CSci102_Module 4.ppt

CSci102_Module 4.ppt
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Free Powerpoint Templates Page 2 Introduction Logic that deals with propositions is incapable of describing most of the statements in mathematics and computer science. Recall that a proposition is a statement that is either true or false, thus, the statement: p: n is an odd integer. is not a proposition because whether p is true or false depends on the value of n. If n = 103, p is true and false if n = 8.
Free Powerpoint Templates Page 3 Introduction Most of the statements in mathematics and computer science use variables. Thus, we must extend the system of logic to include such mathematical statements.
Free Powerpoint Templates Page 4 Domain of Discourse Definition: Let P(x) be a statement involving the variable x and let D be a set. We call P a propositional function (with respect to D) if for each x in D, P(x) is a proposition. We call D the domain of discourse of P.
Free Powerpoint Templates Page 5 Let P(n) be the statement n is an odd integer and let D be the set of positive integers. – P is a propositional function with domain of discourse D since for each n in D, P(n) is a proposition, that is, for each n in D, P(n) is true or false but not both. Domain of Discourse : Example
Free Powerpoint Templates Page 6 – If n = 1, P(1) is true, that is, 1 is an odd integer – If n = 2, P(2) is false, that is, 2 is an odd integer is false. Domain of Discourse : Example
Free Powerpoint Templates Page 7 Propositional Function By itself, it is neither true nor false. However, for each x in its domain of discourse, P(x) is a proposition and is, therefore, either true or false. We can think of a propositional function as defining a class of propositions, one for each element of its domain of discourse.
Free Powerpoint Templates Page 8 Propositional Function If for example, P is a propositional function with domain of discourse equal to the set of positive integers, we obtain the class of propositions P(1), P(2), . . . . in which each of P(1), P(2), . . . is either true or false.
Free Powerpoint Templates Page 9 Examples of Propositional Function • n2 + 2n is an odd integer - D = set of positive integers • x2 - x - 6 = 0 - D = set of real numbers
Free Powerpoint Templates Page 10 Examples of Propositional Function • The student gets a GPA of 2.0 or better during the second semester of SY 2002- 2003 - D = set of students • The actress has won a FAMAS award. - D = set of actresses
Free Powerpoint Templates Page 11 Universal and Existential Quantifiers Most of the statements in mathematics and computer science use terms such as "for every" and "for some." For example in mathematics we have the statements: –For every triangle T, the sum of the angles of T is equal to 180°. –For some triangle S, the sum of two angles is less than 90o.
Free Powerpoint Templates Page 12 Universal Quantifiers Definition: Let P be a propositional function with domain of discourse D. The statement for every x, P(x) is said to be a universally quantified statement. The symbol  means "for every." Thus the statement for every x, P(x) may be written x, P(x).
Free Powerpoint Templates Page 13 Universal Quantifiers The symbol  is called a universal quantifier. The statement for every x, P(x) is true if P(x) is true for every x in D. The statement for every x, P(x) is false if P(x) is false for at least one x in D.
Free Powerpoint Templates Page 14 Existential Quantifiers Definition: The statement for some x, P(x) is said to be an existentially quantified statement. The symbol  means "for some." Thus the statement for some x, P(x) may be written x, P(x).
Free Powerpoint Templates Page 15 Existential Quantifiers The symbol  is called an existential quantifier. The statement for some x, P(x) is true if P(x) is true for at least one x in D. The statement for some x, P(x) is false if P(x) is false for every x in D.
Free Powerpoint Templates Page 16 Remarks on definition of Quantifiers Free or Bound Variable – We call the variable x in the propositional function P(x) a free variable. – The idea is that x is "free" to roam over the domain of discourse. – We call the variable x in the universally quantified statement x, P(x) or in the existentially quantified statement x, P(x) a bound variable.
Free Powerpoint Templates Page 17 Remarks on definition of Quantifiers – The idea is that x is "bound" by the quantifier  or . – We previously pointed out that a propositional function does not have a truth value. On the other hand, quantified statements have corresponding truth values. – In general, a statement with free (unquantified) variables is not a proposition and a statement with no free variables (no unquantified variables) is a proposition.
Free Powerpoint Templates Page 18 The symbol  may be read "for every," "for all," or "for any." The symbol  may be read "for some," "for at least one," or "there exists." Sometimes, to specify the domain of discourse D, we write a universally quantified statement as for every x in D, P(x) and we write an existentially quantified statement as for some x in D, P(x). Remarks on definition of Quantifiers
Free Powerpoint Templates Page 19 The universally quantified statement for every x, P(x) is false if for at least one x in the domain of discourse, the proposition P(x) is false. A value x in the domain of discourse that makes P(x) false is called a counterexample to the statement for every x, P(x). Remarks on definition of Quantifiers
Free Powerpoint Templates Page 20 Quantifiers : Example1 The statement for every real number x, x2  0 is a universally quantified statement. The domain of discourse is the set of real numbers. The statement is true because, for every real number x, it is true that the square of x is positive or zero.
