Foundations of Artificial Intelligence

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Herbrand Expansion, Resolution Wolfram Burgard,

1 Foundations of Artificial Intelligence 9. Predicate Logic Syntax and Semantics, Normal Forms, Herbrand Expansion, Resolution Wolfram Burgard, Bernhard Nebel and Martin Riedmiller Albert-Ludwigs-Universität Freiburg

2 Contents 1 Motivation 2 Syntax and Semantics 3 Normal Forms 4 Reduction to Propositional Logic: Herbrand Expansion 5 Resolution & Unification 6 Closing Remarks (University of Freiburg) Foundations of AI 2 / 57

3 Motivation We can already do a lot with propositional logic. It is, however, annoying that there is no structure in the atomic propositions. Example: All blocks are red There is a block A It should follow that A is red But propositional logic cannot handle this. Idea: We introduce individual variables, predicates, functions,.... First-Order Predicate Logic (PL1) (University of Freiburg) Foundations of AI 3 / 57

4 The Alphabet of First-Order Predicate Logic Symbols: Operators:,,,,, = Variables: x, x 1, x 2,..., x, x,..., y,..., z,... Brackets: (), [], {} Function symbols (e.g., weight(), color()) Predicate symbols (e.g., block(), red()) Predicate and function symbols have an arity (number of arguments). 0-ary predicate: propositional logic atoms 0-ary function: constant We suppose a countable set of predicates and functions of any arity. = is usually not considered a predicate, but a logical symbol (University of Freiburg) Foundations of AI 4 / 57

5 The Grammar of First-Order Predicate Logic (1) Terms (represent objects): 1. Every variable is a term. 2. If t 1, t 2,..., t n are terms and f is an n-ary function, then is also a term. f(t 1, t 2,..., t n ) Terms without variables: ground terms. Atomic Formulae (represent statements about objects) 1. If t 1, t 2,..., t n are terms and P is an n-ary predicate, then P (t 1, t 2,..., t n ) is an atomic formula. 2. If t 1 and t 2 are terms, then t 1 = t 2 is an atomic formula. Atomic formulae without variables: ground atoms. (University of Freiburg) Foundations of AI 5 / 57

6 The Grammar of First-Order Predicate Logic (2) Formulae: 1. Every atomic formula is a formula. 2. If ϕ and ψ are formulae and x is a variable, then ϕ, ϕ ψ, ϕ ψ, ϕ ψ, ϕ ψ, xϕ and xϕ are also formulae., are as strongly binding as. Propositional logic is part of the PL1 language: 1. Atomic formulae: only 0-ary predicates 2. Neither variables nor quantifiers. (University of Freiburg) Foundations of AI 6 / 57

7 Alternative Notation Here Elsewhere ϕ ϕ ϕ ϕ ψ ϕ&ψ ϕ ψ ϕ, ψ ϕ ψ ϕ ψ ϕ; ψ ϕ + ψ ϕ ψ ϕ ψ ϕ ψ ϕ ψ ϕ ψ ϕ ψ xϕ ( x)ϕ xϕ xϕ ( x)ϕ xϕ (University of Freiburg) Foundations of AI 7 / 57

8 Meaning of PL1-Formulae Our example: x[block(x) Red(x)], Block(a) For all objects x: If x is a block, then x is red and a is a block. Generally: Terms are interpreted as objects. Universally-quantified variables denote all objects in the universe. Existentially-quantified variables represent one of the objects in the universe (made true by the quantified expression). Predicates represent subsets of the universe. Similar to propositional logic, we define interpretations, satisfiability, models, validity,... (University of Freiburg) Foundations of AI 8 / 57

9 Semantics of PL1-Logic Interpretation: I = D, I where D is an arbitrary, non-empty set and I is a function that maps n-ary function symbols to functions over D: f I [D n D] maps individual constants to elements of D: a I D maps n-ary predicate symbols to relations over D: P I D n Interpretation of ground terms: (f(t 1,..., t n )) I = f I (t I 1,..., ti n) Satisfaction of ground atoms P (t 1,..., t n ): I = P (t 1,..., t n ) iff t I 1,..., ti n P I (University of Freiburg) Foundations of AI 9 / 57

10 Example (1) D = {d 1,..., d n n > 1} a I = d 1 b I = d 2 c I =... Block I = {d 1 } Red I = D I = Red(b) I = Block(b) (University of Freiburg) Foundations of AI 10 / 57

