First-Order Predicate Logic. First-Order Logic 153/467

Transcription

1 First-Order Predicate Logic First-Order Logic 153/467

2 What Propositional Logic Cannot Express Propositional logic dealt with logical forms of compound propositions It worked well with relationships like not, and, or, if/then We would like to have a way to talk about individuals (also called objects) and in addition to talk about some object, and all objects, without enumerating all objects in a set This requires extensions to Propositional Logic First-Order Logic Introduction and Motivation 154/467

3 Some Example Statements Some example statements: Not all birds can fly Every student is younger than some instructor These refer to things: birds, students, instructors They also refer to properties of things, either as individuals (ability to fly) or in combination (relative age) We would like to make such statements in our logic and to combine them with the connectives of propositional logic Statements like those above require a context, or world, to be meaningful First-Order Logic Introduction and Motivation 155/467

4 Informal Introduction to Predicate Logic We shall briefly discuss some worlds for predicate logic Arithmetic (the inspiration for formal predicate logic) The blocks world (a classical AI example) Graphs (a ubiquitous concept in CS) Databases (the applied CS version of predicate logic) In each, we shall see the following ingredients Domains The set of objects; also called the universe Constants Objects with specific names Relations Properties of objects, alone or in combination Functions Association of objects to others First-Order Logic Introduction and Motivation 156/467

5 Arithmetic, Etc In arithmetic, we have several possible domains N: The natural numbers Z: The integers Q: The rationals Integers modulo n Over these, we have predicates (or relations): <: Less than Has a square root and also functions: +,, etc A sample formula: x ( y (y < x)) First-Order Logic Introduction and Motivation 157/467

6 A Classic AI Example: Blocks Worlds A blocks world consists of a set of blocks, and a table Each block may be on the table, or on one of the other blocks Each block may have a colour In the picture, there are three blocks Two of them are blue (vertical stripes) and one is red (diagonal stripes) B B3 B Set of objects: {B1, B2, B3} We can describe the world with relations: On, OnTable, Red, Blue, First-Order Logic Introduction and Motivation 158/467

7 Describing a Blocks World The domain {B1, B2, B3} is a finite set Therefore, we can list all of the properties, in various ways: OnTable(B1), OnTable(B2), OnTable(B3) On: B1 B2 B3 B1 F F F B2 F F F B3 T F F The set { b Blue(b) } of blue blocks is {B2, B3} First-Order Logic Introduction and Motivation 159/467

8 Properties in the Blocks World Some properties are fundamental to the world No box is on itself x On(x, x) A box on the table is not on any box : x (OnTable(x) ( y On(x, y))) Some properties depend on the situation Every red box has a box on it : x (Red(x) y On(y, x)) Some box is on a box that is on the table : x y (On(x, y) OnTable(y)) First-Order Logic Introduction and Motivation 160/467

9 Graphs A graph is a binary relation A finite graph: An infinite graph: y x 3 2 (The set { x, y y = 1 + x/2 }) A graph is undirected if the relation is symmetric; ie, the formula x y (E(x, y) E(y, x)) holds First-Order Logic Introduction and Motivation 161/467

10 Relational Databases A relational database is a listing of one or more relations Example: Person: The people (or their names) NumberOf : An association between people and their phone numbers Here the domain contains both people and phone numbers the objects about which we have relations A sample statement: Somebody has no phone number x (Person(x) ( y (NumberOf (x, y))) First-Order Logic Introduction and Motivation 162/467

11 A Conundrum Consider the statement, Only people have phone numbers How shall we represent it as a logical formula? Whenever x and y satisfy NumberOf (x, y), then x is a person x ( y (NumberOf (x, y) Person(x))) Whenever x has some phone number y, then x is a person x (( y NumberOf (x, y)) Person(x)) These two formulas are respectively equivalent to x ( y ( NumberOf (x, y) Person(x))) and to x (( ( y NumberOf (x, y))) Person(x)) Are these two formulas equivalent? First-Order Logic Introduction and Motivation 163/467

12 Domains We now turn to general definitions A domain is a non-empty set In principle, any non-empty set can be a domain: the natural numbers, people now alive, {T, F}, etc A constant symbol refers to an object in the domain; eg, 0, B1, Justin Trudeau, etc First-Order Logic Concepts of First-Order Logic 164/467

13 Predicates/Relations A predicate, or relation, represents a property that an individual, or collection of individuals, may (or may not) have In English, we might express a predicate as is a student In symbolic logic, we write S(x) to mean x has property S For example, if S is the property of being a student, then Alex is a student becomes S(Alex) Similarly, we might use I(Sam) for Sam is an instructor and Y (Alex,Sam) for Alex is younger than Sam First-Order Logic Concepts of First-Order Logic 165/467

14 Representing Relations Mathematically, we represent a relation by the set of all things that have the property If S is the set of all students, then x S means x is a student The only restriction on a relation is that it must be a subset of the domain A k-ary relation is a set of k-tuples of domain elements For example, the binary relation less-than, over a domain D, is represented by the set { x, y D 2 x < y } As another example, the On relation in the sample blocks world has just one pair: { B3, B1 } (In a relational database, the listing of such a set is called a table ) First-Order Logic Concepts of First-Order Logic 166/467

15 Variables Variables make statements more expressive You may think of a variable as a place holder, or blank, that can be replaced by a concrete object Alternatively, a variable is a name without a fixed referent Which object the name refers to can vary from time to time A variable lets us refer to an object, without specifying perhaps without even knowing which particular object it is Thus we can express a relation in the abstract S(x): x is a student I(x): x is an instructor Y (x, y): x is younger than y First-Order Logic Concepts of First-Order Logic 167/467

16 Uses of Variables In general, we use variables that range over the domain to make general statements, such as x 2 0, and in expressing conditions which individuals may or may not satisfy, such as x + x = x x This latter condition is satisfied by only two numbers: 0 and 2 The meaning of such an expression will depend on the domain For example, the formula x 2 < x is always false over the domain of integers, but not over the domain of rational numbers First-Order Logic Concepts of First-Order Logic 168/467

17 Quantifiers What about Every student x is younger than some professor y? In math-speak, we say for all to express every and there exists to express some A familiar(?) example from calculus: For all ε > 0, there exists δ > 0 such that for all y, if x y < δ then f(x) f(y) < ε For all is denoted by, the universal quantifier symbol, and there exists is denoted by, the existential quantifier symbol In FOL, the above comes out as the formula ε (ε > 0 δ (δ > 0 y ( x y < δ f(x) f(y) < ε))) First-Order Logic Concepts of First-Order Logic 169/467