Free Powerpoint Templates Page 21 Proof: We must verify that the statement if x > 1, then x + 1 > 1 is true for every real number x. Let x be any real number whatsoever. It is true that for any real number x, either x  1 or x > 1. Prove that the universally quantified statement for every real number x, if x > 1, then x + 1 > 1 is true. Quantifiers : Example2
Free Powerpoint Templates Page 22 Case1: x  1 The conditional proposition If x > 1, then x + 1 > 1 is true because the hypothesis x > 1 is false. (Recall that when the hypothesis is false, the conditional proposition is true regardless of whether the conclusion is true or false.) Quantifiers : Example2
Free Powerpoint Templates Page 23 Case 2: x > 1 x > 1 → x + 1 > 1 + 1 > 1 (API) → x + 1 > 1 (transitivity of >) hence the conditional proposition if x > 1, then x + 1 > 1 is true. Quantifiers : Example 2
Free Powerpoint Templates Page 24 Thus, we have shown that for every real number x, the proposition if x > 1, then x + 1 > 1 is true. Therefore the universally quantified statement for every real number x, if x > 1, then x + 1 > 1 is true. Quantifiers : Example 2
Free Powerpoint Templates Page 25 The universally quantified statement for every real number x, x2 - 1 > 0 is false since, if x = 1, the proposition 12 - 1 > 0 is false. The value 1 is a counterexample to the statement for every real number x, x2 - 1 > 0. Quantifiers : Example 3
Free Powerpoint Templates Page 26 Note: To show that the universally quantified statement for every x, P(x} is false, it is sufficient to find one value x in the domain of discourse for which the proposition P(x} is false, that is, it is sufficient to give a counterexample. Quantifiers : Example 3
Free Powerpoint Templates Page 27 The method of disproving the statement for every x, P(x} is quite different from the method used to prove that the statement is true. To prove that for every x, P(x} is true, we must, in effect, examine every value of x in the domain of discourse and show that for every x, P(x} is true. Quantifiers : Example 3
Free Powerpoint Templates Page 28 The universally quantified statement for every positive integer n, if n is even, then n2 + n + 19 is prime is false. Quantifiers : Example 4
Free Powerpoint Templates Page 29 A counterexample is obtained by taking n = 38. The conditional proposition if 38 is even, then 382 + 38 + 19 is prime is false because the hypothesis 38 is even is true, but the conclusion 382 + 38 + 19 is prime is false. 382 + 38 + 19 is not prime since it can be factored: 382 + 38 + 19 = 38  38 + 38 + 19 = 19(2  38 + 2 + 1) = 19  79. Quantifiers : Example 4
Free Powerpoint Templates Page 30 The existentially quantified statement is true because it is possible to find at least one real number x for which the proposition is true. 5 2 1 , 2   x x x number real some for 5 2 1 2   x x Quantifiers : Example 5
Free Powerpoint Templates Page 31 For example, if x = 2, we obtain the true proposition It is not the case that every value of x results in a true proposition. For example, the proposition is false. 5 2 1 2 2 2   5 2 1 1 1 2   Quantifiers : Example 5
Free Powerpoint Templates Page 32 • The existentially quantified statement for some positive integer n, if n is prime, then n + 1, n + 2, n + 3, and n + 4 are not prime. is true because we can find at least one integer n that makes this conditional proposition true. • For example, if n = 23, we obtain the true proposition if 23 is prime, then 24, 25, 26, and 27 are not prime. Quantifiers : Example 6
Free Powerpoint Templates Page 33 This conditional proposition is true because both the hypothesis “23 is prime" and the conclusion "24, 25, 26, and 27 are not prime" are true. Quantifiers : Example 6
Free Powerpoint Templates Page 34 Some values of n make the conditional proposition for some positive integer n, if n is prime, then n + 1, n + 2, n + 3, and n + 4 are not prime true (e.g., n = 23, n = 47), while others make it false (e.g., n = 2, n = 101). Quantifiers : Example 6
Free Powerpoint Templates Page 35 The point is that we found one value that makes the proposition for some positive integer n, if n is prime, then n + 1, n + 2, n + 3, and n + 4 are not prime true which makes the given existentially quantified statement true. Quantifiers : Example 6
Free Powerpoint Templates Page 36 Generalized De Morgan Laws for Logic Theorem: If P is a propositional function, each pair of propositions in (a) and (b) has the same truth values (i.e., either both are true or both are false). ) ( , ); ( , ) ( x P x x P x a   ) ( , ); ( , ) ( x P x x P x b  
Free Powerpoint Templates Page 37 Generalized De Morgan Laws for Logic Proof of (a): - Suppose that the proposition is true. - Then the proposition x, P(x) is false. - By Definition above, the proposition x, P(x) is false precisely when P(x) is false for at least one x in the domain of discourse. ) ( , x P x 
Free Powerpoint Templates Page 38 Generalized De Morgan Laws for Logic Proof of (a): - But if P(x) is false for at least one x in the domain of discourse, is true for at least one x in the domain of discourse. - Again by definition of quantifiers, when is true for at least one x in the domain of discourse, the proposition x, is true. --- Thus if the proposition is true, the proposition x, is true. ) (x P ) (x P ) (x P ) ( , x P x  ) (x P
Free Powerpoint Templates Page 39 Generalized De Morgan Laws for Logic : Example 7  Verify that the existentially quantified statement is false.  Solution. We must show that is false for every real number x. Now is false precisely when is true. Thus, we must show that is true for every real number x. 1 1 1 2   x 1 1 1 , . 2   x x no real some for 1 1 1 2   x 1 1 1 2   x 1 1 1 2   x
Free Powerpoint Templates Page 40  Let x be any real number whatsoever. Since 0  x2, we may add 1 to both sides of this inequality to obtain 1  x2 + 1. If we divide both sides of this last inequality by x2 + 1, we obtain 1 1 1 2   x Generalized De Morgan Laws for Logic : Example 7
Free Powerpoint Templates Page 41  Therefore the statement is true for every real number x.  Thus is false for every real number x. We have shown that the existentially quantified statement is false. 1 1 1 2   x 1 1 1 2   x 1 1 1 , . 2   x x no real some for Generalized De Morgan Laws for Logic : Example 7
Free Powerpoint Templates Page 42  The example just given illustrates the use of the Generalized de Morgan’s Law for Logic.  In the example, it was shown that an existentially quantified statement for some real number x, P(x) can be proven false by proving that a related universally quantified statement for every real number x, ~P(x) is true. Generalized De Morgan Laws for Logic : Example 7