11 Example (2) D = {1, 2, 3,...} 1 I = 1 2 I = 2 Even I = {2, 4, 6,...} succ I = {(1 2), (2 3),...} I = Even(2) I = Even(succ(2)) (University of Freiburg) Foundations of AI 11 / 57

12 Semantics of PL1: Variable Assignment Set of all variables V. Function α : V D Notation: α[x/d] is the same as α apart from point x. For x : α[x/d](x) = d. Interpretation of terms under I, α: Satisfaction of atomic formulae: x I,α = α(x) a I,α = a I (f(t 1,..., t n )) I,α = f I (t I,α 1,..., ti,α n ) I, α = P (t 1,..., t n ) iff t I,α 1,..., ti,α n P I (University of Freiburg) Foundations of AI 12 / 57

13 Example α = {(x d 1 ), (y d 2 )} I, α = Red(x) I, α[y/d 1 ] = Block(y) (University of Freiburg) Foundations of AI 13 / 57

14 Semantics of PL1: Satisfiability A formula ϕ is satisfied by an interpretation I and a variable assignment α, i.e., I, α = ϕ: I, α = I, α = I, α = ϕ iff I, α = ϕ and all other propositional rules as well as I, α = P (t 1,..., t n ) iff t I,α 1,..., ti,α n P I,α I, α = xϕ iff for all d D, I, α[x/d] = ϕ I, α = xϕ iff there exists a d D with I, α[x/d] = ϕ (University of Freiburg) Foundations of AI 14 / 57

15 Example T = {Block(a), Block(b), x(block(x) Red(x))} D = {d 1,..., d n n > 1} a I = d 1 b I = d 2 Block I = {d 1} Red I = D α = {(x d 1), (y d 2)} Questions: 1. I, α = Block(b) Block(b)? 2. I, α = Block(x) (Block(x) Block(y))? 3. I, α = Block(a) Block(b)? 4. I, α = x(block(x) Red(x))? 5. I, α =? (University of Freiburg) Foundations of AI 15 / 57

16 Free and Bound Variables x[r( y, z ) y{ P (y, x) R(y, z )}] The boxed appearances of y and z are free. All other appearances of x,y,z are bound. Formulae with no free variables are called closed formulae or sentences. We form theories from closed formulae. Note: With closed formulae, the concepts logical equivalence, satisfiability, and implication, etc. are not dependent on the variable assignment α (i.e., we can always ignore all variable assignments). With closed formulae, α can be left out on the left side of the model relationship symbol: I = ϕ (University of Freiburg) Foundations of AI 16 / 57

17 Terminology An interpretation I is called a model of ϕ under α if I, α = ϕ A PL1 formula ϕ can, as in propositional logic, be satisfiable, unsatisfiable, falsifiable, or valid. Analogously, two formulae are logically equivalent (ϕ ψ) if for all I, α: Note: P (x) P (y)! I, α = ϕ iff I, α = ψ Logical Implication is also analogous to propositional logic. Question: How can we define derivation? (University of Freiburg) Foundations of AI 17 / 57

18 Prenex Normal Form Because of the quantifiers, we cannot produce the CNF form of a formula directly. First step: Produce the prenex normal form quantifier prefix + (quantifier-free) matrix Qx 1 Qx 2 Qx 3... Qx n ϕ (University of Freiburg) Foundations of AI 18 / 57

19 Equivalences for the Production of Prenex Normal Form ( xϕ) ψ x(ϕ ψ) if x not free in ψ ( xϕ) ψ x(ϕ ψ) if x not free in ψ ( xϕ) ψ x(ϕ ψ) if x not free in ψ ( xϕ) ψ x(ϕ ψ) if x not free in ψ xϕ xψ x(ϕ ψ) xϕ xψ x(ϕ ψ) xϕ x ϕ xϕ x ϕ... and propositional logic equivalents (University of Freiburg) Foundations of AI 19 / 57

20 Production of Prenex Normal Form 1. Eliminate and 2. Move inwards 3. Move quantifiers outwards Example: x[( xp (x)) Q(x)] x[ ( xp (x)) Q(x)] x[( xp (x)) Q(x)] And now? (University of Freiburg) Foundations of AI 20 / 57

