Unit I LOGIC AND PROOFS. B. Thilaka Applied Mathematics

Transcription

1 Unit I LOGIC AND PROOFS B. Thilaka Applied Mathematics

2 UNIT I LOGIC AND PROOFS Propositional Logic Propositional equivalences Predicates and Quantifiers Nested Quantifiers Rules of inference Introduction to proofs Proof methods and strategy

3 Outline Overview Propositional Logic Definitions, Examples, applications Predicate Logic Definitions, Examples, applications

4 Outline Overview Rules of inference Introduction to Proofs Proof Methods and strategy

5 Preliminaries Set Theory: Universal Set - E Null Set - φ A, B, C subsets of E A, B, C E A B x x A& x B A B x x A or x B A A x x A

6 Preliminaries De Morgan s Laws: B A B A B A B A Distributive Laws: C A B A C B A C A B A C B A

7 Preliminaries Difference: B A B A General formulae: B A&x x x A B B A B A E E A A E A A A A

8 Logic Definition Logic is concerned with all kinds of reasoning based upon a set of rules called rules of inference. Definition Basic units of the object language are called primitive/ primary/ simple or atomic statements.

9 Propositional Logic Definition A proposition is a declarative sentence that is either true or false, but not both. Exercise 1: 1. Canada is a city 2. If x is an integer, then x 2 is positive

10 Propositional Logic 3. Do your homework 4. Where are you going? 5. Inflation will fall next quarter.

11 Logic If the truth value of a proposition is true, it is denoted by T and if the truth value of a proposition is false, it is denoted by F.

12 Logical Operations - Connectives New statements are obtained from atomic statements through the use of connectives. The statements thus formed are called compound / composite statements.

13 Logical Operations - Connectives Examples: AND, OR, NOT, CONDITIONAL, BICONDITIONAL, NAND, NOR, NOR

14 Negation Negation of a statement is formed by using the word NOT It is denoted by / / ~ / / NOT Truth Table p T F p F T

15 Negation Example It will rain tomorrow; it will not rain tomorrow Note : Negation is used even for compound sentences Exercise 1 Write the negation of the following sentences 1. The river is shallow or polluted 2. Jack is tall and thin.

16 Negation 1. The river is shallow or polluted It is not true that the river is shallow or polluted 2. Jack is tall and thin. It is not true that Jack is tall and thin.

17 Conjunction Two statements p and q combined by the word and to form a compound statement is called the conjunction of the original statements and is denoted by p q (٨) Truth Table p q T T T T F F F T F F F F p q

18 Conjunction Example: Roses are red and jasmines are white Exercise 2 Classify the following and 1. Jack and Jill are cousins 2. He opened the book and started to read

19 Disjunction Two statements p and q combined by the word or to form a compound statement is called the disjunction of the original statements p q denoted by (ν) p q and is p q Truth Table T T T T F T F T T F F F

20 Disjunction Exercise 3 Classify the or in the following statements 1. There is something wrong with the bulb or with the wiring. 2. Ram will watch television or go to the beach 3. Either 20 or 30 persons were injured in the landslide.

21 Exclusive OR Exclusive OR of two statements p and q is denoted by p q is the proposition that is true when exactly one p or q is true and is false otherwise. Truth Table p q p q T T F T F T F T T F F F

22 Conditional The conditional of two statements p and q if p then q denoted by p q means (a) q is necessary for p (b) p is sufficient for q (c) q if p (d) p only if q (e) q unless not p (f) p implies q.

23 Conditional Truth Table p q p q T T T T F F F T T F F T

24 Conditional Examples 1. If he studies well, he will get good marks 2. If there is a flood, the crop will be destroyed.

25 Conditional Note : The statement p is called the antecedent / hypothesis / premise, while the statement q is called the consequent / conclusion.

26 Conditional Note : 1. p q is the same as p q. 2. For the conditional p q, q p is called p q is called q p is called converse inverse contrapositive (p q) is called negation

27 Conditional Truth Table for converse p q q p T T T T F T F T F F F T

28 Truth Table for inverse Conditional p q p q T T T T F T F T F F F T

29 Conditional Truth Table for contrapositive p q q p T T T T F F F T T F F T

30 Conditional Truth Table for negation p q (p q) T T F T F T F T F F F F

31 Conditional Exercise 4 Write the converse, inverse, contrapositive and negation of (a) If it is cold, he wears a hat (b) If she works, she will earn money.

32 Conditional (a) If it is cold, he wears a hat Converse: If he wears a hat, it is cold. Inverse: If it is not cold, he does not wear a hat. Contrapositive: If he does not wear a hat, it is not cold. Negation: It is false that if he wears a hat, it is cold.

33 Conditional (b) If she works, she will earn money. Converse: If she earns money, then she will work. Inverse: If she does not earn money, she will not work. Contrapositive: If she does not earn money, she will not work. Negation: She works and she does not earn money.

34 Biconditional If p and q are any two statements, then the statement p q (p iff q) is called a biconditional statement and is equivalent to (p q) Λ ( q p )

35 Biconditional Truth Table p q p q T T T T F F F T F F F T

36 Connectives Note : The connective NOT ( ) is called a unary connective because it acts on one expression to produce a second expression. The connectives Λ,,, are called binary connectives, since they operate on two expressions to produce a third expression.

37 Connectives Exercise 5 Write the negation of He swims iff the water is warm It is not the case that he swims iff the water is warm It is not false that he swims iff the water is warm The water is warm and he does not swim or he swims and the water is not warm.

38 Connectives Order of precedence 1. Connectives within parenthesis, innermost parenthesis first Λ,V 4. 5.

39 Connectives English Word Logical Connective Not, It is not true that, It is false that, Negation Or Disjunction ν Logical expression

40 Connectives English Word Logical Connective And, but, In addition, Moreover, Also Conjunction Logical expression Λ

41 Connectives English Word If p then q, p if only q, p is sufficient for q, q is necessary for p, q if p, p implies q, q unless p Logical Connective Conditional Logical expression p q

42 Connectives English Word p if and only if q, p is necessary and sufficient for q Logical Connective Logical expression Biconditional p q

43 Connectives Problems 1. Using the statement p: Mark is happy, q: Mark is rich, write the following statements in symbolic form. a) Mark is poor but happy b) Mark is rich or unhappy q Λ p q ν p c) Mark is poor or he is both rich and unhappy d) Mark is neither rich nor happy (p ν q) q ν (q Λ p)

44 Connectives Problems 2. While appropriately denoting the simple statements, write the following statement in symbolic form. It is not the case that houses are cold and haunted and it is false that cottages are warm or houses ugly.

