3 Propositional Logic

Transcription

1 3 Propositional Logic 3.1 Syntax 3.2 Semantics 3.3 Equivalence and Normal Forms 3.4 Proof Procedures 3.5 Properties Propositional Logic (25th October 2007) 1

2 3.1 Syntax Definition 3.0 An alphabet Σ consists of a finite or countably infinite set. The elements of this set are calles symbols. Without loss of generality (Wlog) we assume that Λ Σ. The set of words (or strings) over Σ is denoted by Σ and is defined as follows: 1. Λ Σ. 2. If w Σ and a Σ, then aw Σ. 3. There are no other words in Σ. Λ is called empty word or empty string. Example Let Σ = {1, 2}. We can form the following words: Λ, 1Λ, 2Λ, 11Λ, 12Λ, 21Λ, 22Λ, 111Λ,.... Propositional Logic (25th October 2007) 2

3 3.1 Alphabet of Propositional Logic Definition 3.1 An alphabet of propositional logic consists of a (countably) infinite set R = {p 1, p 2,...} of propositional variables, the set { /1, /2, /2, /2, /2} of connectives, and the special characters ( and ). Indices are sometimes omitted. We will occasionally use other letters to denote variables. Propositional variables are also called nullary relation symbols. Different alphabets of propositional logic differ in R. Propositional Logic (25th October 2007) 3

4 Propositional Formulas Definition 3.2 A (propositional) atomic formula, briefly called atom, is a propositional variable. Definition 3.3 The set of propositional formulas is the smallest set L(R) with the following properties: 1. If F is an atomic formula, then F L(R). 2. If F L(R), then F L(R). 3. If /2 is a binary connective, F, G L(R), then (F G) L(R). Does such a smallest set exist? There are sets, which satisfy all properties. The intersection of such sets satisfies the properties too. The smallest set is the intersection of all sets satisfying the properties. Notation A (possibly indexed) denotes an atom, F, G, H (possibly indexed) denote propositional formulas. Propositional Logic (25th October 2007) 4

5 Structural Induction How can we prove properties of sets of formulas? Theorem 3.4 (Principle of structural induction) Every propositional formula has a property E, if the following conditions are met: 1. Induction Base: Every atom has property E. 2. Induction Steps: If F has property E, then F has property E as well. If /2 is a binary connective, and F and G have property E, then (F G) has property E as well. Proof Let E be the set of all formulas satisfying E. E L(R). E satisfies the three conditions of definition 3.3. Because L(R) is the smallest set satisfying these conditions, L(R) E must hold. qed Propositional Logic (25th October 2007) 5

6 Structural Recursion How can we define functions acting on formulas? Theorem 3.5 (Principle of structural recursion) Let M be an arbitrary set and let the following functions be given: foo R : R M, foo : M M, foo : M M M for {,,, }. Then there is exactly one function foo : L(R) M satisfying the following conditions: 1. Recursion base foo(a) = foo R (A) for all A R. 2. Recursion steps foo( G) = foo (foo(g)) for all G L(R). foo((g 1 G 2 )) = foo (foo(g 1 ), foo(g 2 )) for all G 1, G 2 L(R). Proof Exercise Propositional Logic (25th October 2007) 6

7 Structural Recursion Example M = N. foo 1R (A) = 0 for all A R. foo 1 (X) = X for all X M. foo 1 (X, Y ) = X + Y + 1 for all {,,, } and all X, Y M. We obtain: foo 1 (F ) = 8 < : 0 if F is an atom, foo 1 (G) if F is of the form G, foo 1 (G 1 ) + foo 1 (G 2 ) + 1 if F is of the form (G 1 G 2 ). Often foo 1R, foo 1 and foo 1 are not explicitely specified. foo is a function symbol af a meta language. Propositional Logic (25th October 2007) 7

8 Structural Recursion More Examples foo 2 (F ) = 8 < : 1 if F is an atom, foo 2 (G) if F is of the form G, foo 2 (G 1 ) + foo 2 (G 2 ) if F is of the form (G 1 G 2 ). counts the number of occurrences of atoms in F. foo 3 (F ) = 8 < : eliminates parentheses in F. foo 4 (F ) = {Λ} F if F is an atom, foo 3 (G) if F is of the form G, foo 3 (G 1 ) foo 3 (G 2 ) if F is of the form (G 1 G 2 ). 8 >< >: computes the set of positions in F. if F is an atom, {1π π foo 4 (G)} if F is of the form G, {1π 1 π 1 foo 4 (G 1 )} {2π 2 π 2 foo 4 (G 2 )} if F is of the form (G 1 G 2 ) Propositional Logic (25th October 2007) 8

9 Subformulas Definition 3.6 Let F be a propositional formula. The set of subformulas of F is the smallest set of formulas S(F ) satisfying the following conditions: 1. F S(F ). 2. If G S(F ), then G S(F ). 3. If (G 1 G 2 ) S(F ), then G 1, G 2 S(F ). Example S( ((p 1 p 2 ) p 1 )) = { ((p 1 p 2 ) p 1 ), ((p 1 p 2 ) p 1 ), (p 1 p 2 ), p 1, p 2 }. Propositional Logic (25th October 2007) 9

10 3.2 Semantics What is the meaning of ((p 1 p 2 ) p 1 )? Here: formulas may be true or false binary logic. Propositional Logic (25th October 2007) 10