18 Quantifiers: Examples Quantifiers require a variable: x (for all x) or z (there exists z) For example, the statement Not all birds can fly can be written as ( x (B(x) F (x))) Every student is younger than some instructor can become x (S(x) ( y (I(y) Y (x, y)))) Or should that be y (I(y) x (S(x) Y (x, y)))? These two formulas are NOT equivalent! First-Order Logic Concepts of First-Order Logic 170/467

19 Syntax of Predicate Logic Syntax of First-Order Logic 171/467

20 The seven kinds of symbols: The Language of First-Order Logic 1 Constant symbols Usually c, d, c 1, c 2,, d 1, d 2 2 Variables Usually x, y, z, x 1, x 2,, y 1, y 2 3 Function symbols Usually f, g, h, f 1, f 2,, g 1, g 2, 4 Predicate symbols Usually P, Q,, P 1, P 2,, Q 1, Q 2, 5 Connectives:,,,, and 6 Quantifiers: and 7 Punctuation: (, ), and, Function symbols and predicate symbols have an assigned arity the number of arguments required The last three kinds of symbols connectives, quantifiers, and punctuation will have their meaning fixed by the syntax and semantics Constants, variables, functions and predicate symbols are not restricted They may be assigned any meaning, consistent with their kind and arity Syntax of First-Order Logic 172/467

21 Terms In FOL, we need to consider two kinds of expressions: those that can have a truth value, called formulas, and those that refer to an object of the domain, called terms We start with terms Definition The set of terms is defined inductively as follows 1 Each constant symbol is a term, and each variable is a term Such terms are called atomic terms 2 If t 1,, t n are terms and f is an n-ary function symbol, then f(t 1,, t n ) is a term If n = 2 (a binary function symbol), we may write (t 1 f t 2 ) instead of f(t 1, t 2 ) 3 Nothing else is a term Syntax of First-Order Logic 173/467

22 Examples of Terms Example 1 If 0 is a constant symbol, x and y are variables, and s (1) and + (2) are function symbols, then 0, x, and y are terms, as are s(0) and +(x, s(y)) The expressions s(x, y) and s + x are not terms Example 2 Suppose f is a unary function symbol, g is a binary function symbol, and a is a constant symbol Then g(f(a), a) and f(g(a, f(a))) are terms The expressions g(a) and f(f(a), a) are not terms Syntax of First-Order Logic 174/467

23 Atomic Formulas As in propositional logic, a formula represents a proposition (a true/false statement) The relation symbols produce propositions Definition: An atomic formula (or atom) is an expression of the form P (t 1,, t n ) where P is an n-ary relation symbol and each t i is a term (1 i n) If P has arity 2, the atom P (t 1, t 2 ) may alternatively be written (t 1 P t 2 ) Syntax of First-Order Logic 175/467

24 General Formulas We define the set of well-formed formulas of first-order logic inductively as follows 1 An atomic formula is a formula 2 If α is a formula, then ( α) is a formula 3 If α and β are formulas, and is a binary connective symbol, then (α β) is a formula 4 If α is a formula and x is a variable, then each of ( x α) and ( x α) is a formula 5 Nothing else is a formula In case 4, the formula α is called the scope of the quantifier The quantifier keeps the same scope if it is included in a larger formula Syntax of First-Order Logic 176/467

25 Parse Trees Parse trees for FOL formulas are similar to parse trees for propositional formulas Quantifiers x and y form nodes is the same way as negation (ie, only one sub-tree) A predicate P (t 1, t 2,, t n ) has a node labelled P with a sub-tree for each of the terms t 1, t 2,, t n Syntax of First-Order Logic 177/467

26 Examples: Parse trees Example: ( x ((P (x) Q(x)) S(x, y))) Syntax of First-Order Logic 178/467

27 Examples: Parse trees Example: ( x ((P (x) Q(x)) S(x, y))) Example: ( x (F (b) ( y ( z (G(y, z) H(u, x, y)))))) Syntax of First-Order Logic 178/467

28 Examples: Parse trees Example: ( x ((P (x) Q(x)) S(x, y))) Example: ( x (F (b) ( y ( z (G(y, z) H(u, x, y)))))) Ordinarily, one would omit many of the parentheses in the second formula, and write simply x (F (b) y z (G(y, z) H(u, x, y))) Syntax of First-Order Logic 178/467

29 Semantics: Interpretations We cover more on syntax later, but we first start the discussion of semantics Definition: Fix a set L of constant symbols, function symbols, and relation symbols (The language of our formulas) An interpretation (for the set L) consists of A non-empty set dom( ), called the domain (or universe) of For each constant symbol c, a member c of dom( ) For each function symbol f (i), an i-ary function f For each relation symbol R (i), an i-ary relation R Huth and Ryan use the term model instead of interpretation (Not a standard usage) Variable-free semantics 179/467

30 Values of Variable-Free Terms For terms and formulas that contain no variables or quantifiers, an interpretation suffices to specify their meaning The meaning arises in the obvious(?) fashion from the syntax of the term or formula Definition: Fix an interpretation For each term t containing no variables, the value of t under interpretation, denoted t, is as follows If t is a constant c, the value t is c If t is f(t 1,, t n ), the value t is f (t 1,, t n) The value of a term is always a member of the domain of Variable-free semantics 180/467

31 Formulas with Variable-Free Terms Formulas get values in much the same fashion as terms, except that values of formulas lie in {F, T} Definition: Fix an interpretation For each formula α containing no variables, the value of α under interpretation, denoted α, is as follows If α is R(t 1,, t n ), then { α T = { F if t 1,, t n R otherwise If α is ( β) or (β γ), then α is determined by β and γ in the same way as for propositional logic Variable-free semantics 181/467

32 Examples Let 0 be a constant symbol, f (1) a function symbol and E (1) a relation symbol Thus E(f(0)) and E(f(f(0))) are both formulas Consider an interpretation with Domain: N, the natural numbers 0 : zero f : successor; { x, x + 1 x N } E : is even ; { y + y y N } Terms get numerical values: f(0) is 1 and f(f(0)) is 2 Formula E(f(0)) means 1 is even, and E(f(0)) = F Formula E(f(f(0))) means 2 is even, and E(f(f(0))) = T What about some other interpretation? Variable-free semantics 182/467

