Formal Logic: Quantifiers, Predicates, and Validity. CS 130 Discrete Structures

Transcription

1 Formal Logic: Quantifiers, Predicates, and Validity CS 130 Discrete Structures

2 Variables and Statements Variables: A variable is a symbol that stands for an individual in a collection or set. For example, the variable x may stand for one of the days. We may let x = Monday, x = Tuesday, etc. We normally use letters at the end of the alphabet as variables, such as x, y, z. A collection of objects is called the domain of objects. For the above example, the days in the week is the domain of variable x. CS 130 Discrete Structures 55

3 Quantifiers Propositional wffs have rather limited expressive power. E.g., For every x, x > 0. Quantifiers: Quantifiers are phrases that refer to given quantities, such as "for some" or "for all" or "for every", indicating how many objects have a certain property. Two kinds of quantifiers: Universal Quantifier: represented by, for all, for every, for each, or for any. Existential Quantifier: represented by, for some, there exists, there is a, or for at least one. CS 130 Discrete Structures 56

4 Predicates Predicate: It is the verbal statement which describes the property of a variable. Usually represented by the letter P, the notation P(x) is used to represent some unspecified property or predicate that x may have. P(x) = x has 30 days. P(April) = April has 30 days. What is the truth value of ( x)p(x) where x is all the months and P(x) = x has less than 32 days Combining the quantifier and the predicate, we get a complete statement of the form ( x)p(x) or ( x)p(x) The collection of objects is called the domain of interpretation, and it must contain at least one object. CS 130 Discrete Structures 57

5 Truth Values of the Expressions What is the truth value of ( x)p(x) in the following interpretations? P(x): x is yellow; the domain of interpretation is the collection of all flowers. P(x): x is a plant; the domain of interpretation is the collection of all flowers. P(x): x is either positive or negative; the domain of interpretation consists of the integers. Can you find one interpretation in which both ( x)p(x) is true and ( x)p(x) is false? Can you find one interpretation in which both ( x)p(x) is true and ( x)p(x) is false? CS 130 Discrete Structures 58

6 Unary, Binary,, N-ary Predicates Predicates involving properties of a single variable: unary predicates Binary, ternary and n-ary predicates are also possible ( x) ( y)q(x,y) is a binary predicate. This expression reads as for every x there exists a y such that Q(x,y) Assume Q(x,y) is the property that x < y, what re the interpretations? ( x)( y) Q(x, y), ( y) ( x) Q(x, y), and ( x)( x) Q(x, x) We cannot collapse separate variables together into one without changing the nature of the expression Constants are also allowed in expressions, such as a, b, c, 0, 1, 2, etc. CS 130 Discrete Structures 59

7 Interpretation Formal definition: An interpretation for an expression involving predicates consists of the following: A collection of objects, called domain of interpretation, which must include at least one object. An assignment of a property of the objects in the domain to each predicate in the expression. An assignment of a particular object in the domain to each constant symbol in the expression. Predicate wffs can be built similar to propositional wffs using logical connectives with predicates and quantifiers. Must obey the rules of syntax to be considered a wff Examples of predicate wffs ( x)[p(x) Q(x)] ( x) (( y)[p(x,y) V Q(x,y)] R(x)) S(x,y) Λ R(x,y) CS 130 Discrete Structures 60

8 Scope of a Variable in an Expression The parentheses or brackets are used wisely to identify the scope of the variable. ( x) (( y)[p(x,y) V Q(x,y)] R(x)) Scope of ( y) is P(x,y) V Q(x,y) while the scope of ( x) is the entire expression ( x)s(x) V ( y)r(y) Scope of ( x) is S(x) while the scope of ( y) is R(y) ( x)[p(x,y) ( y) Q(x,y)] Scope of variable y is not defined for P(x,y) hence y is called a free variable. Such expressions might not have a truth value at all. P(x): x > 0; P(y)^ P(5), P(y) V P(5). What is the truth of the wff ( x)(a(x) Λ ( y)[b(x,y) C(y)]), where A(x) is x > 0, B(x, y) is x > y, C(y) is y 0, and x is the domain of positive integers and y is the domain of all integers? CS 130 Discrete Structures 61

