Logic and Proof
Jeffrey X. Watt
Logic And Proof
Jeffrey X. Watt, Ph.D. Associate Professor of Mathematical Sciences Indiana University - Purdue University Indianapolis
Copyright ©
ii Contents
Chapter 1 Sets 1
1.1 Definitions and Notations 1 1.2 Operations 19 1.3 Properties of Sets 28 1.4 Theorems of Set Theory 41
Chapter 2 Logic 55
2.1 Propositional Calculus 55 2.2 Tautology, Equivalence, and Conditional 65 2.3 Theorems of Propositional Calculus 78 2.4 Analysis of Arguments for Logical Validity 89
Chapter 3 From Logic to Proof 101
3.1 Predicate Calculus 101 3.2 Quantified Predicates in One Variable 123 3.3 Quantification in Several Variables 141 3.4 Introduction to Proof 156 3.5 Infinite Unions and Intersections 175
Chapter 4 Methods of Mathematical Proof 189
4.1 Proof by Transitivity 189 4.2 Division into Cases 208 4.3 Proof by Induction 226 4.4 Epsilon-Delta Proofs 243 4.5 Indirect Proofs 262
Chapter 5 Advanced Methods of Mathematical Proof 287
5.1 Existence and Uniqueness 287 5.2 Advanced Methods 294 5.3 Equivalence Relations 315 5.4 Equivalence Classes and Partitions 328 5.5 Partial Orderings 346
Appendices 352
A Answers to Selected Problems 352
Glossary 397
Index 403
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3.2 QUANTIFIERS
The first chapter of this book focused on how to determine the truth values of compound statements and how to determine whether an argument is valid; however, the methods learned thus far are often incapable of describing many of the statements in mathematics. Consider the following sentence
x is a prime number.
This sentence is not either true or false. That is, the truth value of this sentence depends on the value of x. For example, if x = 2 then the sentence is false and if x = 3 then the sentence is true. Since the majority of statement in mathematics include variables, we must expand our understanding of logic to include such statements.
Let p(x) be a statement involving the variable x and let U be a set. p(x) is called a propositional function if for each x in U , p(x) is a statement.
A propositional function is a sentence containing one or more variables.
The set U is called the domain of discourse of p(x) .
A propositional function, by itself, is neither true nor false. However, for each x in its domain of discourse, p(x) is a statement, and is either true or false. That is, each value of x will determine the truth value of the proposition.
Let p(x) be a propositional function over a domain of discourse U, the set P contains all x in U that make p(x) true.
The set P is called the truth set of p(x) .
Example A Let p(x) be the statement
x is a prime number,
and let U = {1, 2, 3, 4, 5, 6, 7, 8, 9} . Then p(x) is a propositional function with domain of discourse U.
Determine the truth values of p(x) for x = 1, 2, 3, 4, and 5. Determine the truth set P for p(x) .
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Solution If x = 1, we obtain the proposition, 1 is a prime number, which is false. If x = 2, we obtain the proposition, 2 is a prime number, which is true. If x = 3, we obtain the proposition, 3 is a prime number, which is true. If x = 4, we obtain the proposition, 4 is a prime number, which is false. If x = 5, we obtain the proposition, 5 is a prime number, which is true.
Therefore,
p(1) and p(4) are false statements. p(2), p(3), and p(5) are true statements.
Note: each of p(1), p(2), is either true or false, but not both. …
The truth set for p(x) is the set containing all values of x from the domain of discourse that make p(x) a true statement. Thus,
P = {2, 3, 5, 7, 9}
We can connect propositional functions together to form compound propositional functions in the same manner that we formed compound statements in the first two sections of this chapter. With propositional functions, we want to find the values of x in the domain of discourse that make the compound propositional function true based on the truth values of the individual propositional functions. For example, the truth set for
q(x) → s(x) are those values of x that make q(x) → s(x) a true statement.
1. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9} be the domain of discourse for the following propositional functions.
p(x) : x is a prime number q(x) : x ≤ 3 r(x) : x + 5 > 10 s(x) : x is an even integer t(x) : x < 7
Determine the truth sets for each propositional function.