45 Connectives p 1 : houses are cold p 2 : houses are haunted p 3 : cottages are warm p 4 : houses are ugly (p 1 ν p 2 ) Λ (p 3 ν p 4 ) Ξ p 1 Λ p 2 Λ p 3 Λ p 4

46 Connectives Problems 3. Given the truth values of p and q as T and the truth value of r as F, find the truth value of a) p ν (q Λ r) (T) b) [p Λ (q Λ r) ] ν [ (p ν q) Λ (p Λ r)] (F)

47 Well Formed Formulae A statement formula is an expression which is a string consisting of variables (propositions) parenthesis and connective symbols. An expression that is a legitimate string is called a well formed formula (wff).

48 Well Formed Formulae Examples: p, p, p ν q, p q p Λ q is not a well formed formula.

49 Connectives Problems 4. Construct the truth table for a) ( p Λ q) b) p Λ ( p ν q) c) [ q Λ ( p q) ] p d) ( p ν q) HW e) ( p ν q) ( p ν q) f) ( p Λ q) p ν q HW Discussion

50 Connectives Problems g) p Λ ( q Λ p) HW h) [ p Λ ( p q) q HW i) ( p ν q) ν p Discussion

51 Connectives Solution : The truth tables are given by a) ( p Λ q) p q q pλ q ( p Λ q) T T F F T T F T T F F T F F T F F T F T

52 Connectives Solution : The truth tables are given by b) p Λ ( p ν q) p q q p ν q p Λ (p ν q) T T F T T T F T T T F T F F F F F T T F

53 Connectives Solution : c) [ q Λ ( p q) ] p p q p q qλ(p q) [ q Λ ( p q) ] p T T T T T T F F F T F T T T F F F T F T

54 Substitution Instance A formula A is called a substitution instance of another formula B if A can be obtained from B by substituting formula for some variable of B wherever it occurs.

55 Substitution Instance Note : 1. More than one variable can be substituted by other formula. 2. Substitutions are made for atomic formula and not for compound formula.

56 Connectives Solution : 4 f) ( p Λ q) p ν q p q (p Λq) p ν q ( p Λ q) p ν q T T F F T T F T T T F T T T T F F T T T

57 Definition Tautologies and Contradictions A statement formula which is true regardless of the truth values of the statements in it is called a universally valid formula or a tautology or a logical truth Example : pv p is a tautology

58 Tautology Note : 1. Conjunction of two tautologies is a tautology 2. Disjunction of two tautologies is a tautology 3. A substitution instance of a tautology is a tautology

59 Definition Tautologies and Contradictions A statement formula which is false regardless of the truth values of the statements in it is called a contradiction Example : pλ p is a tautology Can you give a sentence as an example for a contradiction?

60 Contradiction Note : 1. The negation of a tautology is a contradiction 2. The negation of a contradiction is a tautology

61 Tautologies and Contradictions 1. Show that ( p q) { p ( q r)} ( p q) ( p r) is a tautology Solution: Consider

62 Tautologies and Contradictions ( p q) { p ( q r)} ( p q) ( p r) ( p q) {p ( q r)} ( p q) (p r) (De Morgan s Law) ( p q) {p (q r)} ( p q) (p r) (De Morgan s Law)

63 Tautologies and Contradictions ( p q) {(p q) (p r)} ( p q) (p r) ( Distributive Law) ( p q) (p ( p q) (p r) r) ( Associative and Idempotent Laws) (p q) (p r) T ( p q) (p r) ( De Morgan s Law)

64 Tautologies and Contradictions Hence the proof. 2. Show that the following expressions are tautologies a) p (p q) b) q (p q ( p q) c) p (p q) d) ( p q) (p q)

65 Equivalence of Formulae Let A and B be two statement formulae and let p 1, p 2,.., p n be the variables occurring in A and B. If the truth value of A is equal to the truth value of B for every one of the possible 2n set of truth values assigned to p 1, p 2,.., p n, then A and B are said to be logically equivalent / equivalent / equal and denoted by A B (A Ξ B)

66 Equivalence of Formulae Example: p q ( pvq) Can you give more examples? Note: Equivalence is symmetric, transitive

67 Basic Equivalence Formulae / Law of Algebra of Propositions No. Formula Law 1 p p p; p p p Idempotent 2 Commutative p q p q q p; q p 3 Associative ( p q) r p (q r); ( p q) r p (q r)

68 Basic Equivalence Formulae / Law of Algebra of Propositions No. Formula Law 4 Distributive p (q r) (p q) (p r); p (q r) (p q) (p r) 5 p F p; p F F Identity (Fcontradictory statement) p T 6 p T T; Identity (Ttautology) p

69 Basic Equivalence Formulae / Law of Algebra of Propositions No. Formula Law 7 Absorption p (p q) p; p (p q) p 8 ( p q) p q; De Morgan ( p q) p q 9 Involution/ p p Double Negation 10 Complement p p T; p p F

70 Equivalence of Formulae 1. Show that p q r q r p r r Solution: Consider

71 Equivalence of Formulae r p r q r q p p r q r ( Associative and Commutative Laws) r q p p) (q r ( De Morgan and Distributive Laws) r q p

72 Equivalence of Formulae p q r ( p q) r ( Commutative Law) r p q r (p q) ( Commutative Law) r p q p q r T ( Complement Law) ( Distributive Law) r ( Identity Law) Hence the proof.

73 Duality Law Definition: Two statements A and A* are said to be duals of each other if either one can be obtained from the other by replacing Λ by and by Λ

74 Duality Law The connectives and Λ of each other. Note : are called duals If the formula A contains special variables T or F then its dual A* is obtained by replacing T by F and F by T in addition to the above mentioned interchanges.

75 Duality Law Theorem: Let A and A* be dual formulae and let p 1, p 2,.., p n be all the atomic variables that occur in A and A*. Then, A (p 1, p 2,.., p n ) = A* ( p 1, p 2,.., p n ) Similarly, A ( p 1, p 2,.., p n ) = A* (p 1, p 2,.., p n ) ( Proof : De Morgan s Law)

76 Duality Law Theorem: Let p 1, p 2,.., p n be all the atomic variables that appear in A and B. Given that A is equivalent to B means that the biconditional AB is a tautology, then the following are also tautologies: Similarly, A ( p 1, p 2,.., p n ) = A* (p 1, p 2,.., p n ) ( Proof : De Morgan s Law)

77 Duality Law A ( p 1, p 2,.., p n ) B ( p 1, p 2,.., p n ) A* (p 1, p 2,.., p n ) B* (p 1, p 2,.., p n ) A* (p 1, p 2,.., p n ) B* (p 1, p 2,.., p n ) ( Proof : De Morgan s Law)

78 Replacement Process If we replace any part of the statement formula by any other formula that is equivalent to it, then the resultant formula is equivalent to the original formula. By this process, we obtain new formulae which are equivalent to the original formula.

79 Replacement Process Consequently, if we replace any part or parts of a tautology by formulae that are equivalent to these parts, we again obtain a tautology.