11 3.2.1 The Structure of the Truth Values A structure consists of a set of individuals and a set of functions defined on it. The set of truth values W is the set {, }. Unary functions of type W W: Negation : ( ) = and ( ) =. Identity. Projections onto and. Binary functions of type W W W: Conjunction /2. Disjunction /2. Implication /2. NAND /2. NOR /2. Equivalence /2. Etc. Propositional Logic (25th October 2007) 11

12 The Functions over the Set of Truth Values Here negation, conjunction, disjunction, implication, equivalence. Propositional Logic (25th October 2007) 12

13 3.2.2 Interpretations and Models Definition 3.7 A (propositional) interpretation I = (W, I) consists of the set W and a mapping I : L(R) W with: [F ] I = 8 < : w W if F R, [G] I if F is of the form G, ([G 1 ] I [G 2 ] I ) if F is of the form (G 1 G 2 ). An interpretation I = (W, I) is uniquely defined by specifying how I acts on propositional variables. Recall that A denotes an atom. We will sometimes identify an interpretation I = (W, I) with {A [A] I = }. We often write F I instead of [F ] I. I (possibly indexed) denotes an interpretation. Propositional Logic (25th October 2007) 13

14 Satisfiability of Formulas Definition 3.10 Let F L(R). F is called satisfiable if there is an interpretation I = (W, I) with F I =. F is called valid (or F is a tautology) if for all interpretations I = (W, I) we have: F I =. F is called falsifiable if there is an interpretation I = (W, I) with F I =. F is called unsatisfiable if for all interpretations I = (W, I) we have: F I =. Propositional Logic (25th October 2007) 14

15 Models Definition 3.11 An interpretation I = (W, I) is called model for a propositional formula F, in symbols I = F, if F I =. Abbreviations If I is not a model for the formula F, i. e. if F I =, then we write I = F. If F is valid, then we write = F. Propositional Logic (25th October 2007) 15

16 Validity and Unsatisfiability Theorem 3.12 A propositional formula F is valid ( = F ), iff F is unsatisfiable. Proof = F iff all interpretations are models for F iff no interpretation is a model for F iff F is unsatisfiable. qed Propositional Logic (25th October 2007) 16

17 Satisfiability of a Set of Formulas Definition 3.13 Let G be a set of formulas. G is satisfiable, if there is an interpretation mapping each element F G to. An interpretation I is called model for G, in symbols I = G, if I is model for all F G. Propositional Logic (25th October 2007) 17

18 Logical Consequence Definition 3.14 A propositional formula F is a (propositional) consequence of a set of propositional formulas G, in symbols G = F, iff for every interpretation I we have: if I = G, then I = F as well. Exercises Does {p, (p q)} = q hold? Does {(p q)} = q hold? Does {(p p)} = q hold? Theorem 3.15 Let F, F 1,..., F n be propositional formulas. {F 1,..., F n } = F holds iff = ((... (F 1 F 2 )... F n ) F ) holds. Proof Exercise Propositional Logic (25th October 2007) 18

19 Truth Tabling How can we compute the value of a formula F under all possible interpretations? Computing a truth table: 1 Let m = S(F ) be the number of subformulas of F. 2 Let n = {p R p S(F )} be the number of propositional variables occurring in F. 3 Form a table with 2 n rows and m columns, where the first n columns are marked by the n propositional variables occurring in F, the last column is marked by F, and the remaining columns are marked by the other subformulas of F. 4 Fill in the first n columns with und as follows: In the first column fill in alternating downwards..., in the second column..., etc. 5 Calculate the values in the remaining columns using the known functions on the set of truth values. Propositional Logic (25th October 2007) 19

20 Working with Truth Tables Let F be a formula and T a corresponding truth table. Then: F is satisfiable iff T contains a row with in the last column. F is valid iff all rows in T have in the last column. F is falsifiable iff T contains a row with in the final column. F is unsatisfiable iff all rows T have in the final column. We can apply truth tabling to decide whether G = F holds for finite G Exercise. Exercise Write a (Prolog) program which generates a truth table for a given formula F. Propositional Logic (25th October 2007) 20

21 3.3 Equivalence and Normal Forms Semantic Equivalence Definition 3.16 Two propositional formulas F and G are semantically equivalent, in symbols F G, if for all interpretations I we have: I = F iff I = G. Observe is an equivalence relation: is reflexive: F F. is symmetric: If F G, then G F. is transitive: If F G and G H, then F H. This follows immediately from Definition 3.16 Exercise. Propositional Logic (25th October 2007) 21

22 Well-Known Semantic Equivalences Theorem 3.17 The following equivalences hold: (F F ) F (F F ) F idempotency (F G) (G F ) (F G) (G F ) commutativity ((F G) H) (F (G H)) ((F G) H) (F (G H)) associativity ((F G) F ) F ((F G) F ) F absorption (F (G H)) ((F G) (F H)) (F (G H)) ((F G) (F H)) distributivity Propositional Logic (25th October 2007) 22

23 Theorem 3.17 (continued) Theorem 3.17 (continued) The following equivalences hold: F F double negation (F G) ( F G) (F G) ( F G) de Morgan (F G) F, if F is valid (F G) G, if F is valid tautology (F G) G, if F is unsatisfiable (F G) F, if F is unsatisfiable unsatisfiability (F G) ((F G) ( G F )) equivalence (F G) ( F G) implication Proof Exercise Propositional Logic (25th October 2007) 23

