4 Predicate / First Order Logic

Transcription

1 4 Predicate / First Order Logic 4.1 Syntax 4.2 Substitutions 4.3 Semantics 4.4 Equivalence and Normal Forms 4.5 Unification 4.6 Proof Procedures 4.7 Implementation of Proof Procedures 4.8 Properties First Order Logic (24th November 2008) 1

2 4.1 Syntax Definition 4.1 An alphabet of (first order) predicate logic consists of a finite or countably infinite set R of relation symbols, a finite or countably infinite set F of function symbols, a countably infinite set V of variables, the set { /1, /2, /2, /2, /2} of connectives, the set {, } of quantifiers, and the special characters (, ), and,. First Order Logic (24th November 2008) 2

3 Syntax (Continued) is called universal quantifier, is called existential quantifier. An arity n N is assigned to each function and relation symbol. Nullary function symbols are called constant symbols. Notation p, q,... relation symbols, p/n relation symbol with arity n, g, h,... function symbols, g/n function symbol with arity n, a, b,... constant symbols, X, Y,... variables, Agreement In the following we assume that R, F, and V are given. First Order Logic (24th November 2008) 3

4 Terms Definition 4.2 The set T (F, V) of terms is the smallest set satisfying the following conditions: 1. Every variable X V is a term. 2. If g/n F and {t 1,..., t n } T (F, V), then the string g(t 1,..., t n ) is a term. A term is closed or ground (instantiated), if it does not contain any variables. The set of ground terms is denoted by T (F). Notation We write g instead of g(). First Order Logic (24th November 2008) 4

5 Predicate Logic Formulas Definition 4.3 The set of atomic formulas (or, briefly, atoms) of the language L(R, F, V) is the set of strings of the form p(t 1,..., t n ), where p/n R and {t 1,..., t n } T (F, V). Notation We write p instead of p(). Definition 4.4 The set of predicate logic formulas of the language L(R, F, V) is the smallest set satisfying the following conditions: 1. Every atom is a formula. 2. If F is a formula, then F is a formula. 3. If F 1 and F 2 are formulas and /2 is a binary connective, then (F 1 F 2 ) is a formula. 4. If F is a formula, Q is a quantifier and X V, then (QX) F is a formula. Notation A (possibly indexed) denotes an atom, F, G, H (possibly indexed) denote formulas. First Order Logic (24th November 2008) 5

6 Structural Induction Structural induction theorems can be formulated for T (F, V) and L(R, F, V). Every term in T (F, V) has property E, if the following conditions are satisfied: 1. Induction basis: Every variable has property E. 2. Induction step: If f/n is an n-ary function symbol, and the terms t 1,..., t n in T (F, V) have property E, then the term f(t 1,..., t n ) also has property E. Likewise for formulas Exercise. First Order Logic (24th November 2008) 6

7 Structural Recursion Structural recursion theorems can be formulated for T (F, V) and L(R, F, V). There is exactly one function foo over the set of terms satisfying the following conditions: 1. Recursion base: The value of foo for variables and constant symbols is explicitly defined. 2. Recursion step: The value of foo for f(t 1,..., t n ), n > 0, is defined depending only on the values of foo for t 1,..., t n. Example foo(t ) = 8 < : 0 if T is a variable, 1 if T is a constant symbol, P n i=1 foo(t i) if T is of the form f(t 1,..., t n ). Likewise for formulas Exercise. First Order Logic (24th November 2008) 7

8 Subterms and Subformulas Definition 3.8 (propositional subformulas) can be extended to terms and first order formulas Exercise. Example {g(f(x), Y ), f(x), X, Y } is the set of subterms of the term g(f(x), Y ). { ( X)( Y ) (q(x) p(g(a, b), f(f(y )))), ( Y ) (q(x) p(g(a, b), f(f(y )))), (q(x) p(g(a, b), f(f(y )))), q(x), p(g(a, b), f(f(y ))) } is the set of subformulas of the formula ( X)( Y ) (q(x) p(g(a, b), f(f(y )))). First Order Logic (24th November 2008) 8

9 Free and Bound Occurrences of Variables Definition 4.5 The free occurrences of a variable in a predicate logic formula are defined as follows: 1. The free occurrences of a variable in an atomic formula F are the occurrences of a variable in F. 2. The free occurrences of a variable in a formula F are the free occurrences of a variable in F. 3. The free occurrences of a variable in a formula (F 1 F 2 ) are the free occurrences of a variable in F 1 together with the free occurrences of a variable in F The free occurrences of a variable in a formula (QX) F are the free occurrences of a variable in F without the occurrences of X. The occurrence of a variable is called bound, if it is not free. First Order Logic (24th November 2008) 9

10 Closed Terms and Formulas Remember A term is closed if it does not contain any occurrence of a variable. Definition 4.6 A closed formula (or, briefly, a sentence) is a formula, where every occurrence of a variable is bound. First Order Logic (24th November 2008) 10

11 4.2 Substitutions Definition 4.7 A substitution is a mapping σ : V T (F, V), where only a finite number of variables is not mapped to itself. Let σ be a substitution: dom(σ) = {X X V and σ(x) X}. dom(σ) is finite. σ can be represented by a finite set of pairs and vice versa. {X σ(x) X dom(σ)} If dom(σ) =, then σ is called empty substitution. ε denotes an empty substitution. The restriction of σ to a set U V of variables is defined as σ U = {X t X t σ and X U}. First Order Logic (24th November 2008) 11

12 Instances Notation Instead of σ(x) we will write Xσ in the sequel. Definition 4.8 Let σ be a substitution. σ : V T (F, V) is extended to a mapping bσ : T (F, V) T (F, V) as follows: Let t T (F, V), then tbσ = j tσ if t V, f(t 1 bσ,..., t n bσ) if t is of the form f(t 1,..., t n ). tbσ is called instance of t under bσ. If tbσ is closed, then tbσ is called ground instance of t under bσ, and bσ is said to be a ground substitution for t. Proposition 4.9 Let t be a term, µ = {X r} a substitution, and assume that variable X does not occur in t. Then, tbµ = t. First Order Logic (24th November 2008) 12

13 Composition of Substitutions Definition 4.10 Let σ and θ be substitutions. The composition σθ of σ and θ is the substitution {X t b θ X t σ and X t b θ} {Y s Y s θ and Y dom(σ)}. Corollary 4.11 For each substitution σ we find εσ = σ = σε. Proposition 4.12 Let σ and θ be substitutions. For every term t, we have t( c σθ) = (tbσ) b θ. Proof Structural induction over t Exercise. First Order Logic (24th November 2008) 13

14 Composition of Substitutions (continued) Proposition 4.13 Let t T (F, V) and σ, θ and λ be substitutions. Then: 1. t( (σθ)λ) = t( σ(θλ)). 2. (σθ)λ = σ(θλ). Proof 1. t( (σθ)λ) = (t( σθ)) c λ b (Proposition 4.12) = ((tbσ) θ) b λ b (Proposition 4.12) = (tbσ)( θλ) c (Proposition 4.12) = t( σ(θλ)) (Proposition 4.12) 2. Exercise qed Notation From now on we write σ instead of bσ. First Order Logic (24th November 2008) 14

15 More on Substitutions Definition 4.14 Let σ be a substitution. σ X = j σ σ \ {X t} if X dom(σ) if X t σ Proposition 4.15 Let σ be a substition and t a term, and assume that variable X does not occur in t. Then, tσ = tσ X. First Order Logic (24th November 2008) 15

