Section 1.1 Propositions

Transcription

1 Set Theory & Logic Section 1.1 Propositions Fall, 2009 Section 1.1 Propositions In Chapter 1, our main goals are to prove sentences about numbers, equations or functions and to write the proofs. Definition. A proposition or statement is a sentence that is either true or false but not both. No proposition can be true and false at the same time. There are many sentences that are neither true nor false. Example 1. Let us see some examples. If a polygon has three sides, then it is a triangle. This is a proposition with the truth value, true. There is the largest integer. This is a proposition with the truth value, false. He is sad. This is not a proposition, because we cannot tell the truth value, true or false. Big the runs fox This is not a proposition, because it is not even a sentence. We can introduce a connective between two propositions (except for negation). There are seven connectives: conjunction, disjunction, negation, conditional, contrapositive, converse, and biconditional. Conjunction Definition. A conjunction (sometimes called exclusive or) is a proposition combined by two propositions with and. Those two propositions in a conjunction are called conjuncts. A conjunction is true when both conjuncts are true and false otherwise. Example 2. For two propositions, the base angles of an isosceles triangle are congruent and a square has no right angles, we can make a conjunction, the base angles of an isosceles triangle are congruent and a square has no right angles. Page 1 of 3

2 Set Theory & Logic Section 1.1 Propositions Fall, 2009 Disjunction Definition. A disjunction (sometimes called inclusive or) is a proposition combined by two propositions with or. Those two propositions in a disjunction are called disjuncts. A disjunction is true when at least one of its disjuncts is true and false otherwise. Example 3. For two propositions, the base angles of an isosceles triangle are congruent and a square has no right angles, we can make a disjunction, the base angles of an isosceles triangle are congruent or a square has no right angles. Negation Definition. A negation of a proposition is a proposition that has opposite meaning of the original proposition. So the truth value of a negation is opposite of the original proposition s truth value. Example 4. For a proposition, it is raining, its negation is it is not raining. Example 5. For a proposition, 2 3 = 100 its negation is Conditional Definition. A conditional proposition is a proposition of form if P, then Q. Here, P and Q are propositions and the first proposition P is called hypothesis, while the second proposition Q is called conclusion. Example 6. For two propositions, it is raining and Page 2 of 3

3 Set Theory & Logic Section 1.1 Propositions Fall, 2009 I have my umbrella, we can make a conditional proposition, if it is raining, then I have my umbrella. How to determine the truth value of a conditional proposition, if P, then Q? The truth value of a conditional proposition is false only when P is true and Q is false. Otherwise, it is always true. Example 7. Conditional propositions of all possible truth values. If = 9, then = 5. This is a conditional proposition with the truth value, true. If = 9, then = 8. This is a conditional proposition with the truth value, false. If = 8, then 4 10 = 40. This is a conditional proposition with the truth value, true. If = 8, then 4 10 = 21. This is a conditional proposition with the truth value, true. There are other ways to express a conditional proposition without changing the truth value. if P, then Q if P, Q P implies/gives/yields Q P is sufficient for Q Q if P Q when/provided by P Q is necessary for P Sometimes if is dropped in a conditional proposition. Squares have four sides This proposition, in fact, is a conditional proposition fully expressed as follows: if the shape is a square, then it has four sides. Page 3 of 3

4 Set Theory & Logic Section 1.2 Propositional Forms Fall, 2009 Section 1.2 Propositional Forms In this section, we will study the propositional calculus, specifically, discuss on the truth value of a proposition. We use capital letters such as P, Q, and R, to represent a proposition and for this reason, a capital letter representing a proposition is called the propositional variable. The symbol := is used to assign a proposition to a propositional variable. Example 1. For a proposition, we assign a propositional variable P as follows for all real numbers x and y, x + y = y + x, P := for all real numbers x and y, x + y = y + x. Now we introduce symbols to represent connectives. Connective and or not implies if and only if Symbol Note that P Q means P Q and P Q. We use the symbols to connect propositions. Example 2. With the given propositions below, we can consider the followings: P := it is raining, Q := I have my umbrella, R := I am struck by lightning, Combination P Q R P Q Q R R P Translation It is raining and I have my umbrella I am not struck by lightning It is not raining or I do not have my umbrella If I have my umbrella, then I am struck by lightning I am struck by lightning if and only if it is raining The symbols (such as P Q, R, etc.) on the left hand side in the table of the example above are not propositions, but they are called the propositional forms. Small letters such as p, q, and r, are used with the notation := to represent the propositional forms. Example 3. For the given propositional forms, p := R (P Q), q := (R P ) Q, the propositional forms p q and q p represent, respectively, [R (P Q)] [(R P ) Q], [(R P ) Q] [R (P Q)]. Page 1 of 3

5 Set Theory & Logic Section 1.2 Propositional Forms Fall, 2009 For a given propositional form, it is important to interpret it correctly. We follow the rules below. Theorem (Order of Operations). To interpret a propositional form, we read from left to right and use the following precedence: 1. propositional forms within parentheses (innermost first), 2. negations, ( ) 3. conjunctions, ( ) 4. disjunctions, ( ) 5. conditionals, ( ) 6. biconditionals. ( ) Often we don t use parentheses. So P Q R means P (Q R) and P Q R means (P Q) R. Example 4. P Q R means [( P ) Q] R and P Q R means P (Q R). When we translate a propositional form, we need to be careful. 1. (P Q) R is translated as either (P and Q) or R. 2. P (Q R) is translated as P and either Q or R. Example 5. We translate certain propositional forms typically as follows. Propositional Form (P Q) P Q P Q Translation It is not the case that P and Q or It is false that P and Q not P and Q neither P nor Q A propositional form represents a proposition so that we can examine possible truth value of a propositional form. For propositional forms p and q, we have the following table, which is required to be memorized. p q p p q p q p q p q T T F T T T T T F F F T F F F T T F T T F F F T F F T T Example 6. Using the table above, let us find the truth value of the propositional form P (Q P ). Example 7. Find the truth value of First, we define propositions as follows: P Q P Q P P (Q P ) T T F F F T F F F F F T T T T F F T F T if the derivative is zero and the second derivative is positive, then the function has a relative minimum. P := the derivative is zero, Q := the second derivative is positive, R := the function has a relative minimum. Then the given statement is expressed as the propositional form (P Q) R. Now, we set up the truth table of the form. Page 2 of 3