24 Positions Recall foo 4 (F ) = {Λ} 8 >< >: if F is an atom, {1π π foo 4 (G)} if F is of form G, {1π 1 π 1 foo 4 (G 1 )} {2π 2 π 2 foo 4 (G 2 )} if F is of the form (G 1 G 2 ). foo 4 (F ) is the set of positions in a propositional logic formula F. Notation We denote this set also with P(F ). Positions are words over {1, 2}, where Λ denotes the empty word. Propositional Logic (25th October 2007) 24

25 Replacements The subformula of F at position π P(F ), in symbols F π, is defined by: 1. F Λ = F. 2. F 1π = G π, if F is of the form G. 3. F iπ = G i π, if F is of the form (G 1 G 2 ) and i {1, 2}. The replacement of the subformula of F at position π P(F ) by H, in symbols F π H, is defined by: 1. F Λ H = H, 2. F 1π H = (G π H ), if F is of the form G. 3. F 1π H = (G 1 π H G 2 ), if F is otf (G 1 G 2 ). 4. F 2π H = (G 1 G 2 π H ), if F is otf (G 1 G 2 ). Propositional Logic (25th October 2007) 25

26 The Replacement Theorem Theorem 3.18 (Replacement Theorem) Let F L(R), π P(F ), F π = G and G H. Then, F F π H. Proof Structural induction on π. Induction basis Let π = Λ. F = F Λ = G H = F Λ H. Induction hypothesis (IH) The theorem holds for π. Propositional Logic (25th October 2007) 26

27 Proof Replacement Theorem (Continuation) Induction step Let π = iπ. We distinguish two cases: i = 1 π = 1π P(F ), hence F must be otf F or (G 1 G 2 ). F F 1π H = ( F ) 1π H = (F π H ). From IH we conclude: F F π H Hence, for all interpretations I we find: [F ] I = [ F ] I Assumption = [F ] I Def I = [F π H ] I IH and Def = [ (F π H )] I Def I = [( F ) 1π H ] I Def replacement = [F 1π H ] I Assumption Hence, F F 1π H (G 1 G 2 ) analogously i = 2 analogously qed Propositional Logic (25th October 2007) 27

28 Agreement If π P(F ) then F π H := F. Let F π = G. If it is obvious from the context which subformula G is to be replaced, then we write F G H instead of F π H. Theorem 3.18 allows to replace all occurrences of the connectives and. Wlog we consider only formulas over R {,,, (, )}. Propositional Logic (25th October 2007) 28

29 3.3.2 Negation Normal Form Definition 3.19 A propositional formula F is in negation normal form if all negation signs occurring in F appear directly in front of propositional variables. Proposition 3.20 There is an algorithm which transforms any formula into a semantically equivalent formula in negation normal form. Algorithm Given F. While F is not in negation normal form do: Select a subformula of F having the form G with G R. If G = H, then replace H by H. If G = (G 1 G 2 ), then replace (G 1 G 2 ) by ( G 1 G 2 ). If G = (G 1 G 2 ), then replace (G 1 G 2 ) by ( G 1 G 2 ). Propositional Logic (25th October 2007) 29

30 The Rules of the Algorithm The rules will be abbreviated by: H H (G 1 G 2 ) ( G 1 G 2 ) (G 1 G 2 ) ( G 1 G 2 ) The algorithm is (don t care) non-deterministic! Example (r ( p q)) ( r ( p q)) de Morgan ( r ( p q)) double negation Propositional Logic (25th October 2007) 30

31 3.3.3 Clause Form Definition 3.21 A literal is a propositional variable, or a negated propositional variable. Notation L (possible indexed) denotes a literal. Generalized disjunction Generalized conjunction [F 1,..., F n ] = (... ((F 1 F 2 ) F 3 )... F n ) F 1,..., F n = (... ((F 1 F 2 ) F 3 )... F n ) Observe [F ] = F and F = F. Hence, [F ] F and F F. Empty generalized disjunction: [ ] with [ ] I = for all I. Empty generalized conjunction: with I = for all I. Observe (G [ ]) G and (G ) G. Propositional Logic (25th October 2007) 31

32 Clauses Definition 3.22 A clause is a generalized disjunction [L 1,..., L n ], n 0, where every L i, 1 i n, is a literal. A dual clause is a generalized conjunction L 1,..., L n, n 0, where every L i, 1 i n, is a literal. A formula is in conjunctive normal form, or clause form, iff it is of the form C 1,..., C m, m 0, and if every C j, 1 j m, is a clause. A formula is in disjunctive normal form, or dual clause form, iff it is of the form [C 1,..., C m ], m 0, and if every C j, 1 j m, is a dual clause. Notation C (possibly indexed) denotes a clause. Propositional Logic (25th October 2007) 32

33 An Observation Let F be a formula in clause form. The following holds: F is unsatisfiable if a clause occurring in F is unsatisfiable. A clause is unsatisfiable iff it is of the form [ ]. Propositional Logic (25th October 2007) 33

34 Transformation into Clause Form Theorem There is an algorithm which transforms any formula into a semantically equivalent formula in clause form. 2 There is an algorithm which transforms any formula into a semantically equivalent formula in dual clause form. Proof of 1 We have to specify the algorithm. We have to show that the algorithm is correct or sound, i.e., if it outputs G given F, then G must be in clause form and F G must hold. We have to show that the algorithm terminates, i.e., it stops in finite time given F. Propositional Logic (25th October 2007) 34