33 Example, Continued Let be the interpretation with Domain: Q, the rational numbers 0 : two f : halving; { x, x/2 x Q } E : is an integer ; { x x Z } E(f(0)) means 1 is an integer, and E(f(0)) is T E(f(f(0))) means 1/2 is an integer, and E(f(f(0))) is F Exercise: in both and, the formula E(f(f(0))) E(f(0)) receives value F Find another interpretation which gives it the value T Variable-free semantics 183/467

34 Gotchas Two often-overlooked points about interpretations 1 There is NO default meaning for relation, function or constant symbols = 3 might mean that one plus two equals three but only if we specify that interpretation Any interpretation of constants 1, 2, and 3, function symbol + (2) and relation symbol = (2) is possible 2 Functions must be defined at every point in the domain (Ie, they must be total) If we have language with a binary function symbol, we cannot specify an interpretation with domain N and subtraction for Subtraction is not total on N Variable-free semantics 184/467

35 Variables To discuss the evaluation of formulas that contain variables, we need a few more concepts from syntax We shall discuss bound and free variables, substitution of terms for variables Syntax, Continued Free and Bound Variables 185/467

36 Free and Bound Variables Recall: the scope of a quantifier in a sub-formula x α or x α is the formula α An occurrence of a variable in a formula is bound if it lies in the scope of some quantifier of the same variable; otherwise it is free In other words, a quantifier binds its variable within its scope Example In formula x ( y (x + y = z)), x is bound (by x), y is bound (by y), and z is free Example In formula P (x) ( x ( Q(x))), the first occurrence of x is free and the last occurrence of x is bound (The variable symbol immediately after or is neither free nor bound) Syntax, Continued Free and Bound Variables 186/467

37 Free and Bound Variables Formally, a variable occurs free in a formula α if and only if it is a member of the set FV(α) defined as follows 1 If α is P (t 1,, t k ), then FV(α) = { x x appears in some t i } 2 If α is ( β), then FV(α) = FV(β) 3 If α is (β γ), then FV(α) = FV(β) FV(γ) 4 If α is Qx β (for Q {, }), then FV(α) = FV(β) {x} A formula has the same free variables as its parts, except that a quantified variable becomes bound A formula with no free variables is called a closed formula, or a sentence Syntax, Continued Free and Bound Variables 187/467

38 Substitution The notation α[t/x], for a variable x, a term t, and a formula α, denotes the formula obtained from α by replacing each free occurrence of x with t Intuitively, it is the formula that answers the question, What happens to α if x has the value specified by term t? Examples If α is the formula E(f(x)), then α[(y + y)/x] is E(f(y + y)) α[f(x)/x] is E(f(f(x))) E(f(x + y))[y/x] is E(f(y + y)) Substitution does NOT affect bound occurrences of the variable If β is x (E(f(x)) S(x, y)), then β[g(x, y)/x] is β, because β has no free occurrence of x Syntax, Continued Substitution 188/467

39 Examples: Substitution Example Let β be P (x) ( x Q(x)) What is β[y/x]? Syntax, Continued Substitution 189/467

40 Examples: Substitution Example Let β be P (x) ( x Q(x)) What is β[y/x]? β[y/x] is P (y) ( x Q(x)) Only the free x gets substituted Syntax, Continued Substitution 189/467

41 Examples: Substitution Example Let β be P (x) ( x Q(x)) What is β[y/x]? β[y/x] is P (y) ( x Q(x)) Only the free x gets substituted Example What about β[(y 1)/z], where β is x ( y ((x + y) = z)))? Syntax, Continued Substitution 189/467

42 Examples: Substitution Example Let β be P (x) ( x Q(x)) What is β[y/x]? β[y/x] is P (y) ( x Q(x)) Only the free x gets substituted Example What about β[(y 1)/z], where β is x ( y ((x + y) = z)))? At first thought, we might say x ( y ((x + y) = (y 1))) But there s a problem the free variable y in the term (y 1) got captured by the quantifier y We want to avoid this capture Syntax, Continued Substitution 189/467

43 Avoiding Capture Example Formula α = S(x) y (P (x) Q(y)); term t = f(y, y) The leftmost x can be substituted by t since it is not in the scope of any quantifier, but substituting in P (x) puts the variable y into the scope of y We can prevent capture of variables in two ways Declare that a substitution is undefined in cases where capture would occur One can often evade problems by a different choice of variable (Above, we might be able to substitute f(z, z) instead of f(y, y) Or alter α to quantify some other variable) Write the definition of substitution carefully, to prevent capture Huth and Ryan opt for the first method We shall use the second Syntax, Continued Substitution 190/467

44 Substitution Formal Definition Let x be a variable and t a term For a term u, the term u[t/x] is u with each occurrence of the variable x replaced by the term t For a formula α, 1 If α is P (t 1,, t k ), then α[t/x] is P (t 1 [t/x],, t k [t/x]) 2 If α is ( β), then α[t/x] is ( β[t/x]) 3 If α is (β γ), then α[t/x] is (β[t/x] γ[t/x]) 4 (Continued next page ) Syntax, Continued Substitution 191/467

45 Substitution Formal Definition (2) For variable x, term t and formula α: 4 If α is (Qx β), then α[t/x] is α 5 If α is (Qy β) for some other variable y, then (a) If y does not occur in t, then α[t/x] is (Qy β[t/x]) (b) Otherwise, select a variable z that occurs in neither α nor t; then α[t/x] is (Qz (β[z/y])[t/x]) The last case prevents capture by renaming the quantified variable to something harmless (Huth and Ryan specify that the substitution is undefined if capture would occur case 5(b) above With this more complex definition, one never has to add a condition regarding undefined substitutions Substitution always behaves the way it should ) Syntax, Continued Substitution 192/467

46 Example, Revisited Example If α is x ( y (x + y = z)), what is α[(y 1)/z]? This falls under case 5(b): the term to be substituted, namely y 1, contains a variable y quantified in formula α Let β be (x + y = z); thus α is x ( y β) Syntax, Continued Substitution 193/467

47 Example, Revisited Example If α is x ( y (x + y = z)), what is α[(y 1)/z]? This falls under case 5(b): the term to be substituted, namely y 1, contains a variable y quantified in formula α Let β be (x + y = z); thus α is x ( y β) Select a new variable, say w Then β[w/y] is x + w = z, and β[w/y][(y 1)/z] is (x + w) = (y 1) Thus the required formula α[(y 1)/z] is x w((x + w) = (y 1)) Syntax, Continued Substitution 193/467