80 Tautological Implication A statement formula A is said to tautologically imply B iff A B is a tautology and is denoted by A B Note : If each of the formulae A and B imply the other, then A and B are equivalent

81 Tautological Implication Note : In order to show that show that A B, it is sufficient to i) An assignment of the truth value T to the antecedent A leads to the truth value T of the consequent B.

82 Tautological Implication (or) ii) An assignment of the truth value F to the consequent B leads to the truth value F of the antecedent A.

83 Tautological Implication 1. Show that ( p q) (p r) (q r) r Solution: Assume that ( p q) (p r) (q r) is T. We need to show that r is T. Since ( p q) (p r) (q r) is T, we have

84 Tautological Implication ( p q) is T, ( p r) is T and ( q r) is T. Let p be T. Then ( p r) is T implies that r is T. Let p be F. Then ( p q) is T implies that q is T. Further, ( q r) is T implies that r is T. Hence the proof.

85 Tautological Implication Exercise 6 Prove the following implications: 1. p q p,p q q Solution: Consider the Truth Table for p q p

86 Tautological Implication p q pλq (pλq)p T T T T T F F T F T F T Since (pλq)p is a tautology, p q p Hence the proof. F F F T

87 Tautological Implication Exercise 6 Prove the following implications: 2. p p q, q p q 3. p p q 4. q p q 5. ( p q) p

88 Tautological Implication 6. ( p q) q 7. p (p q) q 8. p (p q) q

89 Tautological Implication Remarks: 1. If a formula is equivalent to a tautology, then it must be a tautology 2. If a formula is implied by a tautology, then it is a tautology

90 Tautological Implication 3. Both implication and equivalence are transitive. 4. If 5. If A B and C A then A B C A B and B C then A C

91 Some Basic Tautological Implications No. Formula Law 1 Multiplication 2 Addition 3 p p q p p,p q, q q p p p q q q 4 q p q 5 ( p q) p

92 Some Basic Tautological Implications No. Formula 6 ( p q) q Law 7 p,q p q 8 p,p q q Disjunctive Syllogism 9 Modus p,p q q Ponens 10 q,p q p Modus Tollens

93 Some Basic Tautological Implications No. Formula Law 11 p q,q r p r Hypothetical Syllogism 12 Dilemma p q,p r,q r r

94 Equivalences Involving Conditionals 1. p q p q p q 2. p q p q 3. p q (p q) 4. (p q) p q 5. (p q) (p r) p (q r)

95 Equivalences Involving Conditionals 6. (p r) (q r) (p q) 7. (p q) (p r) p (q r r) 8. (p r) (q r) (p q) r

96 Equivalences Involving Biconditionals 1. p q (p q) q p 2. p q p q 3. p q (p q) ( p q) 4. (p q) p q

97 Tautological Implications and Equivalences Exercise 7 1. Prove the following tautological implications : (a) q (p q) p (b) (c) ( pq) p (p q) ( p (q r)) (p q) (p r)

98 Tautological Implications and Equivalences Solution: (a) Consider q (p q) p q (p q) Assume that is T. Hence q is T and p q) ( is T. In other words, q is F and p q) ( is T. Hence p is F. In other words, p is T.

99 Tautological Implications and Equivalences Hence the proof. (c) Consider ( p (q r)) (p q) (p r) Assume that ( p q) (p r) is F. Hence ( p q) ( p r) is T and is F.

100 Tautological Implications and Equivalences Now, ( p r) is F means that p is T and r is F. Also, p is T and p q is T means that q is T. Hence, we have p is T, q is T and r is F. Thus q r is F. Hence p ( q r ) is F. Hence the proof.

101 Tautological Implications and Equivalences 2. Prove the following equivalences: (a) p q p q p q (b) p (q p) p (p q) (c) p q p q p q

102 Tautological Implications and Equivalences Solution: (a) Consider Now, p q p q p q ( p q) (q p) p q (Definition) ( p q) ( q p) (Equivalence)

103 Tautological Implications and Equivalences ( p q) ( q p) ( De Morgan Law) ( p q) (q p) ( De Morgan Law) p (q p) q (q p) ( Distributive Law)

104 Tautological Implications and Equivalences ( p q) (p p) ( q q) ( q p) ( p q) T ( T ( q p) ( Distributive Law) ( p q) ( p q) ( Complement Law) ( De Morgan and Identity Laws) Hence the proof.

105 Tautological Implications and Equivalences Solution: (c) Consider p Now, q p q p q ( p q) (q p) p q (Definition) ( p q) ( q p) (Equivalence)

106 Tautological Implications and Equivalences ( p q) ( q p) ( p q) (q p) ( De Morgan Law) ( De Morgan Law) Hence the proof.

107 Theory of Inference The main function of logic is to provide rules of inference or principles of reasoning. The theory associated with such rules is known as inference theory because it is concerned with the inferring of a conclusion from certain premises.

108 Theory of Inference Arguments An argument is an assertion that a given set of propositions H 1, H 2,.., H n called premises yields another proposition C called conclusion.

109 Theory of Inference Definition If an implication A B is a tautology, where A and B are statement formulae, we say that B logically follows from A or B is a valid conclusion from the premise A

110 Theory of Inference In other words, we say that the conclusion C follows logically from a set of premises H 1, H 2,.., H n, that is {H 1, H 2,.., H n } C where H 1, H 2,.., H n are functions of the atomic propositions p 1, p 2,.., p m

111 Theory of Inference Truth Table Technique Given a set of premises H 1, H 2,.., H n and a conclusion C, it is possible to determine in a finite number of steps whether the conclusion C logically follows from the given premises by constructing the appropriate truth table.

112 Truth Table Technique Theory of Inference This method of determining whether the conclusion logically follows from the given premises by construction of the relevant truth table is called truth table technique.

113 Truth Table Technique Exercise 8 Using truth table technique, determine whether the conclusion C follows logically from the premises H 1 and H 2 for (a) :p q,h :p; H 2 1 C : q (b) :p q,h : p; H1 2 C : q

114 Truth Table Technique (c) H 2 1 :p q,h : (p q); C: p (d) H 2 1 : p,h :p q; C : (p q) (e) : p,h :q; H 2 1 C : p Solution: (a) Consider H 2 1 :p q,h :p; C : q

115 Truth Table Technique We construct the following Truth Table p (H 2 ) q (C) p q (H 1 ) T T T T F F F T T F F T

116 Truth Table Technique Since from the above table, we see that C takes the value T whenever both the hypotheses H 1 and H 2 are T and at least one of H 1 and H 2 is F whenever C is F, the given conclusion is a valid conclusion.

117 Truth Table Technique (b) Consider :p q,h : p; H1 2 We construct the following Truth Table C : q p q ( C) p q (H 1 ) p (H 2 ) T T T F T F F F F T T T F F T T

118 Truth Table Technique Since in the fourth row of the above table, both the hypotheses H 1 and H 2 are T and C is F, the given conclusion is not a valid conclusion.