16 Applications of Substitutions to Formulas Definition 4.16 Let σ be a substitution. The application of a substitution to a formula is defined as follows: 1. p(t 1,..., t n )σ = p(t 1 σ,..., t n σ) for every atom of the form p(t 1,..., t n ). 2. ( F )σ = (F σ). 3. (F 1 F 2 )σ = (F 1 σ F 2 σ) for every binary connective /2. 4. ((QX) F )σ = (QX) (F σ X ) for every quantifier Q. If F is a formula and F σ is closed, then F σ is called ground instance of F under σ, and σ is said to be a ground substitution for F. First Order Logic (24th November 2008) 16

17 Substitutions and Formulas Definition 4.17 conditions: A substition σ is free for F if it free according to the following 1. σ is free for F if F is an atom. 2. σ is free for F iff σ is free for F. 3. σ is free for (F G) iff σ is free for F and G. 4. σ is free for (QY ) F iff σ Y is free for F and if for all variables X, which occur freely in F and which are different from Y, we find that Y does not occur in Xσ. Theorem 4.18 If σ is free for F and θ is free for F σ, then (F σ)θ = F (σθ). First Order Logic (24th November 2008) 17

18 Proof of Theorem 4.18 (1) Theorem 4.18 If σ is free for F and θ is free for F σ, then (F σ)θ = F (σθ). Proof Structural induction on F. IA If F is an atom of the form p(t 1,..., t n ), then p(t 1,..., t n )(σθ) = p(t 1 (σθ),..., t n (σθ)) = p((t 1 σ)θ,..., (t n σ)θ) = p(t 1 σ,..., t n σ)θ = (p(t 1,..., t n )σ)θ. IH The result holds for F. IS F Let σ be free for F and θ free for ( F )σ. Because σ is free for F, so is σ for F. Because θ is free for ( F )σ and ( F )σ = (F σ), so is θ for F σ. (( F )σ)θ = ( (F σ))θ = ((F σ)θ) = (IH) (F G) Exercise. (F (σθ)) = ( F )σθ. First Order Logic (24th November 2008) 18

19 Proof of Theorem 4.18 (2) ( X) F Let σ be free ( X) F θ be free for (( X) F )σ. Because σ is free for ( X) F, so is σ X for F. Because θ is free for (( X) F )σ = ( X) (F σ X ), so is θ X for F σ X. Conjecture F (σ X θ X ) = F (σθ) X. Then: ((( X) F )σ)θ = (( X) (F σ X ))θ (Definition 4.14(4.)) = ( X) ((F σ X )θ X ) (Definition 4.14(4.)) = ( X) (F (σ X θ X )) (IH) = ( X) (F (σθ) X ) (Conjecture) = (( X) F )(σθ) (Definition 4.14(4.)) ( X) F Exercise. First Order Logic (24th November 2008) 19

20 Proof of Theorem 4.18 (3) Conjecture Under the given assumptions we find F (σ X θ X ) = F (σθ) X. Proof Because only variables occurring freely in F are replaced, it suffices to prove for each Y occurring in F that Y (σ X θ X ) = Y (σθ) X. Y = X Y (σ X θ X ) = Y = Y (σθ) X. Y X Y σ = Y σ X and Y (σθ) = Y (σθ) X. Because σ is free for ( X) F, X does not occur in Y σ. Hence, (Y σ)θ = (Y σ)θ X. Therefore, Y (σ X θ X ) = (Y σ X )θ X (Proposition 4.12) = (Y σ)θ X (X Y ) = (Y σ)θ (s.a.) = Y (σθ) (Proposition 4.12) = Y (σθ) X (X Y ) First Order Logic (24th November 2008) 20

21 Variants Definition 4.19 Let E 1 and E 2 be either two terms or two predicate logic formulas. E 1 and E 2 are called variants, if there are substitutions σ and θ such that E 1 = E 2 σ and E 2 = E 1 θ. In this case, we will call E 1 variant of E 2 and E 2 variant of E 1. If E 1 and E 2 are variants and the variables occurring in E 2 have not been used so far in the context, then E 2 is a new variant of E 1. First Order Logic (24th November 2008) 21

22 4.3 Semantics What is the meaning of predicate logic formulas? Relations and Functions Interpretations Herbrand-Interpretations Models for Closed Formulas Models for Non-Closed Formulas First Order Logic (24th November 2008) 22

23 4.3.1 Relations and Functions Let D be a set. D 2 = D D = {(d 1, d 2 ) d 1 D and d 2 D} D n = D n 1 D = {(t, d n ) t D n 1 and d n D} =: {(d 1,..., d n ) d i D, 1 i n}, n > 2 D 1 = D =: {(d) d D} D 0 = = {( )} Relations R D 2 R = {(n, m) n, m N and n < m} R D 3 R = {(x, y, z) x, y, z N and x 2 + y 2 = z 2 } R D 1 R = {(n) n N and n even} R D 0 R = or R = {( )} First Order Logic (24th November 2008) 23

24 Functions Remember Let D be a set. D 2 = D D = {(d 1, d 2 ) d 1 D and d 2 D} D n = D n 1 D = {(t, d n ) t D n 1 and d n D} =: {(d 1,..., d n ) d i D, 1 i n}, n > 2 D 1 = D =: {(d) d D} D 0 = = {( )} Functions + : N 2 N +(2, 3) 5 succ : N 1 N succ(3) 4 0 : N 0 N 0( ) 0 First Order Logic (24th November 2008) 24

25 4.3.2 Interpretations and Models Definition 4.20 A predicate logic interpretation I for a predicate logic language L(R, F, V) consists of a non-empty set D and a mapping I, which satisfies the following conditions: 1. Every n-ary function symbol g/n F is mapped to an n-ary function g I : D n D. 2. Every n-ary relation symbol p/n R is mapped to an n-ary relation p I D n. D is called domain of the interpretation. First Order Logic (24th November 2008) 25

26 Variable Assignments What is the meaning of a variable? Definition 4.21 A variable assignment wrt an interpretation I = (D, I) is a mapping Z : V D. The image of a variable X under Z is denoted by X Z. Let Z be a variable assignment and d D. {X d}z denotes the variable assignment, where: Y {X d}z = j d if Y = X, otherwise. Y Z First Order Logic (24th November 2008) 26

27 Interpretation of Terms Definition 4.22 Let I = (D, I) be an interpretation and Z a variable assignment wrt I. The meaning t I,Z of a term t T (F, V) is defined as follows: 1 For every variable X V we have X I,Z = X Z. 2 For every term of the form g(t 1,..., t n ) we have [g(t 1,..., t n )] I,Z = g I (t I,Z 1,..., t I,Z n ), where g/n F and t 1,..., t n are terms. First Order Logic (24th November 2008) 27

28 Example Let F = {h/1, g/2, a/0}. D 1 = N with successor function s/1 and addition +/2. D 2 = Z with predecessor function p/1 and subtraction /2. D 3 = set of words over Σ = {o, m} with function add m /1, which appends m to a word at the right, and function conc/2, which concatenates two words. a h g X Z g(h(a), h(h(a))) g(h(x), g(h(a), h(x))) I 1 0 s I 2 1 p I 3 o add m conc omo omomm omomomomom First Order Logic (24th November 2008) 28