6 Set Theory & Logic Section 1.2 Propositional Forms Fall, 2009 P Q R P Q (P Q) R T T T T T T T F T F T F T F T T F F F T F T T F T F T F F T F F T F T F F F F T As we can see, the given statement is false only when P and Q are true and R is false. Suppose that we want to know (1) if (P Q) R is always true or always false, (2) if (P Q) R is false, then at what truth values of P, Q and R we have the false value. The answers to those questions can be obtained from the truth table above. However, it is very painful, because we have to make the big table as above. Is there a better way to find the answer to those questions? Yes, there is. From the truth table of the conditional connectives, we observe (P Q) R is false, when P Q is true and R is false, i.e., (P Q) R can be false (and obviously can be true) and P Q R P Q (P Q) R F T F Since P Q is true only when both P and Q are true, hence, we deduce P Q R P Q (P Q) R T T F T F That is, when both P and Q are true, but R is false, the given statement becomes false. In this way, we don t have to draw the whole table to find the answers to the given questions (1) and (2) above. We end the section with a special propositional forms. Definition. A propositional form p is called a tautology if each entry of the column for p in its truth table is true. If every entry in its column is false, then it is called a contradiction. Example 8. Propositional forms P P and P P are tautologies. Propositional forms P P and P P are contradictions. P P P P P P P P P P T F T T F F F T T T F F Page 3 of 3

7 Set Theory & Logic Section 1.3 Rules of Inference Fall, 2009 Section 1.3 Rules of Inference One of our goals in this section is writing a basic proof. Two rules of logic are taken: Rules of Inference and Rules of Replacement. Consider the case: when all propositional forms h 1, h 2,..., h n are true, a propositional form p is also true. It is expressed as follows h 1, h 2,..., h n p. The collection of propositional forms h 1, h 2,..., h n is called hypotheses or premises. And p is called a theorem or conclusion of h 1, h 2,..., h n. We say p is logically follows from those hypotheses. The symbol is called a turnstile. The expression is read as h 1, h 2,..., h n give/yield p. When all propositional forms h 1, h 2,..., h n are true, a propositional form p is also true. This is called a rule of inference or valid argument. Given propositional forms h 1, h 2,..., h n, we want to find a propositional form p logically following from those forms h 1, h 2,..., h n. When h 1, h 2,..., h n are all true, but p is false, we denote this by and it is called an invalid argument. h 1, h 2,..., h n p Example 1. For propositions P and Q, we observe P, P Q Q because whenever both P and P Q are true, Q must also be true. P P because when P is true, P is false. Example 2. For propositions P and Q, what is the conclusion of P and Q? That is, P, Q? When both P and Q are true, we observe P Q is also true. Hence, we get Similarly, we can get P, Q P Q P P Q and Q P Q and P Q P. How to draw the conclusion? We have two ways: truth table and proof. When we use the truth table, the following definition may be useful. Definition. Let h 1, h 2,..., h n and p be propositional forms. Then h 1, h 2,..., h n p if and only if (h 1 h 2 h n ) p is a tautology. Page 1 of 6

8 Set Theory & Logic Section 1.3 Rules of Inference Fall, 2009 Example 3. Draw the conclusion of (P Q) Q, P, i.e., (P Q) Q, P? By definition above, ([(P Q) Q] P )? should be a tautology. Let us make the truth table. P Q P Q (P Q) Q [(P Q) Q] P ([(P Q) Q] P ) Q T T T T T T T F T F F T F T T T F T F F F T F T As we can see, ([(P Q) Q] P ) Q is a tautology. Hence, we get (P Q) Q, P Q. By the same argument, we can deduce (P Q) Q, P P. Aside: Later in Section 1.4 Rules of Replacement, we will study the following equivalence relations. [(P Q) Q] P [ (P Q) Q] P [( P Q) Q] P [( P Q) ( Q Q)] P [ P Q] P ( P P ) (Q P ) Q P Since Q P Q and also Q P P, hence we get (P Q) Q, P Q and (P Q) Q, P P. Example 4. Does the hypothesis (P Q) Q, P give the conclusion Q? If so, show it. If not, find the truth values of P and Q which does not yield the conclusion Q. Answer. If there is a combination of truth values of P and Q such that ([(P Q) Q] P ) Q (1) has the false truth value, then the hypothesis does not give the conclusion Q. For this reason, we argue backward as follows: Step 1. ([(P Q) Q] P ) Q is false, when Q is false and [(P Q) Q] P is true. Step 2. [(P Q) Q] P is true, when P is true and (P Q) Q is true. Step 3. From Step 1, we deduced Q was false. So (P Q) Q is true, when P Q is true. Step 4. Again from Step 1, we deduced Q was false. So P Q is true, when P is true. Combining Step 1 Step 4, we deduce ([(P Q) Q] P ) Q is false, when P is true, but Q is false. Page 2 of 6

9 Set Theory & Logic Section 1.3 Rules of Inference Fall, 2009 That is, (P Q) Q, P Q when P is true, but Q is false. If there is no combination of truth values of P and Q making the proposition (1) to be false/true, then (1) is obviously a tautology/contradiction. The backward arguing above is better than using the whole truth table, because it saves the time. Example 5. Does the hypothesis (P Q) R give the conclusion P Q? If so, show it. If not, find the truth values of P and Q which does not yield the conclusion P Q. Answer. If there is a combination of truth values of P and Q such that ((P Q) R) (P Q) has the false truth value, then the hypothesis does not give the conclusion P Q. For this reason, we argue backward as follows: Step 1. ((P Q) R) (P Q) is false, when P Q is false, but (P Q) R is true. Step 2. P Q is false, when P is true, but Q is false. Step 3. From Step 1, we deduced P was true, but Q was false. So (P Q) R is true, when R is true. Combining Step 1 Step 3, we deduce ((P Q) R) (P Q) is false, when P is true, but Q is false, and R is true. That is, (P Q) R P Q when P is true, but Q is false, and R is true. As the second way to draw a conclusion, we will use the proof, which will be explained through examples. Example 6. Let us prove the statement, When using symbols, it can be expressed as opposite angles in a parallelogram are congruent 1. if ABCD is a parallelogram, then B = D. Here the hypothesis is ABCD is a parallelogram and conclusion is B Math Composer = D. D C A B 1 congruent means same, identical Page 3 of 6