9 Translation of Verbal Statements to Symbolic Form Using Intermediate Statements Every person is nice can be rephrased as For any thing, if it is a person, then it is nice. So, if P(x) is x is a person and Q(x) be x is nice, the statement can be symbolized as ( x)[p(x) Q(x)] Variations: All persons are nice or Each person is nice. There is a nice person can be rewritten as There exists something that is both a person and nice in symbolic form ( x)[p(x) Λ Q(x)] Variations: Some persons are nice or There are nice persons What would the following forms mean for the example above? ( x) [P(x) Λ Q(x)] is too strong ( x)[p(x) Q(x)] will be true if there are no persons in the world but that is not the case. So almost always, goes with Λ (conjunction) and goes with (implication) Can be confusing, so remember to frame the statement in different forms as possible CS 130 Discrete Structures 62

10 More On Translation The word only can be tricky depending on its presence in the statement: X loves only Y If X loves anything, then that thing is Y Only X loves Y If anything loves Y, then it is X X only loves Y If X does anything to Y, then it is love Example for forming symbolic forms from predicate symbols: D(x) is x is dog R(x) is x is a rabbit C(x,y) is x chases y All dogs chase all rabbits For anything, if it is a dog, then for any other thing, if it is a rabbit, then the dog chases it ( x)[d(x) ( y)(r(y) C(x,y))] Some dogs chase all rabbits There is something that is a dog and for any other thing, if that thing is a rabbit, then the dog chases it ( x)[d(x) Λ ( y)(r(y) C(x,y))] Only dogs chase rabbits For anything, if it is a rabbit then, if anything chases it, that thing is a dog ( y) [R(y) ( x) (C(x, y) D(x))] Or, for any two things, if one is a rabbit and the other chases it, then the other is a dog ( y) ( x)[r(y) Λ C(x,y) D(x)] CS 130 Discrete Structures 63

11 Negation of Statements A(x): Everything is beautiful Negation will be it is false that everything is beautiful, i.e. something is not beautiful In symbolic form, [( x)a(x)] ( x)[a(x)] Similarly, negation of Something is beautiful is Nothing is beautiful or Nothing is beautiful Hence, [( x)a(x)] ( x)[a(x)] What is the negation of Everybody loves somebody sometime Everybody hates somebody sometime Somebody loves everybody all the time Everybody hates everybody all the time Somebody hates everybody all the time CS 130 Discrete Structures 64

12 More Examples on Negation What is the negation of the following statements? Some pictures are old and faded. Every picture is not old or not faded. All people are tall and thin. Someone is short or fat. Some students eat only pizza. Every student eats something which is not a pizza Only students eat pizza. There is a non-student who eats pizza. CS 130 Discrete Structures 65

13 Class Exercises S(x): x is a student I(x): x is intelligent M(x): x likes music Write wffs than express the following statements: All students are intelligent. Some intelligent students like music. Everyone who likes music is a stupid student. Only intelligent students like music. For anything, if it is a student, then it is intelligent ( x)[s(x) I (x)] There is something that is intelligent and it is a student and it likes music ( x)[i(x) Λ S(x) Λ M(x)] For anything, if that thing likes music, then it is a student and it is not intelligent ( x)(m(x) S(x) Λ [I (x)] ) For any thing, if it likes music, then it is a student and it is intelligent ( x)(m(x) S(x) Λ I(x)) CS 130 Discrete Structures 66

14 Validity Analogous to a tautology of propositional logic Truth of a predicate wff depends on the interpretation A predicate wff is valid if it is true in all possible interpretations just like a propositional wff is true if it is true for all rows of the truth table A valid predicate wff is intrinsically true Truth Values Intrinsic truth Propositional Wffs True or false depends on the truth value of statement letters Tautology- true for all truth values of its statements Predicate Wffs True, false or neither(if the wff has a free variable) Valid wff- true for all interpretations Methodology Truth table to determine if it is a tautology No algorithm to determine validity CS 130 Discrete Structures 67