29 Interpretation of Formulas Definition 4.23 Let I = (D, I) be an interpretation and Z a variable assignment wrt I. I and Z assign to every formula F L(R, F, V) a truth value F I,Z as follows: 1. [p(t 1,..., t n )] I,Z = iff (t I,Z 1,..., t I,Z n ) pi. 2. [ F ] I,Z = (F I,Z ). 3. [(F 1 F 2 )] I,Z = (F I,Z 1 F I,Z 2 ) for all binary connectives /2. 4. [( X) F ] I,Z = iff for all d D: F I,{X d}z =. 5. [( X) F ] I,Z = iff for some d D: F I,{X d}z =. First Order Logic (24th November 2008) 29

30 Interpretations of Closed Formulas Proposition 4.24 If F L(R, F, V) is closed, then F I,Z = F I,Z for all interpretations I and variable assignments Z and Z wrt I. Proof Exercise If formulas are closed, then their meaning is independent of a variable assignment. Notation If F is closed, then we write F I instead of F I,Z. First Order Logic (24th November 2008) 30

31 Substitutions and Variable Assignments Lemma 4.25 Let s, t be terms, G a formula, Y a variable, I = (D, I) an interpretation and Z a variable assignment wrt I and d D. If [t] I,Z = d, then: 1. [s{y t}] I,Z = [s] I,{Y d}z, 2. [G{Y t}] I,Z = [G] I,{Y d}z if t is free for Y in G. Proof induction on the structure of s and G Exercise. First Order Logic (24th November 2008) 31

32 4.3.3 Herbrand-Interpretations In the following we assume that F contains at least one constant. Otherwise we add a symbol a/0 to F. Definition 4.26 An interpretation I = (D, I) for a predicate logic language L(R, F, V) is a Herbrand-interpretation, if the following conditions are satisfied: 1. D = T (F); it is called Herbrand-universe. 2. For every t T (F) we have t I = t. F = {a/0, b/0, h/1, g/2}. T (F) = {a, b, h(a), h(b), g(a, a), g(a, b), g(b, a), g(b, b), h(h(a)),...}. First Order Logic (24th November 2008) 32

33 Herbrand-Interpretations and Formulas L({p/1, q/1, r/1}, {g/1, a/0}, V) F = (( X) (p(x) q(x)) ( Y ) r(g(y ))) T (F) = {a, g(a), g(g(a)),...} I 1 : p I 1 = q I 1 = r I 1 = T (F). I 1 is a model for F. I 2 : p I 2 = q I 2 = r I 2 =. I 2 is not a model for F. I 3 : p I 3 = {g(g(a)), g(g(g(g(a)))),...}, q I 3 = {a, g(a)}, r I 3 I 3 is not a model for F. = {g(g(a))}. Notation I 1 = {p(a), q(a), r(a), p(g(a)), q(g(a)), r(g(a)),...} I 2 = I 3 = {p(g(g(a))), p(g(g(g(g(a))))),..., q(a), q(g(a)), r(g(g(a)))} First Order Logic (24th November 2008) 33

34 4.3.4 Models for Closed Formulas Definition 4.27 Let I = (D, I) be an interpretation and F L(R, F, V) be closed. I is called model for F, in symbols I = F, if F I = holds. Many notions and results known from propositional logic can be extended to predicate logic. For example: Validity, satisfiability, falsifiability, unsatisfiability. E.g., a closed formula F is valid iff all interpretations are models for F. Theorem 3.12 (extended) A closed formula F is valid iff F is unsatisfiable. Theorem 3.15 (extended) Let F, F 1,..., F n be closed formulas. {F 1,..., F n } = F iff = ( F 1,..., F n F ). First Order Logic (24th November 2008) 34

35 Logical Consequence Definition 4.28 A closed formula F is a (predicate logic) consequence of a set G of closed formulas, in symbols G = F, iff every model for G is also a model for F. First Order Logic (24th November 2008) 35

36 Propositional vs. First Order Logic What is the relation between propositional and first order logic? If all relation symbols in R are nullary, then first order logic is equivalent to propositional logic. If no variables are occurring in the formulas, then first order logic is equivalent to propositional logic. First Order Logic (24th November 2008) 36

37 4.3.5 Models for Non-Closed Formulas We want to extend the notion of models for all formulas. How shall we treat free variables? Let fv : L(R, F, V) 2 V be a function, which assigns to each formula G the set of variables which occur free in G. Definition 4.29 Let G L(R, F, V) and fv(g) = {X 1,..., X n }. 1. ucl(g) := ( X 1,..., X n ) G is the universal closure of G. 2. ecl(g) := ( X 1,..., X n ) G is the existential closure of G. Definition 4.30 Let G L(R, F, V) and I be an interpretation. 1. I = u G if I = ucl(g). 2. I = e G if I = ecl(g). First Order Logic (24th November 2008) 37

38 Closures of Closed Formulas Propositition 4.31 For all sentences G we find: 1. ucl(g) = G = ecl(g). 2. I = G iff I = u G iff I = e G. In the sequel I will discuss = u. First Order Logic (24th November 2008) 38

39 Some Properties of Universal Closures = u (( X) p(x) p(x)) To show = ( X) (( X) p(x) p(x)). Suppose = ( X) (( X) p(x) p(x)). Then we find (D, I) with [( X) (( X) p(x) p(x))] I =. Then we find Z and d D with [(( X) p(x) p(x))] I,{X d}z =. But this holds only if [( X) p(x)] I = and [p(x)] I,{X d}z =, In other words, for all e D we find e p I and d p I, which is impossible. = u (p(x) ( X) p(x)) To show = ( X) (p(x) ( X) p(x)). Consider ({a, b}, I) and Z with p I = {a}. Then we find [p(x)] I,{X a}z = and [( X) p(x)] I =. Hence, [( X) (p(x) ( X) p(x))] I =. First Order Logic (24th November 2008) 39

40 4.4 Equivalence and Normal Forms Semantic Equivalence Prenex Normal Form Skolem Normal Form Clause Form First Order Logic (24th November 2008) 40

41 4.4.1 Semantic Equivalence In this subsection, formulas need not to be closed. Two formulas F and G are called semantically equivalent, in symbols F G if F I,Z = G I,Z holds for all interpretations I and all variable assignments Z wrt I. Note For closed formulas we find G F if G I = F I for all interpretations I. Theorem 3.19 (propositional semantic equivalences) holds also for first-order formulas. First Order Logic (24th November 2008) 41

42 Some Semantic Equivalences Theorem 4.32 Let F and G be formulas. The following equivalences hold: ( X) F ( X) F ( X) F ( X) F (( X) F ( X) G) ( X) (F G) (( X) F ( X) G) ( X) (F G) ( X) ( Y ) F ( Y ) ( X) F ( X) ( Y ) F ( Y ) ( X) F (( X) F G) ( X) (F G), if X does not occur free in G. (( X) F G) ( X) (F G), if X does not occur free in G. (( X) F G) ( X) (F G), if X does not occur free in G. (( X) F G) ( X) (F G), if X does not occur free in G. First Order Logic (24th November 2008) 42

43 Proof for ( X) F ( X) F Let I = (D, I) be an interpretation and Z a variable assignment wrt I. We find: [ ( X) F ] I,Z = iff [( X) F ] I,Z = iff [( X) F ] I,Z = iff F I,{X d}z = does not hold for all d D iff F I,{X d}z = holds for some d D iff F I,{X d}z = holds for some d D iff [( X) F ] I,Z =. qed First Order Logic (24th November 2008) 43