35 An Algorithm for the Transformation into Clause Form Input: A propositional formula F. Output: A formula, which is in conjunctive normal form and is equivalent to F. G := [F ]. (G is a conjunction of disjunctions.) While G is not in conjunctive normal form do: Select a non-clausal element H from G. Select a non-literal element K from H. Apply the rule among the following ones which is applicable. D D (D 1 D 2 ) (D 1 D 2 ) (D 1 D 2 ) (D 1 D 2 ) D 1 D 2 D 1, D 2 D 1, D 2 D 1 D 2 A rule D D is applicable to K if K is of the form D. If applied, then K is replaced by D. D A rule D 1 D is applicable to K if K is of the form D. 2 If applied, H is replaced by two disjunctions: The first one is obtained from H by replacing the occurrence of D by D 1. The second one is obtained from H by replacing the occurrence of D by D 2. Propositional Logic (25th October 2007) 35

36 An Example and a Lemma Example [ (p ( (p q) r))] [ p], [ ( (p q) r)] [ p], [ (p q), r] [ p], [(p q), r] [ p], [(p q), r] [ p], [p, r], [q, r] Lemma 3.24 If a propositional formula G is a conjunction of disjunctions and G is obtained from G by applying one of the rules of the algorithm, then G is a conjunction of disjunctions and G G. Proof Exercise. Propositional Logic (25th October 2007) 36

37 Induction Principle for While-Loops A loop invariant for a while-loop W is a statement E having the following property: If E holds, then E still holds after one execution of the body of W. Theorem 3.25 (Induction Principle for While-Loops) If a statement E holds before entering a loop W, and is a loop invariant for W, then E holds after termination of W as well (if W terminates). Proof Literature. Propositional Logic (25th October 2007) 37

38 Soundness of the Algorithm Let F be the given propositional formula and let E be the statement: G ist a conjunction of disjunctions and F G holds. Initially G is defined to be [F ]. Because F [F ] holds, E holds before entering the while-loop. From Lemma 3.24 we learn that E is a loop invariant for the while-loop. Hence, by Theorem 3.25 E holds when the loop has terminated. The loop terminates only when G is in conjunctive normal form. The algorithm is sound! Propositional Logic (25th October 2007) 38

39 Termination of the Algorithm When does a non-deterministic algorithm terminate? Weak Termination There is at least one way to execute the algorithm such that it terminates. Strong Termination The algorithm terminates independently of which choice we make in the non-deterministic situations. Definition 3.26 The function r, called rank of a propositional formula, is defined over L(R) as follows: r(h) = 8 >< >: 0 if H is a literal, r(f ) + 1 if H is of the form F, r(f ) + r(g) + 1 if H is of the form (F G) or (F G), r( F ) + r( G) + 1 if H is of the form (F G) or (F G). Propositional Logic (25th October 2007) 39

40 Multisets A multiset is comparable to a set, but the individual elements may occur multiple times in it. Here multisets of natural numbers. {1, 1, 2, 2, 2, 3 }. M 1 M 2 holds iff M 2 is obtained from M 1 by deleting a number n from M 1 and replacing it by finitely many numbers which are all less than n. {1, 1, 2, 2, 2, 3 } {1, 1, 1, 1, 2, 2, 2, 2 }. Observation there cannot be infinitely long sequences (M i i 0) of finite multisets of natural numbers such that M 0 M Proof see Dershowitz, Manna: Proving Termination with Multiset Orderings, CACM 22, (1979). Propositional Logic (25th October 2007) 40

41 Proof of Termination Let G = [H 11,..., H 1,n1 ],..., [H m1,..., H mn m]. We define: f(g) = { n 1 X j=1 r(h 1j ),..., nmx j=1 r(h mj ) } Observation With every execution of the while-loop, the associated multisets become smaller with respect to /2 (Proof Exercise) Termination. qed Example [ (p ( (p q) r))] {5 } [ p], [ ( (p q) r)] {0, 4 } [ p], [ (p q), r] {0, 3 } [ p], [(p q), r] {0, 2 } [ p], [(p q), r] {0, 1 } [ p], [p, r], [q, r] {0, 0, 0 } Observation If G is in clause form, then f(g) = {0,..., 0 } and the number of clauses occurring in G is equal to the number of occurrences of 0 in f(g). Propositional Logic (25th October 2007) 41

42 3.4 Proof Methods {F 1,..., F n } = F iff iff iff ( F 1,..., F n F ) is valid ( F 1,..., F n F ) is unsatisfiable F 1,..., F n, F is unsatisfiable. Truth tabling Resolution Semantic tableaus Calculus of natural deduction Sequent calculus Connection method Hilbert systems Propositional Logic (25th October 2007) 42

43 3.4.1 Resolution Resolution is a method to show unsatisfiability of formulas. Here: Wlog we assume that formulas are in clause form. A formula in clause form in unsatisfiable if it contains the empty clause, or it can be transformed into a semantically equivalent formula which contains the empty clause. Resolution is search for the empty clause using an appropriate inference rule. Propositional Logic (25th October 2007) 43