21 Renaming of Variables ϕ[ x t ] is obtained from ϕ by replacing all free appearances of x in ϕ by t. Lemma: Let y be a variable that does not appear in ϕ. Then it holds that xϕ yϕ[ x y ] and xϕ yϕ[ x y ] Theorem: There exists an algorithm that calculates the prenex normal form of any formula. (University of Freiburg) Foundations of AI 21 / 57

22 Derivation in PL1 Why is prenex normal form useful? Unfortunately, there is no simple law as in propositional logic that allows us to determine satisfiability or general validity (by transformation into DNF or CNF). But: we can reduce the satisfiability problem in predicate logic to the satisfiability problem in propositional logic. In general, however, this produces a very large number of propositional formulae (perhaps infinitely many) Then: apply resolution. (University of Freiburg) Foundations of AI 22 / 57

23 Skolemization Idea: Elimination of existential quantifiers by applying a function that produces the right element. Theorem (Skolem Normal Form): Let ϕ be a closed formula in prenex normal form such that all quantified variables are pair-wise distinct and the function symbols g 1, g 2,... do not appear in ϕ. Let ϕ = x 1 x i yψ, then ϕ is satisfiable iff [ ϕ = x 1 x i ψ is satisfiable. y g i (x 1,...,x i ) Example: x y[p (x) Q(y)] x[p (x) Q(g(x))] ] (University of Freiburg) Foundations of AI 23 / 57

24 Skolem Normal Form Skolem Normal Form: Prenex normal form without existential quantifiers. Notation: ϕ* is the SNF of ϕ. Theorem: It is possible to calculate the Skolem normal form of every closed formula ϕ. Example: x(( xp (x)) Q(x)) y(( xp (x)) Q(y)) y( x(p (x) Q(x))) x(p (x) Q(g 0 )) develops as follows: Note: This transformation is not an equivalence transformation; it only preserves satisfiability! Note:... and is not unique. (University of Freiburg) Foundations of AI 24 / 57

25 Ground Terms & Herbrand Expansion The set of ground terms (or Herbrand Universe) over a set of SNF formulae θ* is the (infinite) set of all ground terms formed from the symbols of θ* (in case there is no constant symbol, one is added). This set is denoted by D(θ*). The Herbrand expansion E(θ*) is the instantiation of the Matrix ψ i of all formulae in θ* through all terms t D(θ*): E(θ*) = {ψ i [ x 1 t 1,..., xn t n ] ( x 1,..., x n ψ i ) θ*, t j D(θ*)} Theorem (Herbrand): Let θ* be a set of formulae in SNF. Then θ* is satisfiable iff E(θ*) is satisfiable. Note: If D(θ*) and θ* are finite, then the Herbrand expansion is finite finite propositional logic theory. Note: This is used heavily in AI and works well most of the time! (University of Freiburg) Foundations of AI 25 / 57

26 Infinite Propositional Logic Theories Can a finite proof exist when the set is infinite? Theorem (compactness of propositional logic): A (countable) set of formulae of propositional logic is satisfiable if and only if every finite subset is satisfiable. Corollary: A (countable) set of formulae in propositional logic is unsatisfiable if and only if a finite subset is unsatisfiable. Corollary: (compactness of PL1): A (countable) set of formulae in predicate logic is satisfiable if and only if every finite subset is satisfiable. (University of Freiburg) Foundations of AI 26 / 57

27 Recursive Enumeration and Decidability We can construct a semi-decision procedure for validity, i.e., we can give a (rather inefficient) algorithm that enumerates all valid formulae step by step. Theorem: The set of valid (and unsatisfiable) formulae in PL1 is recursively enumerable. What about satisfiable formulae? Theorem (undecidability of PL1): It is undecidable, whether a formula of PL1 is valid. (Proof by reduction from PCP) Corollary: The set of satisfiable formulae in PL1 is not recursively enumerable. In other words: If a formula is valid, we can effectively confirm this fact. Otherwise, we can end up in an infinite loop. (University of Freiburg) Foundations of AI 27 / 57

28 Derivation in PL1 Clausal Form instead of Herbrand Expansion. Clauses are universally quantified disjunctions of literals; all variables are universally quantified ( x 1,..., x n )(l 1... l n ) l 1... l n {l 1,..., l n } written as or (University of Freiburg) Foundations of AI 28 / 57