44 Standardizing Apart Replacement Theorem 3.23 holds also for predicate logic formulas. Definition 4.33 The variables occurring in a formula F are standardized apart, if no two quantifiers occurring in F bind the same variable, and no variable occurs both free and bound. Proposition 4.34 For every formula, there is a semantically equivalent formula where the variables are standardized apart. Auxiliary Proposition Let Q be a quantifier, F a formula of the form (QX) G, and Y a variable not occurring in F. Then F (QY ) G{X Y } holds. Proof Exercise. First Order Logic (24th November 2008) 44

45 Proof Sketch of Proposition 4.34 Proof Let F be a predicate logic formula. We find only finitely many, say n, occurrences of subformulas of the form (QX)G in F. We prove by induction on n that each of these occurrences can be replaced according to the auxiliary proposition. Let G be the formula obtained from F in this way. The variables occurring in G are standardized apart. qed Agreement From now on we assume wlog that variables are standardized apart. First Order Logic (24th November 2008) 45

46 4.4.2 Prenex Normal Form Definition 4.35 A formula F is in prenex normal form, if it is of the form (Q 1 X 1 )... (Q n X n ) G, where Q i {, } and n 0 holds, X 1,..., X n are variables, and G itself does not contain any further quantifier. G is called matrix of F. Proposition 4.36 There is an algorithm which transforms each sentence F into a semantically equivalent sentence F in prenex normal form. First Order Logic (24th November 2008) 46

47 An Algorithm for Transformation into Prenex Normal form While F is not in prenex normal form apply one of the following rules: ( X) G ( X) G ( X) G ( X) G ((QX) G H) (QX) (G H) (G (QX) H) (QX) (G H) ((QX) G H) (QX) (G H) (G (QX) H) (QX) (G H) Example ( ( X) ( Y ) p(x, Y ) ( Z) q(z)) (( X) ( Y ) p(x, Y ) ( Z) q(z)) (( X) ( Y ) p(x, Y ) ( Z) q(z)) ( X) (( Y ) p(x, Y ) ( Z) q(z)) ( X) ( Y ) ( p(x, Y ) ( Z) q(z)) ( X) ( Y ) ( Z)( p(x, Y ) q(z)). Soundness and termination Exercise. First Order Logic (24th November 2008) 47

48 4.4.3 Skolem Normal Form Idea We eliminate all existential quantifiers. Definition 4.37 Let L(R, F, V) be a predicate logic language. Let F S be a countable set of function symbols such that F S F = and F S contains countably many function symbols for each arity. The elements of F S are called Skolem function symbols. We now consider L(R, F F S, V). Notation Nullary Skolem function symbols are often called Skolem constant symbols. Definition 4.38 A formula is in Skolem normal form if it is of the form ( X 1 )... ( X n ) G, where n 0 holds, X 1,..., X n are variables, and G itself does not contain any further quantifier. First Order Logic (24th November 2008) 48

49 Transformation into Skolem Normal Form Let F be a formula in prenex normal form (variables are standardized apart). While F is not in Skolem normal form apply the following rule: ( X 1 )... ( X n ) ( Y ) G ( X 1 )... ( X n ) (G{Y g(x 1,..., X n )}) Theorem 4.39 If G is a Skolem normal form of the sentence F, then F is satisfiable iff G is satisfiable. The transformation into Skolem normal form preserves satisfiability. Alternatively we could have eliminated the universal quantifiers: dual Skolem Normal Form, validity preserving. First Order Logic (24th November 2008) 49

50 Proof Sketch of Theorem 4.39 Suppose F is in prenex normal form and variables are standardized apart. Auxiliary Proposition Let F be a formula in prenex normal form, where the variables are standardized apart, and F be the formula obtained from F by a single application of the replacement rule. Then the following holds: F is satisfiable iff F is satisfiable. Proof of Auxiliary Proposition Exercise. Let E be the proposition: F is a formula in prenex normal form, where all variables are standardized apart, and is satisfiable iff F is satisfiable. With F = F proposition E holds before entering the while-loop. From the Auxiliary Proposition we learn that E is a loop invariant. From Theorem 3.30 we learn that E holds after the loop has been left. The loop is left only if F is in Skolem normal form. qed First Order Logic (24th November 2008) 50

51 4.4.4 Clause Form Let F be a sentence and H a Skolem normal form of F. F is satisfiable iff H is satisfiable. H is of the form G = ( X 1 )... ( X n ) G, where X 1,..., X n are all the variables occurring in F. There are no quantifiers occurring in the matrix G. Every variable occurring in H is universally quantified. We can transform G into clause form. Let G be a formula in clause form which is semantically equivalent to G. F is satisfiable F is unsatisfiable iff G is satisfiable. iff G is unsatisfiable. First Order Logic (24th November 2008) 51

52 Example: Proving Mathematical Theorems by Machines Theorem: Let G be a group with as binary operator in infix notation and e as identity element. If, for all elements X of G, the property X X = e holds, then G is a commutative group. A group G satisfies the following axioms: A 1 If X, Y G, then also X Y G. A 2 For all X, Y, Z G holds (X (Y Z)) = ((X Y ) Z). A 3 For all X G holds X e = e X = X. A 4 For every X G there exists a unique (inverse) element X 1 G with property X X 1 = X 1 X = e. We represent X Y = Z by p(x, Y, Z) and X 1 by f(x). First Order Logic (24th November 2008) 52

53 Proving Mathematical Theorems by Machines (continued) The group axioms can be represented as follows: A 1 ( X 1 )( Y 1 )( Z 1 ) p(x 1, Y 1, Z 1 ) A 2 (( X 2 )( Y 2 )( Z 2 )( U 2 )( V 2 )( W 2 ) (p(x 2, Y 2, U 2 ), p(y 2, Z 2, V 2 ), p(u 2, Z 2, W 2 ) p(x 2, V 2, W 2 )) ( X 3 )( Y 3 )( Z 3 )( U 3 )( V 3 )( W 3 ) ( p(x 3, Y 3, U 3 ), p(y 3, Z 3, V 3 ), p(x 3, V 3, W 3 ) p(u 3, Z 3, W 3 ))) A 3 (( X 4 ) p(x 4, e, X 4 ) ( X 5 ) p(e, X 5, X 5 )) A 4 (( X 6 ) p(x 6, f(x 6 ), e) ( X 7 ) p(f(x 7 ), X 7, e)) First Order Logic (24th November 2008) 53

54 Proving Mathematical Theorems by Machines (continued) The conclusion of the theorem can be represented as follows: C (( X 8 ) p(x 8, X 8, e) (( U 9 )( V 9 )( W 9 ) (p(u 9, V 9, W 9 ) p(v 9, U 9, W 9 )))) Hence, the theorem is represented by the formula: F = ( A 1, A 2, A 3, A 4 C ) The theorem is proven if F is valid, or, equivalently, if is unsatisfiable. F = A 1, A 2, A 3, A 4, C First Order Logic (24th November 2008) 54

55 Proving Mathematical Theorems by Machines (continued) As clause form of F we obtain: [p(x 1, Y 1, g(x 1,Y 1 ))], [ p(x 2, Y 2, U 2 ), p(y 2, Z 2, V 2 ), p(u 2, Z 2, W 2 ), p(x 2, V 2, W 2 )], [ p(x 3, Y 3, U 3 ), p(y 3, Z 3, V 3 ), p(x 3, V 3, W 3 ), p(u 3, Z 3, W 3 )], [p(x 4, e, X 4 )], [p(e, X 5, X 5 )], [p(x 6, f(x 6 ), e], [p(f(x 7 ), X 7, e)], [p(x 8, X 8, e)], [p(a, b, c)], [ p(b, a, c)]. How can we show unsatisfiablity of a predicate logic formula? First Order Logic (24th November 2008) 55