10 Set Theory & Logic Section 1.3 Rules of Inference Fall, 2009 Proof. Proof. 1. ABCD is a parallelogram Given 2. Have line AC Postulate 3. DAC = ACB Alternate interior angles 4. CAB = ACD Alternate interior angles 5. AC = AC Reflexive 6. ACD = CAB ASA 2 7. B = D Corresponding parts The example above tells us how to draw a conclusion from the given hypotheses. In general, Proof of a propositional form p (conclusion) from the propositional forms (hypotheses), h 1, h 2,..., h k, means a list of propositional forms, p 1, p 2,..., p n, such that p n = p and every propositional form is either a hypothesis or follows by a rule of logic. We use the rules of inferences to draw the conclusion. Before we give the rules, let us introduce the terminology. 1. Theorem: statement that is proven. 2. Lemma: technical theorem used to prove another theorem. 3. Corollary: theorem that is quickly proven from another. 4. Axiom/Postulate: statement that is assumed to be true. Theorem (Rules of Inference). Let p, q, r and s be propositional forms. Modus Pones [MP] p q, p q Disjunctive Syllogism [DS] p q, p q Modus Tolens [MT] p q, q p Addition [Add] p p q Simplification [Simp] Transitivity [Trans] 3 p q p p q, q r p r Conjunction [Conj] p, q p q Constructive Dilemma [CD] Destructive Dilemma [DD] (p q) (r s), p r q s (p q) (r s), q s p r 2 Angle Side Angle: If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent 3 Also known as Hypothetical Syllogism or the Chain Rule Page 4 of 6

11 Set Theory & Logic Section 1.3 Rules of Inference Fall, 2009 We can use the truth table to prove each of the results in the theorem above. However, we will focus on how to use them to prove a propositional form form the hypotheses, which is one of main concerns. Example 7. Simple usage of the rules. (1) Let us draw a conclusion from (P Q) R, P Q, i.e., (P Q) R, P Q? Let p := P Q and q := R. Then the left hand side of the statement above becomes p q, p. By the rule Modus Ponens, we have p q, p q i.e., (P Q) R, P Q R. (2) It is easy to see that the rule Addition gives P P (Q R). (3) Let us draw a conclusion from (Q R) P, P, i.e., (Q R) P, P? Let p := Q R and q := P. Then the left hand side of the statement above becomes p q, q. By the rule Modus Tolens, we have p q, q p i.e., (Q R) P, P (Q R). Example 8. Invalid uses of the rules of inference, although each are valid arguments. (1) P Q P. When we use the truth table, we find that it is a tautology. That is, it is a valid argument. However, the rule of inference, i.e., Modus Ponens, says P P Q. (2) P Q, Q P. When we use the truth table, we find that it is a tautology. That is, it is a valid argument. However, the rule inference, i.e., Disjunctive Syllogism, says P Q, P Q. Now let us see examples on how to apply the rules of inferences to prove. Example 9. Prove (P Q) (Q R), P Q Proof. We make up the following table. Proof. 1. (P Q) (Q R) Given 2. P Given Show Q 3.?? 4.?? 5.?? Page 5 of 6

12 Set Theory & Logic Section 1.3 Rules of Inference Fall, 2009 Step 1: We list all the hypotheses and put Given to each of them as a reason. Step 2: We put the conclusion as above. The line containing the conclusion is called a comment. Step 3: We apply the rule of inference, Addition with p := P and q := Q, to the line 2. Then we get p p Q, i.e., P P Q. So on the line 3 in the table, we put P Q with the reason 2 Add. Step 4: Since we have (P Q) (Q R) on line 1 and P Q on line 3, so we use the rule of inference, Modus Ponens with p := P Q and q := Q R, to get p q, p q, i.e., (P Q) (Q R), P Q Q R. So on the line 4 in the table, we put Q R with the reason 1, 3 MP. Step 5: Now since we have Q R on the line 4 and want to deduce Q, so we apply the rule of inference, Simplification with p := Q and q := R, to the line 4 and get p q p, i.e., Q R Q. So on the line 5 in the table, we put Q with the reason 4 Simp. Combining the results in all steps, we have Proof. 1. (P Q) (Q R) Given 2. P Given Show Q 3. P Q 2 Add 4. Q R 1, 3 MP 5. Q 4 Simp Example 10. Show Example 11. Prove Example 12. Prove P Q, Q R, S R, S P P Q, (P Q) (T S), P T, Q S P Q, Q R, R Q P Page 6 of 6

13 Set Theory & Logic Section 1.4 Rules of Replacement Fall, 2009 Section 1.4 Rules of Replacement It is often important to be able to substitute one proposition for another if we are sure that their truth values coincide. This brings us to the second type of logic rule: rules of replacement. Definition. Any two propositional forms p and q are (logically) equivalent if their columns in their truth tables are the same. The notation for this is p q. Example 1. Any implication is equivalent to its contrapositive, but not equivalent to its converse. That is, for propositions P and Q, P Q Q P, but P Q Q P. Proof. Let us make the truth table: P Q P Q Q P Q P Q P T T T F F T T T F F T F F T F T T F T T F F F T T T T T As we can see, P Q and Q P have the same truth values. But P Q and Q P does not have the same value. Example 2. All tautologies are logically equivalent. Also all contradictions are logically equivalent. Example 3. The following equivalence is very important and useful. So please memorize it. P Q P Q. Example 4. Using the equivalence in the previous example, we have Q (P Q) Q ( P Q). Example 5. Since P Q is equivalent to Q P, so the propositional form R (P Q) is equivalent to R ( Q P ), i.e., R (P Q) R ( Q P ). Theorem. Let p, q, and r be propositional forms. 1. Commutative Laws [Com] p q q p and p q q p. 2. De Morgan s Laws [DeM] (p q) p q and (p q) p q. Page 1 of 3

14 Set Theory & Logic Section 1.4 Rules of Replacement Fall, Associative Laws [Assoc] (p q) r p (q r) and (p q) r p (q r). 4. Tautology [Taut] p p p and p p p. 5. Distributive Laws [Distr] p (q r) (p q) (p r) and p (q r) (p q) (p r). 6. Material Equivalence [Equiv] p q (p q) (q p) and p q (p q) ( p q). 7. Contrapositive Law [Contra] 8. Material Implication [Impl] 9. Double Negation [DN] 10. Exportation [Exp] p q q p. p q p q. p p. (p q) r p (q r). We can prove the rules above by using the truth tables, which is not our concerns. Just as we did with the rules of inference, we will utilize the rules above. Example 6. Show that P Q, R S R P. Proof. We introduce the table as we proved with the rules of inference. Proof. 1. P Q Given 2. R S Given Show R P Show R P 3. R 2 Simp 4. R P 3 Add 5. P 4 DN 6. R P 5 Impl Page 2 of 3

15 Set Theory & Logic Section 1.4 Rules of Replacement Fall, 2009 In this proof, (1) the rule of replacement is used on the last line 5 Impl, (2) the second comment Show R P is a just translation of the first comment Show R P, (3) we did not use the first hypothesis P Q. This happens very rarely! What we have actually proven is R S R P. Example 7. Show that P Q, R Q (P R) Q. Proof. We make the table as follows: Proof. 1. P Q Given 2. R Q Given Show (P R) Q 3. (P Q) (R Q) 1, 2 Conj 4. ( P Q) ( R Q) 3 Impl 5. (Q P ) (Q R) 4 Com 6. Q ( P R) 5 Dist 7. ( P R) Q 6 Com 8. (P R) Q 7 DeM 9. (P R) Q 8 Impl Example 8. Prove (P Q) (R S) (P S) (Q R). The proof is given in the textbook. We skip. Page 3 of 3