Free Powerpoint Templates Page 43  A universally quantified proposition generalizes the compound proposition P1  P2  . . .  Pn (1) in the sense that the individual propositions P1 , P2 , . . . , Pn are replaced by an arbitrary family P(x), where x is a member of the domain of discourse, and (1) is replaced by for every x, P(x) (2) Generalized De Morgan Laws for Logic : Example 7
Free Powerpoint Templates Page 44  The proposition (1) is true if and only if Pi is true for every i = 1, . . . , n.  The truth value of proposition (2) is defined similarly, that is (2) is true if and only if P(x) is true for every x in the domain of discourse. Generalized De Morgan Laws for Logic : Example 7
Free Powerpoint Templates Page 45  Similarly, an existentially quantified proposition generalizes the compound proposition P1  P2  . . .  Pn (3) in the sense that the individual propositions P1  P2  . . .  Pn are replaced by an arbitrary family P(x), where x is a member of the domain of discourse, and (3) is replaced by for some x, P(x). Generalized De Morgan Laws for Logic : Example 7
Free Powerpoint Templates Page 46  The preceding observations explain how the above theorem generalizes De Morgan's laws for logic (Law 2.5). Recall that the first De Morgan law for logic states that the propositions and have the same truth values.  In the above theorem part (b), is replaced by and is replaced by n P P P    ... 2 1 n P P P    ... 2 1 n P P P    ... 2 1 ) ( , x P x  n P P P    ... 2 1 . ) ( , x P x  Generalized De Morgan Laws for Logic : Example 7
Free Powerpoint Templates Page 47 Statements in words often have more than one possible interpretation. Consider the well-known quotation from Shakespeare All that glitters is not gold. One possible interpretation of this quotation is: Nothing that glitters is gold (i.e., a gold object never glitters). GDML : Example 8
Free Powerpoint Templates Page 48 However, this is surely not what Shakespeare intended. The correct interpretation is: Not all that glitters is gold or Something that glitters is not gold. GDML : Example 8
Free Powerpoint Templates Page 49 If we let P(x) be the propositional function "x glitters" and Q(x) be the propositional function "x is gold," the first interpretation becomes for all x, and the second interpretation becomes for all x, ) ( ) ( x Q x P  ) ( ) ( x Q x P  GDML : Example 8
Free Powerpoint Templates Page 50 From Chapter 2 [Law 2.4], we have seen that p  q  ~p  q  ~(p  q)  ~(~p  q)  ~(p  q)  p  ~q. Hence, the truth values of the propositional function for some x, and for some x, are the same. ) ( ) ( x Q x P    ) (x Q x P  GDML : Example 8
Free Powerpoint Templates Page 51 Using the Generalized De Morgan Laws for Logic, the truth values of for some x, and are the same. Thus an equivalent way to represent the second interpretation is .   ) (x Q x P  ) ( ) ( , x Q x P x all for  ) ( ) ( , x Q x P x all for  GDML : Example 8
Free Powerpoint Templates Page 52 Comparing the symbolic representations of the two interpretations, we see that the ambiguity results from whether the negation applies to Q(x) (the first interpretation) or to the entire statement for all x, P(x) → Q(x) (the second interpretation). GDML : Example 8
Free Powerpoint Templates Page 53 A Generalization For positive statements, the words "any," "all," "each," and "every" have the same meaning.
Free Powerpoint Templates Page 54 A Generalization In negative statements, the situation changes. The statements Not all C1 is C2 Not each C1 is C2 Not every C1 is C2 are considered to have the same meaning as Some C1 is not C2. whereas Not any C1 is C2 means No C1 is C2 .
Free Powerpoint Templates Page 55 Suppose that the domain of discourse is the set of real numbers. Consider the statement for every x, for some y, x + y = 0. The meaning of this statement is that for any x whatsoever, there is at least one y, which may depend on the choice of x, such that x + y = 0. We can show that the statement for every x, for some y, x + y = 0 is true. GDML : Example 9
Free Powerpoint Templates Page 56 For any x, we can find at least one y, namely y = - x, such that x + y = 0 is true. Suppose that we revise the above statement to read for some y, for every x, x + y = 0. If this statement is true, then it is possible to select some value of y such that the statement for every x, x + y = 0 is true. GDML : Example 9
Free Powerpoint Templates Page 57 However, we can disprove this last statement with a counterexample. For example, we might take x = 1 – y. We then obtain the false statement 1 – y + y = 0. Therefore the statement for some y, for every x, x + y = 0 is false. GDML : Example 9
Free Powerpoint Templates Page 58 Let P(x, y) be the statement if x2 < y2, then x < y. This statement has its domain of discourse the set of real numbers. Now, the statement for every x, for every y, P(x, y) is false. A counterexample is x = 1, y = -2. In this case, we obtain the false proposition if 12 < (-2) 2, then 1 < -2. GDML : Example 10
Free Powerpoint Templates Page 59 The statement for every x, for some y, P(x, y) is true. To prove this, we need to show that for every x, the proposition for some y, if x2 < y2, then x < y is true by exhibiting a value of y for which if x2 < y2, then x < y is true. GDML : Example 10
Free Powerpoint Templates Page 60 Indeed, if we set y = 0, we obtain the true proposition if x2 < 0, then x < 0. The conditional proposition is true, because the hypothesis x2 < 0 is false. GDML : Example 10
Free Powerpoint Templates Page 61 The statement for every y, for some x, P(x, y) is true. To justify this, we show that for every y, the proposition for some x, if x2 < y2, then x < y is true by exhibiting a value of x for which if x2 < y2, then x < y is true. GDML : Example 10
Free Powerpoint Templates Page 62 Indeed, if we set x = |y| + 1, we obtain the true proposition if (|y| + 1)2 < y2, then |y| + 1 < y. The conditional proposition is true, because the hypothesis is false. GDML : Example 10
Free Powerpoint Templates Page 63 Rule 1: Proving / Disproving quantified statements To prove that the universally quantified statement for every x, P(x) is true, show that for every x in the domain of discourse, the proposition P(x) is true. Showing that P(x) is true for a particular value x does not prove that for every x, P(x) is true.