56 4.5 Unification We consider terms from T (F, V). Definition 4.40 An equation is an expression of the form s t, where s and t are terms. Definition 4.41 A unification problem consists of a multiset of equations {s 1 t 1,..., s n t n } and is the question, whether there exists a substitution σ such that s i σ = t i σ holds for all 1 i n. If such a substitution σ exists, then the terms s i and t i, 1 i n, are said to be simultaneously unifiable and σ is said to be a unifier for the unification problem. First Order Logic (24th November 2008) 56

57 Example {X a, Y Z, g(w, X) g(w, a) } Unifiers: σ 1 = {X a, Y Z} σ 2 = {X a, Z Y } σ 3 = {X a, Y a, Z a} σ 4 = {X a, Y g(a, U), Z g(a, U), W a} Do you prefer unifiers over other unifiers? First Order Logic (24th November 2008) 57

58 Most General Unifiers Definition 4.42 Let σ and θ be substitutions. σ is more general than θ, in symbols σ θ, if there is a substitution λ such that σλ = θ holds. Definition 4.43 Two substitutions σ and θ are said to be variants, in symbols σ θ, if σ θ and θ σ hold. Definition 4.44 Let U be a unification problem. A substitution σ is a most general unifier for U, if σ is a unifier for U and σ θ holds for every unifier θ for U. Theorem 4.45 (Unification Theorem) Let U be a solvable unification problem. There exists a most general unifier for U. Proof Specification of a unification algorithm next slide. Proof of termination and soundness see lecture Foundations of LP. Corollary 4.46 Let U be a unfication problem and σ as well as θ mgus for U. Then, θ σ. First Order Logic (24th November 2008) 58

59 Unification Algorithm Input A unification problem U. Output A most general unifier θ for U, if U is solvable, or not unifiable, otherwise. θ := ε. While U is non-empty do: Select an equation s t from U. U := U \ {s t }. Apply one of the following rules: (1) If s t is of the form X r or r X and variable X does not occur in the term r, then θ := θ{x r} and U := U{X r}. (2) If s t is of the form X X, then do nothing. (3) If s t is of the form X r or r X, variable X occurs in the term r and r X holds, then terminate with not unifiable. (4) If s t is of the form f(s 1,..., s n ) f(t 1,..., t n ), then add the equations s 1 t 1,..., s n t n to U. (5) If s t is of the form f(s 1,..., s n ) g(t 1,..., t m ) where f g, then terminate with not unifiable. First Order Logic (24th November 2008) 59

60 4.6 Proof Methods Resolution Semantic Tableaus Calculus of Natural Deduction Other Methods Hilbert Systems Sequent Calculus Connection Method First Order Logic (24th November 2008) 60

61 4.6.1 Resolution Formulas are negated and an attempt is made to show unsatisfiability. The alphabet is the alphabet of predicate logic. The language is the set of predicate logic formulas in clause form. The only axiom is the empty clause. The derivation rules are the resolution and the factorization rule. First Order Logic (24th November 2008) 61

62 The Resolution Rule Definition 4.47 Let and be clauses, where k, m, n 0. C 1 = [p(s 1,..., s k ), L 1,..., L m ] C 2 = [ p(t 1,..., t k ), L m+1,..., L n ] If {s i t i 1 i k } is unifiable with mgu σ then C = [L 1,..., L n ]σ is called resolvent of C 1 and C 2 wrt p(s 1,..., s k ) and p(t 1,..., t k ). C has been obtained by applying the resolution rule to C 1 and C 2, where p(s 1,..., s k ) and p(t 1,..., t k ) are said to be the literals resolved upon. First Order Logic (24th November 2008) 62

63 The Factoring Rule Definition 4.48 Let C = [p(s 1,..., s k ), p(t 1,..., t k ), L 1,..., L m ] or C = [ p(s 1,..., s k ), p(t 1,..., t k ), L 1,..., L m ] be clauses, where k, m 0. If {s i t i 1 i k } is unifiable with mgu σ then C = [p(t 1,..., t k ), L 1,..., L m ]σ and C = [ p(t 1,..., t k ), L 1,..., L m ]σ, are factors of C, respectively. C has been obtained by applying the factoring rule to C. First Order Logic (24th November 2008) 63

64 Resolution Derivations and Refutations Definition 4.49 Let F = C 1,..., C n be a sentence in clause form, where C i, 1 i n, are clauses. 1. The sequence (C i 1 i n) is a resolution derivation for C. 2. If (C i 1 i m) is a resolution derivation for F, and C m+1 is obtained by applying the resolution or the factoring rule to new variants of 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. Note A resolvent can be computed from two variants of one clause. F may contain already [ ]. It suffices to consider refutations in which [ ] occurs only once. We may assume that [ ] is the last clause in a refutation. First Order Logic (24th November 2008) 64

65 Example Is the following formula unsatisfiable? Clause form: ((p(0) ( X)(p(X) p(h(x)))) p(h(h(h(h(0)))))) ( X) [p(0)], [ p(x), p(h(x))], [ p(h(h(h(h(0)))))]. Resolution refutation of the matrix: 1 [p(0)] 2 [ p(x), p(h(x))] 3 [ p(h(h(h(h(0)))))] 4 [ p(x 1 ), p(h(h(x 1 )))] res(2,2) 5 [ p(x 3 ), p(h(h(h(h(x 3 )))))] res(4,4) 6 [p(h(h(h(h(0)))))] res(1,5) 7 [ ] res(3,6) First Order Logic (24th November 2008) 65

66 Another Example The barber shaves all male inhabitants, who do not shave themselves. Clause form: ( X)( p(x, X) p(b, X)) ( X)( Y ) [p(x, X), p(b, X)], [ p(b, Y ), p(y, Y )]. Resolution refutation of the matrix: 1 [p(x, X), p(b, X)] 2 [ p(b, Y ), p(y, Y )] 3 [p(b, b)] fac(1) 4 [ p(b, b)] fac(2) 5 [ ] res(3,4) First Order Logic (24th November 2008) 66

67 Resolution Proof Definition 4.50 Let F = C 1,..., C n be a predicate logic formula in clause form and S = (C i 1 i m) a resolution derivation for F. The length of S is m n. Definition 4.51 Let F be a sentence and G a clause form of F. A resolution proof for F is a resolution refutation for G. F is called theorem of the resolution calculus, if there exists a resolution proof for F. We denote with r F that there exists a resolution proof for F. First Order Logic (24th November 2008) 67