16 Set Theory & Logic Section 2.1 Predicates and Sets Fall, 2009 Section 2.1 Predicates and Sets In this chapter, we will study the sentences with variables and see how they can become propositions, both by substitution and by using words such as for all and there exists. Also we will use them in proofs. The collection of rules that govern these predicates is known as first order predicate logic, which is our subject of this chapter. Predicates A sentence that has variables into which we can substitute values is called a predicate. How to represent a predicate? See the example below. Example 1. A predicate y + 2 = 7 is represented by P (y) := y + 2 = 7. Although the predicate y + 2 = 7 does not have x, we can add another variable x as follows: Q(x, y) := y + 2 = 7. When we put y = 3 into P (y), we have P (3) := = 7, which becomes a false proposition. But when we put y = 5, we have P (5) := = 7, which becomes a true proposition. Also Q(x, 3) is always false (,i.e., contradiction) for any x. Q(x, 5) is always true (,i.e., tautology) for any x. Example 2. We can think of another predicate, P (f) := the linear function f is increasing. Here the variable is f. So for g(x) = 3x 5, P (g) := the linear function g is increasing, which is a true proposition. But for h(x) = 3x + 10, P (h) := the linear function h is increasing, which is a false proposition, because h(x) = 3x + 10 is not increasing. Sets In mathematics, an object can be a number, a function, an ordered pair, or almost anything that we would want to study. Let us collect some or all objects. The collection is called the set of the objects and the object is called Page 1 of 3

17 Set Theory & Logic Section 2.1 Predicates and Sets Fall, 2009 the set s element or member. 1. For a set A (set is written by a capital letter) and its element a, we write a A. 2. If a is not an element of A, we write a / A. 3. If a, b A means a A and b A. 4. If a set does not have any element, then it is called the empty set denoted by. 5. For a set A of numbers from 1 to 10, we write A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, or A = {1, 2,..., 9, 10}, where the list of the elements is called a roster. 6. A set of only one element is called the singleton. 7. Two sets A and B are equal means A and B have the same elements. 8. The repetition of elements in a set does not change the set, i.e., { 92, 4x + 23, school } = { 92, 4x + 23, school, 92, 92, 92, school }. Notations: Symbol N or Z + Z Z Q R C Name the set of natural numbers the set of integers the set of negative integers the set of rational numbers the set of real numbers the set of complex numbers Example Z +, 0 / Z +, 4 N, 5 / N, 5 Z, 0.65 / Z, 0.65 Q, 1/2 Q, π / Q, π R, 3 2i / R, 3 2i C Notations: If a, b R so that a < b, then we have the following: closed interval [a, b] closed ray [a, ) open interval (a, b) closed ray (, a] half open interval [a, b) open ray (a, ) half open interval (a, b] open ray (, a) Example 4. (4, 7) means the interval between 4 and 7, explicitly, set of all real numbers greater than 4 and less than 7, in another way, set of all real numbers x such that 4 < x < 7. Page 2 of 3

18 Set Theory & Logic Section 2.1 Predicates and Sets Fall, 2009 When the objects of a set are the elements that are allowed to be substituted into a predicate, the set is called the domain of the predicate. Example 5. Let P (x) := x + 2 = 7. Since P (5) is true and no other real number satisfies P (x), exactly one element of R satisfies P (x). However, no element of Z or (, 5) satisfies P (x). Example 6. If Q(x) := x 10, then all elements of [20, 100] satisfy Q(x), some but not all elements of Q satisfy Q(x), and no elements of {1, 2, 3} satisfy Q(x). Page 3 of 3

19 Set Theory & Logic Section 2.2 Quantification Fall, 2009 Section 2.2 Quantification We will introduce another way to form a proposition from a predicate. I. Two Quantifiers: Existential and Universal Let us start with a predicate x + 2 = 7. Assuming the domain is the set R of real numbers, this predicate is true when x = 5 R. It means there exists x R such that x + 2 = 7 is a true proposition. The notation for this is ( x R)(x + 2 = 7). Such a statement is known as an existential proposition. When no element in the domain satisfies the predicate, the predicate becomes a false proposition. In general, if P (x) is a predicate with domain D, the propositional form ( x D)P (x) means there exists x D such that P (x). Consider x + 5 = 5 + x. This is true for any real number x. So we can say for all x R, x + 5 = 5 + x. The notation for this is ( x R)(x + 5 = 5 + x). Such a statement is known as a universal proposition. It would be false if one element of the domain did not satisfy the predicate. In general, if P (x) is a predicate with domain D, the propositional form ( x D)P (x) means for all x D such that P (x). The symbols and are called quantifiers. is the universal quantifier and is the existential quantifier. Example 1. Other expressions 1. for all y R, y + 0 = y Page 1 of 5

20 Set Theory & Logic Section 2.2 Quantification Fall, 2009 is a true proposition. There are other expressions with the same meaning. For each real number y, y + 0 = y For every y, y + 0 = y y + 0 = y for all y R y + 0 = y for each real y 2. there exists y R such that 8 = 2y is a true proposition. There are other expressions with the same meaning. There is a real number y so that 8 = 2y For some y R, 8 = 2y 8 = 2y for some real number y 8 is equal to 2y for a particular y R Example 2. When we put multiple quantifiers, typically we put all of them in front or in back. But universal and existential quantifiers are exceptional. The proposition ( x R)( y R)(x + y = 0) can be translated and for all x R, there exists y R such that x + y = 0 for every real number x, x + y = 0 for some y R. There are certain cases that we prefer to put the quantifiers in back. Example 3. We want to express d divides one of the integers in the list, a 1, a 2,..., a k. Then we may write d divides a i for some i = 1, 2,..., k. Similarly, if d divides them all, then we write d divides a i for all i = 1, 2,..., k. Now we introduce a very useful theorem. Theorem. If D is a domain for a predicate P (x), then: ( x D)P (x) ( x) [x D P (x)], ( x D)P (x) ( x) [x D P (x)]. The example below shows how useful the theorem is. Page 2 of 5