2. Using the results of exercise 1, find the truth sets of the compound propositional functions.
[a] ¬p(x) [b] p(x) ∧ q(x) [c] q(x) ∨ r(x) [d] t(x) ∧ ¬s(x) [e] q(x) → s(x) [f] p(x) ↔ t(x)
124 In exercise 2[e], you must find the values in set U that make the compound propositional function q(x) → s(x) true. Remember, a conditional statement is true unless the premise is true and the conclusion is false (i.e., for a given value of x, q(x) is true and s(x) is false). The only values of x in U = {1, 2, 3, 4, 5, 6, 7, 8, 9} that make the conditional statement false are 1 and 3. Therefore, the truth set of q(x) → s(x) is {2, 4, 5, 6, 7, 8, 9} .
Universal and Existential Quantifiers
You most likely have noticed that many of the statements in mathematics use terms such as “for every” and “for some.” For example:
For every rectangle, the sum of the angles is equal to 360 .
For some number x, 5 + x = 8 .
We now extend our understanding of logic to include “for every” and “for some” in statements.
If p(x) is a propositional function with variable x and domain of discourse U, then:
The sentence for all x, p(x) , symbolized (∀x)( p(x)) , is a proposition that is true if and only if the truth set of p(x) equals U.
The sentence for some x, p(x) , symbolized (∃x)( p(x)) , is a proposition that is true if and only if the truth set of p(x) is nonempty.
It is important to understand that an expression (∀x)( p(x)) or (∃x)( p(x)) is a statement (either true or false), and not a propositional function, even though it contains a variable. Unlike a propositional function, its truth value does not depend on which value is substituted for the variable x.
Example B Let the domain of discourse be U = {0,1, 2, 3, 4, 5, 6, 7, 8, 9} . For each of the following statements, convert the statement into symbolic form and determine its truth value.
[a] For every number x, x is a prime number.
[b] For some number x, x is a prime number.
[c] For every number x, there is a number y, such that x − y = x .
Solution [a] (∀x)( p(x)) where p(x) : x is a prime number, and U = {0,1, 2, 3, 4, 5, 6, 7, 8, 9} . This is a false statement because not every number in U is a prime number.
[b] (∃x)( p(x)) where p(x) : x is a prime number, and U = {0,1, 2, 3, 4, 5, 6, 7, 8, 9} .
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This is a true statement because at least one number in the domain of discourse is a prime number.
[c] (∀x)(∃y)(s(x, y)) where s(x, y) : x − y = x , and U = {0,1, 2, 3, 4, 5, 6, 7, 8, 9} . This is a true statement because for every number in the domain of discourse, there is a number, 0, in the domain of discourse such that x − y = x .
3. Let the domain of discourse be all students at IUPUI in the year 2004. Define three propositional functions over U by:
p(x) : x is a male q(x) : x is an athlete r(x) : x is younger than 25 years old
For each of the following statements, convert into symbolic form and determine its truth value.
[a] All students are male. [b] Some students are athletes. [c] Not all students are male. [d] Some students are male and are not younger than 25 years old. [e] All students are male athletes. [f] Some students are male or athletes.
4. Let s(x, y) be a propositional function “x is taller than y,” and let the domain of discourse be all high school students in Perry Township this year. Express each statement in an English sentence and determine its truth value.
[a] (∀x)(∀y)(s(x, y)) [b] (∀x)(∃y)(s(x, y)) [c] (∃x)(∀y)(s(x, y)) [d] (∃x)(∃y)(s(x, y))
In exercise 3[c], you should have found the statement to be true, but did you find ¬(∀x)( p(x)) or (∃x)(¬p(x)) to be the symbolic form? It turns out that both forms are correct. The first symbolic form reads, “it is not the case that all students are male,” and the second form reads, “some students are not male.” These two forms are logically equivalent.
In exercise 3[e], you should have found this statement to be false, since IUPUI has a female basketball team. You may have found one of two symbolic forms for this statement:
(∀x)( p(x) ∧ q(x))
(∀x)( p(x)) ∧ (∀x)(q(x))
These two symbolic forms are logically equivalent and leads us to our next definition.