44 Example Some Facts K 1 If it is hot and humid, then it will rain. K 2 If it is humid, then it is hot. It is humid now. K 3 Question K 4 Will it rain? Introducing propositional variables p q r it is hot it is humid it will rain Formalization F 1 ((p q) r) F 2 (q p) F 3 q r F 4 Propositional Logic (25th October 2007) 44

45 Example (continued) {((p q) r), (q p), q} = r iff ( ((p q) r), (q p), q r) is valid iff ( ((p q) r), (q p), q r) is unsatisfiable iff ((p q) r), (q p), q, r is unsatisfiable iff [ p, q, r], [ q, p], [q], [ r] is unsatisfiable Propositional Logic (25th October 2007) 45

46 How to Show Unsatisfiability? Recall [ p, q, r], [ q, p], [q], [ r] Observation {[ p, q, r], [q]} = [ p, r] Hence [ p, q, r], [ q, p], [q], [ r] [ p, q, r], [ q, p], [q], [ r], [ p, r] Propositional Logic (25th October 2007) 46

47 Resolution Rule Definition 3.27 Let C 1 be a clause where the atom A occurs, and C 2 a clause where the negated atom A occurs. Let C be the result of performing the following steps: 1. Remove all occurrences of A from C Remove all occurrences of A from C Combine the clauses obtained in this way disjunctively. C was generated applying the (propositional) resolution rule to C 1 and C 2, where A is the literal that was resolved upon. We also call C the resolvent of C 1 and C 2 wrt A. C may be the resolvent of C 1 and C 1. Propositional Logic (25th October 2007) 47

48 Derivations and Refutations Definition 3.28 Let F = C 1,..., C n be a formula in clause form. 1. The sequence (C i 1 i n) is a resolution derivation for F. 2. If (C i 1 i m) is a resolution derivation for F, and C m+1 is obtained by applying the resolution rule to two elements from (C i 1 i m), then (C i 1 i m + 1) is a resolution derivation for F. 3. A resolution derivation for F which contains the empty clause [ ] is called resolution refutation for F. F may contain [ ]. It suffices to consider refutations in which [ ] occurs once. We may assume that [ ] is the last clause in a refutation. Propositional Logic (25th October 2007) 48

49 Example (continued) 1 [ p, q, r] 2 [ q, p] 3 [q] 4 [ r] 5 [ q, q, r] res(1, 2) 6 [r] res(3, 5) 7 [ ] res(4, 6) Propositional Logic (25th October 2007) 49

50 Resolution Proofs Definition 3.29 Let F be a propositional formula, and G a formula in clause form equivalent to F. A propositional resolution proof for F is a resolution refutation for G. F is called theorem of the resolution calculus, if there is a resolution proof for F. We denote by r F, that F has a resolution proof. Requirements Soundness If r F, then = F. Completeness If = F, then r F. Observation Let F be a formula with n propositional variables. (F (p p)) is unsatisfiable. The proof consists of a single resolution step! Propositional Logic (25th October 2007) 50

51 3.4.4 Further Proof Methods and Calculi Definition 3.35 A calculus consists of an alphabet, a language, a set of axioms, i. e., a set of formulas of the language, and a set of inference rules defining the provability relation. Definition 3.36 Let L be a logic and C a calculus, both defined over the same alphabet and the same language. Let = be the logical consequence relation of L and the provability relation of C. C is sound wrt L, if for every formula F of the language we find: if F holds then so does = F. C is complete wrt L, if for every formula F of the language we find: if = F holds then so does F. Propositional Logic (25th October 2007) 51

52 A Characterization of Calculi Definition 3.37 A calculus is positive, if its axioms are valid; negative, if its axioms are unsatisfiable; generating, if it operates by proving a formula starting out from the axioms; analyzing, if it operates by reducing the formula to be proved to the axioms. Propositional Logic (25th October 2007) 52

53 Additional Calculi and Methods Calculi Semantic Tableaus Hilbert Systems Calculus of Natural Deduction Sequent Calculus Davis-Putman-Logeman-Loveland (DPLL) Procedure Connection Method Model Generation Greedy Satisfiability Testing Binary Decision Diagrams Propositional Logic (25th October 2007) 53

54 Natural Deduction Gentzen 1935: How to represent logical reasoning in mathematics? Calculus of natural deduction. It s alphabet is the alphabet of propositional logic. Remember [ ] denotes some unsatisfiable formula. It s language is the language of propositional logic. Propositional Logic (25th October 2007) 54

55 An Introductory Example Suppose we want to show the validity of A proof could look like the following: ((p (q r)) ((p q) (p r))). Assume that (p (q r)) holds. We distinguish two cases: (i) Assume that p holds. Hence, (p q) must hold. (ii) Assume that (q r) holds. Hence, q and, consequently, (p q) must hold. Thus, (p q) must hold in any case. (1) Likewise, we conclude that (p r) must hold. (2) From (1) and (2) we conclude that ((p q) (p r)) holds. We can now cancel the assumption and, thus, obtain a proof of ((p (q r)) ((p q) (p r))). Propositional Logic (25th October 2007) 55

56 Inference Rules Example Elimination of the implication G (G F ) F ( E) Schema which needs to be instantiated, e.g., (p q) ((p q) (q p)) (q p) ( E) Propositional Logic (25th October 2007) 56