15 Validity Examples ( x)p(x) ( x)p(x) This is valid because if every object of the domain has a certain property, then there exists an object of the domain that has the same property. Therefore, whenever the antecedent is true, so is the consequent, and the implication is therefore true. ( x)p(x) P(a) Valid quite obvious since a is a member of the domain of X. ( x)p(x) ( x)p(x) Not valid since the property cannot be valid for all objects in the domain if it is valid for some objects of than domain. Can use a mathematical context to check as well. Say P(x) = x is even, then there exists an integer that is even but not every integer is even. How about ( x)[p(x) V Q(x)] ( x)p(x) V ( x)q(x) Invalid, can prove by mathematical context by taking P(x) = x is even, Q(x) = x is odd. In that case, the hypothesis is true but not the conclusion is false because it is not the case that every integer is even or that every integer is odd. CS 130 Discrete Structures 68

16 More Examples What s the validity? ( x)[p(x) ^ Q(x)] ( x)p(x) ^ ( x)q(x) P(x) [Q(x) P(x)] CS 130 Discrete Structures 69

17 Class Exercises What is the truth of the following wffs where the domain consists of integers: ( x)[l(x) O(x)] where O(x) is x is odd and L(x) is x < 10 ( y)( x)(x + y = 0) ( y)( x)(x 2 = y) ( x)[x < 0 ( y)(y > 0 Λ x + y = 0)] Using predicate symbols and appropriate quantifiers, write the symbolic form of the following English statement: D(x) is x is a day ; M is Monday ; T is Tuesday. S(x) is x is sunny ; R(x) is x is rainy. Some days are sunny and rainy It is always a sunny day only if it is a rainy day It rained both Monday and Tuesday Every day that is rainy is not sunny CS 130 Discrete Structures 70

18 Answers D(x) is x is a day ; M is Monday ; T is Tuesday. S(x) is x is sunny ; R(x) is x is rainy. Some days are sunny and rainy ( x) S(x) Λ R(x) Λ D(x) It is always a sunny day only if it is a rainy day ( x) [S(x) Λ D(x) R(x) Λ D(x)] It rained both Monday and Tuesday R(M) Λ R(T) Every day that is rainy is not sunny ( x) [R(X) Λ D(x) S (x)] CS 130 Discrete Structures 71

19 CS 130 Discrete Structures Formal Logic -- Predicate Logic

20 Valid Argument (P 1 Λ P 2 Λ... Λ P n ) Q where the wffs are built from predicates and quantifiers as well as logical connectives and grouping symbols no equivalent of the truth table exists to prove validity the meaning and the structure of the quantifiers and predicates determines the interpretation and the validity of the arguments Predicate logic: 4 more new derivation rules to build a proof sequence leading from the hypothesis to the conclusion The equivalence rules and inference rules still applies Why predicate logic? More rules? A valid argument for predicate logic need not be a tautology to be valid Example: ( x)p(x) ( x)p(x) Hence: we need to learn 4 new rules CS 130 Discrete Structures 73

21 Steps To Prove the Validity Basic approach to prove arguments: Strip off quantifiers Manipulate the unquantified wffs Reinsert the quantifiers Four new inference rules Two rules to strip the quantifiers Two rules to reinsert the quantifiers CS 130 Discrete Structures 74

22 Inference Rules in Predicate Logic From Can Derive Name / Abbreviation ( x)p(x) ( x)p(x) P(t) where t is a variable or constant symbol P(t) where t is a variable or constant symbol not previously used in a proof sequence Universal Instantiation- ui Existential Instantiation- ei P(x) ( x)p(x) Universal Generalization- ug P(x) or P(a) ( x)p(x) Existential Generalization- eg Restrictions on Use If t is a variable, it must not fall within the scope of a quantifier for t Must be the first rule used that introduces t P(x) has not been deduced from any hypotheses in which x is a free variable nor has P(x) been deduced by ei from any wff in which x is a free variable To go from P(a) to ( x)p(x), x must not appear in P(a) CS 130 Discrete Structures 75

23 Universal Instantiation (1) This rule says if P is true for every element of the domain, we can name such an element by an arbitrary variable name like x, y, or z, or we can specify a particular constant in the domain, and P is still true for all these things. Example: All flowers are plants. Sunflower is a flower. Therefore, sunflower is a plant. P(x) is x is a plant a is a constant symbol (Sunflower) F(x) is x is a flower The argument is ( x)[f(x) P(x)] Λ F(a) P(a) The proof sequence is as follows: 1. ( x)[f(x) P(x)] hyp 2. F(a) hyp 3. F(a) P(a) 1, ui 4. P(a) 2, 3, mp CS 130 Discrete Structures 76