21 Set Theory & Logic Section 2.2 Quantification Fall, 2009 Example 4. Determined the truth values of the following two propositions: 1. ( x )P (x). (1) By the theorem, it is equivalent to ( x) [x P (x)]. (2) Since x is always false, so the predicate x P (x) is always true, i.e., true for any x. Hence, the latter proposition (2) is true, i.e., the given proposition (1) is also true. 2. ( x )P (x). (3) By the theorem, it is equivalent to ( x) [x P (x)]. (4) Since x is always false, so the predicate x P (x) is always false, i.e., false for any x. Hence, the latter proposition (4) is false, i.e., the given proposition (3) is also false. II. Proposition with Many Quantifiers Next we consider propositions with many quantifiers. For instance, the Associative Law for addition: x + (y + z) = (x + y) + z for all real numbers x, y and z. Using logic notation, we can rewrite this as ( x R)( y R)( z R) [x + (y + z) = (x + y) + z]. (5) Step 1. We observe the predicate ( y R)( z R) [x + (y + z) = (x + y) + z] does not have a quantifier on x. So in this predicate, we say that the occurrence of x is free. However, since it has quantifiers on y and z, we say that the occurrence of y and z is bound. Since the occurrence of x is free in the predicate, we can regard x as a variable in the predicate and rewrite this as P (x) := ( y R)( z R) [x + (y + z) = (x + y) + z]. (6) When we use this, the proposition (5) can be rewritten by ( x R)( y R)( z R) [x + (y + z) = (x + y) + z] ( x R)P (x). Step 2. We observe the predicate ( z R) [x + (y + z) = (x + y) + z] Page 3 of 5

22 Set Theory & Logic Section 2.2 Quantification Fall, 2009 does not have a quantifiers on x and y. So in this predicate, the occurrence of x and y is free. However, since it has quantifier on z, the occurrence of z is bound. Since the occurrence of x and y is free in the predicate, we can regard x and y as variables in the predicate and rewrite this as Q(x, y) := ( z R) [x + (y + z) = (x + y) + z]. (7) When we use this, the proposition (6) can be rewritten by P (x) := ( y R)( z R) [x + (y + z) = (x + y) + z] ( y R)Q(x, y). and thus the proposition (5) can be rewritten by ( x R)( y R)( z R) [x + (y + z) = (x + y) + z] ( x R)P (x) ( x R)( y R)Q(x, y). Step 3. When we let R(x, y, z) := [x + (y + z) = (x + y) + z], the proposition (7) can be rewritten by Q(x, y) := ( z R) [x + (y + z) = (x + y) + z] ( z R)R(x, y, z). and thus the proposition (5) can be rewritten by ( x R)( y R)( z R) [x + (y + z) = (x + y) + z] ( x R)P (x) ( x R)( y R)Q(x, y) ( x R)( y R)( z R)R(x, y, z). Example 5. From the graph of the equation y = 2x 2 + 1, we know for all x R, there exists y R such that y = 2x In logic notation, it is written by ( x R)( y R)(y = 2x 2 + 1). (8) Following the argument above, we let P (x) := ( y R)(y = 2x 2 + 1), then the proposition (8) can be rewritten by ( x R)( y R)(y = 2x 2 + 1) ( x R)P (x). Page 4 of 5

23 Set Theory & Logic Section 2.2 Quantification Fall, 2009 When we substitute any value to x, P (x) becomes a true proposition, because for example, 33 = 2(4) and 33 is a real number, i.e., P (4) is true. Therefore, the proposition (8) is true. Page 5 of 5

24 Set Theory & Logic Section 2.3 Negating Quantifiers Fall, 2009 Section 2.3 Negating Quantifiers Let P (x) be a predicate and D = { 4, 5, 6 } a domain. If ( x D ) P (x) is true, then we have P (4) T, P (5) T, P (6) T, where T means true. If ( x D ) P (x) is false, then at least one between P (4), P (5) and P (6) should be false. In other words, there exists an element in D such that P (x) is false. In symbols, if ( x D ) P (x) is false, then ( x D ) P (x). In general, the negation of ( x D ) P (x) is ( x D ) P (x). When we use symbols, we have ( x D ) P (x) ( x D ) P (x). (1) If ( x D ) P (x) is true, then at least one between P (4), P (5) and P (6) should be true. Under the circumstance, if ( x D ) P (x) is false, then there should be no element in D such that P (x) is true, i.e., for all elements in D, P (x) should be false. In symbols, if ( x D ) P (x) is false, then ( x D ) P (x). In general, the negation of ( x D ) P (x) is ( x D ) P (x). When we use symbols, we have The relations (1) and (2) should be memorized. Example 1. Consider the negation of ( x ) [ P (x) Q(x) ]: ( x D ) P (x) ( x D ) P (x). (2) ( x ) [ P (x) Q(x) ] ( x ) [ P (x) Q(x) ] (by (1)) ( x ) [ P (x) Q(x) ] (by De Morgan s Law) When we negate a proposition with a quantifier, we can simplify the negated proposition as much as we can by using the Rules of Replacement. When the negation is as far into the predicate as possible, the form is called a positive form. That is, the last proposition ( x ) [ P (x) Q(x) ] is the positive form of ( x ) [ P (x) Q(x) ]. Consider the negation of ( x ) [ R(x) S(x) ]: ( x ) [ R(x) S(x) ] ( x ) [ R(x) S(x) ] (by (2)) ( x ) [ R(x) S(x) ] ( x ) [ R(x) S(x) ] ( x ) [ R(x) S(x) ] (by Material Implication) (by Distributive Laws) (by Double Negation) The last proposition is the positive form of the first proposition. Example 2. Write the negation of ( x ) ( y ) [ P (x) Q(y) ] in positive form. ( x ) ( y ) [ P (x) Q(y) ] ( x ) ( y ) [ P (x) Q(y) ] (by (1)) ( x ) ( y ) [ P (x) Q(y) ] (by (2)) ( x ) ( y ) [ P (x) Q(y) ] (by Material Implication) ( x ) ( y ) [ P (x) Q(y) ] (by De Morgan s Laws) ( x ) ( y ) [ P (x) Q(y) ] (by Double Negation) Page 1 of 3