119 Truth Table Technique Remark: If the number of atomic statements present in the formula representing the premises and the conclusion is large, the truth table technique is tedious.

120 Theory of Inference Valid Arguments When a conclusion is derived from a set of premises by accepted rules of reasoning/inference, then such a process of derivation is called a deduction or a formal proof and the argument is called a valid argument.

121 Theory of Inference Note: An argument which is not valid is called a fallacy Rules of Inference There are three rules of inference called rule P, rule T and rule CP.

122 Theory of Inference Rules of Inference 1. Rule P : A premise may be introduced at any point of the derivation. 2. Rule T : A formula S may be introduced in a derivation if S is tautologically implied by one or more of the preceding formulae in the derivation.

123 Rules of Inference Theory of Inference 3. Rule CP : If we can derive S from R and a set of premises P, then we may derive R S, from the set of premises P alone.

124 Theory of Inference Note: If the conclusion is a conditional of the form R S then we may add r as an additional premise and derive S alone.

125 Theory of Inference Rules of Inference Exercise 9 Determine the validity of the following arguments. 1. p q, q r, s p, r s ( ( p r) ( q s) s r 2. p q)

126 Theory of Inference 3. p q, p m, q r, m r (p q) 4. p (q r), q (r s) p (q s) p q,q r, 5. ( p r), p r r

127 Theory of Inference Solution: 1. Consider p q,q r, s p, r s (1) q r Rule P (2) r Rule P

128 Theory of Inference (3) q Rule T ( Disjunctive Syllogism of (1) and (2) ) ( p,p q (4) p q Rule P (5) p Rule T q) (Modus Tollens of (3) and (4) )

129 Theory of Inference (6) s p Rule P (7) s Rule T (Modus Tollens of (5) and (6) ) Hence the given argument is a valid argument.

130 Theory of Inference 2. Consider ( p q) (p r) (q s) s r (1) p q Rule P (2) p q Rule T (Equivalence of (1) ) (3) q s Rule P (4) p s Rule T ( Hypothetical Syllogism of (2) and (3) )

131 Theory of Inference (5) p r Rule P r p (6) Rule T (Equivalence of (5) ) (7) r s Rule T ( Hypothetical Syllogism of (4) and (6) ) (8) r s Rule T (Equivalence of (7) ) (9) s r Rule T (Commutative Law )

132 Theory of Inference Hence the given argument is a valid argument. 3. Consider p q, p m,q r, m r (p q) (1) p q Rule P (2) p q Rule T (Equivalence of (1) )

133 Theory of Inference (3) q r Rule P (4) p r Rule T ( Hypothetical Syllogism of (2) and (3) ) p m (5) Rule P (6) m p Rule T (Equivalence of (5) )

134 Theory of Inference (7) m r Rule T ( Hypothetical Syllogism of (4) and (6) ) (8) m Rule P (9) r Rule T (Modus Ponens of (7) and (8) ) Rule T (10) r (p q) (Combining (1) and (9) )

135 Theory of Inference Hence the given argument is a valid argument. 4. Consider p (q r),q (r s) p (q s) Since the RHS is a conditional p (q s) we add p as an additional premise and conclude q s (Rule CP).

136 Theory of Inference Therefore, the given argument reduces to p (q r),q (r s),p (q s) Again, since the RHS is a conditional q s we add q as an additional premise and conclude s alone (Rule CP).

137 Theory of Inference Hence, the given argument is equivalent to p (q r),q (r s),p, q s (1) p Rule P (2) q Rule P (3) p (q r) Rule P

138 Theory of Inference (4) q r Rule T (Modus Ponens of (7) and (3) ) (5) r Rule T (Modus Ponens of (2) and (4) ) (6) q (r s) Rule P (7) r s Rule T (Modus Ponens of (2) and (6) ) (8) s Rule T (Modus Ponens of (5) and (7) )

139 Theory of Inference Hence the given argument is a valid argument. 5. Consider p q,q r, (p q), p r r p q (1) Rule P (2) q r Rule P

140 Theory of Inference (3) p r Rule T ( Hypothetical Syllogism of (1) and (2) ) (4) ( p r) Rule P (5) p r Rule T ( De Morgan s Law ) (6) p r Rule T (Equivalence of (3) )

141 Theory of Inference (7) (8) p r Rule P ( p r) ( p r) Rule T (Combining (6) and (7) ) (9) ( p p) r Rule T ( Distributive Law ) (10) F r Rule T (Complement Law ) (11) r Rule T (Identity Law )

142 Theory of Inference (12) r ( p r) Rule T (Combining (5) and (11) ) (13) ( r p) (r r) Rule T ( Distributive Law ) (14) (15) ( r p) F Rule T (Complement Law ) r p Rule T ( Identity Law )

143 Theory of Inference (16) r Rule T (Multiplication) Hence the given argument is a valid argument. Problems: Determine whether the following arguments are valid. 1. q, p q p

144 Theory of Inference 2. p q, q r, p r 3. c d, ( c d) h, h (a b), ( a b) (r s) r s 4. ( p q) (r s), ( t u), p r ( q t) (s p u),

145 Theory of Inference 5. p q, q r, r s p s 6. p (q s), r p, q r s

146 Theory of Inference Definition: A set of formulae H 1, H 2,.., H n is said to be inconsistent if their conjunction implies a contradiction. That is, H 1, H 2,.., H n R R for some formula R

147 Theory of Inference Note: A set of formulae H 1, H 2,.., H n is said to be be consistent if it is not inconsistent.

148 Indirect Method of Proof Proof by Contradiction This technique comprises the following steps: 1. Introduce the negation of the desired conclusion as a new premise. 2. From the new premise, together with the given premises, derive a contradiction.

149 Indirect Method of Proof Proof by Contradiction 3. Assert the desired conclusion as a logical inference from the premises

150 Indirect Method of Proof Proof by Contradiction Note: This method is applicable in two cases: 1. to determine whether an argument is valid or a fallacy 2. to determine the consistency of a set of statements

151 Exercise 10 Indirect Method of Proof Using indirect method of proof verify whether ( p q) 1. follows from p q q 2. follows from p q, p 3. p follows from r q,r s,s q,p q

152 Indirect Method of Proof 4. p s follows from p q r, q p,s r,p 5. q,p q,p r imply r 6. p q,q r, (p r),p r imply r

153 Indirect Method of Proof Solution: 2. Consider p q, p Assume that the conclusion q is false. Hence (q) q ( De Moivre) is true. The premises now become p q, p,q

154 Indirect Method of Proof (1) q Rule P (2) (3) p p q Rule P Rule P (4) p q Rule T ( Equivalence of (3)) Since all the above are true statements, we do

155 Indirect Method of Proof not arrive at a contradiction. Hence, the given argument is valid. 4. Consider p q r, q p,s r, p Assume that the conclusion is false.