Free Powerpoint Templates Page 64 To prove that the existentially quantified statement for some x, P(x) is true, find one value of x in the domain of discourse for which P(x) is true. One value suffices. Rule 2: Proving / Disproving quantified statements
Free Powerpoint Templates Page 65 Rule 3 : Proving/Disproving quantified statements To prove that the universally quantified statement for every x, P(x) is false, find one value of x (a counterexample) in the domain of discourse for which P(x) is false.
Free Powerpoint Templates Page 66 Rule 4: Proving/Disproving quantified statements To prove that the existentially quantified statement for some x, P(x) is false, show that for every x in the domain of discourse, the proposition P(x) is false. Showing that P(x) is false for a particular value of x does not prove that for some x, P(x) is false.
Free Powerpoint Templates Page 69 Mathematical System  A mathematical system consists of axioms, definitions, and undefined terms.  Axioms are assumed true.  Definitions are used to create new concepts in terms of existing ones .  Theorem is a proposition that has been proved to be true. Special kinds of theorems are referred to as lemmas and corollaries.
Free Powerpoint Templates Page 70  Special kinds of theorems are referred to as lemmas and corollaries. -- A lemma is a theorem that is usually not too interesting in its own right but is useful in proving another theorem. -- A corollary is a theorem that follows quickly from another theorem.  Proof is an argument that establishes the truth of a theorem. Mathematical System
Free Powerpoint Templates Page 71 Example 1  Euclidean geometry is an example of a mathematical system. Among the axioms are: 1. Given two distinct points, there is exactly one line that contains them. 2. Given a line and a point not on the line, there is exactly one line parallel to the line through the point.
Free Powerpoint Templates Page 72  The terms point and line are undefined terms that are implicitly defined by the axioms that describe their properties. Example 1
Free Powerpoint Templates Page 73 Among the definitions in this system are:  Two triangles are congruent if their vertices can be paired so that the corresponding sides and corresponding angles are equal.  Two angles are supplementary if the sum of their measures is 180°.
Free Powerpoint Templates Page 74 • If two sides of a triangle are equal, then the angles opposite them are equal. • If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. Examples of theorems in this system are:
Free Powerpoint Templates Page 75 An example of a corollary in Euclidean geometry is  If a triangle is equilateral, then it is equiangular. -- This corollary follows immediately from the first theorem given above for this example.
Free Powerpoint Templates Page 77 Direct Proof Theorems are often of the form For all x1, x2, . . . , xn, if p(x1, x2, . . . ,xn), then q(x1, x2, . . . , xn). This universally quantified statement is true provided that the conditional proposition if p(x1, x2, . . . , xn), then q(x1, x2, . . ., xn) (*)
Free Powerpoint Templates Page 78 is true for all x1, x2, . . . , xn in the domain of discourse. To prove (*), we assume that x1, x2, . . . , xn are arbitrary members of the domain of discourse. If p(x1, x2, . . . , xn) is false, by definition of the truth value for implication, (*) is true. Thus, we need only consider the case that p(x1, x2, . . . , xn) is true. Direct Proof
Free Powerpoint Templates Page 79 A direct proof assumes that p(x1, x2, . . . , xn) is true and then, using p(x1, x2, . . . , xn) as well as other axioms, definitions, and previously derived theorems, shows directly that q(x1, x2, . . . , xn) is true. Direct Proof
Free Powerpoint Templates Page 80 Example For all real numbers d, d1, d2, and x, prove that If d = min {d1, d2} and x  d, then x  d1 and x  d2 . Proof. Assume that d, d1, d2, and x are arbitrary real numbers. Then it suffices to assume that d = min { d1, d2} and x  d is true and then show that x  d1 and x  d2 is also true.
Free Powerpoint Templates Page 81 From the definition of min, it follows that d  d1 and d  d2. From x  d and d  d1, we may derive x  d1 from a theorem on real numbers (Transitive Property of Inequality). From x  d and d  d2, we may derive x  d2 for the same reason. Therefore, x  d1 and x  d2. Example
Free Powerpoint Templates Page 83 Proof by contradiction A proof by contradiction establishes the proposition if p(x1, x2, . . . , xn), then q(x1, x2, . . . , xn) (*) by assuming that the hypothesis p is true and that the conclusion q is false and then, using p and as well as other axioms, definitions, and previously derived theorems, derives a contradiction.