25 Set Theory & Logic Section 2.3 Negating Quantifiers Fall, 2009 Example 3. Now we try to negate propositions written in English: there exists a real number x so that x 2 = 4. In symbols, it is written by Its negation follows ( x R ) ( x 2 = 4 ). which means, in English, ( x R ) ( x 2 = 4 ) ( x R ) ( x 2 = 4 ) (by (2)) ( x R ) ( x 2 4 ), for every real number x, x 2 4, or the square of every real number is not four. Let us discuss another one: for every x R, there exists y R such that x > 0 and y = cos x. In symbols with P (x) := x > 0 and Q(x, y) := y = cos x, it is written by Its negation follows ( x R ) ( y R ) [ P (x) Q(x, y) ]. ( x R ) ( y R ) [ P (x) Q(x, y) ] ( x R ) ( y R ) [ P (x) Q(x, y) ] (by (1)) which means, in English, ( x R ) ( y R ) [ P (x) Q(x, y) ] (by (2)) ( x R ) ( y R ) [ P (x) Q(x, y) ], there exists x R so that for every y R, x 0 or y cos x. (by DeM) As the last topic in this section, we discuss the counterexample. In mathematics, we want to show It is equivalent that its negation is true, i.e., ( x D ) P (x) is false. ( x D ) P (x) is true, i.e., ( x D ) P (x) is true. The last one is about the existence of an element in D such that P (x) is not true. Hence, by finding such an element a D, we can prove the last statement, eventually, the given original statement. This element a D is called a counterexample of the original statement ( x D ) P (x). Example 4. Show that is false. It has the negation: ( x R ) ( x + 2 = 7 ) ( x R ) ( x + 2 = 7 ) ( x R ) ( x + 2 = 7 ) ( x R ) ( x ), which is true for x = 3, because x = 3 R does not satisfy x + 2 = 7. Since x = 3 makes the negation of the original proposition to be true, hence x = 3 is the counterexample of the original proposition and the original proposition becomes false. Page 2 of 3

26 Set Theory & Logic Section 2.3 Negating Quantifiers Fall, 2009 Example 5. As another example, consider the proposition every quadratic function has a real root, which is false. Let us prove that the proposition is false. In symbols, the proposition is written by It has the negation: ( f ) ( f is quadratic ( x R ) [ f(x) = 0 ] ). ( f ) ( f is quadratic ( x R ) [ f(x) = 0 ] ) ( f ) ( f is quadratic ( x R ) [ f(x) = 0 ] ) ( f ) ( f is quadratic ( x R ) [ f(x) = 0 ] ) ( f ) ( f is quadratic ( x R ) [ f(x) = 0 ] ) ( f ) ( f is quadratic ( x R ) [ f(x) 0 ] ), which is true for f(x) = x 2 + 3, because f(x) = x is a quadratic function and for any real number x, f(x) = x Hence, f(x) = x is a counterexample of the original proposition and the original proposition becomes false. Page 3 of 3

27 Set Theory & Logic Section 2.4 Proofs with Quantifiers Fall, 2009 Section 2.4 Proofs with Quantifiers We study four rules of inference involved with the quantifiers and used in the proof: Universal Instantiation (UI), Universal Generalization (UG), Existential Generalization (EG), Existential Instantiation (EI). Theorem (Universal Instantiation (UI)). If a is a constant from a nonempty domain D, then ( x D ) P (x) P (a). It is easy to understand. When ( x D ) P (x) is true, every element of the domain satisfies P (x). Example 1. If a D, then the following are legitimate uses of Universal Instantiation: 1. ( x D ) [ P (x) Q(x) ] P (a) Q(a) 2. ( x D ) [ P (x) ( y D ) Q(y) ] P (a) ( y D ) Q(y) 3. ( x D ) ( y D ) [ Q(x) R(y) ] ( y D ) [ Q(a) R(y) ] Before we discuss further, we need to specify a constant: arbitrary and particular. A constant a is arbitrary if it can represent a randomly selected element of the domain. For example, the expression let a be a real number means that a represents any real number without restriction. If a constant is not arbitrary, then it is particular. A particular element has some extra characteristic other than simply being a member of the domain. Example 2. Consider the predicate P (x, y) := y = x + 5. Let us pick an element a R. This a is arbitrary, because it is chosen randomly. When we put x = a into P (x, y), we get P (a, y) := y = a + 5. If y = b is another constant satisfying y = a + 5, then this constant b is particular. Example 3. An arbitrary real number will satisfy x + 7 = 7 + x, but only a particular real number will satisfy 2 + x = 10. One way to see if a symbol is arbitrary is to check if it has a free occurrence in a hypothesis of a proof. If the symbol does not have a free occurrence, then it is arbitrary. If the symbol has a free occurrence, then it is particular, because the assumption is declaring that the symbol has a specific property. Example 4. ( x ) P (x) P (a). Here a is arbitrary. But a in the following is not arbitrary, because the occurrence of a in the hypothesis P (a) is free: Proof. 1. P (a) Given 2. P (a) Q(a) From now on, the arbitrary elements will usually be presented by a and sometimes b. Theorem (Universal Generalization (UG)). Let a be an arbitrary constant symbol from a nonempty domain D. If P (a) contains no particular constants from D, then P (a) ( x D ) P (x). legal, lawful; acceptable; justified; reasonable, logical; correct, sound; authorized Page 1 of 6

28 Set Theory & Logic Section 2.4 Proofs with Quantifiers Fall, 2009 Since a D is arbitrary, P (a) means that P (x) is satisfied by every element of the domain D. Thus, the conclusion ( x D ) P (x) is valid. Example 5 (Illegal Uses of Universal Generalization). 1. The following is invalid: Proof. 1. P (c) Given 2. ( x ) P (x) 1 UG Since c in line 1 is a particular constant, we cannot use the Universal Generalization to the line The following is an attempt to prove ( x R ) ( y R ) ( x + y = 2x ) from ( x R ) ( x + x = 2x ). Proof. 1. ( x R ) ( x + x = 2x ) Given 2. a + a = 2a 1 UI 3. ( y R ) ( a + y = 2a ) 2 UG 4. ( x R ) ( y R ) ( x + y = 2x ) 3 UG In line 3, an illegal substitution was made. When applying Universal Generalization (UG), we should get ( y R ) ( y + y = 2y ) rather than ( y R ) ( a + y = 2a ). Example 6 (Valid Proofs). 1. Prove ( x ) ( y ) P (x, y) ( y ) ( x ) P (x, y). Proof. Proof. 1. ( x ) ( y ) P (x, y) Given Show ( y ) ( x ) P (x, y) 2. ( y ) P (a, x) 1 UI 3. P (a, b) 2 UI 4. ( x ) P (x, b) 3 UG 5. ( y ) ( x ) P (x, y) 4 UG 2. Prove ( x ) [ P (x) Q(x) ], ( x ) [ Q(x) R(x) ] ( x ) P (x). Proof. Proof. 1. ( x ) [ P (x) Q(x) ] Given 2. ( x ) [ Q(x) R(x) ] Given Show ( x ) P (x) 3. P (a) Q(a) 1 UI 4. [ Q(a) R(a) ] 2 UI 5. Q(a) R(a) 4 DeM 6. Q(a) 5 Simp 7. P (a) 3, 5 MT 8. ( x ) P (x) 7 UG Notice that the a in line 3 did not occur free in a hypothesis. Hence a is arbitrary throughout the proof. Page 2 of 6

