence, by definition of I, I(p i ) = 0. Logic in Computer Science 55

56 Proof for Point 2 Assume Claim 1 holds for P 1 and P 2. Suppose (P 1 P 2 ) S n. Then, by non-applicability of = to S n, P 1 S n and P 2 S n ; By induction hypothesis, I(P 1 ) = 1 and I(P 2 ) = 1; Hence I(P 1 P 2 ) = 1. Suppose (P 1 P 2 ) S n. Then, by non-applicability of = to S n, P 1 S n or P 2 S n ; By induction hypothesis, I(P 1 ) = 0 or I(P 2 ) = 0; Hence I(P 1 P 2 ) = 0. Logic in Computer Science 56

57 Proof for Point 3 Assume Claim 1 holds for P 1 and P 2. Suppose (P 1 P 2 ) S n. Then, by non-applicability of = to S n, P 1 S n or P 2 S n ; By induction hypothesis, I(P 1 ) = 1 or I(P 2 ) = 1; Hence I(P 1 P 2 ) = 1. Suppose (P 1 P 2 ) S n. Then, by non-applicability of = to S n, P 1 S n and P 2 S n ; By induction hypothesis, I(P 1 ) = 0 and I(P 2 ) = 0; Hence I(P 1 P 2 ) = 0. Logic in Computer Science 57

58 Proof for Point 4 Assume Claim 1 holds for Q. We show Claim 1 for Q. Suppose Q S n. By induction hypothesis, I(Q) = 0. Hence I( Q) = 1. Suppose Q S n. Then, by non-applicability of = to S n, Q S n. By induction hypothesis, I(Q) = 1; Hence I( Q) = 0. Logic in Computer Science 58

59 Completeness of the tableau algorithm Assume that P is satisfiable. We have to construct a complete tableau path {P } = S 0, S 1,..., S n such that S n does not contain a clash. Let I be an interpretation with I(P ) = 1. We construct the tableau path as follows: Let S 0 = {P } and assume that S 0 S 1 S i have already been defined such that I(Q) = 1 for all Q S i. Then S i does not contain a clash. If no completion rule is applicable, then the path is complete and we are done. Now assume that a completion rule is applicable. We show that we can apply the rule in such a way that S i = S i+1 and I(Q) = 1 for all Q S i+1. Logic in Computer Science 59

60 Construction of S i+1 1. If P 1 P 2 S i and = is applicable, then set S i+1 = S i {P 1, P 2 }. Then I(P 1 ) = I(P 2 ) = 1 since I(P 1 P 2 ) = 1. Thus I(Q) = 1 for all Q S i Otherwise, if (P 1 P 2 ) S i and = is applicable, then set S i+1 = S i { P 1, P 2 }. Then I( P 1 ) = I( P 2 ) = 1 since I( (P 1 P 2 )) = 1. Thus I(Q) = 1 for all Q S i Otherwise, if P 1 S i and = is applicable, then set S i+1 = S i {P 1 }. Then I(P 1 ) = 1 since I( P 1 ) = 1. Thus I(Q) = 1 for all Q S i+1. Logic in Computer Science 60

61 Construction of S i+1 1. Otherwise, if (P 1 P 2 ) S i and = is applicable, then I( P 1 ) = 1 or I( P 2 ) = 1 since I( (P 1 P 2 )) = 1. In the first case let S i+1 = S i { P 1 }. In the second case let S i+1 = S i { P 2 }. In both cases I(Q) = 1 for all Q S i Otherwise, if (P 1 P 2 ) S i and = is applicable, then I(P 1 ) = 1 or I(P 2 ) = 1 since I(P 1 P 2 ) = 1. In the first case let S i+1 = S i {P 1 }. In the second case let S i+1 = S i {P 2 }. In both cases I(Q) = 1 for all Q S i+1. Logic in Computer Science 61

62 Modern SAT solvers High performance SAT solvers are not tableau based. They are based on modern versions of the Davis-Putnam-Logemann-Loveland algorithm (DPLL) developed in the 1960s (which is based on a very general proof method called resolution); and on stochastic local search algorithms. Many solvers are available as free and open source software. Logic in Computer Science 62

63 Tautology Definition A tautology is a formula which is true under all interpretations. Example All formulas of the form P P are tautologies, because for all interpretations I: I(P P ) = 1 P P P P Observation: A formula P is a tautology if, and only if, P is not satisfiable. Logic in Computer Science 63

64 Contradiction Definition A contradiction is a formula which is false under all interpretations. Example All formulas of the form P P are tautologies, because for all interpretations I: I(P P ) = 0 P P P P Observation: A formula P is a contradiction if, and only if, P is not satisfiable Logic in Computer Science 64

65 Semantic consequence Definition Suppose X is a finite set of formulas and P is a formula. Then P follows from X (is a semantic consequence of X) if the following holds for every interpretation I: If I(Q) = 1 for all Q X, then I(P ) = 1. This is denoted by X = P. Logic in Computer Science 65

66 Example 1 Show {p 1 p 2 } = p 1 p 2. Solution: p 1 p 2 p 1 p 2 p 1 p The statement follows, because in any row where the column for p 1 p 2 contains 1 the column for p 1 p 2 also contains 1. Logic in Computer Science 66

67 Example 2 Show {p 1 } = p 2. Solution: Take the interpretation I with I(p 1 ) = 1 and I(p 2 ) = 0. Logic in Computer Science 67

68 Example 3 We show that = P if, and only if, P is a tautology. ( ) Assume P is not a tautology. Take interpretation I with I(P ) = 0. Then I(Q) = 1 for all Q, but I(P ) 1, Hence = P. ( ) Assume = P. Take interpretation I with I(Q) = 1 for all Q and I(P ) 1. Then P is not a tautology. Logic in Computer Science 68