57 Derivations Some Notation Derivations have the form of trees, where the nodes are labeled by formulas. Derivations will be denoted by A derivation where the root is labeled by F is denoted by This will be called a derivation of F. A formula labeling a leaf node in a derivation is called hypothesis or assumption. If we want to single out a hypothesis G in a derivation of F, then we will write F G F This will be called a derivation of F from G. Propositional Logic (25th October 2007) 57

58 Derivations More Notation Some inference rule will cancel hypotheses. If a hypothesis has been cancelled, then this will be denoted by marking the hypothesis with. The hypotheses cancelled by the application a of a rule as well as a itself will be marked by a unique index. Propositional Logic (25th October 2007) 58

59 Natural Deduction - Inference Rules Negation [ ] G (f) F [ ] F (raa) F [ ] F ( I) F F [ ] ( E) Conjunction F G (F G) ( I) (F G) F ( E) (F G) G ( E) Disjunction F (F G) ( I) G (F G) ( I) (F G) F H H G H ( E) Propositional Logic (25th October 2007) 59

60 Natural Deduction Inference Rules (Continued) Implication G F (G F ) ( I) F G ( I) (G F ) G (G F ) F ( E) Equivalence G F (G F ) F G ( I) G (G F ) F ( E) F (G F ) G ( E) Propositional Logic (25th October 2007) 60

61 Derivations (1) The set of derivations in the calculus of natural deduction is the smallest set X with the following properties 1 L(R) X. 2 If is a derivation in X and H 1 H 1 H 2 is the instance of a rule r {(f), ( E), ( I)}, then H 1 r H 2 is a derivation in X. Propositional Logic (25th October 2007) 61

62 Derivations (2) 3 If H 1 H 2 is a derivation in X, j is a new index, and H 1 H 2 H 3 is an instance of a rule r {(raa), ( I), ( I)}, then H 1 j H 2 is a derivation in X. The hypothesis H 1 was cancelled by the application of the rule. H 3 r j Propositional Logic (25th October 2007) 62

63 Derivations (3) 4 If are derivations in X and if 1 2 H 1, H 2 1 H 1 2 H 2 H 3 is an instance of a rule r {( E), ( E), ( E)}, then 1 H 1 2 H 2 r H 3 is a derivation in X. Propositional Logic (25th October 2007) 63

64 Derivations (4) 5 If (F G), H, H are derivations in X and j is a new index, then F G 1 (F G) F j 2 H H G j 3 H ( E) j is a derivation in X. The hyotheses F and G are cancelled by the application of ( E). Propositional Logic (25th October 2007) 64

65 Derivations (5) 6 If 1 2 F, G are derivations in X and j is a new index, then G F G j 1 F (G F ) F j 2 G ( I) j is a derivation in X. The hypotheses G and F are cancelled by the application of ( I). Propositional Logic (25th October 2007) 65

66 Proofs An element of X is called derivation of F in the calculus of natural deduction if in its tree representation the root node is labeled by F. A proof of F in the calculus of natural deduction is a derivation of F in the calculus of natural deduction, where each hypothesis is cancelled. This will be denoted by n F. A Simple Example A proof of (p p) p 1 ( I) 1 (p p) Propositional Logic (25th October 2007) 66

67 The Introductory Example Revisited (p (q r)) 1 p 2 (q r) 2 ( E) q p 3 ( I) ( I) ( I) (p q) (p q) (p (q r)) 1 (p r) ( E) 2 (p q) (p r) ( I) ((p q) (p r)) ( I) 1 ((p (q r)) ((p q) (p r))) (q r) 3 ( E) r ( I) (p r) ( E) 3 Please compare this to the informal proof given at the beginning. Propositional Logic (25th October 2007) 67

68 Lemmas (1) Subderivations may occur over and over again, e.g., F F 1 ( E) [ ] F (raa) 1 Idea Generate such derivations once and for all, specify corresponding inference rules called lemmas, and apply the lemmas in addition to the usual inference rules. In the example we obtain F F (l1) Propositional Logic (25th October 2007) 68

69 Lemmas (2) The lemma (F G) G F (l2) and its derivation F G 1 ( I) (F G) [ ] G (F G) ( I) 1 ( E) Propositional Logic (25th October 2007) 69

70 Lemmas (3) The lemma (F G) F G (l3) and its derivation F 1 G ( I) (F G) (F G) ( E) [ ] ( I) 1 F Propositional Logic (25th October 2007) 70

71 Lemmas (4) The lemma (F G) ( F G) (l4) and its derivation (F G) [ ] F F 1 ( I) (F G) (F G) ( E) [ ] ( I) 1 G ( F G) G 2 (F G) ( I) 2 ( I) ( I) ( E) Propositional Logic (25th October 2007) 71

72 Another Example Demonstrating the Use of Lemmas A proof of (p p) (p p) 1 (l4) ( p p) ( E) p (l1) p [ ] (p p) (p p) 1 (l4) ( p p) ( E) p ( E) (raa) 1 Propositional Logic (25th October 2007) 72

73 Some Remarks on Natural Deduction Lemmas shorten proofs, but enlarge the search space. The calculus of natural deduction is sound and complete. Human readable proofs versus machine generated proofs. Sequent calculus. Proof theory. Propositional Logic (25th October 2007) 73

74 Sequent Calculus A sequent is an expression of the form F G, where F and G are multisets of formulas. Notation We use the following abbreviations: {F 1,..., F n } F1,..., F n. F {G } F, G. The inference rules are of the form S 1... S n r or S S r Propositional Logic (25th October 2007) 74