48 Semantics of Predicate Logic Semantics 194/467

49 FOL Adds to Propositional Logic In propositional logic, semantics was described in terms of valuations to propositional atoms FOL includes more ingredients (ie, predicates, functions, variables, terms, constants, etc) and, hence, the semantics for FOL must account for all of the ingredients We already saw the concept of an interpretation, which specifies the domain and the identities of the constants, relations and functions Formulas that include variables, and perhaps quantifiers, require additional information, known as an environment (or assignment) Semantics 195/467

50 Environments A first-order environment is a function that assigns a value in the domain to each variable Example With the domain N, we might have environment E 1 given by E 1 (x) = 9 and E 1 (y) = 2 If the interpretation specifies < is less-than, then x < y gets value false Example With the domain of fictional animals, we might have E 2 (x) = Tweety and E 2 (y) = Nemo If the interpretation specifies < is was created before, then x < y gets value true Semantics Environments 196/467

51 Constants Vs Variables Example: Let α 1 be P (c) (where c is a constant), and let α 2 be P (x) (where x a variable) Let be the interpretation with domain N, c = 2 and P = is even Then α 1 = T, but α 2 is undefined To give α 2 a value, we must also specify an environment For example, if E(x) = 2, then α (,E) 2 = T If we wish, we can consider a formula such as α 2 that contains a free variable x as expressing a function: the function that maps E(x) to α (,E) 2 Semantics Environments 197/467

52 Meaning of Terms The combination of an interpretation and an environment supplies a value for every term Definition: Fix an interpretation and environment E For each term t, the value of t under and E, denoted t (,E), is as follows If t is a constant c, the value t (,E) is c If t is a variable x, the value t (,E) is x E If t is f(t 1,, t n ), the value t (,E) is f (t (,E) 1,, t (,E) n ) To extend this definition to formulas, we must consider quantifiers But first, a few examples Semantics Environments 198/467

53 Meaning of Terms Example Example Suppose a language has constant symbol 0, a unary function s, and a binary function + We shall write + in infix position: x + y instead of +(x, y) The expressions s(s(0) + s(x)) and s(x + s(x + s(0))) are both terms The following are examples of interpretations and environments dom{ } = {0, 1, 2, }, 0 = 0, s is the successor function and + is the addition operation Then, if E(x) = 3, the terms get values (s(s(0) + s(x))) (,E) = 6 and (s(x + s(x + s(0)))) (,E) = 9 Semantics Environments 199/467

54 Meaning of Terms Example 2 dom{ } is the collection of all words over the alphabet {a, b}, 0 = a, s appends a to the end of a string, and + is concatenation Let E(x) = aba Then (s(s(0) + s(x))) (,E) = aaabaaa and (s(x + s(x + s(0)))) (,E) = abaabaaaaa Semantics Environments 200/467

55 Quantified Formulas To evaluate the truthfulness of a formula x α (resp x α), we should check whether α holds for every (resp, for some) value a in the domain How can we express this precisely? Definition: For any environment E and domain element d, the environment E with x re-assigned to d, denoted E[x d], is given by { d if y is x E[x d](y) = { E(y) if y is not x Semantics Environments 201/467

56 Values of Quantified Formulas Definition: The values of x α and x α are given by { ( x α) (,E) T = { F { ( x α) (,E) T = { F if α (,E[x d]) = T for every d in dom( ) otherwise if α (,E[x d]) = T for some d in dom( ) otherwise Note: The values of ( x α) (,E) and ( x α) (,E) do not depend on the value of E(x) The value E(x) only matters for free occurrences of x Semantics Environments 202/467

57 Examples: Value of a Quantified Formula Example Let dom( ) = {a, b} and R = { a, a, a, b, b, b } Let E(x) = a and E(y) = b We have R(x, x) (,E) = T, since E(x), E(x) = a, a R R(y, x) (,E) = F, since E(y), E(x) = b, a R ( y R(y, x)) (,E) = T, since R(y, x) (,E[y a]) = T (That is, E[y a](y), E[y a](x) = a, a R ) What is ( x y R(x, y)) (,E)? Semantics Environments 203/467

58 Examples: Continued Example Let dom( ) = {a, b} and R = { a, a, a, b, b, b } Let E(x) = a and E(y) = b What is ( x y R(x, y)) (,E)? Since b, a R, we have R(x, y) (,E[x b][y a]) = F, and thus ( x y R(x, y)) (,E) = F Semantics Environments 204/467

59 Examples: Continued Example Let dom( ) = {a, b} and R = { a, a, a, b, b, b } Let E(x) = a and E(y) = b What is ( x y R(x, y)) (,E)? Since b, a R, we have R(x, y) (,E[x b][y a]) = F, and thus ( x y R(x, y)) (,E) = F What about ( x y R(x, y)) (,E)? Semantics Environments 204/467

60 A Question of Syntax In the previous example, we wrote R(x, y) (,E[x b][y a]) = F Why did we not write simply R(b, a) = F or perhaps R(b, a) (,E) = F? Semantics Environments 205/467

61 A Question of Syntax In the previous example, we wrote R(x, y) (,E[x b][y a]) = F Why did we not write simply R(b, a) = F or perhaps R(b, a) (,E) = F? Because R(b, a) is not a formula The elements a and b of dom( ) are not symbols in the language; they cannot appear in a formula Semantics Environments 205/467

62 Satisfaction of Formulas An interpretation and environment E satisfy a formula α, denoted E α, if α (,E) = T; they do not satisfy α, denoted E α, if α (,E) = F Form of α R(t 1,, t k ) β β γ β γ β γ x β x β Condition for E α t (,E) 1,, t (,E) k R E β both E β and E γ either E β or E γ (or both) either E β or E γ (or both) for every a dom( ), E[x a] β there is some a dom( ) such that E[x a] β If E α for every E, then satisfies α, denoted α Semantics Satisfaction of Formulas 206/467

63 Example: Satisfaction Example Consider the formula y R(x, y y) (For R a binary relation and a binary function) Suppose dom( ) = {1, 2, 3, }, is the addition operation, and R is the equality relation Then E y R(x, y y) iff E(x) is an even number Semantics Satisfaction of Formulas 207/467

64 Validity and Satisfiability Validity and satisfiability of formulas have definitions analogous to the ones for propositional logic Definition: A formula α is valid if every interpretation and environment satisfy α; that is, if E α for every and E, satisfiable if some interpretation and environment satisfy α; that is, if E α for some and E, and unsatisfiable if no interpretation and environment satisfy α; that is, if E α for every and E (The term tautology is not used in predicate logic) Semantics Satisfaction of Formulas 208/467

