1.3 Predicates and Quantifiers

Transcription

1 1.3 Predicates and Quantifiers INTRODUCTION Statements x>3, x=y+3 and x + y=z are not propositions, if the variables are not specified. In this section we discuss the ways of producing propositions from such statements. In the statement x is greater than 3 : x variable, is greater than 3 - predicate and it is donated by P. So the statement x is greater than 3 can be denoted by P(x) and it is said to be the value of the propositional function P at x. Once x gets a value, the statement P(x) becomes a proposition and has a truth value. Example 1. Let P(x) be the statement x>3. What are the truth values of P(4) and P(2)? Solution: P(4) is true and P(2) is false, since 4>3 is true and 2>3 is false. Example 2. Let Q(x,y) denote the statement x=y+3. What are the truth values of Q(1,2) and Q(3,0)? Solution: Q(1,2) is false and Q(3,0) is true, since 1=2+3 is false and 3=0+3 is true. Example 3. Let R(x,y,z) be the statement x + y=z. What are the truth values of R(1,2,3) and R(0,0,1)? Solution: R(1,2,3) is true and R(0,0,1) is false, since 1+2=3 is true and 0+0=3 is false. In general a statement involving the n variables x,, xn can be denoted by P(x,, xn). Propositional functions occur in computer programs, as the following shows. Example 4. Consider the statement If x>0, then x:=x+1. Solution: If the stored value of x is positive, then assignment x:=x+1 is executed, that is value of x increased by one. Otherwise assignment x:=x+1 is executed, x remains with the original stored value. 1

2 QUANTIFIERS For a propositional function P(x), if the variable is assigned a value, then P(x) becomes a proposition. However there is another way, called quantification, creating a proposition from P(x). We see two types of quantification: universal and existential. The area of logic that deals with predicates and quantifiers is called predicate calculus. THE UNIVERSAL QUANTIFIER Many mathematical statements assert that a property is true for all values of a variable in a particular domain, called the universe of discourse or the(definition) domain. The universal quantification of a propositional function is the proposition that asserts that P(x) is true for all values of x in universe of discourse. The universe of discourse specifies the possible values of the variable x. Definition 1 The universal quantification of P(x) is the proposition P(x) for all values of x in the domain. The notation x P(x) denotes the universal quantification of P(x). Here is called the universal quantifier. The proposition x P(x) is read as for all x P(x), for every x P(x). Example 5.a) Let P(x) be the statement x>x+1. What is the truth value of the quantification x P(x), where the domain is R(all real numbers)? b) What is if P(x) is the statement x+1>x. Solution: a) x P(x) is false, since x>x+1 is false for all x in R. b) x P(x) is true, since x+1>x is true for all x in R. Example 6. Let Q(x) be the statement x<2. What is the truth value of the quantification x Q(x), where the domain is R? Solution: the quantification x Q(x) is false, since x<2 is false for x=3. Example 7. What is the truth value of x P(x), where P(x) is the statement x 2 <10 the domain is {1,2,3,4}? 2

3 Solution: x P(x) is false, since 4 2 <10 is false and so P(1) P(2) P(3) P(4) is false. Example 8. What does the statement xt(x) mean if T(x) is x has two parents and the domain consists of all people? Example 9. What is the truth value of x (x x 2 ) if the domain is R and what is its truth value if the domain is Z (all integers)? Solution: x (x x 2 ) is false, since x x 2 is false, if 0<x<1. If the domain is Z, then x (x x 2 ) is true, since x x 2 is true for all x in Z. Counter Example: one value for x from the universe of discourse which makes P(x) false, is enough to show that statement x P(x) is false. Example 10. Suppose that P(x) is x 2 >0. Give a counterexample that indicates the statement x P(x) is false when the domain is Z. Solution: x P(x) is false for x=0, since 0 2 >0 is false. Definition 2 The existential quantification of P(x) is the proposition There exists an element x in the domain such that P(x). We use the notation x P(x) for the existential quantification of P(x). Here is called existential quantifier. The existential quantification x P(x) is read as There is an x such that P(x), There is at least one x such that P(x) or For some x P(x). Example 11. Let P(x) be x>3. What is the truth value of x P(x), where the domain is R. Solution: x P(x) is true, since x>3 is true for x=4. Example 12. Let Q(x) be the statement x=x+1. What is the truth value of x Q(x), where the domain is R. Solution: x Q(x) is false, since x=x+1 is false for any x in R. Remark: If the the domain is {x1, x2,, xn}, then x P(x) is the same as P(x1) P(x2) P(xn) and x P(x) is the same as P(x1) P(x2) P(xn). 3