156 Indirect Method of Proof Assume that p s is false. ( p s) ( p s) p s (De Morgan) is true. Hence, we add p s as an additional premise. The premises now become

157 Indirect Method of Proof p q r, q p,s r,p, p s (1) p Rule P (2) p q r Rule P (3) q r (Modus ponens of (1) and (2)) Rule T (4) p s Rule P

158 Indirect Method of Proof (5) s Rule T (Multiplication of (4)) (6) s r Rule P (7) r Rule T (Modus ponens of (5) and (6)) (8) q p Rule P

159 Indirect Method of Proof (9) p q Rule T (Contrapositive of (8) ) (10) q Rule T (Modus ponens of (5) and (9)) (11) r Rule T (Disjunctive syllogism of (3) and (10)) (12) r r Rule T (Combining of (7) and (11) ) which is a contradiction.

160 Indirect Method of Proof Hence, the assumption p s is T is incorrect. p s is T. Hence, the given premises together imply the required conclusion.

161 Validity of Verbal Arguments Exercise 11 Check whether the following statements are consistent: 1. (a) If Jack misses many classes through illness, then he fails high school. (b) If Jack fails high school, then he is uneducated.

162 Validity of Verbal Arguments (c) If Jack reads a lot of books, then he is not uneducated. (d) Jack misses many classes through illness and he reads a lot of books. Solution:

163 Validity of Verbal Arguments We denote the statements by the following variables: A - Jack misses many classes through illness B Jack fails high school C Jack is uneducated D Jack reads a lot of books

164 Validity of Verbal Arguments The given argument is now translated as follows: (a) (b) (c) A B B C D C (d) A Λ D

165 Validity of Verbal Arguments (1) A B Rule P (2) B C Rule P (3) (4) A C D C Rule T (Hypothetical Syllogism of (1) and (2)) Rule P (5) C D Rule T (Contrapositive of (4))

166 Validity of Verbal Arguments (6) A D Rule T (Hypothetical Syllogism of (3) and (5)) (7) A Λ D Rule P (8) A Rule T (Multiplication of (7)) (9) D Rule T (Multiplication of (7))

167 Validity of Verbal Arguments (10) D Rule T (Modus ponens of (6) and (8)) (11) D Λ D Rule T (Combining (9) and (10)) which is a contradiction. Hence, the given set of statements is inconsistent.

168 Validity of Verbal Arguments 2. (a) If A works hard, then either B or C will enjoy himself. (b) If B enjoys himself, then A will not work hard. (c) Therefore, if A works hard, D will not enjoy himself.

169 Validity of Verbal Arguments Solution: hide We denote the statements by the following variables: P A works hard Q B enjoys himself R C enjoys himself S D enjoys himself

170 Validity of Verbal Arguments The given argument is now translated as follows: (a) (b) (c) P ( Q R) Q P S R (d) P ( Q R), Q P, S R P S

171 Validity of Verbal Arguments Since the conclusion is in the form of a conditional, we use Rule CP to add P as an additional premise to obtain the following equivalent argument: P ( Q R), Q P, S R, P S (1) P ( Q R) Rule P (2) P Rule P

172 Validity of Verbal Arguments (3) Q V R Rule T (Modus ponens of (1) and (2)) (4) R Q Rule T (Equivalence of (3)) (5) Q P Rule P (6) R P Rule T (Hypothetical Syllogism of (4) and (5))

173 Validity of Verbal Arguments (7) S R Rule P (8) S P Rule T (Hypothetical syllogism of (6) and (7)) (9) S Rule T (Modus tollens of (1) and (8)) Hence, the given statements are consistent.

174 Some More Connectives NAND The connective NAND denoted by is a combination of NOT and AND. Truth Table : p q T T F T F T F T T F F T p q( ( p q))

175 NOR Some More Connectives The connective NOR denoted by is a combination of NOT and OR. Truth Table : p q T T F T F F F T F F F T p q( ( p q))

176 Functionally Complete Set of Connectives Definition A set of connectives is said to be a functionally complete set of connectives if every formula can be expressed in terms of an equivalent formula containing the connectives only from the set.

177 Functionally Complete Set of Connectives Note The connectives and are duals of each other. If A is a formula such that the connectives that occur in A is a subset of { Λ,,,, } then the dual A* is obtained from A by replacing the symbols { Λ,,,, } and the special variables

178 Functionally Complete Set of Connectives T and F that occur in A by their respective duals. Example If A : P ( Q Λ ( R P) Λ T ), then the dual of A is A* : P ( Q V ( R P) V F ).

179 Functionally Complete Set of Connectives Exercise 12 Verify whether the following sets are functionally complete : 1. {, } 3. { V, Λ } 2. { Λ, } 4. { } 5. { }

180 Functionally Complete Set of Connectives Solution: 1. To prove that a statement formula containing any of the connectives can be replaced in terms of the connectives V and. Consider P P V P P P

181 Functionally Complete Set of Connectives P Λ Q ( P Q ) P Q P V Q P Q ( P Q ) Λ ( Q P ) ( P V Q ) Λ ( Q V P ) { ( P V Q ) V ( Q V P ) }

182 Functionally Complete Set of Connectives P Q ( P Λ Q ) P Q P Q ( P V Q ) Hence the set {, } is functionally complete. 4. Consider the set { } Since the conditional and biconditional can be expressed in equivalent forms containing negation, conjunction and disjunction alone, to prove that {}

183 Functionally Complete Set of Connectives is functionally complete, it is enough to show that negation, conjunction and disjunction can be expressed in terms of Consider P P V P ( P Λ P) P P

184 Functionally Complete Set of Connectives P Λ Q ( P Q ) ( P Q ) ( P Q ) P V Q ( P Λ Q ) P Q ( P P ) ( Q Q ) Hence the set { } is functionally complete.

185 Functionally Complete Set of Connectives 2. Consider the set { Λ, } To prove that a statement formula containing any of the connectives can be replaced in terms of the connectives Λ and. Consider P P P Λ P P P V Q ( P Λ Q )

186 Functionally Complete Set of Connectives P Q P V Q ( P Λ Q ) P Q ( P Q ) Λ ( Q P ) ( P Λ Q ) Λ ( Q Λ P ) P Q ( P Λ Q ) P Q ( P V Q ) P Λ Q

187 Functionally Complete Set of Connectives Hence the set { Λ, } is functionally complete. 3. Consider the set { Λ, V } To verify whether a statement formula containing any of the connectives can be replaced in terms of the connectives Λ and. Consider P P P P Λ P The above statement cannot be written using

188 Functionally Complete Set of Connectives Λ and V alone. Hence the set { Λ, V } is functionally complete.

189 Normal Forms If A(p 1, p 2,.., p n ) is a statement formula (where p 1, p 2,.., p n are the atomic variables) which has the truth value T for at least one combination of the truth values assigned to p 1, p 2,.., p n then, A is said to be satisfiable The problem of determining in a finite number of steps, whether a given statement formula is a tautology or a contradiction is known as a decision problem

190 Normal Forms As the construction of truth tables may not be practical, procedures known as reduction to normal forms are considered. Elementary Product : A product of the variables and their negations in a formula is called an elementary product. Example: P, Q, P Λ Q, P Λ Q

191 Elementary Sum : Normal Forms A sum of the variables and their negations is called an elementary sum. Example: Note: P, Q, P V Q, P V Q Any part of an elementary sum or product which is itself an elementary sum or product is called a factor of the original elementary sum or product

192 Normal Forms Theorem: A necessary and sufficient condition for an elementary product to be identically false is that it contains at least one pair of factors in which one is the negation of the other. Note: If P Λ P appears in the elementary product, then the product is identically false.