65 Example: Satisfiability and Validity Let α be the formula P (f(g(x), g(y)), g(z)) The formula is satisfiable: dom( ): N f : summation g : squaring P : equality E(x) = 3, E(y) = 4 and E(z) = 5 α is not valid (Why?) Semantics Satisfaction of Formulas 209/467

66 Quantifiers Over Finite Domains The universal and existential quantifiers may be understood respectively as generalizations of conjunction and disjunction If the domain D = {a 1,, a k } is finite then: For all x, R(x) iff R(a 1 ) and and R(a k ) There exists x, R(x) iff R(a 1 ) or or R(a k ) where R is a property Semantics Satisfaction of Formulas 210/467

67 Relevance Lemma Lemma: Let α be a first-order formula, be an interpretation, and E 1 and E 2 be two environments such that E 1 (x) = E 2 (x) for every x that occurs free in α Then E1 α if and only if E2 α Proof by induction on the structure of α Semantics Satisfaction of Formulas 211/467

68 Logical Consequence Suppose Σ is a set of formulas and α is a formula We say that α is a logical consequence of Σ, written as Σ α, iff for any interpretation and environment E, we have E Σ implies E α α means that α is valid Semantics Satisfaction of Formulas 212/467

69 Example: Entailment Example: Show that ( x(α β)) (( x α) ( x β)) Proof by contradiction Suppose there are and E such that E ( x(α β)) (( x α) ( x β)) Then we must have E x(α β) and E ( x α) ( x β); the second gives E x α and E x β Using the definition of for formulas with, we have for every a dom( ), E[x a] α β and E[x a] α Thus also E[x a] β for every a dom( ) Thus E x β, a contradiction Semantics Satisfaction of Formulas 213/467

70 Example II: Entailment Example Show that x( γ) ( x γ) Semantics Satisfaction of Formulas 214/467

71 Example II: Entailment Example Show that x( γ) ( x γ) Suppose that E x( γ) By definition, this means for every a dom( ), E[x a] γ Again by definition (for a formula with ), this is equivalent to for every a dom( ), E[x a] γ and also to there is no a dom( ) such that E[x a] γ This last is the definition of E ( x γ), as required Semantics Satisfaction of Formulas 214/467

72 Example Example: Show that, in general, ( x α) ( x β) x(α β) (That is, find α and β such that consequence does not hold) Semantics Satisfaction of Formulas 215/467

73 Example Example: Show that, in general, ( x α) ( x β) x(α β) (That is, find α and β such that consequence does not hold) Key idea: φ 1 φ 2 yields true whenever φ 1 is false Let α be R(x) Let have domain {a, b} and R = {a} Then ( x α) ( x β) for any β (Why?) Semantics Satisfaction of Formulas 215/467

74 Example Example: Show that, in general, ( x α) ( x β) x(α β) (That is, find α and β such that consequence does not hold) Key idea: φ 1 φ 2 yields true whenever φ 1 is false Let α be R(x) Let have domain {a, b} and R = {a} Then ( x α) ( x β) for any β (Why?) To obtain x(α β), we can use R(x) for β (Why?) Thus ( x α) ( x β) x(α β), as required (Why?) Semantics Satisfaction of Formulas 215/467

75 Example Example: for any formula α and term t, ( x α) α[t/x] Recall that functions must be total! Semantics Satisfaction of Formulas 216/467

76 Proofs in First-Order Logic Using Natural Deduction Natural Deduction 217/467

77 Natural Deduction for FOL Natural Deduction for FOL extends Natural Deduction for propositional logic by including rules for introduction and elimination of quantifiers Other proof techniques and tricks remain the same as natural deduction for propositional logic Natural Deduction 218/467

78 e and i Elimination of and introduction of are fairly straightforward Name -notation inference notation -elimination ( e) -introduction ( i) If Σ x α then Σ α[t/x] If Σ α[t/x], then Σ x α x α α[t/x] α[t/x] x α Given that a formula is true for every value of x, conclude it is true for any particular value, such as that of t Given that a formula is true for a particular value (denoted by t), conclude it is true for some value Natural Deduction -Elimination and -Introduction 219/467

79 Example: e All fish can swim Nemo is a fish Therefore, Nemo can swim In FOL: show that { x(f (x) S(x)), F (Nemo)} S(Nemo) Proof: 1 x(f (x) S(x)) Premise 2 F (Nemo) Premise 3 F (Nemo) S(Nemo) e: 1 4 S(Nemo) e: 2, 3 The proof doesn t care what F and S mean Fishiness and swimming ability really have nothing to do with the argument Natural Deduction -Elimination and -Introduction 220/467

80 Example: i Example Show P (y) x(p (x) Q(y)) 1 P (y) Premise 2 P (y) Assumption 3 e: 2, 1 4 Q(y) e: 3 5 P (y) Q(y) i: x(p (x) Q(y)) i: 5 Natural Deduction -Elimination and -Introduction 221/467

81 Note to the example The general form of rule i: α[t/x] x α Use in the previous example: P (y) Q(y) x(p (x) Q(y)) We took P (x) Q(y) for α However, knowing what α[t/x] is, does not determine what α is We could also take P (x) Q(x) for α; thus the derivation step would be P (y) Q(y) x(p (x) Q(x)) But the formula x(p (x) Q(x)) is not what we wanted to prove Natural Deduction -Elimination and -Introduction 222/467

82 Soundness of -Elimination and -Introduction Claim: For any formula φ, variable x and term t, x φ φ[t/x] and φ[t/x] xφ Proof: Suppose E Qx φ; ie, for (every/some) d dom( ), φ (,E[x d]) = T Since t (,E) is value in the domain, it suffices to show Claim II: For every formula φ, variable x and term t, φ[t/x] (,E) = φ (,E[x t(,e) ]) To prove this second claim, we use the definition of substitution to do an induction on the structure of formula φ Natural Deduction -Elimination and -Introduction 223/467

83 Soundness of -Elimination, Cont d One of the cases: φ is a quantified formula, say y α We have ( y α) (,E[x t(,e) ]) = T iff for some d dom( ), α (,E[x t(,e) ][y d]) = T Likewise, we have (( y α)[t/x]) (,E) = T iff for some d dom( ), (α[t/x]) (,E[y d]) = T In the first case, t is evaluated under environment E In the second, t is evaluated under environment E[y d] (or a further modification) If y is free in t, the difference matters! Also, we have a problem if y is the same variable as x Natural Deduction -Elimination and -Introduction 224/467