4 Example 13. Let P(x) be x 2 >10. What is the truth value of x P(x), where the domain is {1,2,3,4}. Solution: x P(x) is true, since x 2 >10 is true for x=4. Other Quantifiers Universal and existential quantifiers are the most important quantifiers. However many other quantifiers can be defined such as the uniqueness quantifier, quantifiers with restricted domains. Actually using two essential quantifiers and propositional logic we can express the other quantifiers. Precedence of Quantifiers The quantifiers and have higher precedence than all logical operators of logical calculus. Binding Variables When a quantifier is used on the variable x, we say that this occurrence of the variables is bound. An occurrence of a variable that is not bound to particular value is said to be free. To turn a propositional function into a proposition, all variables in a proposition must be bound or a particular value must be assigned to them. This can be done using a combination of quantifiers (universal, existential) and value assignments. The part of a logical expression to which a quantifier is applied is called the scope of this quantifier. Thus a variable is free if it is outside the scope of all quantifiers. Example: In x(x+y=1), the variable x is bound by existential quantification, but y is free, because it is not bound by a quantifier and no value is assigned. Logical Equivalences Involving Quantifiers We have seen the notion of logical equivalences of compound propositions. We can extend this notion to expressions involving predicate and quantifiers. Definition 3 Statements involving predicates and quantifiers are logically equivalent if and only if they have the same truth value independent of predicates and the domain. 4

5 Example19. Show that x(p(x) Q(x)) and x P(x) x Q(x) are logically equivalent. Solution: If x(p(x) Q(x)) is true, then for every a in the domain P(a) Q(a) is true. Then of course P(a) is true and Q(a) is true. Since P(a) is true and Q(a) is true for every element in the domain, x P(x) and x Q(x) are both true. This means x P(x) x Q(x) is true. If x P(x) x Q(x) is true, then x P(x) is true and x Q(x) is true. So if a is in the domain, then P(a) is true and Q(a) is true. This means then P(a) Q(a) is true for all a in the domain. This is x(p(x) Q(x)) is true. So x(p(x) Q(x)) x P(x) x Q(x). Negating Quantified Expressions Some times we need to negate a quantified expression. Negation of universal quantifier: Every student in your class has taken a course in calculus. Let P(x) be x has taken a course in calculus and the domain be all students in the class. So the statement is x P(x). Negation: Not every student in your class has taken a course in calculus. This means There is at least one student in the class who has not taken a class in calculus. This is simply the existential quantification of the negation of the original propositional function: x P(x). Therefore x P(x) x P(x). Verification: x P(x) is true if and only if x P(x) is false. This can happen if and only if there is an x in the domain for which P(x) is false. This happens if and only if there is an x in the domain for which P(x) is true. Therefore there is an x in the domain for which P(x) is true if and only if x P(x) is true. Thus x P(x) and x P(x) are logically equivalent. Negation of existential quantifier: There is a student in the class who has taken a course in calculus. 5

6 The statement is the existential quantification x Q(x), where Q(x) is the statement x has taken a course in calculus. So negation is It is not the case that there is a student in the class who has taken a course in calculus. This is equivalent to Every student in this class has not taken a course in calculus. This is just the universal quantification of the negation of the original propositional function: x Q(x) x Q(x). Proof: Let Q(x)=T(x). Since x T(x) x T(x) we get x T(x) x T(x) and so x T(x) x T(x). Thus x Q(x) x Q(x). Example 20 What are the negations of the statements There is an honest politician And All Americans eat cheeseburgers? Solution: Let H(x) be x is honest. So the statement There is an honest politician is represented by x H(x), where the domain consists of all politicians. The negations is x H(x) x H(x) and expressed as Every politician is dishonest. Let C(x) be x eats cheeseburgers. Then All Americans eat cheeseburgers is represented by x C(x), where the domain consists of all Americans. The negation is x C(x) x C(x) and means There is an American who does not eat cheeseburgers. Example 21 What are the negations of the statements x (x 2 >x) and x(x 2 =2)? Solution: x (x 2 >x)= x (x 2 >x)= x(x 2 =<x), x(x 2 =2)= x (x 2 =2)= x (x 2 2). Example 22 Show that x (P(x) Q(x)) x(p(x) Q(x)). Solution: By De Morgan s Law for quantifiers, x (P(x) Q(x))= x (P(x) Q(x))= x ( P(x) Q(x))= x(p(x) Q(x)). Translating from English into logical expressions 6

7 Translating sentences in English into logical expressions is a crucial task in: Mathematics, Logic programming, Artificial intelligence, Software engineering, Many other disciplines. We used propositions and logical operators to translate English sentences to logical expressions. With quantifiers involving, translation is more complex and there might be many ways of translating a particular sentence. Example 23 Express Every student in this class has studied calculus using predicates and quantifiers. Solution: Every: ; student in this class: x; x has studied calculus: C(x). So the statement is x C(x). If we change the domain to consist of all people, then the statement is expressed as For every person x, if person x is student in this class, then x has studied calculus. Denoting person x is student in this class with S(x), the statement is then x (S(x) C(x)). Simplest way is the best way! Example 24 Express Some student in this class has studied calculus and Every student in this class has studied calculus or music using predicates and quantifiers. Solution: (I) Some student: there is/exists a student -- x; students in this class: domain; x has studied calculus C(x). So the statement is x C(x). (II) Students in this class: domain; every student: x; x studied calculus: C(x); x studied music: M(x). So x(c(x) or M(x))= x(c(x) M(x)). 7