193 Normal Forms Theorem: A necessary and sufficient condition for an elementary sum to be identically false is that it contains at least one pair of factors in which one is the negation of the other. Note: If P V P appears in the elementary sum, then the product is identically true.

194 Normal Forms Disjunctive Normal Forms A formula which consists of a sum of elementary products is called a disjunctive normal form. Example: The disjunctive normal form of p ( p q) ( p p) ( p q) is WHY??? p ( p q) p ( p q) ( p p) ( p q)

195 Normal Forms Conjunctive Normal Forms A formula which consists of a product of elementary sums is called a conjunctive normal form. Example: The conjunctive normal form of p ( p q) is p ( p q) WHY??? p ( p q) p ( p q)

196 Normal Forms Extended Distributive Law ( p q) ( r s) Exercise 13 ( p r) ( p s) ( q r) ( q s) Obtain a disjunctive normal form and a conjunctive normal form of the following statement formulae 1. ( p q) p q 2. p ( p q) ( q p)

197 3. 4. Normal Forms p ( q r) ( p q) r q ( p r) ( p r) q p Solution: 1. Consider ( p q) p q We know that p q ( p q) ( p q)

198 ) ( )} ( { q p q p ) ( ) ( ( q p q p Normal Forms q p q p ) ( ) ( ) ( q p q p ) ( ) ( q p q p q p q p ) ( ) ( q p q p F ) ( ) ( q p q p ) ( ) ( q p q p The RHS is a conjunctive normal form of the LHS.

199 ) ( )} ( { q p q p ) ( ) ( ( q p q p Normal Forms q p q p ) ( ) ( ) ( q p q p ) ( ) ( q p q p q p q p ) ( ) ( q p q p F ) ( ) ( q p q p We now obtain the disjunctive normal form of the given expression. Consider

200 Normal Forms ( p q) p q ( p q) ( p q) ( p q) p ( p q) q) ( p) ( p q) p ( q p) ( q q) F ( p q ( q p) F ) p q p q The RHS is a disjunctive normal form of the LHS.

201 Principal Normal Forms The normal forms are not unique, although different normal forms of the same formulae are equivalent. The need to arrive at a unique normal form gives rise to the concept of principal normal forms

202 Principal Normal Forms Definition: For a given number of variables, a minterm consists of a conjunction in which each variable or its negation, but not both, appears exactly once. In other words, if p 1, p 2,.., p n are n variables, the expression x x... be a minterm (of p 1, p 2,.., p n ) if x 1 2 n is said to be a x i { pi, pi} i 1,2,.., n

203 Principal Normal Forms Definition: For a given number of variables, a maxterm consists of a disjunction in which each variable or its negation, but not both, appears exactly once. In other words, if p 1, p 2,.., p n are n variables, the expression x x... 2 be a maxterm (of p 1, p 2,.., p n ) if xn 1 is said to be a x i { pi, pi} i 1,2,.., n

204 Principal Normal Forms Definition: For a given formula φ, an equivalent formula consisting of disjunctions of only minterms is called principal disjunctive normal form (PDNF/ sum-of-products canonical form) of φ. Definition: For a given formula φ, an equivalent formula consisting of conjunctions of maxterms alone is

205 Principal Normal Forms called principal conjunctive normal form (CDNF/ product-of-sums canonical form) of φ. Remark: 1. Given n variables, p 1, p 2,.., p n there can be only 2 n minterms x x... maxterms x x... x, 1 2 n 1 2 n and 2 n x xi { pi, pi} i 1,2,.., n

206 Principal Normal Forms 2. Each minterm as a formula has its truth value T in only one row of its truth table. Hence no two minterms are equivalent. 3. Every formula which is not a contradiction has a unique principal disjunctive normal form; every formula which is not a tautology has a unique principal conjunctive normal form.

207 Exercise 14 Principal Normal Forms Obtain the principal disjunctive normal form and principal conjunctive normal form of the following formulae by using truth tables. 1. p q 2. p q 3. ( p q) ( p r) ( q r)

208 Principal Normal Forms Solution: 1. Consider the formula p q We know that p q ( p q) ( q p) The truth table of p q is given by

209 Principal Normal Forms p T T F F q T F T F minterm p q p q p q p q T F F T p q Hence the PDNF of p q is given by ( p q) ( p q)

210 The PCNF of Principal Normal Forms p q ( p q) ( p q) is now obtained as ( p q) ( p q) 2. Consider the formula p q The truth table of p q is given by

211 Principal Normal Forms p q minterms p p V q T T p ᴧ q F T T F p ᴧ q F T F T p ᴧ q T T F F p ᴧ q T F Hence the PDNF of p q is given by ( p q) ( p q) ( p q) The PCNF of p q is now obtained as

212 Principal Normal Forms ( p q) p q 3. Consider the formula S ( p q) The truth table of is given by ( p r) ( q r) ( p q) ( p r) ( q r)

213 Principal Normal Forms p q r minterm p Λ q q Λ r p Λ r S T T T p ᴧ q Λ r T T F T T T F p ᴧ q Λ r T F F T T F T p ᴧ q Λ r F F F F T F F p ᴧ q Λ r F F F F F T T p ᴧ q Λ r F T T T F T F p ᴧ q Λ r F F F F F F T p ᴧ q Λ r F F T T F F F p ᴧ q Λ r F F F F

214 Principal Normal Forms Hence the PDNF of S is given by ( p q r) ( p q r) The PCNF of S is given by ( p q r) ( p q r) ( p q r) ( p q r) ( p q r) ( p q r) Hence the PCNF of S is given by

215 Principal Normal Forms ) ( ) {( r q p r q p )} ( ) ( r q p r q p ) ( ) ( r q p r q p ) ( ) ( r q p r q p

216 Exercise 15 Principal Normal Forms Obtain the principal disjunctive normal form and principal conjunctive normal form of the following formulae without using truth tables. 1. p q 2. p q 3. ( p q) ( p r) ( q r)