29 Set Theory & Logic Section 2.4 Proofs with Quantifiers Fall, 2009 Let us introduce a terminology. A paragraph proof is an argument written in English or some other language. When we write a paragraph proof for a proposition with universal quantifier, we may write as follows, for example, In logic notation, Let a D. or Take a D. or Suppose a is in D. These types of proofs are called universal proofs. P (a) is true. ( x D ) P (x) Let a D Show P (a) Example 7. Prove that for all real numbers x, (x 1) 3 = x 3 3x 2 + 3x 1. Proof. Let a R. Then (a 1) 3 = (a 1)(a 1) 2 = (a 1)(a 2 2a + 1) = a 3 3a 2 + 3a 1. Definition (Even and Odd Integers). Let c be an integer. Then Example 8. Prove the proposition It can be rewritten by c is even if and only if c = 2l for some l Z. c is odd if and only if c = 2k + 1 for some k Z. the square of every even integer is even. for all even integer n, n 2 is even. Proof. Let n be an even integer. Then there exists k Z such that n = 2k. We compute n 2 = (2k) 2 = 4k 2 = 2(2k 2 ), which implies that n 2 is even, because 2k 2 Z. In the example above, we used the existential quantifier (there exists k Z such that n = 2k). Next, we discuss the rules involved with the existential quantifier. Theorem (Existential Generalization (EG)). Let D be a nonempty domain. If a is an element from D, then P (a) ( x D ) P (x). The rule says that if P (a) is true for some constant a in the domain, then it is also true that ( x D ) P (x). Example P (a) R(a) ( x ) [ P (x) R(x) ] 2. Q(a) T (b) ( y ) [ Q(a) T (y) ] Page 3 of 6

30 Set Theory & Logic Section 2.4 Proofs with Quantifiers Fall, 2009 We introduce a notation â to represent a particular element a of the domain. Theorem (Existential Instantiation (EI)). If a is a constant symbol from a nonempty domain D that has no prior occurrence in the proof, then ( x D ) P (x) P (â). The propositional form ( x D ) P (x) means that P (x) is satisfied for some element of D. That element is denoted by â. Example ( x ) [ P (x) Q(x) ] P (â) Q(â) 2. ( y ) [ R(a, y, c) R(a, y, c) ] R(a, ˆb, c) R(a, ˆb, c) 3. ( x ) ( y ) ( z ) Q(x, y, z) ( y ) ( z ) Q(â, y, z) Example ( z ) [ P (z) Q(z) ] P (â) Q(z), where the substitution was made incorrectly and can be fixed by ( z ) [ P (z) Q(z) ] P (â) Q(â). 2. ( x ) ( y ) T (a, x, y) ( y ) T (a, a, y), where the new variable was not used with the application of Existential Instantiation and can be fixed by ( x ) ( y ) T (a, x, y) ( y ) T (a, â, y). Example 12 (Illegal Usage). Prove that ( x ) ( y ) ( x + y = 2 ) logically implies ( y ) ( x ) ( x + y = 2 ). Proof. Attempted Proof. 1. ( x ) ( y ) ( x + y = 2 ) Given Show ( y ) ( x ) ( x + y = 2 ) 2. ( y ) ( a + y = 2 ) 1 UI 3. a + ˆb = ( 2 ) 2 EI 4. ( x ) x + ˆb = 2 3 UG 5. ( y ) ( x ) ( x + y = 2 ) 4 EG Since ˆb is particular, the application of Universal Generalization in line 4 is invalid. Example 13 (Valid Proofs). 1. ( x ) [ P (x) Q(x) ], ( x ) [ P (x) R(x) ] ( x ) R(x) Proof. Proof. 1. ( x ) [ P (x) Q(x) ] Given 2. ( x ) [ P (x) R(x) ] Given Show ( x ) R(x) 3. P (ĉ) Q(ĉ) 1 EI 4. P (ĉ) R(ĉ) 2 UI 5. P (ĉ) 3 Simp 6. R(ĉ) 4, 5 MP 7. ( x ) R(x) 6 EG Page 4 of 6

31 Set Theory & Logic Section 2.4 Proofs with Quantifiers Fall, ( x ) ( y ) [ Q(x) T (y) ] ( x ) [ Q(x) ( y ) T (y) ] Proof. Proof. 1. ( x ) ( y ) [ Q(x) T (y) ] Given Show ( x ) [ Q(x) ( y ) T (y) ] 2. ( y ) [ Q(a) T (y) ] 1 UI 3. Q(a) T (ĉ) 2 EI 4. T (ĉ) Q(a) 3 Com 5. T (ĉ) 4 Simp 6. ( y ) T (y) 5 EG 7. Q(a) 3 Simp 8. Q(a) ( y ) T (y) 6, 7 Conj 9. ( x ) [ Q(x) ( y ) T (y) ] 8 UG 3. P (a) ( x ) [ Q(x) R(x) ], P (a) ( x ) R(x) Proof. In the proof, a is a particular symbol because it appears free in the hypothesis in lines 1 and 2. That is why it has a hat from the beginning of the proof. Proof. 1. P (â) ( x ) [ Q(x) R(x) ] Given 2. P (â) Given Show ( x ) R(x) 3. ( x ) [ Q(x) R(x) ] 1, 2 MP 4. Q(ˆb) R(ˆb) 3 EI 5. R(ˆb) Q(ˆb) 4 Com 6. R(ˆb) 5 Simp 7. ( x ) R(x) 6 EG As we did after the universal quantifier, we introduce a terminology. When we write a paragraph proof for a proposition with existential quantifier, we may write as follows, for example, ( x D ) P (x) Find c D such that P (c) Choose a candidate Choose a candidate These types of proofs are called existential proofs. Example 14. Prove that there exists x Z such that x 2 + 2x 3 = 0. Proof. A basic factorization shows 0 = x 2 + 2x 3 = (x + 3)(x 1), that is, either x = 3 or x = 1 will satisfy the predicate. Example 15. Prove that there is a function f such that the derivative of f is 20x 4. Page 5 of 6