8 If we are also interested in people who is not student, then (III) All person: domain; person x is student in this class: S(x); every person: x; x has studied calculus: C(x); x has studied music: M(x). So the statement is x(s(x) (C(x) M(x)), in words For every x, if x is in this class, x has studied calculus or x has studied music. Using Quantifiers in System Specifications To represent system specifications not only propositions are used, but also predicates and quantifications can be used. Example 25. Use predicates and quantifiers to express the system specifications Every mail message larger than one megabyte will be compressed and If a user is active, at least one network link will be available. Solution: Let S(m, y) be Mail message m is larger than y megabyte, where the variable m has the domain of all mail messages and the variable y is a positive real number, and let C(m) denote Mail message m will be compressed. So the first message can be expressed as x(s(m,1) C(m)). Let A(u) represent User u is active, where the variable u has the domain of all users, let N(n, s) be Network link n is in state s, where the variable n has the domain of all network links and s has the domain of all possible states for a network link. Then the expression is: u A(u) n N(n, available). Examples by Lewis Carroll Example 26 Consider these statements. The first two are premises and the third is the conclusion. All lions are fierce. Some lions do not drink coffee. Some fierce creatures do not drink coffee. Solution: Let P(x), Q(x) and R(x) be the statements x is a lion, x is fierce and x drinks coffee. respectively. Let the domain consists of all creatures. Now the statements are: 8

9 x (P(x) Q(x)). (For an arbitrary animal x, if x is a lion, then x is fierce.) x (P(x) R(x)). There is an animal x which is lion and x does not drink coffee. x (Q(x) R(x)). There is an animal x which is fierce and does not drink coffee. Not okay: x (P(x) R(x)). There is an animal x, if x is a lion, then x does not drink coffee. x (Q(x) R(x)). There is an animal x, if x is fierce, then x does not drink coffee. Not exact -- both are true even if P(x) and Q(x) both are not true! The two implications claims more than the original statements. Example 27 Consider these statements. The first three are premises and the fourth is a valid conclusion. All hummingbirds are richly colored. No large birds live on honey. Birds that do not live on honey are dull in color. Hummingbirds are small. Solution: Let H(x): x is a hummingbird, B(x): x is large, (large=big) L(x): x lives on honey, C(x): x is richly colored. Let the domain consists of all birds. So the statements are: x (H(x) C(x)). For any bird x, if x is a hummingbird, then x is richly colored. x (B(x) L(x)) or x (B(x) L(x)). It is not the case that there is a bird x which is large/big and lives on honey. For any bird x, if x is large/big, then x does not live on honey. 9

10 Warning: No large birds live on honey. Some/all small birds live on honey. The later one is If the bird x is small (not big), then x lives on honey., i.e, x ( B(x) L(x))= x (B(x) L(x)) and this is saying that For any bird x, x is either large or lives on honey. But this could mean The bird x is large and lives on honey. and clearly this is not equal to No large birds live on honey. x ( L(x) C(x)). For any bird x, if x does not live on honey, then x is not rich in color. x (H(x) B(x)). For any bird x, if x is a hummingbird, then x is small(not big). Logic Programming An important type of programming language is designed to reason using the rules of predicate logic. Prolog. Prolog programs include a set of declarations consisting of two types of statements, Prolog facts and Prolog rules. Prolog facts define predicates by specifying the elements that satisfy these predicates. Prolog rules are used to define new predicates using those already defined by Prolog facts. Example 28 Consider a Prolog program given facts telling it the instructor of each class and in which classes students are enrolled. The program uses these facts to answer queries concerning the professors who teach particular students. Such a programs could use the predicates instructor(p, c) and enrolled(s, c) to represent that professor p is the instructor of course c and that the student s is enrolled in course c, respectively. For example, the Prolog facts in such a program might include: instructor(chan, math273) instructor(patel, ee222) instructor(grossman, cs301) enrolled(kevin, math273) enrolled(juana, ee222) enrolled(juana, cs301) enrolled(kiko, math273) enrolled(kiko, cs301) A new predicate teaches(p, s), representing that professor p teaches student s, can be defined using the prolog rule 10

11 teaches (P, S): - instructor(p, C), enrolled(s, C) which means that teaches(p, s) is true if there exists a class c such that professor p is the instructor of class c and student s is enrolled in class c. Remark: in Prolog: conjunction is represented by,, disjunction is represented by ; and query is represented by?. The query Is Kevin enrolled in math273?,? enrolled(kevin, math273) the answer: yes. The query who is enrolled in math273?,? enrolled(x, math273) the answer: kevin kiko The query who are the instructors of classes taken by Juana?,? teaches(x, juana) the answer: patel grossman 11