217 Principal Normal Forms Solution: 1. Consider the formula p q We know that p q ( p q) ( q p) ( p q) ( q p) which is the PCNF of the given formula The PDNF of p q is given by

218 Principal Normal Forms ( p q) ( p q) ( p q) ( p q) 2. Consider the formula p q p q is itself the PCNF. The PDNF of p q is given by

219 Principal Normal Forms {( p q) ( p q) ( p q)} ( p q) ( p q) ( p q) 3. Consider the formula ( p q) ( p r) ( q r) The formula is in DNF but not in PDNF since in the first and second terms neither r nor r appears and

220 Principal Normal Forms third term neither p nor p appears. Now, ( p q) ( p q) ( q r) [( p q) ( r r)] [( p q) ( r r)] [( q r) ( p p)] [( p q r) ( p q r)] [( p q r) ( p q r)] [( q r p) ( q r p)]

221 Principal Normal Forms ) ( ) ( r q p r q p ) ( ) ( r q p r q p which is the required PDNF of the given formula. The PCNF is obtained as follows: ) ( ) ( ) ( r q q p q p ) ( ) {( r q p r q p )} ( ) ( r q p r q p

222 Principal Normal Forms ) ( ) ( r q p r q p ) ( ) ( r q p r q p Problems 1. Find the minterm normal form (PDNF) and hence the PCNF of ) ( ] ) [( r p r q p Solution ) ( ] ) [( r p r q p S Let

223 Principal Normal Forms Then S [ ( p q) r] ( p r) [( p q) r] ( p r) [{( p q) r} p] [{( p q) r} r] [{ p ( p q)} ( p r [{ r ( p q)} ( r r)] )]

224 Principal Normal Forms [( p p q) ( p r )] [{ r ( p q)} F] [{( p p) q} ( p r ( r p q) ( p q r) ( p r) )] [ F ( p r)] ( r p q) ( p q r) [( p r) ( q q)]

225 Principal Normal Forms ( p q r) [( p r q) ( p r q)] ( p q r) ( p r q) ( p r q) which is the PDNF of S. Now, the PDNF of S is given by S ( p q r) ( p q r) ( p q r) ( p q r) ( p q r)

226 Principal Normal Forms Hence the PCNF of S is given by S [( p q r) ( p q r) ( p q r) ( p q r) ( p q r)] ( p q r) ( p q r) ( p q r) ( p q r) ( p q r)

227 Principal Normal Forms which is the required PCNF of S. Now, the PDNF of S is obtained as ) ( ) ( ) ( [ r q p r q p r q p ) ( ) ( r q p r q p )] ( ) ( r q p r q p

228 ) ( ) ( r q p r q p Principal Normal Forms ) ( ) ( r q p r q p ) ( ) ( r q p r q p ( r) q p

229 Principal Normal Forms 2. Obtain the PCNF and PDNF of Solution ) ( ) ( p q r p S Let ] ) ( [ r p S ( r) p )] ( ) ( [ p q p q ) ( ) ( p q r p )] ( ) ( [ p q p q

230 Principal Normal Forms ( p r) [{( q p) q} {( q p) p}] ( p r) [( q q) ( q p)] [( p q) ( p p)] ( p r) ( q q) ( q p) ( p q) ( p p) ( p r) ( q p) ( p q)

231 Principal Normal Forms ] ) ( [ F r p ] ) ( [ F p q ] ) ( [ F q p )] ( ) ( [ q q r p )] ( ) ( [ r r p q )] ( ) ( [ r r q p )] ( ) ( [ q r p q r p )] ( ) ( [ r p q r p q )] ( ) ( [ r q p r q p

232 Principal Normal Forms ) ( ) ( q r p q r p ( r) p q )] ( ) ( [ r q p r q p The PDNF of S is obtained as ) ( ) ( [ q r p q r p ( r)] q p

233 Principal Normal Forms ( p q r) ( p q r) ( p q r) Problems Obtain the PCNF and PDNF of 1. p ( p q r) 2. ( p r s) ( r s)

234 Applications Digital Logic, Logic Gates, NP hard, NP - complete

235 Predicate Logic Definition: Let A be a given set. A propositional function (open sentence) P(x) defined on A is an expression which has the property that P(a) is true or false for every a in A. P(x) is called a predicate.

236 Predicate Logic Note: A predicate which involves one variable is called one place predicate, two variables is called two place predicate,.., n variables is called n place predicate. Example: 1. Let A={1,2,3,4,5} P(x) : x+4 > 7, x ε A

237 Predicate Logicc 2. Let A = {students in a class} P(x,y) : x is taller than y. 3. P(x) : x+3<8, x ε N P(1), P(2), P(3), P(4) True P(5), P(6), P(7),... - False

238 Statement Function: Predicate Logic A simple statement function of one variable is defined to be an expression consisting of a predicate symbol and an individual variable. Note: A statement function becomes a statement when the variable is replaced by the name of any object.

239 Predicate Logic Substitution Instance: The statement resulting from a replacement is called a substitution instance of the statement function and is a formula of statement calculus Universe of Discourse: The collection of variables which can replace a variable in the predicate is called the universe of discourse.

240 Predicate Logic Equivalent Predicates: Two predicates are said to be equivalent if they have the same truth values for all possible values of the variables in their discourse Note: The universe of discourse must be the same for both the predicates.

241 Quantifiers Universal Quantifier: Whenever a predicate is such that the statement obtained by substituting the values of the variables is always true for all elements of the universe, we say that the variables are quantified universally. It is denoted by or ( ). The term for all (for every, for each) is called a universal quantifier.

242 Universal Quantifier: Example: Quantifiers All maths majors are intelligent. A={maths majors} P(x) : x is intelligent The given statement is x P(x)

243 Existential Quantifier: Quantifiers Whenever a predicate is such that the statement obtained is true for some value of the variable from the universe, we say that the variables are quantified existentially. It is denoted by

244 Quantifiers Existential Quantifier: Example: There exists a positive integer x such that x+2>10 A={positive integers} P(x) : x+ 2 > 10 The given statement is x P( x), x A

245 Quantifiers Negation of Quantified Statements: De Morgan s Laws 1. xp( x) ( x) P( x) 2. xp( x) ( x) P( x) It is not true that for all a in A, P(a) is true is equivalent to there exists an a in A such that P(a) is false.

246 Quantifiers Example: 1. All maths majors are intelligent There exists a maths major who is not intelligent. 2. There exists a person who is 150 years old. Every living person is not 150 years old. Let the universe of discourse be A = {living persons}. The given statement becomes xp(x) Its negation is ( x) P( x)

247 Quantifiers 3. Every student has at least one course where the lecturer is a research associate. Let A = { students }, B = { courses }. P(x,y) Lecturer is a research associate. The given statement is equivalent to ( x) ( y) P( x, y), x A, y The negation of the above formula is ( x) ( y) P( x, y), x A, B yb There exists a student, such that for all courses, the lecturers are not research associates.