32 Set Theory & Logic Section 2.4 Proofs with Quantifiers Fall, 2009 Proof. We choose f(x) = 4x 5 as a candidate. Now check, which will satisfy the predicate. d dx ( 4x 5 ) = 20x 4, We end this section with the paragraph proofs involved with both the universal and the existential quantifier. Example Prove that for every x R, there exists a y R such that x + y = 2. Proof. In logic notation, ( x R ) ( y R ) ( x + y = 2 ), where we use the quantifier diagram (will be explained in class). Let x R. Then we need find y R so that x + y = 2. We choose y = 2 x and compute x + y = x + (2 x) = Prove that there exists a x R such that for all y R, x + y = y. Proof. In logic notation, ( x R ) ( y R ) ( x + y = y ). We claim that 0 is the sought after element. Let y R. Then 0 + y = y. Hence we have such an element x = 0 R. 3. Prove that there is a function f and a real number x such that the derivative of f at x is 2. Proof. In logic notation, ( f ) ( x R ) [ f (x) = 2 ]. We define f(x) = x 2. This means f (x) = 2x. Thus, let x = 1, we get f (1) = 2(1) = 2. Page 6 of 6

33 Set Theory & Logic Section 2.5 Direct and Indirect Proof Fall, 2009 Section 2.5 Direct and Indirect Proof In this section, we will study three topics: direct proof, uniqueness rule and indirect proof. Most propositions in mathematics are implications, i.e., if p, then q. In many cases, it is not easy to prove by using the Rules of Inferences and Replacement. So we study more rules. Topic I. Direct Proof Theorem (Direct Proof (DP)). For propositional forms, h 1,..., h k, p, q, h 1,..., h k p q if and only if h 1,..., h k, p q. This theorem is also called the Deduction Theorem. Rough Proof of Theorem. The Exportation (Rules of Inference) implies [ h 1... h k ] ( p q ) [ h 1... h k p ] q. The left hand side means h 1,..., h k p q and the right hand side means h 1,..., h k, p q. Example 1. Proof. 1. ( P Q ) ( R S ) Given Show P R 2. P Assumption Show R 3. P Q 2 Add 4. R S 1, 3 MP 5. R 4 Simp 6. P R 2 5 DP Topic II. Uniqueness Rule Topic III. Indirect Proof Theorem (Indirect Proof (IP)). For propositional forms, h 1,..., h k, p, h 1,..., h k p if and only if h 1,..., h k, p q q for some propositional form q. This method is also known as proof by contradiction or reductio ad absurdum. Page 1 of 1

34 Section 2.6 More Methods We study proofs involved with Material Equivalence, disjunctions and cases. Each method will rely on Direct Proof. Topic I. Material Equivalence Theorem (Biconditional Proof (BP)). Let h 1,..., h k and p and q be propositional forms. Then, h 1,..., h k p q if and only if h 1,..., h k p q and h 1,..., h k q p. Since p q is equivalent to p q and q p, the theorem holds. Example Prove P Q ( P Q ) P. Proof. 1. P Q Given Show ( P Q ) R ( ) 2. P Q Assumption Show P 3. P 2 Simp ( ) 4. P Assumption Show P Q 5. Q 1, 4 MP 6. P Q 4, 5 Conj 7. ( P Q ) P 2 6 BP Example Prove for all n Z, n is even if and only if n 3 is even. In logic notation, it can be rewritten by ( n Z ) [ P (n) Q(n) ], where P (n) := n is even and Q(n) := n 3 is even. To prove Q(n) P (n), we use the BP and have the following structure of proof. ( ) Assume n is even Then n = 2k, k Z Show n 3 is even Show n 3 = 2l, l Z. n 3 is even ( ) Assume n is odd Then n = 2k + 1, k Z Show n 3 is odd Show n 3 = 2l + 1, l Z. n 3 is odd 1

35 Proof. Let n Z. ( ) Assume n is even. Then, n = 2k for some k Z. We must show n 3 is even. To do this, we calculate: n 3 = ( 2k ) 3 = 2 ( 4k 3 ), which means that n 3 is even. ( ) Suppose n is odd. This means that n = 2k + 1 for some k Z. To show n 3 is odd, we again calculate: n 3 = ( 2k + 1 ) 3 = 8k k 2 + 6k + 1 = 2 ( 4k 3 + 6k 2 + 3k ) + 1. Hence, n 3 is odd. Using the rules of replacement, we can have an better expression equivalent to the biconditional. Example Prove ( P Q ) ( P Q ) P. Proof. P Q P Q ( P P ) Q P ( P Q ) ( P Q ) P ( P Q ) P ( P Q ) P ( P Q ) P Sometimes, we need to prove a sequence of biconditionals. Definition The propositional forms p 1, p 2,..., p k are pairwise equivalent if for all i, j, p i p j. In other words, p 1 p 2, p 1 p 3,..., p 2 p 3, p 2 p 4,..., p k 1 p k. To prove the pairwise equivalence, we make use of Transitivity. This result is the Equivalence Rule. Theorem (Equivalence Rule). To prove that the propositional forms p 1, p 2,..., p k equivalent, it is enough to prove are pairwise p 1 p 2, p 2 p 3,..., p k 1 p k, p k p 1. The Equivalence Rule is typically useful in proving propositions that include the phrase the following are equivalent. Example For a polynomial f(x) = a n x n + a n 1 x n a 1 x + a 0, prove that the following are equivalent. 1. r is a root of f(x), 2. r is a solution to f(x) = 0, 2

36 3. x r is a factor of f(x). Proof. By the Equivalence Rule, it is enough to prove that 1 2, 2 3 and : Let r be a root of f(x). This means f(r) = 0 and so r is a solution to f(x) = : Suppose r is a solution to f(x) = 0. Then by the Polynomial Division Algorithm (will be discussed in class), there are polynomials q(x) and s(x) such that f(x) = q(x) ( x r ) + s(x). Here the degree of s(x) should be less than that of x r, i.e., s(x) is a constant, say s(x) = c. Now, 0 = f(r) = q(r) ( r r ) + c = 0 + c = c, i.e., c = 0, i.e., f(x) = q(x) ( x r ). Hence, x r is a factor of f(x). 3 1: Assume x r is a factor of f(x). This means there is a polynomial q(x) such that f(x) = q(x) ( x r ). Thus, we get which means r is a root of f(x). f(r) = q(r) ( r r ) = 0, Topic II. Disjunctions The second topic is to prove a disjunction. We observe p q p q p q. So to prove p q, we prove p q, because they are equivalent and p q is much easier to be proven by the Direct Proof. Theorem (Proof of Disjunctions). For any propositional forms p and q: p q p q. Two points on the theorem should be emphasized. 1. When proving p q, because of the hypothesis p, one may think of the Indirect Proof. But it is Direct Proof. 2. When we have p in proof, one may conclude p q by the Addition (Rule of Inference). But this is not reasonable. We can deduce p q by Addition in the middle of proof, but the rule cannot be used to give the conclusion. Example Prove for all a, b in Z, if ab = 0, then a = 0 or b = 0. The structure of the proof is given by the following: Assume ab = 0 Show a = 0 or b = 0 Suppose a 0. b = 0 a = 0 or b = 0 3