29 Production of Clausal Form from SNF Skolem Normal Form quantifier prefix + (quantifier-free) matrix x 1 x 2 x 3 x n ϕ 1. Put Matrix into CNF using distribution rule 2. Eliminate universal quantifiers 3. Eliminate conjunction symbol 4. Rename variables so that no variable appears in more than one clause. Theorem: It is possible to calculate the clausal form of every closed formula ϕ. Note: Same remarks as for SNF (University of Freiburg) Foundations of AI 29 / 57

30 Conversion to CNF (1) Everyone who loves all animals is loved by someone: x[ yanimal(y) Loves(x, y)] [ yloves(y, x)] 1. Eliminate biconditionals and implications x [ y Animal(y) Loves(x, y)] [ yloves(y, x)] 2. Move inwards: xp x p, xp x p x[ y ( Animal(y) Loves(x, y))] [ yloves(y, x)] x[ y Animal(y) Loves(x, y)] [ yloves(y, x)] x[ yanimal(y) Loves(x, y)] [ yloves(y, x)] (University of Freiburg) Foundations of AI 30 / 57

31 Conversion to CNF (2) 3. Standardize variables: each quantifier should use a different one x[ yanimal(y) Loves(x, y)] [ zloves(z, x)] 4. Skolemize: a more general form of existential instantiation. Each existential variable is replaced by a Skolem function of the enclosing universally quantified variables: x[animal(f (x)) Loves(x, F (x))] [Loves(G(x), x)] 5. Drop universal quantifiers: [Animal(F (x)) Loves(x, F (x))] [Loves(G(x), x)] 6. Distribute over : [Animal(F (x)) Loves(G(x), x)] [ Loves(x, F (x)) Loves(G(x), x)] (University of Freiburg) Foundations of AI 31 / 57

32 Clauses and Resolution Assumption: All formulae in the KB are clauses. Equivalently, we can assume that the KB is a set of clauses. Due to commutativity, associativity, and idempotence of, clauses can also be understood as sets of literals. The empty set of literals is denoted by. Set of clauses: Set of literals: C, D Literal: l Negation of a literal: l (University of Freiburg) Foundations of AI 32 / 57

33 Propositional Resolution C 1 {l}, C 2 {l} C 1 C 2 C 1 C 2 are called resolvents of the parent clauses C 1 {l} and C 2 {l}. l and l are the resolution literals. Example: {a, b, c} resolves with {a, d, c} to {a, b, d}. Note: The resolvent is not equivalent to the parent clauses, but it follows from them! Notation: R( ) = {C C is a resolvent of two clauses from } (University of Freiburg) Foundations of AI 33 / 57

34 What Changes? Examples {{Nat(s(0)), Nat(0)}, {Nat(0)}} {Nat(s(0))} {{Nat(s(0)), Nat(x)}, {Nat(0)}} {Nat(s(0))} {{Nat(s(x)), Nat(x)}, {Nat(0)}} {Nat(s(0))} We need unification, a way to make literals identical. Based on the notion of substitution, e.g., { x 0 }. (University of Freiburg) Foundations of AI 34 / 57

35 Substitutions A substitution s = { v 1 t1,..., vn t n } substitutes variables v i for terms t i (t i does NOT contain v i ). Applying a substitution s to an expression ϕ yields the expression ϕs which is ϕ with all occurrences of v i replaced by t i for all i. (University of Freiburg) Foundations of AI 35 / 57

36 Substitution Examples P (x, f(y), B) P (z, f(w), B) s = { x z, y w } P (x, f(a), B) s = { y A } { x P (g(z), f(a), B) s = g(z), y } A P (C, f(a), A) (University of Freiburg) Foundations of AI 36 / 57

37 Composing Substitutions Composing substitutions s 1 and s 2 gives s 1 s 2 which is that substitution obtained by first applying s 2 to the terms in s 1 and adding remaining term/variable pairs (not occurring in s 1 ) to s 1. Example: { z g(x,y)} { x A, y B, w C, z D } = { z g(a,b), x A, y B, w C Application example: P (x, y, z) P (A, B, g(a, B)) } (University of Freiburg) Foundations of AI 37 / 57

38 Properties of substitutions For a formula ϕ and substitutions s 1, s 2 (ϕs 1 )s 2 = ϕ(s 1 s 2 ) (s 1 s 2 )s 3 = s 1 (s 2 s 3 ) associativity s 1 s 2 s 2 s 1 no commutativity! (University of Freiburg) Foundations of AI 38 / 57

