Introduction EXAMPLE 1 Express xx yy(xx + yy = 0) without the existential quantifier. Solution: xx yy(xx + yy = 0) is the same as xxxx(xx) where QQ(xx) is yyyy(xx, yy) and PP(xx, yy) = xx + yy = 0 EXAMPLE 2 Translate xx yy((xx > 0) (yy < 0) xxxx < 0)) into English. The Order of Quantifiers Solution: This statement says that for every real number x and for every real number y, if xx > 0 and yy < 0, then xxxx < 0. That is, this statement says that for real numbers x and y, if x is positive and y is negative, then xy is negative. The can be stated more succinctly as The product of a positive real number and a negative real number is always a negative real number. EXAMPLE 3 Let PP(xx, yy) be the statement xx + yy = yy + xx. " What are the truth values of the quantifications xx yyyy(xx, yy) and yy xxxx(xx, yy) where the domain for all variables consists of all real numbers. Solution: The quantification xx yyyy(xx, yy) denotes the proposition For all real numbers x, for all real numbers y, xx + yy = yy + xx. Because PP(xx, yy) is true for all real numbers x and y (it is the commutative law for addition which is an axiom for the real numbers), the proposition xx yyyy(xx, yy) is true. Note that the statement yy xxxx(xx, yy) says For all real numbers y, for all real numbers x, xx + yy = yy + xx. This has the same meaning as the statement For all real numbers x, for all real numbers y, xx + yy = yy + xx. That is, xx yyyy(xx, yy) and yy xxxx(xx, yy) have the same meaning, and both are true. This illustrates the principle that the order of nested universal quantifiers in a statement without other quantifiers can be changed without changing the meaning of the quantified statement. 1
EXAMPLE 4 Let QQ(xx, yy) denote xx + yy = 0. What are the truth values of the quantifications yy xxxx(xx, yy), where the domain for all variables consists of all real numbers. Solution: The quantification yy xxxx(xx, yy) denotes the proposition There is a real number y such that for every real number x, QQ(xx, yy). No matter what value of y is chosen, there is only one value of x for which xx + yy = 0. Because there is no real number y such that xx + yy = 0 for all real numbers x, the statement yy xxxx(xx, yy) is false. The quantification denotes the proposition xx yyyy(xx, yy) For every real number x there is a real number y such that QQ(xx, yy). Given a real number x, there is a real number y such that xx + yy = 0; namely, yy = xx. Hence, the statement xx yyyy(xx, yy) is true. Example 4 illustrates that the order in which quantifiers appear makes a difference. The statements yy xxxx(xx, yy) and xx yyyy(xx, yy) are not logically equivalent. The statement yy xxxx(xx, yy) is true if and only if there is a y that makes PP(xx, yy) true for every x. So, for this statement to be true, there must be a particular value of y for which PP(xx, yy) is true regardless of the choice of x. On the other hand, xx yyyy(xx, yy) is true if and only if for every value of x there is a value of y for which PP(xx, yy) is true. So, for this statement to be true, no matter which x you choose, there must be a value of y (possibly depending of the x you choose) for which PP(xx, yy) is true. In other words, in the second case, y can depend on x, whereas in the first case y is a constant independent of x. From these observations it follows that if yy xxxx(xx, yy) is true, then xx yyyy(xx, yy) must also be true. However, if xx yyyy(xx, yy) is true, it is not necessary for yy xxxx(xx, yy) to be true. 2
TABLE 1 Quantifications of Two Variables. Statement When True? When False? xx yyyy(xx, yy) PP(xx, yy) is true for every pair x, y. There is a pair x, y for which PP(xx, yy) is yy xxxx(xx, yy) false. xx yyyy(xx, yy) For every xx there is a yy for which PP(xx, yy) is true. There is an x such that PP(xx, yy) is false for every y. xx yyyy(xx, yy) There is an x for which PP(xx, yy) is For every x there is a y for which PP(xx, yy) xx yyyy(xx, yy) yy xxxx(xx, yy) true for every y. There is a pair xx, yy for which PP(xx, yy) is true. is false. PP(xx, yy) is false for every pair xx, yy. EXAMPLE 5 Let QQ(xx, yy, zz) be the statement xx + yy = zz. What are the truth values of the statements xx yy zzzz(xx, yy, zz) and zz xx yyyy(xx, yy, zz), where the domain of all variables consists of all real numbers? Solution: Suppose that xx and yy are assigned values. Then, there exists a real number zz such that xx + yy = zz. Consequently, the quantification xx yy zzzz(xx, yy, zz) which is the statement For all real numbers x and for all real numbers y there is a real number z such that xx + yy = zz. " is true. The order of the quantification is important, because the quantification zz xx yyyy(xx, yy, zz) which is the statement There is a real number z such that for all real numbers x and for all real numbers y it is true that xx + yy = zz. is false, because there is no value of z that satisfies the equation xx + yy = zz for all values of x and y. 3