Free Powerpoint Templates Page 84 Contradiction A contradiction is a proposition of the form r Λ ~ r (r may be any proposition whatever). A proof by contradiction is sometimes called an indirect proof since to establish (*) using proof by contradiction, one follows an indirect route: derive r , then conclude that (*) is true.      r
Free Powerpoint Templates Page 85 The only difference between the assumptions in a direct proof and a proof by contradiction is the negated conclusion. In a direct proof the negated conclusion is not assumed, whereas in a proof by contradiction the negated conclusion is assumed. Contradiction
Free Powerpoint Templates Page 86 Proof by contradiction may be justified by noting that the propositions and are equivalent. The equivalence is immediate from the following truth table. q p  r r q p    Contradiction
Free Powerpoint Templates Page 87 p q r p ~ q r  ~ r T T T T F F T T T F T F F T T F T F T F F T F F F T F F F T T T F F T F T F T F F T F F T T F F T F F F T F F T q p  r r q p    Contradiction
Free Powerpoint Templates Page 88 Example Prove, by contradiction, the following statement: For all real numbers x and y, if x + y  2, then either x  1 or y  1. Proof. Suppose that the conclusion is false. Then, x < 1 and y < 1. (Remember that negating an or results in an and (De Morgan’s laws for logic). From the properties of inequality in Algebra, we may add these inequalities to obtain x + y < 2.
Free Powerpoint Templates Page 89 At this point, we have derived the contradiction p , where p: x + y  2. Thus we conclude that the statement is true. Suppose that we give a proof by contradiction of (*) in which, as in the above example, we deduce . In effect, we have proved. (1.4.2) This special case of proof by contradiction is called proof by contrapositive. p q  Example
Free Powerpoint Templates Page 91 Deductive Reasoning In constructing a proof, we must be sure that the arguments used are valid. In this section we make precise the concept of a valid argument and explore this concept in some detail.
Free Powerpoint Templates Page 92 Deductive Reasoning  Consider the following sequence of propositions. The bug is either in module 17 or in module 81. The bug is a numerical error. Module 81 has no numerical error. (1)  Assuming these statements are true, it is reasonable to conclude: The bug is in module 17. (2)
Free Powerpoint Templates Page 93  This process of drawing a conclusion from a sequence of proposition is called deductive reasoning.  The given propositions, such as (1), are called hypotheses or premises.  The proposition that follows from the hypotheses, like (2), is called the conclusion.  A (deductive) argument consists of hypotheses together with a conclusion. Many proofs in mathematics and computer science are deductive arguments. Deductive Reasoning
Free Powerpoint Templates Page 94  Any argument has the form If p1 and p2 and . . . and pn, then q. (3)  Argument (3) is said to be valid if the conclusion follows from the hypotheses: that is, if p1 and p2 and . . . and pn are true, then q must also be true. This discussion motivates the following definition. Deductive Reasoning
Free Powerpoint Templates Page 95  An argument is a sequence of propositions written p1 p2 : pn or p1 and p2 , . . . , pn / q  Deductive Reasoning
Free Powerpoint Templates Page 96  The propositions p1 and p2 , . . . , pn are called the hypotheses (or premises) and the proposition q is called the conclusion.  The argument is valid provided that if p1 and p2 and p3. . . and pn are all true, then q must also be true; otherwise, the argument is invalid (or a fallacy). Deductive Reasoning
Free Powerpoint Templates Page 97  In a valid argument, we sometimes say that the conclusion follows from the hypotheses. Notice that we are not saying that the conclusion is true; we are only saying that if you grant the hypotheses, you must also grant the conclusion. An argument is valid because of its form, not because of its content. Deductive Reasoning
Free Powerpoint Templates Page 98  Determine whether the argument is valid. q p  p q  Example 1
Free Powerpoint Templates Page 99 Example 1  [First solution.] We construct a truth table for all the propositions involved: p q p q T T T T T T F F T F F T T F T F F T F F q p 
Free Powerpoint Templates Page 100 We observe that whenever the hypotheses p q and p are true, the conclusion q is also true; therefore, the argument is valid.  [Second solution.] We can avoid writing the truth table by directly verifying that whenever the hypotheses are true, the conclusion is also true. Suppose that p q and p are true. Then q must be true, for otherwise p q would be false. Therefore, the argument is valid. Example 1
Free Powerpoint Templates Page 101 Example 2 Represent the argument If 2 = 3, then I ate my hat. I ate my hat. 2 = 3 symbolically and determine whether the argument is valid. 
Free Powerpoint Templates Page 102 Solution. If we let p: 2 =3, q: I ate my hat. the argument may be written q p  q p  Example 2
Free Powerpoint Templates Page 103 If the argument is valid, then whenever and q are both true, p must also be true. Suppose that and q are true. This is possible if p is false and q is true. In this case, p is not true; thus the argument is invalid. q p  q p  Example 2
Free Powerpoint Templates Page 104 We can also determine the validity of the argument in this example by examining the truth table of Example 1. In the third row of the table, the hypotheses are true and the conclusion is false; thus the argument is invalid. Example 2
Free Powerpoint Templates Page 105 If arguments contain more than two or three different simple statement as components, it is cumbersome and tedious to use truth tables to test their validity. A more convenient method of establishing the validity of some arguments is to construct a formal proof of validity. Example 2
Free Powerpoint Templates Page 106 Formal proof of validity A formal proof of validity for a given argument is defined to be a sequence of statements, each of which is either a premise of that argument or follows from preceding statements by an elementary valid argument, and such that the last statement in the sequence is the conclusion of the argument whose validity is being proved. This definition must be completed and made definite by specifying what is to count as an elementary valid argument.
Free Powerpoint Templates Page 107 Elementary valid argument An elementary valid argument is any argument that is a substitution instance of an elementary valid argument form.