68 Proving Mathematical Theorems by Machines (continued) 1 [p(x 1, Y 1, g(x 1, Y 2 ))] 2 [ p(x 2, Y 2, U 2 ), p(y 2, Z 2, V 2 ), p(u 2, Z 2, W 2 ), p(x 2, V 2, W 2 )] 3 [ p(x 3, Y 3, U 3 ), p(y 3, Z 3, V 3 ), p(x 3, V 3, W 3 ), p(u 3, Z 3, W 3 )] 4 [p(x 4, e, X 4 )] 5 [p(e, X 5, X 5 )] 6 [p(x 6, f(x 6 ), e)] 7 [p(f(x 7 ), X 7, e)] 8 [p(x 8, X 8, e)] 9 [p(a, b, c)] 10 [ p(b, a, c)] 11 [p(x 9, V 2, e), p(y 9, U 9, V 9 ), p(x 2, Y 2, U 2 )] res(2, 8) 13 [p(x 10, V 10, W 10 ), p(e, Z 10, W 10 ), p(x 10, Z 10, V 10 )] res(2, 8) 14 [p(x 11, V 11, U 11 ), p(y 11, e, V 11 ), p(x 11, Y 11, U 11 )] res(2, 4) 40 [p(x 12, e, W 12 ), p(e, X 12, W 12 )] res(13, 8) 43 [p(x 13, V 13, Z 13 ), p(x 13, Z 13, V 13 )] res(13, 5) 67 [p(x 14, g(x 14, Z 14 ), Z 14 )] res(43, 1) 3239 [p(e, V 15, Y 15 ), p(y 15, e, V 15 )] res(14, 5) 3325 [ p(y 16, e, a), p(b, Y 16, c)] res(10, 14) 3627 [ p(g(b, c), e, a)] res(3325, 67) 4630 [p(e, Z 17, X 17 ), p(x 17, Z 17, e)] res(3239, 43) 5199 [p(z 18, e, X 18 ), p(x 18, Z 18, e)] res(40, 4630) 5800 [ p(a, g(b, c), e)] res(3627, 5199) [p(a, V 19, e), p(b, c, V 19 )] res(11, 9) [p(a, g(b, c), e)] res(518609, 1) [ ] res(645778, 5800) First Order Logic (24th November 2008) 68

69 4.6.3 First-Order Calculus of Natural Deduction We extend the propositional calculus of natural deduction by adding introduction and elimination rules for the quantifiers. We cannot restrict ourselves to closed formulas, but must consider formulas with free variables. There are several possibilities to define models for non-closed formulas. The new inference rules depend on the these possibilities. Here we consider the universal closure of a formula. Recall R n = R {[ ]}. [ ] is nullary with [ ] I = for all interpretations I. Now we consider L(R n, F, V). First Order Logic (24th November 2008) 69

70 A Note on Substitutions Let t T (F, V). If t Y then {Y t} represents a substitution. However, if t = Y then {Y t} does not represent a substitution. In this section we agree to use {Y Y } as an acronym for ε. Note that ε and, hence, {Y Y } is free for all G L(R, F, V). First Order Logic (24th November 2008) 70

71 Additional Inference Rules Universal Quantification F ( Y ) F ( I) ( Y ) F F {Y t} ( E) Existential Quantification F {Y t} ( Y ) F ( I) ( Y ) F G F. G ( E) First Order Logic (24th November 2008) 71

72 Derivation Trees Definition 4.56 The set of derivation trees in the first-order calculus of natural deduction is the smallest set X with the following properties: 1. Each tree consisting of a single node labeled by some F L(R n, F, V) is in X as in the propositional case. 7. If F X and Y does not occur free in any uncancelled hypothesis of F, then F ( I) ( Y ) F is a derivation tree in X. 8. If ( Y ) F X, t T (F, V) and {Y t} is free for F, then ( Y ) F F {Y t} ( E) is a derivation tree in X. First Order Logic (24th November 2008) 72

73 Derivation Trees (Continued) 9. If F {Y t} X, t T (F, V) and {Y t} is free for F, then is a derivation tree in X. F {Y t} ( I) ( Y ) F 10. If ( Y ) F, F G X, Y is not free in G or in an uncancelled hypothesis of F G other than in F, and j is a new index, then ( Y ) F G F j G ( E) j is a derivation tree in X. First Order Logic (24th November 2008) 73

74 Why do we need the restriction on ( I)? Consider But p(x, 0) 1 ( I) ( X) p(x, 0) ( I) 1 (p(x, 0) ( X) p(x, 0)) ( I) ( X) (p(x, 0) ( X) p(x, 0)) ( E) (p(0, 0) ( X) p(x, 0)) = (p(0, 0) ( X) p(x, 0)). Let I = ({a, b}, I) such that 0 I = a p I = {(a, a), (b, b)} Then [p(0, 0)] I =, but [( X) p(x, 0)] I =. First Order Logic (24th November 2008) 74

75 Why do we need the restriction on ( E)? Consider ( X) ( Y ) p(x, Y ) 1 ( E) ( Y )p(y, Y ) (( X) ( Y ) p(x, Y ) ( Y ) p(y, Y )) ( I) 1 But = (( X) ( Y ) p(x, Y ) ( Y ) p(y, Y )). Let I = ({a, b}, I) with p I = {(a, a), (b, b)}. Then [( X) ( Y ) p(x, Y )] I =, but [ ( Y ) p(y, Y ))] I =. First Order Logic (24th November 2008) 75

76 Some Examples If X fv(f ) then n (( X) F F ) ( X) F 1 F 1 ( E) F ( X) F (( X) F F ) ( I) ( I) 1 Note that {X X} is free for F in ( E) and that X fv(f ) in ( I). If X fv(f ) then n (( X) F F ) ( X) F 1 F 2 ( E) 2 F 1 F ( X) F (( X) F F ) ( I) ( I) 1 Note that {X X} is free for F in ( I) and that X fv(f ) in ( E). First Order Logic (24th November 2008) 76

77 Another Example n (( X) F ( X) F ) ( X) F 4 ( E) F 1 F [ ] ( E) F 3 ( X) F 2 ( X) F 2 ( X) F ( X) F ( E) (f) ( E) 1 ( X) F 4 [ ] (raa) 3 ( X) F ( I) ( E) [ ] ( I) 2 F ( I) ( X) F ( X) F ( I) 4 (( X) F ( X) F ) First Order Logic (24th November 2008) 77

78 Comments The first-order calculus of natural deduction is sound and complete wrt = u, i.e., n F iff = u F (see e.g. van Dalen: Logic and Structure. Springer 1997). The first-order calculus of natural deduction is incomplete wrt = e : Consider F = (p(a) p(x)). We find = e F and = u F. Because the calculus is sound and complete wrt = u we conclude n F. First Order Logic (24th November 2008) 78

79 4.8 Properties Herbrand interpretations For each interpretation there is a corresponding Herbrand interpretation Soundness and completeness = F iff r F Compactness F is satisfiable iff each finite subset of F is satisfiable Undecidability The question whether F = G is undecidable in first-order logic. First Order Logic (24th November 2008) 79

80 Completeness of the Resolution Method Overview {F 1,..., F n } = F Theorem 3.17 (extended) ( F 1,..., F n F ) is valid Theorem 3.14 (extended) ( F 1,..., F n F ) is unsatisfiable Theorem 4.39 G is unsatisfiable, where G is a clause form of ( F 1,..., F n F ) Theorem 4.62 {Cσ C is clause in G and σ is ground substitution for C} is unsatisfiable Bijection ground atoms propositional variables H is unsatisfiable, where H is the set of propositional clauses corresponding to {Cσ...} Corollary 3.46 finite subset H H is unsatisfiable Theorem 3.49 there is a propositional resolution refutation for H Bijection ground atoms propositional variables there is a first-order resolution refutation for H Lemma 4.66 there is a first-order resolution refutation for G Definition 4.51 there is a first-order resolution proof for ( F 1,..., F n F ) First Order Logic (24th November 2008) 80

81 4.8.1 Herbrand Interpretations F = {a/0, f/1} and R = {p/1, q/1, r/1}. G = ( X)( Y ) [p(x), q(x)], [r(f(y ))] Herbrand universe: T (F) = {a, f(a), f(f(a))...}. Herbrand interpretations, e.g.: I 1 = {p(a), q(a), r(f(a)), p(f(a)), q(f(a)), r(f(f(a))),...} I 2 =. Different interpretations, e.g.: I 3 = (D, I3) where D = {0, 1}, a 0, f/1 f /1 : D D with f (0) = 1 and f (1) = 0, p I = {0}, q I = {0}, r I = {0}. We can construct a corresponding Herbrand interpretation! First Order Logic (24th November 2008) 81