84 Defining Substitution The definition of substitution included the following For a variable x and a term t: 4 If α is (Qx β), then α[t/x] is α 5 If α is (Qy β) for some other variable y, then If y does not occur in t, then α[t/x] is (Qy β[t/x]) Otherwise, let z be a variable that occurs in neither α nor t; then α[t/x] is (Qz (β[z/y])[t/x]) With this definition, we always get that, as required, Proof left to you (Qy α) (,E[x t(,e) ]) = ((Qy α)[t/x]) (,E) (Note: ((Qy α)[t/x]) (,E) may differ from (Qy(α[t/x])) (,E)!) Natural Deduction -Elimination and -Introduction 225/467

85 Proving a Universal To motivate the rule of -introduction, we consider ordinary mathematical usage In order to prove that a property holds for all integers, one often starts with Let x be an integer This means the same as Assume that the variable x refers to an integer Then one proves that x has the property Since we know nothing about the value x, except that it is an integer, this justifies that every integer has the property One could also start the proof with Let x be anything If x is an integer, then The conclusion is essentially the same Natural Deduction -Introduction 226/467

86 Rule -Introduction Definition: a variable is fresh in a subproof if it occurs nowhere outside the box of the subproof Freshness captures the notion of know nothing about Name -notation inference notation -introduction ( i) If Σ α[y/x] and y not free in Σ or α, then Σ x α y fresh α[y/x] x α In words: in order to prove x α(x), prove α(y) for arbitrary y Natural Deduction -Introduction 227/467

87 Rule i Is Sound To further clarify the rule i, we show that it is sound That is, Suppose that Σ α[y/x] and y is not free in Σ or α Then Σ x α Proof: Fix an arbitrary and E with E Σ The supposition Σ α[y/x] thus requires E α[y/x] We need to show that E[x a] α for every a dom( ) Consider an arbitrary a dom( ) Since y is not free in Σ, the Relevance Lemma yields E[y a] Σ Since y is not free in α, we have α[y/x] (,E[y a]) = α (,E[x a]) Therefore E[x a] α for every a, and thus E x α as required Natural Deduction -Introduction 228/467

88 Example: Use of i Example Show that x α x α, for any α 1 x α Premise 2 u fresh α[u/x]?? n x α i: 2 6 Note: u fresh means we choose any variable not in α Natural Deduction -Introduction 229/467

89 Example: Use of i Example Show that x α x α, for any α 1 x α Premise 2 u fresh 3 α[u/x] Assumption 4 x α i: 3 5 e: 1, 4 6 α[u/x] i: x α i: 2 6 Note: u fresh means we choose any variable not in α Natural Deduction -Introduction 229/467

90 Example: Another use of i Show that x(α β) ( x α) ( x β) 1 x(α β) Premise ( x α) ( x β) i?? Note: do not apply rule e until you know which term to use Natural Deduction -Introduction 230/467

91 Example: Another use of i Show that x(α β) ( x α) ( x β) 1 x(α β) Premise 2 x α Assumption x β i??? 8 ( x α) ( x β) i?? Note: do not apply rule e until you know which term to use Natural Deduction -Introduction 230/467

92 Example: Another use of i Show that x(α β) ( x α) ( x β) 1 x(α β) Premise 2 x α Assumption 3 u fresh 6 β[u/x]?? 7 x β i??? 8 ( x α) ( x β) i?? Note: do not apply rule e until you know which term to use Natural Deduction -Introduction 230/467

93 Example: Another use of i Show that x(α β) ( x α) ( x β) 1 x(α β) Premise 2 x α Assumption 3 u fresh 4 α[u/x] β[u/x] e: 1 5 α[u/x] e: 2 6 β[u/x] e: 4, 5 7 x β i: ( x α) ( x β) i: 2 7 Note: do not apply rule e until you know which term to use Natural Deduction -Introduction 230/467

94 Elimination of an Existential Quantifier Name -notation inference notation -elimination ( e) If Σ, α[u/x] β, with u fresh, then Σ, x α β x α α[u/x], u fresh β β In e, the variable u should not occur free in Σ, α, or β (Of course, u will normally be free in α[u/x]) Natural Deduction -Elimination 231/467

95 Rule e Is Sound The rule e is sound That is, Suppose that Σ, α[u/x] β and u is not free in Σ, α, or β Then Σ, x α β Proof: Exercise Follow the proof of soundness of i Natural Deduction -Elimination 232/467

96 Example: Use of e Example Show that x R(x) y R(y) 1 x R(x) Premise 2 R(u), u fresh Assumption 3 y R(y) i: 2 (term u) 4 y R(y) e: 1, 2 3 Natural Deduction -Elimination 233/467

97 Extending the example? Clearly, the previous proof did not depend on the particular relation R that we used Can we do the same proof for arbitrary formulas? Does x α y α[y/x] hold? 1 x α Premise 2 α[u/x], u fresh Assumption 3 α[y/x][u/y]???? 4 y α[y/x] i: 3 (term u) 5 y α[y/x] e: 1, 2 4 Is the formula on line 2 the same as the one on line 3? Natural Deduction -Elimination 234/467

98 Extending the example? Clearly, the previous proof did not depend on the particular relation R that we used Can we do the same proof for arbitrary formulas? Does x α y α[y/x] hold? 1 x α Premise 2 α[u/x], u fresh Assumption 3 α[y/x][u/y]???? 4 y α[y/x] i: 3 (term u) 5 y α[y/x] e: 1, 2 4 Is the formula on line 2 the same as the one on line 3? If y is free in α, then no the derivation fails But otherwise, it works Natural Deduction -Elimination 234/467

99 Example: and together Example Show that x α x α 1 x α Premise x α e?? Natural Deduction -Elimination 235/467

100 Example: and together Example Show that x α x α 1 x α Premise 2 α[u/x], u fresh Assumption x α i?? 7 x α e?? Natural Deduction -Elimination 235/467

101 Example: and together Example Show that x α x α 1 x α Premise 2 α[u/x], u fresh Assumption 3 x α Assumption 4 α[u/x] e: 3 5 e: 4, 2 6 x α i: x α e: 1, 2 6 Natural Deduction -Elimination 235/467

102 Example: e and i together, again We can interchange the quantifiers in the previous deduction Example Show x α x α 1 x α Premise 2 x α Assumption 3 α[u/x] (u fresh) Assumption 4 α[u/x] e: 1 5 e: 3, 4 6 e: 2, x α i: 2 6 Natural Deduction -Elimination 236/467