69 Example 4 (ex falso quodlibet) We show that {(P P )} = Q holds for all formulas Q. Let Q be arbitrary. There is no interpretation I such that I(P P ) = 1. Thus, if I is an interpretation such that I(P P ) = 1, then I(Q) = 1. Thus {(P P )} = Q. Logic in Computer Science 69

70 Reduction to Satisfiability Recall that we call a finite set S of formulas satisfiable if there exists an interpretation I such that I(Q) = 1 for all Q S. Note that the set S = {Q 1,..., Q n } is satisfiable if, and only if, the formula obtained by taking the conjunction of all Q 1,..., Q n, (Q 1... Q n ), is satisfiable. Observation For every finite set S of formulas and every formula P : S = P if, and only if, S { P } is not satisfiable. Thus, we can use the tableau algorithm to check semantic consequence: to check whether S = P check that S { P } is not satisfiable. Logic in Computer Science 70

71 Example We check {p q} = p q. To this end, we have to show that S 0 = {p q, (p q)} is not satisfiable. We do this using the tableau algorithm: an application of = to S 0 gives S 1 = S 0 {p, q, (p q)} an application of = to S 1 gives S 2 = S 1 { p, q} S 2 contains a clash: {p, p} S 2. Thus, all tableau paths starting with S 0 contain a clash. Hence S 0 is not satisfiable. Logic in Computer Science 71

72 Logical equivalence Definition Two formulas P and Q are called equivalent if they have the same truth value under every possible interpretation. In other words, P and Q are equivalent if I(P ) = I(Q) for every interpretation I. This is denoted by P Q. Observation For any two formulas P and Q: P Q if, and only if, neither (P Q) nor (Q P ) are satisfiable. Thus, we can use the tableau algorithm to check logical equivalence: to check whether P Q check that (P Q) is not satisfiable and (Q P ) is not satisfiable. Logic in Computer Science 72

73 Laws for equivalences The following equivalences can be checked using the tableau algorithm or by truth tables: Associative laws: P (Q R) (P Q) R, P (Q R) (P Q) R Commutative laws: P Q Q P, P Q Q P Logic in Computer Science 73

74 Laws for Equivalence Distributive laws: P (Q R) (P Q) (P R), P (Q R) (P Q) (P R) Complement law: P P De Morgan s laws: (P Q) P Q, (P Q) P Q. Logic in Computer Science 74

75 Proof of {(p (q r))} = ((p q) (p r)) We have to show that S 0 = {(p (q r)), ((p q) (p r))} is not satisfiable. an application of = to S 0 gives S 1 = S 0 {p, (q r)} an application of = to S 1 gives S 2 = S 1 { (p q), (p r)} Logic in Computer Science 75

76 Continue by decomposing (q r) S 2 (1) First option: an application of = to (q r) S 2 gives S 3 = S 2 {q} an application of = to (p q) S 3 gives S 4 contains the clash {p, p}. = S 3 { p} which the other application of = to (p q) S 3 gives S 5 which contains the clash {q, q}. = S 3 { q} Thus, every complete tableau path starting with S 0, S 1, S 2, S 3 contains a clash. Logic in Computer Science 76

77 Continue by decomposing (q r) S 2 (2) Second option: an application of = to (q r) S 2 gives S 6 = S 2 {r}. an application of = to (p r) S 6 gives S 7 contains the clash {p, p}. = S 6 { p} which the other application of = to (p r) S 6 gives S 8 which contains the clash {r, r}. = S 6 { r} Thus, every complete tableau path starting with S 0, S 1, S 2, S 6 contains a clash. We can conclude that all complete tableau paths starting with S 0 contain a clash. Logic in Computer Science 77

78 Conjunctive and disjunctive normal form A formula (P 1 P 2 P n ) is called a disjunction of P 1,..., P n ; Similarly, (P 1 P 2 P n ) is called a conjunction of P 1,..., P n ; A formula which is either an atomic formula or its negation is called a literal; A formula is in conjunctive normal form (CNF) if it is a conjunction of disjunctions of literals. A formula is in disjunctive normal form (DNF) if it is a disjunction of conjunctions of literals. Logic in Computer Science 78

79 Examples p 1, p 1, p 5 are literals. They are also in CNF and in DNF. (p q) is in CNF and in DNF. ((p 1 p 2 ) ( p 1 p 3 )) and ((p 1 p 2 ) p 1 ) are in CNF and not in DNF. ((p 1 p 2 ) ( p 1 p 3 )) and (p (p p)) are in DNF and not in CNF. Logic in Computer Science 79

80 CNF Theorem (1) Every formula is equivalent to a formula in CNF. (2) Every formula is equivalent to a formula in DNF. Proof of (1) Suppose a formula P is given. We transform P to a formula in CNF using the Laws of equivalence: Step 1. Apply De Morgan s laws and Complement law P P until negation ( ) occurs in front of atomic formulas only. Step 2. Apply Distributive law P (Q R) (P Q) (P R) and Commutative laws until the formula is in CNF. Logic in Computer Science 80

81 Example Transform ( (p 0 p 1 ) (p 2 p 1 )) into CNF. ( (p 0 p 1 ) (p 2 p 1 )) is equivalent to (de Morgan s Law) is equivalent to (Distributive law) is equivalent to (Distributive law) is equivalent to (Distributive law) (( p 0 p 1 ) (p 2 p 1 )) ((( p 0 p 1 ) p 2 ) (( p 0 p 1 ) p 1 )) (( p 0 p 2 ) ( p 1 p 2 ) ( p 0 p 1 ) p 1 )) (( p 0 p 2 ) ( p 1 p 2 ) ( p 0 p 1 ) ( p 1 p 1 )). Logic in Computer Science 81