Free Powerpoint Templates Page 108 Rules of Inference The following list of argument forms are regarded as elementary valid argument forms and are considered as the Rules of Inference. 1. Modus Ponens (M.P.) q p  p q 
Free Powerpoint Templates Page 109 2. Modus Tollens (M.T.) q p  q ~ p ~  3. Hypothetical Syllogism (H.S.) q p  r q  r p   Rules of Inference
Free Powerpoint Templates Page 110 4. Disjunctive Syllogism (D.S.) q p  p ~ q  5. Conditional Proof r  Rules of Inference q p    r q p  
Free Powerpoint Templates Page 111 6. Proof by Cases Rules of Inference r p  r q    r q p    7. Destructive Dilemma (D.D.) ) ( ) ( s r q p    s q ~ ~  r p ~ ~  
Free Powerpoint Templates Page 112 9. Simplification (Simp.) q p  p  Rules of Inference 8. Constructive Dilemma (C.D.) ) ( ) ( s r q p    r p  s q  
Free Powerpoint Templates Page 113 10. Conjunction (Conj.) q q p   p 11. Addition (Add.) p q p  Rules of Inference
Free Powerpoint Templates Page 114 These eleven Rules of Inference are elementary valid argument forms, whose validity is easily established by truth tables. They can be used to construct formal proofs of validity for a wide range of more complicated arguments. The names listed are standard for the most part, and the use of their abbreviations permits formal proofs to be set down with a minimum of writing. Rules of Inference
Free Powerpoint Templates Page 115 Example 3 Establish the validity of the following arguments: Either the Attorney General has imposed a strict censorship or if Black mailed the letter then Davis received a warning. If our lines of communication have not broken down completely, then if Davis received a warning then Emory was informed about the matter.
Free Powerpoint Templates Page 116 If the Attorney General has imposed a strict censorship, then our lines of communication have broken down completely. Our lines of communication have not broken down completely. Therefore, if Black mailed the letter, then Emory was informed about the matter. Example 3
Free Powerpoint Templates Page 117 Example 3 Solution: Symbolically, the above arguments may be written as: C ~ ) ( ~ E D C   C A  ) ( D B A   E B  
Free Powerpoint Templates Page 118 Example 3 where A: Attorney General imposes a strict censorship. B: Black mails the letter. C: Our lines of communication breaks down completely. D: Davis receives a warning. E: Emory was informed about the matter.
Free Powerpoint Templates Page 119 To establish the validity of this argument by means of a truth table would require a table with thirty-two rows. We can prove the given argument valid, however, by deducing its conclusion from its premises by a sequence of just four arguments whose validity has already been established. Example 3
Free Powerpoint Templates Page 120 o From the third and fourth premises, A  C and ~ C, we validly infer ~ A by Modus Tollens. o From ~ A and the first premises, A v (B  D), we validly infer B  D by a Disjunctive Syllogism. Example 3
Free Powerpoint Templates Page 121 o From the second and fourth premises, ~ C  (D  E) and ~ C, we validly infer D  E by Modus Ponens. o And finally, from these last two conclusions (or subconclusions), B  D and D  E, we validly infer B  E by a Hypothetical Syllogism. Example 3
Free Powerpoint Templates Page 122 o That its conclusion can be deduced from its premises using valid arguments exclusively, proves the original argument to be valid. o Here the elementary valid argument forms Modus Ponens (M.P.), Modus Tollens (M.T.), Disjunctive Syllogism (D.S.), and Hypothetical Syllogism (H.S.) are used as Rules of Inference by which conclusions are validly deduced from premises. Example 3
Free Powerpoint Templates Page 123 o A more formal and more concise way of writing out this proof of validity is to list the premises and the statements deduced from them in one column, with “justifications” for the latter written beside them. In each case the “justifications” for a statement specifies the preceding statements from which, and the Rule of Inference by which, the statements in question was deduced. It is convenient to put the conclusion to the right of the last premise, separated from it by a slanting line which automatically marks all of the statements above it to be premises.
Free Powerpoint Templates Page 124 The formal proof of validity for the given argument can be written as follows: 1.A v (B  D) 2. ~ C  (D  E) 3. A  C 4. ~ C /B  E 5. ~ A 3, 4, M.T. 6. B  D 1, 5, D.S. 7. D  E 2, 4, M.P. 8. B  E 6, 7, H.S. Example 3
Free Powerpoint Templates Page 125 One matter that needs to be emphasized here is that any substitution instance of an elementary valid argument form is an elementary valid argument. Thus the argument ~ C  (D  E) ~ C  D  E Example 3
Free Powerpoint Templates Page 126 is an elementary valid argument because it is a substitution instance of the elementary valid argument form Modus Ponens (M.P.). It results from p  q p  q Example 3
Free Powerpoint Templates Page 127 by substituting ~C for p and D  E for q; therefore, it is of that form even though Modus Ponens is not the specific form of the given argument. Example 3
Free Powerpoint Templates Page 128 Rule of Replacement There are many valid truth-functional arguments that cannot be proved valid using only the nine Rules of Inference. However, if we can replace any part of a compound statement by an expression that is logically equivalent to the part replaced, the truth value of the resulting statement is the same as that of the original statement. This is sometimes called the Rule of Replacement or the Principle of Extensionality.