Translating Mathematical Statements into Statements Involving Nested Quantifiers EXAMPLE 6 Translate the statement The sum of two positive integers is always positive into a logical expression. Solution: 1. To translate this statement into a logical expression, we first rewrite it so that the implied quantifiers and a domain are shown. For every two integers, if these integers are both positive, then the sum of these integers is positive. 2. Next, we introduce the variables x and y to obtain For all positive integers x and y, xx + yy is positive. 3. We can express the foregoing as xx yy((xx > 0) (yy > 0) (xx + yy > 0)) where the domain for both variables consists of all integers. 4. Note that we could also translate this using the positive integers as the domain. Then the statement The sum of two positive integers is always positive becomes For every two positive integers, the sum of these integers is positive. We can express this as xx yy(xx + yy > 0). 4
Translating from Nested Quantifiers into English EXAMPLE 9 Translate the statement xx(cc(xx) yyyy(yy) FF(xx, yy))) into English, where CC(xx) is the x has a computer, FF(xx, yy) is x and y are friends, and the domain for both x and y consists of all students in your school. Solution: The statement says that for every student x in your school, x has a computer or there is a student y such that y has a computer and x and y are friends. In other words, every student in your school has a computer or has a friend who has a computer. Translating English Sentences into Logical Expressions EXAMPLE 11 Express the statement If a person is female and is a parent, then this person is someone s mother as a logical expression involving predicates, quantifiers with a domain consisting of all people, and logical connectives. Solution: 1. FF(xx) x is female. 2. PP(xx) x is a parent. 3. MM(xx, yy) x is the mother of y. xx( FF(xx) PP(xx) yyyy(xx, yy)) or xx yy( FF(xx) PP(xx) MM(xx, yy)) EXAMPLE 13 Use quantifiers to express the statement There is a woman who has taken a flight on every airline in the world. Solution: Let PP(ww, ff) be w has taken f and QQ(ff, aa) be f is a flight on a. We can express the statement as ww aa ff(pp(ww, ff) QQ(ff, aa)), where the domains of discourse for w, f, and a consist of all the women in the world, all airplane flights, and all airlines respectively. 5
Negating Nested Quantifiers TABLE 2 De Morgan s Laws for Quantifiers (from section 1.3) Negation Equivalent Statement When Is Negation True? When False? xxxx(xx) xx PP(xx) For every x, PP(xx) is false There is an xx, for which PP(xx) is true. xxxx(xx) xx PP(xx) There is an xx, for which PP(xx) is false. PP(xx) is true for every x. EXAMPLE 14 Express the negation of the statement xx yy(xxxx = 1) so that no negation precedes a quantifier. Solution: By successively applying De Morgan s laws for quantifiers in Table 2 of Section 1.3, we can move the negation in xx xx(xxxx = 1) inside all the quantifiers. We find that xx yy(xxxx = 1) is equivalent to xx yy(xxxx = 1), which is equivalent to xx yy (xxxx = 1). Because (xxxx = 1) can be expressed more simply as xxxx 1, we conclude that our negated statement can be expressed as xx yy(xxxx 1) EXAMPLE 15 Use quantifiers to express the statement that There does not exist a woman who has taken a flight on every airline in the world. Solution: This statement is the negation of the statement There is a woman who has taken a flight on every airline in the world. from Example 13. By Example 13, our statement can be expressed as ww aa ff PP(ww, ff) QQ(ff, aa), where PP(ww, ff) be w has taken f and QQ(ff, aa) be f is a flight on a. By successively applying De Morgan s laws for quantifiers in Table 2 of Section 1.3 to move the negation inside successive quantifiers and by applying De Morgan s low for negating a conjunction in the last step, we find that our statement is equivalent to each of the sequence of statements: Statement Justification ww aa ff PP(ww, ff) QQ(ff, aa) Original statement ww aa ff PP(ww, ff) QQ(ff, aa) Table 2 Section 1.3 xxxx(xx) xx PP(xx) ww aa ff PP(ww, ff) QQ(ff, aa) Table 2 Section 1.3 xxxx(xx) xx PP(xx) ww aa ff PP(ww, ff) QQ(ff, aa) Table 2 Section 1.3 xxxx(xx) xx PP(xx) ww aa ff PP(ww, ff) QQ(ff, aa) Table 2 Section 1.2 (pp qq) pp qq This last statement states For every woman there is an airline such that for all flights, this woman has not taken that flight or that flight is not on this airline. 6