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Let p(x) and q(x) be propositional functions over a domain of discourse. Then the following statements are logically equivalent:
¬(∀x)( p(x)) ⇔ (∃x)(¬p(x))
¬(∃x)( p(x)) ⇔ (∀x)(¬p(x))
(∀x)( p(x) ∧ q(x)) ⇔ (∀x)( p(x)) ∧ (∀x)(q(x))
(∃x)( p(x) ∨ q(x)) ⇔ (∃x)( p(x)) ∨ (∃x)(q(x))
Exercises
5. Let p(x) be the propositional function x2 + 10 = 14 . The domain of discourse is the set of all real numbers. Determine the truth value of each. [a] p(0) [b] p(−2) [c] p(2) [d] p(4) [e] For some n, p(n) [f] For all n, p(n)
For exercises 6 through 9, determine whether the sentence is a propositional function. For each statement that is a propositional function, give a domain of discourse.
6. Let x be an integer.
7. It rained and John saw a movie.
8. There exists a person who has won the State Math Contest and the Science Fair contest.
9. Every animal eats other animals for food.
10. Let the domain of discourse be all of the problems on an IQ test. Define three propositional functions over U by:
p(x) : x is a geometry problem q(x) : x is a probability problem r(x) : x is an unsolvable problem
Convert each of the symbolic forms into an English sentence. [a] (∀x)( p(x)) [b] (∀x)(¬r(x)) [c] (∀x)( p(x) ∨ q(x)) [d] (∃x)( p(x) ∨ r(x))
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11. Let the domain of discourse be all of the problems on an IQ test. Define three propositional functions over U by:
p(x) : x is a geometry problem q(x) : x is a probability problem r(x) : x is an unsolvable problem
Convert each of the symbolic forms into an English sentence. [a] (∃x)(q(x)) [b] ¬(∃x)(r(x)) [c] (∀x)( p(x) → q(x)) [d] (∃x)( p(x) ∧ ¬r(x))
12. Let the domain of discourse be the set of all positive integers. Define three propositional functions over U by:
p(x) : x is a prime q(x) : x is an odd number r(x) : x is divisable by 3
Determine the truth value of each statement. [a] (∀x)( p(x) → q(x)) [b] ¬(∀x)(r(x)) [c] (∀x)(r(x) ∨ ¬r(x)) [d] (∃x)( p(x) ∧ ¬r(x))
13. Let s(x, y) be a propositional function “x is a sibling of y,” and let the domain of discourse be all living people. Express each statement in symbolic form and determine its truth value. [a] Everybody is a sibling of everybody. [b] Everybody is a sibling of somebody. [c] Somebody is a sibling of everybody. [d] Somebody is a sibling of somebody.
14. Let f (x, y) be a propositional function “x is a friend of y,” and let the domain of discourse be all living people. Express each statement in an English sentence and determine its truth value. [a] (∀x)(∀y)( f (x, y)) [b] (∀x)(∃y)( f (x, y)) [c] (∃x)(∀y)( f (x, y)) [d] (∃x)(∃y)( f (x, y))
For exercises 15 through 18, let t(x, y) : x ≤ y , and the domain of discourse be all nonnegative real numbers. Determine the truth value of each statement.
15. (∀x)(∀y)(t(x, y)) 16. (∀x)(∃y)(t(x, y))
17. (∃x)(∀y)(t(x, y)) 18. (∃x)(∃y)(t(x, y))
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For exercises 19 through 22, rephrase the statement and determine its meaning.
19. Every pot does not contain gold.
20. Some men are not mortal.
21. Some gun owner are not criminal.
22. All carpet stains can’t be cleaned.
23. Determine the truth value of the statement. The domain of discourse is all natural numbers.
(∀x)(∀y)(∃z)((x + z) ≤ (y + z))
24. Determine the truth value of the statement. The domain of discourse is all natural numbers.
(∀x)(∀y)(∀z)((x > y) → (x + z > y + z))
25. Determine the truth value of the statement. The domain of discourse is all natural numbers.
(∀x)(∀y)(∀z)((x > y) → (x ⋅ z > y ⋅ z))
26. Which of the following are logically equivalent to ¬(∀x)(∃y)( p(x, y)) ? [a] (∃x)(∀y)(¬p(x, y)) [b] (∃x)¬(∀y)( p(x, y)) [c] (∃y)(∀x)(¬p(x, y)) [d] (∃x)(∀y)( p(x, y))
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