Free Powerpoint Templates Page 129 Using this rule we can employ the laws of logic that we have seen in Chapter 2 to replace all or part of a statement with a logically equivalent statement without changing the truth value of the original statement. Rule of Replacement
Free Powerpoint Templates Page 130 Let us recall some of the commonly used laws and introduce a few new ones. We also write down their names and abbreviations for ease in writing justifications for our proofs. These logically equivalent expressions can replace each other wherever they occur. We number them consecutively after the first eleven rules already stated. Rule of Replacement
Free Powerpoint Templates Page 131 12. De Morgan’s Laws (De M.) ~(p  q)  (~p  ~q) ~(p  q)  (~p  ~q) 13. Commutation (Com.) p  q  q  p p  q  q  p Rule of Replacement
Free Powerpoint Templates Page 132 14. Association (Assoc.) p  (q  r)  (p  q)  r p  (q  r)  (p  q)  r 15. Distribution (Dist.) p  (q  r)  (p  q)  (p  r) p  (q  r)  (p  q)  (p  r) Rule of Replacement
Free Powerpoint Templates Page 133 16. Double Negation (D.N.) p  ~~p 17. Transportation (Trans.) p  q  ~q  ~p 18. Material Implication (Impl.) p  q  ~p  q Rule of Replacement
Free Powerpoint Templates Page 134 19. Material Equivalence (Equiv.) p  q  (p  q)  (q  p) p  q  (p  q)  (~p  ~q) 20. Exportation (Exp.) (p  q)  r  p  (q  r) 21. Tautology (Taut.) p  p  p p  p  p Rule of Replacement
Free Powerpoint Templates Page 135 Example 4 Example 3 can also be solved using rule of replacement for some steps as follows: 1. A v (B  D) 2. ~ C  (D  E) 3. A  C 4. ~ C /B  E 5. ~ A 3, 4, M.T. 6. B  D 1, 5, D.S. 7. ~~ C  (D  E) 2, Impl. 8. D  E 7, 4, D.S. 9. B  E 6, 8, H.S.
Free Powerpoint Templates Page 137 Axioms of Peano Arithmetic 1 is a natural number If x is a natural number, then x + 1 is also a natural number There is no natural number, Z, such that Z + 1 = 0. Given natural numbers x and y, if x + 1 = y + 1, then x = y. If one can prove that if a property holds for some natural number x then it holds for x + 1, and if it can further be proven that the same property holds for 1, then the property holds for all natural numbers.
Free Powerpoint Templates Page 138 Principle of Mathematical Induction Suppose that for each positive integer n we have a statement S(n) that is either true or false. Suppose that S(1) is true; and (1) if S(i) is true, for all i  n, then S(n + 1) is true. (2) Then S(n) is true for every positive integer n. Condition (1) is sometimes called the Basis Step and condition (2) is sometimes called the Inductive Step.
Free Powerpoint Templates Page 139 Some Remarks on the Principle of Mathematical Induction The axiom allows us to prove that a certain property holds for all natural numbers. The idea is that, in order to prove that a formula (or theorem) about natural numbers is true for all natural numbers, we first verify (by actual substitution) that the formula is true for the number 1.
Free Powerpoint Templates Page 140 Then, we prove that whenever the formula is true for a natural number n, it must also be true for n + 1. Since it was verified to be true for 1, then it had to be true for 2; and since it was true for 2, it had to be true for 3, and so on for all natural numbers. Some Remarks on the Principle of Mathematical Induction
Free Powerpoint Templates Page 141 Principle of Mathematical Induction: Example 1 Prove by mathematical induction that Solution. Let Sn denote the sum of the first n positive integers Sn = =1 + 2 + 3 + . . . + n. Here, a sequence of statements is actually being made, namely, 2 ) 1 ( 1     n n i n i 2 ) 2 ( 1 1 1   S 3 2 ) 3 ( 2 2 1 2     S 6 2 ) 4 ( 3 3 2 1 3      S
Free Powerpoint Templates Page 142 The formula is true for the first 3 natural numbers. Assuming that all of the equations preceding the (n + 1)st equation are true, then, in particular, the nth equation is true, that is, We must show that the (n + 1)st equation is true. Take note that Sn+1 = 1 + 2, + . . . + n + (n + 1). 2 ) 1 ( ... 3 2 1        n n n Sn 2 ) 2 )( 1 ( 1     n n Sn Principle of Mathematical Induction: Example 1
Free Powerpoint Templates Page 143 Use mathematical induction to show that n!  2n-1 for n = 1, 2, . . . . Proof: Basis Step. We must show that the inequality is true if n = 1. This is easily accomplished, since 1! = 1  1 = 21-1. Inductive Step. We must show that if i !  2i-1 for i = 1, . . . , n, then (n + 1)!  2n. Principle of Mathematical Induction: Example 2
Free Powerpoint Templates Page 144 To show that if i !  2i-1 for i = 1, . . . , n, then (n + 1)!  2n we assume that i!  2i-1 for i = 1, . . . , n. Then, in particular, for i = n, we have n!  2n-1. We can relate the 2 inequalities above by observing that (n + 1)! = (n + 1)(n!). Now (n + 1)! = (n + 1)(n!)  (n + 1)2n-1  2 . 2n-1 since n + 1  2 = 2n. Principle of Mathematical Induction: Example 2
Free Powerpoint Templates Page 145 Since we have shown that if i !  2i-1 for i = 1, . . . , n, then (n + 1)!  2n therefore, n!  2n-1 for n = 1, 2, . . . . is true. We have completed the Inductive Step. Since the Basis Step and the Inductive Step have been verified, the Principle of Mathematical Induction tells us that n!  2n-1 is true for every positive integer n. Principle of Mathematical Induction: Example 2
Free Powerpoint Templates Page 146 Strong Form of Mathematical Induction The verification of the Inductive Step where we assume that S (i ) is true for all i < n + 1 and then prove that S (n + 1) is true. This formulation of mathematical induction is called strong form of mathematical induction. Often, as was the case in the preceding examples, we can deduce S (n + 1) assuming only S (n ). indeed, the Inductive Step is often stated: If S (n) is true, then S (n + 1) is true.