103 Quantifiers and Negation: The final case So far, we have shown x α x α, x α x α, and x α x α Example Show that x α x α 1 x α Premise α[t/x]?? x α i:?? Natural Deduction -Elimination 237/467

104 Quantifiers and Negation: The final case So far, we have shown x α x α, x α x α, and x α x α Example Show that x α x α 1 x α Premise α[t/x]?? x α i:?? For what term t can we prove α[t/x]? Natural Deduction -Elimination 237/467

105 Quantifiers and Negation: The final case So far, we have shown x α x α, x α x α, and x α x α Example Show that x α x α 1 x α Premise α[t/x]?? x α i:?? For what term t can we prove α[t/x]? There is no such t! We need to try something cleverer Natural Deduction -Elimination 237/467

106 The Final Case: A full proof Example Show that x α x α 1 ( x α) Premise 2 ( x( α)) Assumption 3 u fresh 4 α[u/x] Assumption 5 x( α) i: 4 6 e: 5, 2 7 ( α[u/x]) i: α[u/x] e: 7 9 x α i: e: 9, 1 11 ( ( x( α))) i: x( α) e: 11 Natural Deduction -Elimination 238/467

107 Repeated Quantifiers The rules for elimination and introduction of quantifiers can be generalized to multiple quantifiers Let x 1,, x n be n distinct variables If Σ x 1 x n α, then Σ α[t 1 /x 1 ] [t n /x n ] If Σ α[t 1 /x 1 ] [t n /x n ], for terms t 1,, t n, then Σ x 1 x n α If Σ α[u 1 /x 1 ] [u n /x n ], with variables u 1,, u n fresh, then Σ x 1 x n α If Σ x 1 x n α and Σ {α[u 1 /x 1 ] [u n /x n ] β, with u 1,, u n fresh, then Σ β Natural Deduction -Elimination 239/467

108 Example: Repeated universal quantifiers Example Show that x y A(x, y) y x A(x, y) 1 x y A(x, y) Premise 2 u, v fresh 3 A(u, v) e ( 2): 1 4 y x A(x, y) i ( 2): 3 Natural Deduction -Elimination 240/467

109 Exercise on Quantifier Rules Exercise Show that { x(q(x) R(x)), x(p (x) Q(x)) } x(p (x) R(x)) Left to you Natural Deduction -Elimination 241/467

110 FOL with Equality Generally, relation symbols have no mandated interpretation Sometimes, however, one makes an exception for the symbol = Definition: First-Order Logic with Equality is First-Order Logic with the restriction that the symbol = must be interpreted as equality on the domain: (=) = { d, d d dom( ) } There are two ways to account for this restriction in proofs 1 Add deduction rules for symbol =: Equals-Introduction: t = t =i Equals-Elimination: t 1 = t 2 α[t 1 /x] α[t 2 /x] 2 Alternatively, use axioms rather than deduction rules Logic With Equality 242/467 =e

111 Axioms for Equality Instead of deduction rules for =, we can choose to use axioms for equality An axiom is a formula that it is implicitly a premise for any proof It may be used at any time, the same as an explicit premise EQ1: EQ2: x (x = x) is an axiom For each formula α and variable z, x y (x = y (α[x/z] α[y/z])) is an axiom (EQ2, which says that that multiple formulas are axioms, is an axiom schema) These axioms imply Symmetry of =: ND= x y(x = y y = x) Transitivity of =: ND= w x y(x = y (y = w x = w)) Logic With Equality 243/467

112 Symmetry of Equality: Proof Lemma (EqSymm): ND= x y(x = y y = x) 1 u, v fresh u = v v = u??? x y(x = y y = x) i ( 2): 1? Logic With Equality 244/467

113 Symmetry of Equality: Proof Lemma (EqSymm): ND= x y(x = y y = x) 1 u, v fresh 2 u = v Assumption v = u?? u = v v = u i: 2? x y(x = y y = x) i ( 2): 1? Logic With Equality 244/467

114 Symmetry of Equality: Proof Lemma (EqSymm): ND= x y(x = y y = x) 1 u, v fresh 2 u = v Assumption 3 x y(x = y (x = u y = u)) EQ2 [ z = u ] 4 u = v (u = u v = u) e ( 2) [u, v]: 3 5 u = u v = u e: 2, 4 v = u?? u = v v = u i: 2? x y(x = y y = x) i ( 2): 1? Logic With Equality 244/467

115 Symmetry of Equality: Proof Lemma (EqSymm): ND= x y(x = y y = x) 1 u, v fresh 2 u = v Assumption 3 x y(x = y (x = u y = u)) EQ2 [ z = u ] 4 u = v (u = u v = u) e ( 2) [u, v]: 3 5 u = u v = u e: 2, 4 6 x(x = x) EQ1 7 u = u e [u]: 6 8 v = u e: 7,5 9 u = v v = u i: x y(x = y y = x) i ( 2): 1 10 Logic With Equality 244/467

116 Transitivity of Equality: Proof Lemma (EqTrans) ND= w x y(x = y (y = w x = w)) 1 w, u, v fresh 2 x y((x = y) (x = w y = w)) EQ2 [(z = w)] 3 v = u (v = w u = w) e ( 2) [v, u]: 2 4 u = v Assumption 5 v = u EQsymm: 4 6 u = w v = w e: 5, 3 7 u = v (v = w u = w) i: w x y (x = y (y = w x = w)) i ( 3): 1 7 Logic With Equality 245/467

117 Derived Proof Rules for Equality Equality satisfies the following derived rules EQtrans(k): t 1 = t 2 t 2 = t 3 t k = t k+1 t 1 = t k+1 for any t 1,, t k+1 EQtrans(k) results from k 1 uses of transitivity EqSubs[r]: t 1 = t 2 r[t 1 /z] = r[t 2 /z] for any variable z and terms r, t 1 and t 2 Prove as an exercise Logic With Equality 246/467

118 Soundness and Completeness of Natural Deduction for First-Order Logic Soundness and Completeness 247/467

119 Soundness and Completeness of Natural Deduction Theorem Natural Deduction is sound for FOL: if Σ α, then Σ α Natural Deduction is complete for FOL: if Σ α, then Σ α Proof outline: Soundness: Each application of a rule is sound By induction, any finite number of rule applications is sound Completeness: We shall show the contrapositive: if Σ α, then Σ α We shall not give the full proof, but we will sketch the main points Soundness and Completeness 248/467