39 Unification Unifying a set of expressions {w i } Find substitution s such that w i s = w j s for all i, j Example {P (x, f(y), B), P (x, f(b), B)} s = { y B, z A } not the simplest unifier s = { y B } most general unifier (mgu) The most general unifier, the mgu, g of {w i } has the property that if s is any unifier of {w i } then there exists a substitution s such that {w i }s = {w i }gs Property: The common instance produced is unique up to alphabetic variants (variable renaming) (University of Freiburg) Foundations of AI 39 / 57

40 Subsumption Lattice a) b) (University of Freiburg) Foundations of AI 40 / 57

41 Disagreement Set The disagreement set of a set of expressions {w i } is the set of sub-terms {t i } of {w i } at the first position in {w i } for which the {w i } disagree Examples {P (x, A, f(y)), P (v, B, z)} gives {x, v} {P (x, A, f(y)), P (x, B, z)} gives {A, B} {P (x, y, f(y)), P (x, B, z)} gives {y, B} (University of Freiburg) Foundations of AI 41 / 57

42 Unification Algorithm Unify(T erms): Initialize k 0; Initialize T k = T erms; Initialize s k = {}; *If T k is a singleton, then output s k. Otherwise continue. Let D k be the disagreement set of T k. If there exists a var v k and a term t k in D k such that v k does not occur in t k, continue. Otherwise, exit with failure. s k+1 s k { v k tk }; T k+1 T k { v k tk }; k k + 1; Goto *. (University of Freiburg) Foundations of AI 42 / 57

43 Example {P (x, f(y), y), P (z, f(b), B)} (University of Freiburg) Foundations of AI 43 / 57

44 Binary Resolution C 1 {l 1 }, C 2 {l 2 } [C 1 C 2 ]s where s = mgu(l 1, l 2 ), the most general unifier [C 1 C 2 ]s is the resolvent of the parent clauses C 1 {l 1 } and C 2 {l 2 }. C 1 {l 1 } and C 2 {l 2 } do not share variables l 1 and l 2 are the resolution literals. Examples: {{Nat(s(0)), Nat(0)}, {Nat(0)}} {Nat(s(0))} {{Nat(s(0)), Nat(x)}, {Nat(0)}} {Nat(s(0))} {{Nat(s(x)), Nat(x)}, {Nat(0)}} {Nat(s(0))} (University of Freiburg) Foundations of AI 44 / 57

45 Some Further Examples Resolve P (x) Q(f(x)) and R(g(x)) Q(f(A)) Standardizing the variables apart gives P (x) Q(f(x)) and R(g(y)) Q(f(A)) Substitution s = { x A } Resolvent P (A) R(g(y)) Resolve P (x) Q(x, y) and P (A) R(B, z) Standardizing the variables apart Substitution s = { x A } and Resolvent Q(A, y) R(B, z) (University of Freiburg) Foundations of AI 45 / 57

46 Factoring C 1 {l 1 } {l 2 } [C 1 {l 1 }]s where s = mgu(l 1, l 2 ) is the most general unifier. Needed because: {{P (u), P (v)}, { P (x), P (y)}} = but cannot be derived by binary resolution Factoring yields: {P (u)} and { P (x)} whose resolvent is. (University of Freiburg) Foundations of AI 46 / 57

47 Derivations Notation: R( ) = {C C is a resolvent or a factor of two clauses from } We say D can be derived from, i.e., D, if there exist C 1, C 2, C 3,..., C n = D such that C i R( {C 1,..., C i 1 }) for 1 i n. (University of Freiburg) Foundations of AI 47 / 57

48 Example From Russell and Norvig: The law says it is a crime for an American to sell weapons to hostile nations. The country Nono, an enemy of America, has some missiles, and all of its missiles were sold to it by Colonel West, who is American. Prove that Col. West is a criminal. (University of Freiburg) Foundations of AI 48 / 57

49 Example... it is a crime for an American to sell weapons to hostile nations: American(x) weapon(y) Sells(x, y, z) Hostile(z) Criminal(x) Nono... has some missiles, i.e., xowns(nono, x) Missile(x): Owns(Nono, M 1) and Missile(M 1)... all of its missiles were sold to it by Colonel West. xmissiles(x) Owns(Nono, x) Sells(West, x, Nono) Missiles are weapons: Missile(x) Weapon(x) An enemy of America counts as hostile : Enemy(x, America) Hostile(x) West, who is American... American(West) The country Nono, an enemy of America Enemy(Nono, America) (University of Freiburg) Foundations of AI 49 / 57