Free Powerpoint Templates Page 147 Use induction to show that if r  1, for n = 0, 1, . . . . The sum on the left is called a geometric sum. In a geometric sum, the ratio of consecutive terms (ar i + 1/ar i = r) is constant. Proof: Basis Step. The Basis Step, which in this case is obtained by setting n = 0, is which is true. 1 ) 1 ( ... 1 2 1         r r a ar ar ar a n n Principle of Mathematical Induction: Example 3
Free Powerpoint Templates Page 148 Inductive Step . Assume that statement is true for n. Now, 1 1 1 2 1 1 ) 1 ( ...             n n n n ar r r a ar ar ar ar a 1 ) 1 ( 1 ) 1 ( 1 1         r r ar r r a n n 1 ) 1 ( 2     r r a n Principle of Mathematical Induction: Example 3
Free Powerpoint Templates Page 149 Since the modified Basis Step and the Inductive step have been verified, the Principle of Mathematical Induction tells us that 1 ) 1 ( ... 1 2 1         r r a ar ar ar a n n is true for n = 0, 1, … Principle of Mathematical Induction: Example 3
Free Powerpoint Templates Page 150 As an example of the use of the geometric sum, if we take a = 1 and r = 2 in the above expression, we obtain the formula . 1 2 1 2 1 2 2 ... 2 2 2 1 1 1 3 2             n n n Principle of Mathematical Induction: Example 3
Free Powerpoint Templates Page 151 The reader has surely noticed that in order to prove the previous formulas one has to be given the correct formulas in advance. A reasonable question is: How does one come up with the formulas? There are many answers to this question. One technique to derive a formula is to experiment with small values and try to discover a pattern. For example, consider the sum 1 + 3 + . . . + (2n - 1). The following table gives the values of this sum for n = 1, 2, 3, 4. Principle of Mathematical Induction: Example 3
Free Powerpoint Templates Page 152 n 1 + 3 + . . . + (2n –1) 1 1 2 4 3 9 4 16 Since the second column consists of squares, we conjecture that 1 + 3 + . . . + (2n - 1) = n2 for every positive integer n. The conjecture is correct and the formula can be proved by mathematical induction (see Exercise Set 3.4 #1). Principle of Mathematical Induction: Example 3
Free Powerpoint Templates Page 153 Use induction to show that 5n - 1 is divisible by 4 for n = 1, 2, . . . . Proof: Basis Step. If n = 1, 5n - 1 = 51 - 1 = 4, which is divisible by 4. Note that also 4 = 50 - 1 + 4  50 If n = 2, 52 - 1 = 25 - 1 = 24, which is divisible by 4. Also 24 = 51 - 1 + 4  51 Principle of Mathematical Induction: Example 4
Free Powerpoint Templates Page 154 If n = 3, 53 - 1 = 125 - 1 = 124, which is divisible by 4. Again 124 = 52 - 1 + 4  52 : : If we look at the pattern, we can say that 5 n - 1 = 5 n-1 - 1 + 4  5 n-1. Principle of Mathematical Induction: Example 4
Free Powerpoint Templates Page 155 Inductive Step. Assume that 5n - 1 is divisible by 4 and that the above formula holds. We must show that 5n+1 - 1 is divisible by 4. To relate the (n + 1)st case to the nth case, we write 5 n+1 - 1 = (5 n - 1) + 4 . 5 n. Principle of Mathematical Induction: Example 4
Free Powerpoint Templates Page 156 By assumption, 5 n - 1 is divisible by 4 and, since 4 . 5 n is divisible by 4, the sum (5 n - 1) + 4. 5 n = 5 n+1 - 1 is divisible by 4. Since the Basis Step and the Inductive Step have been verified, the Principle of Mathematical Induction tells us that 5n - 1 is divisible by 4 for n = 1, 2, . . . Principle of Mathematical Induction: Example 4
Free Powerpoint Templates Page 157 For any set X with n elements, the cardinality of the power set of X has 2n elements, that is, n((X)) = 2n. Or if |X| = n, then |(X)| = 2n Proof. The proof is by induction on n. Basis Step. If n = 0, X is the empty set. The only subset of the empty set is the empty set itself; thus (X)  = 1 = 20 = 2n. Thus the formula is true for n = 0. Principle of Mathematical Induction: Example 5
Free Powerpoint Templates Page 158 Inductive Step. Assume that formula holds for n. Let X be a set with X= n + 1. Choose x  X. We divide (X) into two classes. The first class consists of those subsets of X that include x, and the second class consists of those subsets of X that do not include x. If we list the subsets of X that do not include x, X1, X2, . . . , Xk , then Principle of Mathematical Induction: Example 5
Free Powerpoint Templates Page 159 X1  {x}, X2  {x}, . . . , Xk  {x} is a list of the subsets of X that do include x. Thus the number of subsets of X that include x is equal to the number of subsets of X that do not include x. If Y = X – {x}, (Y) consists precisely of those subsets of X that do not include x. Thus (X)= 2(Y). Since Y = n, by the inductive assumption, (Y)= 2n. Principle of Mathematical Induction: Example 5
Free Powerpoint Templates Page 160 Therefore, (X) = 2(Y)= 2  2n = 2n+1. Thus the formula holds for n + 1 and the inductive step is complete. By the Principle of Mathematical Induction, Law 3.6.2 holds for all n  0. Principle of Mathematical Induction: Example 5

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