75 Sequent Calculus: Axiom, Cut and Structural Rules Axiom Cut H, F G, H ax F G, H F G H, F G cut Structural rules H, H, F G H, F G cl F G, H, H F G, H cr Propositional Logic (25th October 2007) 75

76 Sequent Calculus: Logical Rules Rules for the left hand side Rules for the right hand side F G, H 1 H 2, F G (H 1 H 2 ), F G l H 1, F G, H 2 F G, (H 1 H 2 ) r H 1, H 2, F G (H 1 H 2 ), F G l F G, H 1 F G, H 2 r F G, (H 1 H 2 ) H 1, F G H 2, F G (H 1 H 2 ), F G l F G, H 1, H 2 F G, (H 1 H 2 ) r F G, H H, F G l H, F G F G, H r Propositional Logic (25th October 2007) 76

77 Sequent Calculus: Soundness and Completeness A proof of the sequent S in the sequent calculus is defined as follows: An axiom of the form r is a proof of S. S If T 1,..., T n are proofs of S 1,..., S n respectively and the sequent calculus contains a rule of the form S 1... S n r, then T 1... T n r S S of S. is a proof The propositional sequent calculus is sound and complete. Subformula property: all formulas occurring in the condition of a rule occur as subformulas in the conclusion of the rule. Only the cut rule violates this property. Gentzen s Hauptsatz : Cuts can be removed in sequent calculus proofs. Propositional Logic (25th October 2007) 77

78 3.5 Properties Compactness Theorem Soundness and Completeness Theorems Propositional Logic (25th October 2007) 78

79 3.5.1 Compactness Theorem Introduction How can we decide whether a possibly infinite set of formulas is satisfiable? Idea We test all finite subsets. Remember is an equivalence relation on L(R). Let L(R, n) L(R) be the set of formulas in which only the variables p 1,..., p n occur. There are 2 n different interpretations for L(R, n). How many different equivalence classes are defined by on L(R, n)? 2 2n. Propositional Logic (25th October 2007) 79

80 Compactness Theorem Theorem 3.39 (Compactness Theorem) A set F of propositional formulas is satisfiable iff every finite subset G F is satisfiable. Proof Let F be a (possibly infinite) set of formulas. We have to show: (a) If F is satisfiable, so is each finite G F. (b) If each G F is satisfiable, so is F. (a) follows immediately from the definition of satisfiability for sets. To show (b) we have to prove the existence of a model for F starting from the models for each finite G F. Propositional Logic (25th October 2007) 80

81 Proof of Part (b) of the Compactness Theorem Suppose, each finite G F is satisfiable. Let G n = F L(R, n). There are at most k 2 2n different equivalence classes on G n defined by. Let G 1,..., G k be representatives of these equivalence classes. Hence, for each G G n there is an i, 1 i k, with G G i. Hence, each model for {G 1,..., G k } is also a model for G n. Because {G 1,..., G k } is a finite subset of F, it has a model. Let I n be this model. Because G 1 G 2... G n, I n is also a model for all G j, 1 j n. Propositional Logic (25th October 2007) 81

82 Proof of Part (b) of the Compactness Theorem (continued) We construct a model I for F as follows: Set I :=, K 0 := N and n = 1. Do the following: If there are infinitely many j K n 1 with [p n ] I j =, then set I := I {p n } and K n := K n 1 \ {j [p n ] I j = }, else set K n := K n 1 \ {j [p n ] I j = }. Set n := n + 1. Goto Do the following. ( ) ( ) I is an interpretation. By induction the following proposition E(n) can be proven: K n contains infinitely many elements and for all m K n with m > n we find: [p i ] Im = [p i ] I for all 1 i n. Proof Exercise. Propositional Logic (25th October 2007) 82

83 Proof of Part (b) of the Compactness Theorem (continued) To show: I is a model for F. Let F be an arbitrary but fixed element of F. There are at most finitely many propositional variables occurring in F. Let p n be the variable with the highest index. Then, F G n G n+1... and each interpretation I n, I n+1,... is a model for F. Because of E(n) we find: K n contains infinitely many elements. For all m K with m > n we find: [p i ] Im = [p i ] I for all 1 i n. Select m K n with m > n. Because I m is a model for F and F contains only the variables p 1,..., p n, I must be a model for F as well. Because F is an arbitrary element from F, I must also be a model for F. qed Propositional Logic (25th October 2007) 83

84 Implications of the Compactness Theorem The proof is not constructive. We may negate both sides of the compactness theorem. Corollary 3.40 A set of propositional formulas is unsatisfiable iff one of its finite subsets is unsatisfiable. Procedure to prove unsatisfiability of F: We generate all finite subsets of F in a systematic way and test them for unsatisfiability. Propositional Logic (25th October 2007) 84

85 3.5.2 Soundness and Completeness Theorems Lemma 3.41 (Propositional Resolution Lemma) Let F = C 1,..., C n be a propositional formula in clause form with the clauses C i, 1 i n, and let C be a resolvent of two clauses from F. Then F (F C) holds. Proof To show I = (F C) iff I = F. immediate. Suppose I = F. Let C be the resolvent of C 1 and C 2 wrt A. Let C 1 = C 1 without A and C 2 = C 2 without A. Then C = (C 1 C 2 ). Case 1 I = A. Then, I = A. Because I = C 2 we conclude I = C 2. Hence, I = C. Case 2 I = A. Because I = C 1 we find I = C 1. Hence I = C. qed Propositional Logic (25th October 2007) 85