50 An Example American(x) Weapon(y) Sells(x,y,z) Hostile(z) Criminal(x) American(West) Missile(x) Weapon(x) Missile(M 1 ) Missile(x) Owns(Nono,x) Sells(West,x,Nono) Missile(M 1 ) Owns(Nono,M 1 ) Criminal(West) American(West) Weapon(y) Sells(West,y,z) Hostile(z) Weapon(y) Sells(West,y,z) Hostile(z) Missile(y) Sells(West,y,z) Hostile(z) Sells(West,M 1,z) Hostile(z) Missile(M 1 ) Owns(Nono,M 1 ) Hostile(Nono) Owns(Nono,M 1 ) Hostile(Nono) Enemy(x,America) Hostile(x) Enemy(Nono,America) Hostile(Nono) Enemy(Nono,America) (University of Freiburg) Foundations of AI 50 / 57

51 Another Example Cat(Tuna) Cat(x) Animal(x) Kills(Jack, Tuna) Kills(Curiosity, Tuna) Kills(Curiosity, Tuna) Animal(Tuna) Loves(y, x) Animal(z) Kills(x, z) Kills(Jack, Tuna) Loves(x,F(x)) Loves(G(x), x) Animal(x) Loves(Jack, x) Loves(y, x) Kills(x, Tuna) Animal(F(Jack)) Loves(G(Jack), Jack) Animal(F(x)) Loves(G(x), x) Loves(y, Jack) Loves(G(Jack), Jack) (University of Freiburg) Foundations of AI 51 / 57

52 Properties of Resolution Lemma: (soundness) If D, then = D. Lemma: resolution is refutation-complete: is unsatisfiable implies. Theorem: is unsatisfiable iff. Technique: to prove that = C negate C and prove that { C}. (University of Freiburg) Foundations of AI 52 / 57

53 The Lifting Lemma Lemma: Let C 1 and C 2 be two clauses with no shared variables, and let C 1 and C 2 be ground instances of C 1 and C 2. If C is a resolvent of C 1 and C 2, then there exists a clause such that (1) C is a resolvent of C 1 and C 2 (2) C is a ground instance of C Can be easily generalized to derivations (University of Freiburg) Foundations of AI 53 / 57

54 The General Picture Any set of sentences S is representable in clausal form Assume S is unsatisfiable, and in clausal form Herbrand s theorem Some set S of ground instances is unsatisfiable Ground resolution theorem Resolution can find a contradiction in S Lifting lemma There is a resolution proof for the contradiction in S (University of Freiburg) Foundations of AI 54 / 57

55 Closing Remarks: Processing PL1-Resolution: forms the basis of most state of the art theorem provers for PL1 the programming language Prolog only Horn clauses considerably more efficient methods. not dealt with : search/resolution strategies Finite theories: In applications, we often have to deal with a fixed set of objects. Domain closure axiom: x[x = c 1 x = c 2... x = c n ] Translation into finite propositional theory is possible. (University of Freiburg) Foundations of AI 55 / 57

56 Closing Remarks: Possible Extensions PL1 is definitely very expressive, but in some circumstances we would like more... Second-Order Logic: Also over predicate quantifiers x, y[(x = y) { p[p(x) p(y)]}] Validity is no longer semi-decidable (we have lost compactness) Lambda Calculus: Definition of predicates, e.g., λx, y[ zp (x, z) Q(z, y)] defines a new predicate of arity 2 Reducible to PL1 through Lambda-Reduction Uniqueness quantifier:!xϕ(x) - there is exactly one x... Reduction to PL1: x[ϕ(x) y{ϕ(y) x = y}] (University of Freiburg) Foundations of AI 56 / 57

57 Summary PL1 makes it possible to structure statements, thereby giving us considerably more expressive power than propositional logic. Formulae consist of terms and atomic formulae, which, together with connectors and quantifiers, can be put together to produce formulae. Interpretations in PL1 consist of a universe and an interpretation function. The Herbrand Theory shows that satisfiability in PL1 can be reduced to satisfiability in propositional logic (although infinite sets of formulae can arise under certain circumstances). Resolution is refutation complete Validity in PL1 is not decidable (it is only semi-decidable) (University of Freiburg) Foundations of AI 57 / 57

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