120 Completeness of ND for FOL: Getting started To show: if Σ α, then Σ α Lemma I: If Σ α, then Σ { α} α By rule i, if Σ { α} α, then Σ α α Thus Σ α Lemma II: If there are and E st E Σ { α}, then Σ α and E satisfy Σ but not α Lemma III (the big one): If Σ { α} α, then there are and E such that E Σ { α} Soundness and Completeness Sketch of Proof of Completeness 249/467

121 Whence a Domain? Given: Σ { α} α Required: interpretation and environment E that satisfy Σ { α} To start, we need a domain Where can we get one? Soundness and Completeness Sketch of Proof of Completeness 250/467

122 Whence a Domain? Given: Σ { α} α Required: interpretation and environment E that satisfy Σ { α} To start, we need a domain Where can we get one? We use the set of terms That is, let the domain be { t t is a term } (The notation indicates that we refer to the domain element, rather than to the expression) Soundness and Completeness Sketch of Proof of Completeness 250/467

123 Interpretation of Terms For a set Σ of premises, we want an interpretation I and an environment E, over the domain of terms Constants, variables, and functions are easy to handle For a constant symbol c, we define c I = c For a variable x, we define x E = x For a k-ary function symbol f, we define f I ( t 1,, t k ) = f(t 1,, t k ) Relations pose a problem, since they depend on Σ For a relation symbol R (k), we must determine, for each tuple t 1,, t k, whether to put t 1,, t k into the set R I The basic idea is to consider each possible formula, one by one For each, restrict the set of interpretations to account for it Since we assume no proof of contradiction exists, we can guarantee that we always have some interpretations left as possible Soundness and Completeness Sketch of Proof of Completeness 251/467

124 Checking a Formula Suppose we have a list of all atomic formulas: α 1, α 2, α 3, We will create a list of sets: Σ 0, Σ 1, Σ 2, etc, with Σ 0 = Σ and Σ i+1 determined from Σ i and α i+1 : Σ 0 Σ for i 1, 2, 3, if Σ i 1 {α i } α i set Σ i Σ i 1 {α i } else set Σ i Σ i 1 { α i } Let β k denote the formula added at step k; thus β k is either α k or α k Let Σ denote the union of all Σ i : Σ = Σ {β 1, β 2, β 3, } Soundness and Completeness Sketch of Proof of Completeness 252/467

125 Once we have Σ, it The Satisfying Interpretation Almost is consistent (there is no γ st Σ { γ} γ), and contains either R(t 1,, t k ) or R(t 1,, t k ) for every relation symbol R and terms t 1,, t k Thus Σ defines an interpretation and environment: R = { t 1,, t k R(t 1,, t k ) Σ } This almost works, but we must handle two subtleties We need a list of all formulas, in some order Existential formulas (ie, x ) may never get a term to satisfy them Thus they evaluate to F even if x α is true Soundness and Completeness Sketch of Proof of Completeness 253/467

126 Ensuring Satisfaction of Existential Formulas We use the following trick to ensure that each existential formula in Σ has a term that satisfies it Let c 1, c 2, be a list of fresh constant symbols, that do not occur in any formula of Σ For each formula γ i, add the formula ( x γ i ) γ i [c i /x] to the set Σ 0 at the start of the construction (And include their terms in the domain!) At the end of the construction, if Σ x γ i, then also Σ γ i [c i /x] Soundness and Completeness Sketch of Proof of Completeness 254/467

127 One Piece of the Puzzle : Listing all formulas As one part of the construction, we require a list of all possible formulas Since we may have arbitrarily many constant, variable, functions and relation symbols, of any arity, we must take care that everything gets onto the list at some point For example, if we take the ith formula to be R(c i ), then many formulas (R(x), Q 4 (f 7 (y 66 )), etc) never appear on the list We do the listing in stages, starting from stage 1 At stage j, consider the first j constants, variables, and function symbols Form all terms that combine these, using at most j applications of a function Apply each of the first j relation symbols to each of these terms, and then form all formulas from these that use at most j connectives or quantifiers The set of formulas formed this way is large, but finite After all have been listed, continue to stage j + 1 Soundness and Completeness Sketch of Proof of Completeness 255/467

128 Extended Example: Peano Arithmetic Arithmetic 256/467

129 Axioms Thus far, we have used symbols for functions, relations, etc, without a fixed meaning An interpretation can specify them arbitrarily Often, however, we want to use one or more symbols in a specific way The usual approach to this is to use axioms Definition: An axiom is a formula that is assumed as a premise in any proof An axiom schema is a set of axioms, defined by a pattern or rule Axioms often behave like additional inference rules Arithmetic Axioms for Arithmetic 257/467

130 Example: Axioms for equality Recall the axioms of equality: EQ1: EQ2: x x = x is an axiom For each choice of formula α and variable z, x 1 x 2 (x 1 = x 2 (α[x 1 /z] α[x 2 /z])) is an axiom We shall keep these, and add new ones Notation: in place of writing k x(x = x) Axiom EQ1 k+1 t = t e[term t]: k we shall simply write k t = t EQ1+ e and similarly for other axioms Arithmetic Axioms for Arithmetic 258/467

131 Natural Numbers Fix the domain as N, the natural numbers Interpret the constant symbol 0 as zero and the unary function symbol s as successor Thus each number in N has a term: 0, s(0), s(s(0)), s(s(s(0))), Zero and successor satisfy the following axioms PA1: x(s(x) 0) Zero is not a successor PA2: x y(s(x) = s(y) x = y) Nothing has two predecessors ( PA stands for Peano Axioms, named for Giuseppe Peano) Arithmetic Axioms for Arithmetic 259/467

132 Addition and Multiplication Further axioms characterize + (addition) and (multiplication) PA3: x(x + 0 = x) Adding zero to any number yields the same number PA4: x y x + s(y) = s(x + y) Adding a successor yields the successor of adding the number PA5: x x 0 = 0 Multiplying by zero yields zero PA6: x y(x s(y) = (x y) + x) Multiplication by a successor Arithmetic Axioms for Arithmetic 260/467

133 Induction in Peano Arithmetic The six axioms above define + and for any particular numbers They do not, however, allow us to reason adequately about all numbers For that, we use an additional axiom: induction PA7: For each formula φ and variable v, φ[0/v] (( v(φ φ[s(v)/v])) ( v φ)) is an axiom The formula φ represents the property to be proved To prove φ for every x, we can prove the base case φ[0/x] and the inductive case x(φ φ[s(x)/x]) Arithmetic Axioms for Arithmetic 261/467