24 Universal Instantiation (2) Restriction: ( x)p(x) derives P(t), t must not fall within the scope of a quantifier for t Without this restriction: a hypothesis of the form ( x) ( y) P(x, y) could lead to the wff ( y) P(y, y) this is invalid considering when P(x, y) means y > x CS 130 Discrete Structures 77

25 Example Prove the argument: ( x)[p(x) Q(x)] Λ [Q(y)] [P(y)] Proof sequence: 1. ( x)[p(x) Q(x)] hyp 2. [Q(y)] hyp 3. P(y) Q(y) 1, ui 4. [P(y)] 2, 3, mt CS 130 Discrete Structures 78

26 Existential Instantiation It says that from ( x)p(x) we can derive P(a), P(b) or P(c), provided that these are new constant symbols. The following proof sequence is legitimate: 1. ( x)[p(x) -> Q(x)] hyp 2. ( y)p(y) hyp 3. P(a) 2, ei 4. P(a)->Q(a) 1, ui 5. Q(a) 3, 4, mp However, step 3 and 4 cannot be reversed So, look at your hypotheses, if you plan to use ei on any of them, do it first. CS 130 Discrete Structures 79

27 Universal Generalization P(x) derives ( x)p(x) when x is absolutely arbitrary Example: ( x)[p(x) Q(x)] Λ ( x)p(x) ( x)q(x) Proof sequence: 1. ( x)[p(x) Q(x)] hyp 2. ( x)p(x) hyp 3. P(x) Q(x) 1, ui 4. P(x) 2, ui : no restriction on UI about reusing a name 5. Q(x) 3, 4, mp 6. ( x)q(x) 5, ug Note: step 6 is legitimate since x is not a free variable in any hypothesis nor was ei used before CS 130 Discrete Structures 80

28 Two Restrictions on UG Free variable in hypothesis P(x) hyp ( x)p(x) 1, incorrect ug; x is free in the hypothesis Deduced from ei, and free variable ( x)( y)q(x, y) hyp ( y)q(x, y) 1, ui Q(x, a) 2, ei ( x)q(x, a) 3, incorrect ug; Q(x, a) is deduced by ei from the wff in step 2, in which x is free Example: Q(x, y): x + y = 0 CS 130 Discrete Structures 81

29 Example Prove the argument ( x)[p(x) Λ Q(x)] ( x)p(x) Λ ( x)q(x) Proof sequence: 1. ( x)[p(x) Λ Q(x)] hyp 2. P(x) Λ Q(x) 1, ui 3. P(x) 2, sim 4. Q(x) 2, sim 5. ( x)p(x) 3, ug 6. ( x)q(x) 4, ug 7. ( x)p(x) Λ ( x)q(x) 5, 6, con CS 130 Discrete Structures 82

30 Existential Generalization It allows insertion of an. From P(x) or P(a) we can derive ( x)p(x). Restriction: x must not appear in P(a). Prove the argument ( x)p(x) ( x)p(x) Proof sequence: 1. ( x)p(x) hyp 2. P(x) 1, ui 3. ( x)p(x) 2, eg Without the restriction: from P(a, y) one could derive ( y) P(y, y) incorrect: since y already appeared in the wff in which eg was applied why? P(x, y): y>x CS 130 Discrete Structures 83

31 More Examples Are the following proof sequence legitimate? 1. ( x)p(x) V ( x)q(x) hyp 2. P(a) V Q(a) 1, ei 1. ( x)( y)q(x, y) hyp 2. ( x)q(x, a) 1, ei CS 130 Discrete Structures 84

32 More Examples Prove the argument ( y)[p(x) Q(x,y)] [P(x) ( y)q(x,y)] Using the deduction method, we can derive ( y)[p(x) Q(x,y)] Λ P(x) ( y)q(x,y) Proof sequence: 1. ( y)[p(x) Q(x,y)] hyp 2. P(x) hyp 3. P(x) Q(x,y) 1, ui 4. Q(x,y) 2, 3, mp 5. ( y)q(x,y) 4, ug CS 130 Discrete Structures 85