248 Quantifiers Statement When is it true? When is it false? ( x) P( x) ( x) P( x) ( x) P( x) ( x) P( x) For some a in A, P(a) is true For every a in A, P(a) is true For at least one a in A, P(a) is false, so its negation is true. For every a in A, P(a) is false and its negation is true. For every a in A, P(a) is false There is at least one a in A, for which P(a) is false For every a in A, P(a) is true There is at least one a in A for which P(a) is true

249 Quantifiers Equivalent formula for Quantifiers: Let A x x,..., 1, 2 x n x x P( x) P( x ) P( x2 P( x) P( x ) P( x2 )... P( x 1 n )... P( x 1 n ) )

250 Quantifiers Exercise 16 : Write the negation of the following statements : 1. All the students live in hostels 2. Some of the students are 25 years or older. 3. Every apple is red. 4. Every city in India is clean. 5. If the teacher is absent, then some students do not complete their home work.

251 Quantifiers All the students completed their homework and the teacher is present. Some students have not completed their homework or the teacher is absent. Solution 1. Consider the statement All the students live in hostels Let A = { students } P(x) = x lives in a hostel.

252 Quantifiers The given statement becomes ( x) P( x) Its negation is ( x) P( x) There is a student who does not live in hostel. 2. Consider the statement Some of the students are 25 years or older. Let A = { students } P(x) = x is 25 years or older. The given statement becomes ( x) P( x)

253 Quantifiers Its negation is ( x) P( x) All the students are less than 25 years. 3. Consider the statement Every apple is red. Let A = { apples } P(x) = x is red. The given statement becomes ( x) P( x) Its negation is ( x) P( x) There exists some apples that are not red.

254 Quantifiers (OR) Let A = { fruits } P(x) = x is an apple. Q(x) = x is red. The given statement is equivalent to ( x)[ P( x) Q( x)] Its negation is given by {( x)[ P( x) Q( x)]} ( x) [ P( x) Q( x)]

255 Quantifiers ( x) [ P( x) Q( x)] ( x)[ P( x) Q( x)] There exists fruits which are apples and not red. Some apples are not red.

256 Quantifiers Definition: Let P(x) and Q(x) be open statements defined for a given universe. The statements P(x) and Q(x) are called (logically) equivalent denoted by ( x)( P( x) Q( x)) when the biconditional P( a) Q( a) is true for each a in the universe. If the implication P( a) Q( a) is true for each a in in the universe, then we say that P(x) logically implies Q(x) and denote it by ( x)( P( x) Q( x))

257 Quantifiers Consider the implication ( x)( P( x) Q( x)) Then, we define contrapositive ( x)( Q( x) P( x)) converse inverse ( x)( Q( x) P( x)) ( x)( P( x) Q( x)) Note : The converse and inverse are logically equivalent.

258 Quantifiers Logical equivalences and Logical Implications for Quantified Statements in One Variable 1. ( x)[ p( x) q( x)] [( x) p( x) ( x) q( x)] 2. ( x)[ p( x) q( x)] [( x) p( x) ( x) q( x)] 3. ( x)[ p( x) q( x)] [( x) p( x) ( x) q( x)] 4. {( x) p( x) ( x) q( x)} ( x)[ p( x) q( x)] 5. ( x)[ p( x) { q( x) r( x)}] ( x)[ p( x) q( x) r( x)]

259 Quantifiers Logical equivalences and Logical Implications for Quantified Statements in One Variable 6. ( x)[ p( x) q( x)] ( x)[ p( x) q( x)] 7. ( x) p( x) ( x) p( x) 8. ( x) [ p( x) q( x)] ( x)[ p( x) q( x)] 9. ( x) [ p( x) q( x)] ( x)[ p( x) q( x)]

260 Quantifiers Logical equivalences and Logical Implications for Quantified Statements in One Variable 10. ( x) p( x) ( x) p( x) 11. ( x) [ p( x) q( x)] ( x)[ p( x) q( x)] 12. ( x) [ p( x) q( x)] ( x)[ p( x) q( x)]

261 Free and Bound Variables Given a formula containing a part of the form ( x)( P( x)) or ( x)( Q( x)), such a part is called the x-bound part of the formula. Any occurrence of the variable which is not a bound occurrence is called the free occurrence of the variable. Further, the formula P(x) in ( x)( P( x)) ( x)( Q( x)) or is called the scope of the quantifier.

262 Free and Bound Variables In other words, the scope of a quantifier is the formula immediately following the quantifier. Example: 1. ( x) P( x) Q( x) The first occurrence of x in P(x) is bounded, while the second occurrence of x in Q(x) is free. 2. ( x)[ P( x) Q( x)] Both the occurrences of x in P(x) and Q(x) are bounded.

263 Free and Bound Variables Exercise 16 : Determine the free variables, bound variables and scope in the following predicate formulae ( x) P( x, y) ( x)[ P( x) Q( x)] ( x)[ P( x) ( y) Q( x, y)] ( x)[ P( x) Q( x)] ( y) R( y) ( x) P( x) Q( x)

264 Free and Bound Variables No. Predicate formula Bound variable; scope Free variable ( x ) P( x, y) x; P( x, y) 1 Nil 2 Nil ( x)[ P( x) Q( x)] x; P( x) Q( x) 3 Nil ( x)[ P( x) ( y) Q( x, y)] x; P( x) ( y) Q( x, y) y; ( y) Q( x, y) 4 Nil ( x)[ P( x) Q( x)] ( y) R( y) x; [ P( x) Q( x)] y; R( y) 5 First x; P(x) Second x ( x) P( x) Q( x)

265 Inference Theory of Predicate Calculus If it becomes necessary to eliminate quantifiers during the course of derivation we require two rules of specification called US and ES rules. Once the quantifiers are eliminated, the derivation or inference theory is similar to the statement calculus If it becomes necessary to quantify the desired conclusion, we require two rules of generalisation called UG and EG rules.

266 Inference Theory of Predicate Calculus 1. Rule US : Universal Specification is the rule of inference which states that one can conclude that P(c) is true, if ( x)( P( x)) is true, where c is an arbitrary member of the universe of discourse. This rule is also called universal instantiation. ( x) P( x) P( y)

267 Inference Theory of Predicate Calculus 2. Rule ES : Existential specification is the rule which allows us to conclude that P(c) is true if ( x) P( x) is true, where c is not an arbitrary member of the universe, but one for which P(c) is true. This rule is also called existential instantiation. (Usually c will not be known, but its existence will be assured)

268 Inference Theory of Predicate Calculus Rule ES : ( x) P( x) P( y) provided y is not free in the given premises and also not free in the prior steps of derivation. Note: Choose a new variable every time ES is used.