82 Corresponding Herbrand Interpretations Definition 4.58 Let I = (D, I) be an interpretation and {t 1,..., t n } T (F). A Herbrand interpretation J is called corresponding to I if the following condition is satisfied: for all relation symbols p/n holds: p(t 1,..., t n ) I = iff p(t 1,..., t n ) J. First Order Logic (24th November 2008) 82

83 Example Consider again I 3 = (D, I3) with D = {0, 1}, a 0, f/1 f /1 : D D mit f (0) = 1 and f (1) = 0, p I = {0}, q I = {0}, r I = {0}. Set of ground atoms: {p(a), q(a), r(a), r(f(a)), p(f(a)), q(f(a)), r(f(f(a))),...} Meaning under I 3 : p(a) q(a) r(a) r(f(a)) p(f(a)) q(f(a)) r(f(f(a))) Corresponding Herbrand interpretation: J 3 = {p(a), q(a), r(a), p(f(f(a))),...} First Order Logic (24th November 2008) 83

84 Interpretation vs. Herbrand Interpretation Lemma 4.59 If an interpretation I is a model for a predicate logic sentence F in Skolem normal form, then every Herbrand interpretation corresponding to I is also a model for F. Proof Exercise. Implication The real numbers cannot be characterized within predicate logic. Suppose, the real numbers can be characterized. Then there exists a sentence F such that for all I = (D, I) we find: F I = iff D is uncountable. Select I with F I =. Let J be the Herbrand interpretation corresponding to I. Then F J =. But the domain of Herbrand interpretations is countable. Contradiction qed First Order Logic (24th November 2008) 84

85 Interpretations vs. Herbrand Interpretations (continued) Theorem 4.60 A predicate logic sentence F in Skolem normal form is unsatisfiable iff F is mapped to under each Herbrand interpretation. Proof immediate. Suppose, F is mapped to under each Herbrand interpretation, but is satisfiable. Then there exists an I with F I =. Let J be the Herbrand interpretation corresponding to I. By Lemma 4.59 we find F J =. Contradiction qed First Order Logic (24th November 2008) 85

86 Interpretations vs. Herbrand Interpretations (cont.) Lemma 4.59 and Theorem 4.60 hold only for formulas in Skolem normal form. Example Let F = {a/0}, R = {p/1} and X V. Consider F = (p(a) ( X) p(x)). Let I = ({1, 2}, I) with a I = 1 and p I = {1}. Then, (p(a) ( X) p(x)) I = p(a) I (( X) p(x)) I = a I p I (( X) p(x)) I = 1 {1} (( X) p(x)) I = (( X) p(x)) I = (( X) p(x)) I = because 2 {1} T (F) = {a} I = {p(a)} is the Herbrand interpretation corresponding to I. F I = Lemma 4.59 does not carry over. For I = we find F I = Theorem 4.60 does not carry over. First Order Logic (24th November 2008) 86

87 Further Properties Proposition 4.61 Let F be a predicate logic formula, I a Herbrand interpretation, Z a variable assignment wrt I, and t 1,..., t n T (F). Then, Proof Exercise. F I,{X 1 t 1 }...{X n tn}z = [F {X 1 t 1,..., X n t n }] I,Z. First Order Logic (24th November 2008) 87

88 Further Properties (continued) Theorem 4.62 Let F = C 1,..., C n be a sentence in clause form. F is satisfiable iff {C i σ 1 i n, σ is a ground substitution for C i } is satisfiable. Proof Let X 1,..., X k be all variables occurring in C 1,..., C n. F is satisfiable iff there exists a Herbrand model for F (Lemma 4.59). I = F iff [ C 1,..., C n ] I,{X 1 t 1 }...{X k t k }Z = for all t 1,..., t k T (F) and arbitrary Z iff [ C 1,..., C n {X 1 t 1,..., X k t k }] I = for all t 1,..., t k T (F) (Proposition 4.61) iff [ C 1,..., C n σ] I = for all ground substitutions σ for C 1,..., C n iff [C i σ] I = for 1 i n and all ground substitutions σ for C i iff I = {C i σ 1 i n, σ ground substitution for C i }. qed First Order Logic (24th November 2008) 88

89 Example Let F = {a/0, s/1}, R = {p/1, q/1} and X, Y V. Consider: We obtain: F = C 1, C 2 = ( X)( Y ) [p(x)], [ q(s(y ))]. G = {C i σ 1 i 2, σ ground substitution for C i } = {[p(a)], [p(s(a))],...} {[ q(s(a))], [ q(s(s(a)))],...} Obviously: I = F iff I = G. First Order Logic (24th November 2008) 89

90 4.8.2 Soundness and Completeness Theorems First Order Resolution Lemma 4.63 Let F = C 1,..., C n be a first order sentence in clause form with clauses C i, 1 i n, and let C be a resolvent or a factor of new variants of clauses from F. Then F C 1,..., C n, C. Proof Observe: C 1,..., C n, C (F C). To show: for all I: I = F iff I = (F C). If I = (F C) then I = F for all I. Suppose, I = F. 1st C is a factor Exercise. 2nd Let C = [L 1,..., L n ]σ resolvent of C = [p(s 1,..., s k ), L 1,..., L m ] and C = [ p(t 1,..., t k ), L m+1,..., L n ], where C and C are new variants of clauses from F and σ is a mgu for {s i t i 1 i k }. First Order Logic (24th November 2008) 90

91 Proof of Lemma 4.63 (continued) Suppose, I is not a model for C. Then we find a variable assignment Z wrt. I with [L 1 σ] I,Z... [L n σ] I,Z =. (1) Because I = F we find: I = (C σ) and I = (C σ). Hence: [p(s 1,..., s k )σ] I,Z [L 1 σ] I,Z... [L m σ] I,Z = (2) and [ p(t 1,..., t k )σ] I,Z [L m+1 σ] I,Z... [L n σ] I,Z = (3) Because of (1) and (2) we conclude: [p(s 1,..., s k )σ] I,Z =. Because of (1) and (3) we conclude: [ p(t 1,..., t k )σ] I,Z =. Contradiction, because s i σ = t i σ for all 1 i k. qed First Order Logic (24th November 2008) 91

92 Example Consider the sentence: (( X)p(X) ( Y ) (p(y ) q(y ))). It s clause form is: ( X)( Y ) [p(x)], [ p(y ), q(y )]. As a resolvent of new variants of the two clauses we obtain: [q(z)]. We find: [p(x)], [ p(y ), q(y )] [p(x)], [ p(y ), q(y )], [q(z)]. First Order Logic (24th November 2008) 92

93 Generalized Resolution Lemma Lemma 4.64 Let F = C 1,..., C n be a predicate logic sentence in clauseform and (C i 1 i m) a resolution derivation for F. Then: F C 1,..., C m. Proof Structural induction on the length k = m n of the resolution derivation. IB k = 0: immediate because n = m. IH Suppose the proposition holds for k. IS Let (C i 1 i m + 1) be a resolution derivation of length k + 1. From IH we conclude: F C 1,..., C m. From Lemma 4.63 we conclude: C 1,..., C m C 1,..., C m, C m+1. Because is transitive, we conclude F C 1,..., C m+1. The proposition follows by an application of the induction theorem. qed First Order Logic (24th November 2008) 93