86 An Implication of the Resolution Lemma Corollary 3.41(a) Let F = C 1,..., C n be a propositional logic formula in clause form with clauses C i, 1 i n, and let D 1,..., D m be the computed resolvents in a resolution derivation from F. Then, F F, D 1,..., D n. Proof Induction over n Exercise. Propositional Logic (25th October 2007) 86

87 Resolution Method Theorem 3.42 (Soundness and Completeness of the Prop. Resolution Method) Let F be a propositional formula. = F iff r F. Proof Soundness To show If r F, then = F. Suppose r F. By Definition 3.29 we find a resolution refutation for F, i.e. by Definition 3.28, a resolution derivation for F containing [ ]. By Theorem 3.23 we may assume wlog that F is in clause form. Let C 1,..., C n, [ ] be all resolvents computed in the refutation. By the Corollary 3.41(a) we find F F, C 1,..., C n, [ ]. Because [ ] is unsatisfiable we conlude by Theorem 3.17 that F [ ], i.e. F ist unsatisfiable. Applying Theorem 3.12 we find = F. Propositional Logic (25th October 2007) 87

88 Completeness of the Resolution Method Proof Completeness To show If = F, then r F. Suppose = F. By Theorem 3.23 we find a formula G in clause form with G F. By Theorem 3.12 we find that G is unsatisfiable. To show: there is a resolution refutation for G The proof is by induction on the number n of propositional variables occurring in G. n = 0 Then G = [ ],..., [ ] and we are done. n n + 1 Assume that the proposition holds for n, i.e. if G is unsatisfiable and in G occur only the variables p 1,..., p n, then there is a resolution refutation for G. Wlog we may assume that p 1,..., p n are the n variables occurring in G Exercise. Propositional Logic (25th October 2007) 88

89 Completeness of the Resolution Method (continued) Let G be a formula in clause form in which the variables p 1,..., p n+1 occur. Let G be the formula obtained from G by deleting each clause, in which p n+1 occurs, and deleting each occurrence of p n+1. Let G be the formula obtained from G by deleting each clause, in which p n+1 occurs, and deleting each occurrence of p n+1. We find: G and G are unsatisfiable if G is unsatisfiable Exercise. Propositional Logic (25th October 2007) 89

90 Completeness of the Resolution Method (continued) Applying the induction assumption we find refutations for G and G. resolvents G resolvents G. Case 1 resolution refutation for G. Case 2 resolution derivation with [ p n+1,... p n+1 ] as last entry. resolvents G resolvents G. Case 1 resolution refutation for G. Case 2 resolution derivation with [p n+1,..., p n+1 ] as last entry. If case 2 was selected twice, then compute the resolvent of [p n+1,..., p n+1 ] and [ p n+1,..., p n+1 ] resolution refutation for G. An application of the induction principle yields the desired result. qed Propositional Logic (25th October 2007) 90

91 Satisfiability Up to now Does = F hold? Now Is F satisfiable? Goal To find a model for F. Applications optimization problems, planning problems, etc. Method greedy satisfiability testing (gsat) Propositional Logic (25th October 2007) 91

92 Some Definitions Let F be a formula in clause form. C F iff clause C occurs in F. F u (I) = {C C F and I = C}. flip(i, p) denotes the interpretation obtained from I by reversing the truth value assigned to p. rank(i, F, p) = F u (I) F u (flip(i, p)) denotes the rank of F wrt I and p. Propositional Logic (25th October 2007) 92

93 Satisfiability Testing maxtries, maxloops N. initial selects an arbitrary interpretation I for F. pick selects an arbitrary element from a set of propositional variables. flip implements flip. rank impements rank. Propositional Logic (25th October 2007) 93

94 A Generic Algorithm Input: Output: A propositional formula F in clause form. A model for F, if one is found, or no model found. procedure gensat(f) for J := 1 to maxtries do I := initial(f); for K := 1 to maxloops do if I = F then return I; I := flip(i,pick(hillclimb(f,i))); end for; end for; return no model found ; end gensat; Propositional Logic (25th October 2007) 94

95 Hillclimbing Input: A propositional formula F in clause form and an interpretation I. Output: A set of variables occurring in F. procedure hillclimb(f,i); r := rnd(0,1); return {p select(f,i,p,r) = }; end hillclimb; gsat select(f, I, p, r) = j if rank(f, I, p) = max{rank(f, I, p ) p R}, otherwise. gwsat select(f, I, p, r) = 8 >< >: if r r and rank(f, I, p) = max{rank(f, I, p ) p R}, if r < r and p occurs in F u (I), otherwise. where r is some fixed value in [0, 1]. Propositional Logic (25th October 2007) 95

96 Some Results 3-CNF formulas (formulas in conjunctive normal form, each clause containing 3 literals). maxtries = 1, maxloops = N 100, where N is the number of variables occurring in F. maxtries = 15 : 26.3% 98.8%. N variant forced unforced 100 gsat 76.1% 23.5% gwsat 100% 82.2% 200 gsat 75.3% 8.4% gwsat 99% 48.7% 300 gsat 77.0% 2.7% gwsat 100% 26.3% 500 gsat 70.2% gwsat 100% Propositional Logic (25th October 2007) 96