33 Temporary Hypotheses An extension to the deduction method: we can insert a temporary hypothesis into a proof If some wff T is introduced into the proof of sequence as a temporary hypothesis, and eventually a wff W is deduced from T and other hypotheses, then the wff T -> W has been deduced from the other hypotheses and can be inserted in the proof sequence CS 130 Discrete Structures 86

34 Example of Temporary Hypothesis Prove the argument [P(x) ( y)q(x,y)] ( y)[p(x) Q(x,y)] Proof sequence: 1. P(x) ( y)q(x,y) hyp. 2. P(x) temporary hypothesis (T) 3. ( y)q(x,y) 1, 2, mp 4. Q(x,y) 3, ui (W) 5. P(x) Q(x,y) temp. hyp discharged (T->W) 6. ( y)[p(x) Q(x,y)] 5, ug Universal quantifier can slide over subwffs that do not contain the quantified variable Same for existential quantifier CS 130 Discrete Structures 87

35 More Example Prove the argument ( x)[(b(x) V C(x)) A(x)] ( x)[b(x) A(x)] Proof sequence: 1. ( x)[(b(x) V C(x)) A(x)] hyp 2. (B(x) V C(x)) A(x) 1, ui 3. B(x) temp. hyp. 4. B(x) V C(x) 3, add 5. A(x) 2, 4, mp 6. B(x) A(x) temp. hyp. Discharged 7. ( x)[b(x) A(x)] 6, ug CS 130 Discrete Structures 88

36 More: [( x)a(x)] ( x)[a(x)] [( x)a(x)] ( x)[a(x)] Proof sequence: 1. [( x)a(x)] hyp. 2. A(x) temp. hyp. 3. ( x)a(x) 2, eg 4. A(x) ( x)a(x) temp. hyp. discharged 5. [A(x)] 1, 4, mt 6. ( x)[a(x)] 5, ug ( x)[a(x)] [( x)a(x)] Proof sequence: 1. ( x)[a(x)] hyp. 2. ( x)a(x) temp. hyp. 3. A(a) 2, ei 4. [A(a)] 1, ui 5. [( x)[a(x)] ] 3, 4, inc 6. ( x)a(x) [( x)[a(x)] ] temp. hyp. discharged 7. [(( x)[a(x)] ) ] 1, dn 8. [( x)a(x)] 6, 7, mt inc: inconsistency in exercise 27 in section 1.2 CS 130 Discrete Structures 89

37 More Examples Is the following wff a valid argument? Prove or disprove. ( x)[p(x) v Q(x)] ( x)p(x) v ( x)q(x) ( x)p(x) Λ ( x)q(x) ( x)[p(x) Λ Q(x)] ( x)p(x) Λ [( x)(p(x) Λ Q(x))] ( x)[q(x)] First, consider whether the wff seems valid if yes, try to find a proof sequence for it otherwise, try to find an interpretation in which it is not true CS 130 Discrete Structures 90

38 Verbal Arguments Every crocodile is bigger than every alligator. Sam is a crocodile. But there is a snake, and Sam isn t bigger than that snake. Therefore, something is not an alligator. Use C(x), A(x), B(x,y), s, S(x) ( x) ( y)[c(x) Λ A(y) B(x,y)] Λ C(s) Λ ( x)(s(x) Λ [B(s,x)] ) ( x)[a(x)] CS 130 Discrete Structures 91

39 Practice All rock music is loud music. Some rock music exists; therefore some loud music exists. Use R(x) and L(x). CS 130 Discrete Structures 92

40 Review of Chapter 1 Be able to: construct truth tables for compound wffs recognize tautologies and contradictions translate compound statements into symbolic notations negate a statement apply derivation rules for propositional logic use propositional logic to prove the validity of a verbal argument determine the truth value of a predicate wff in a given interpretation translate statements into predicate wffs apply derivation rules for predicate logic use predicate logic to prove the validity of a verbal argument CS 130 Discrete Structures 93