94 The Lifting-Lemma Lemma 4.65 (Lifting-Lemma) Let C 1 and C 2 be clauses and σ a ground substitution for C 1 and C 2. (i) If C is a resolvent of C 1 σ and C 2 σ, then there exists a resolvent C of C 1 and C 2 as well as a substitution λ, such that C λ = C holds. (ii) If C is a factor of C 1 σ, then there exists a factor C of C 1 as well as a substitution λ, such that C λ = C holds. First Order Logic (24th November 2008) 94

95 Proof of the Lifting-Lemma Proof (i) C 1 = [p(s 1,..., s k ), L 1,..., L m ], C 2 = [ p(t 1,..., t k ), L m+1,..., L n ]. s i σ = t i σ for 1 i k, because σ is a ground substitution for C 1 and C 2. C = [L 1,..., L n ]σ. σ is a unifier for U = {s 1 t 1,..., s k t k }. We find an mgu θ for U and λ, such that: θλ = σ (Theorem 4.45). C = [L 1,..., L n ]θ is resolvent for C 1 and C 2. C λ = [L 1,..., L n ]θλ = [L 1,..., L n ]σ = C. (ii) Exercise qed First Order Logic (24th November 2008) 95

96 Example Consider: C 1 = [ p(w )], C 2 = [p(f(x)), q(x)], σ = {X a, W f(a)}. Then: C 1 σ = [ p(f(a))], C 2 σ = [p(f(a)), q(a)]. As resolvent of C 1 σ and C 2 σ we obtain: C = [ q(a)] As resolvent of C 1 and C 2 with mgu θ = {W f(x)} we obtain C = [ q(x)] With λ = {X a} we find C λ = C and θλ = σ. First Order Logic (24th November 2008) 96

97 The Generalized Lifting Lemma Lemma 4.66 (Generalized Lifting Lemma) Let C j be clauses and σ j ground substitutions for C j, 1 j n. If there is a resolution derivation (B i 1 i m) for C 1 σ 1,..., C n σ n, then there is a resolution derivation (B i 1 i m) for C 1,..., C n and a sequence (λ i 1 i m) of substitutions such that B i λ i = B i for all 1 i m. Proof Induction on the length l = m n of the derivation (B i 1 i m). IB l = 0: In this case n = m and the proposition follows with B i λ i = σ i for all 1 i n. IH Suppose the proposition holds for l. = C i and IS Let (B i 1 i m + 1) be a resolution derivation of length l + 1. Because of IH we find a resolution derivation (B i 1 i m) of length l for C 1,..., C n and a sequence (λ i 1 i m) of substitutions with B i λ i = B i for all 1 i m. ( ) 1st B m+1 is a factor Exercise 2nd B m+1 is a resolvent next slide First Order Logic (24th November 2008) 97

98 Proof Generalized Lifting Lemma (Cont.) Let B m+1 be the resolvent of B j and B k with j, k m. Because of ( ) we find B j and λ j with B j λ j = B j. Let B j be a new variant of B j. Then there is a substitution θ j with B j θ j = B j. Alltogether: B j = B j λ j = (B j θ j)λ j = B j (θ jλ j ). Analogously we find B k, λ k and θ k with B k = B k λ k = (B k θ k)λ k = B k (θ kλ k ). Because B j dom((θ j λ j ) V ar(b and B k are new variants, we conclude: j )) dom((θ kλ k ) V ar(b )) =. k Let σ = (θ j λ j ) V ar(b j ) (θ kλ k ) V ar(b k ). Hence we obtain: B j σ = B j and B k σ = B k. Because of the Lifting Lemma we find a resolvent B m+1 a substitution λ m+1 with B m+1 λ m+1 = B m+1. of B j and B k as well as Together with ( ) we obtain a resolution derivation (B i 1 i m + 1) and a sequence (λ i 1 i m + 1) with B i λ i = B i for all 1 i m + 1. An application of the induction theorem yields the desired result. qed First Order Logic (24th November 2008) 98

99 Example 1 Consider the clauses with [ p(w )], [p(f(x)), q(x)], [q(g(y )), r(y )], [r(z)] σ = {W f(g(a)), X g(a), Y a, Z a}. We obtain the following derivations: B 1 [ p(f(g(a)))] B 1 [ p(w )] B 2 [p(f(g(a))), q(g(a))] B 2 [p(f(x)), q(x)] B 3 [q(g(a)), r(a)] B 3 [q(g(y )), r(y )] B 4 [r(a)] B 4 [r(z)] B 5 [ q(g(a))] B 5 [ q(x 2 )] res(1,2) B 6 [ r(a)] B 6 [ r(y 3 )] res(3,5) B 7 [ ] B 7 [ ] res(4,6) First Order Logic (24th November 2008) 99

100 Example 2 Consider the clauses [p(a)], [ p(x), p(f(x))], [ p(y ), p(f(y ))], [ p(f(f(a)))] with σ = {X a, Y f(a)}. We obtain the derivations: B 1 [p(a)] B 1 [p(a)] B 2 [ p(a), p(f(a))] B 2 [ p(x), p(f(x))] B 3 [ p(f(a)), p(f(f(a)))] B 3 [ p(y ), p(f(y ))] B 4 [ p(f(f(a)))] B 4 [ p(f(f(a)))] B 5 [ p(a), p(f(f(a)))] B 5 [ p(x 1 ), p(f(f(x 1 )))] res(2,3) B 6 [p(f(f(a)))] B 6 [p(f(f(a)))] res(1,5) B 7 [ ] B 7 [ ] res(4,6) First Order Logic (24th November 2008) 100

101 Closed Formulas and Propositional Logic Let F be a generalized conjunction of ground-instantiated clauses. F = [p(a)], [ p(f(f(a)))], [ p(a), p(f(a))], [ p(f(a)), p(f(f(a)))] Let G be the set of ground atoms occurring in F. G = {p(a), p(f(a)), p(f(f(a)))} Let V be a sufficiently large set of propositional variables. V = {p 1, p 2, p 3 } We find a bijection between G and V. p(a) p 1, p(f(a)) p 2, p(f(f(a))) p 3 All propositional results are applicable to F. First Order Logic (24th November 2008) 101

102 The Resolution Theorem Theorem 4.67 (Resolution Theorem) Let F = C 1,..., C l be a sentence in clause form. F is unsatisfiable iff there is a resolution refutation for F. Proof Let (C i 1 i m) be a resolution refutation for F. Lemma 4.64: F C 1,..., C m. Because of C m = [ ] and Theorem 3.19 (unsatisfiability) we conclude F [ ]. Hence, F is unsatisfiable. First Order Logic (24th November 2008) 102

103 Proof of the Resolution Theorem (cont.) Suppose, F is unsatisfiable. Then, G = {C i σ 1 i l, σ ground substitution for C i } is unsatisfiable (Theorem 4.62). To each ground atom occurring in G we can assign a propositional variable such that any two atoms are mapped to different variables. Then, we find a finite subset {C 1 σ 1,..., C m σ m } G, which is propositionally unsatisfiable (Corollary 3.46). Then, we find a propositional resolution refutation for C 1 σ 1,..., C m σ m (Theorems 3.14 and 3.49). Then, we find a predicate logic resolution refutation for C 1,..., C m (Lemma 4.66). By construction each C j, 1 j m, is a clause occurring in F. Hence, we obtain a resolution refutation for F (by eliminating variants of {C 1,..., C m }). qed First Order Logic (24th November 2008) 103