Chapter 2. Logic

1. Chapter 2. Logic Weiqi Luo (骆伟祺) School of Software Sun Yat-Sen University Email：weiqi.luo@yahoo.com Office：A309

2. Chapter two: Logic • 2.1. Propositions and Logical Operation • 2.2. Conditional Statements • 2.3. Methods of Proof • 2.4. Mathematical Induction • 2.5. Mathematical Statements • 2.6. Logic and Problem Solving

3. 2.1. Propositions and Logical Operation • Statement (Proposition) A statement or proposition is a declarative sentence that is either true or false but not both Example 1: (a) The earth is round. (b) 2+3=5 (c) Do you speak English? (d) 3-x=5 (e) Take two aspirins. (f) The temperature on the surface of the planet Venus is 800 F (g) The sun will come out tomorrow Yes. Yes. No. This is a question. No. May true or false No. This is a command. Yes. yes

4. 2.1. Propositions and Logical Operation • Paradox A male barber shaves all and only those men who do not shave themselves Paradox: A paradox is a seemingly true statement or group of statements that lead to a contradiction or a situation which seems to defy logic or intuition . Paradox is not a statement. Refer to http://en.wikipedia.org/wiki/List_of_paradoxes Others “I am lying”, “this sentence is wrong”, so on and so forth

5. 2.1. Propositions and Logical Operation • Propositional variables In logic, the letters p, q, r … denote propositional variables, which are replaced by statements p: 1+2 = 5 q: It is raining. • Compound statements Propositional variables can be combined by logical connectives to obtain compound statements. E.g. p and q : 1+2 =5 and it is raining.

6. 2.1. Propositions and Logical Operation • Negation (a unary operation) If p is a statement, the negation of p is the statement not p, denoted by ~ p, meaning “it is not the case that p”. if p is true, then ~p is false, and if p is false, then ~p is true. Truth Table : List the truth value of a compound statement in terms of its component parts.

7. 2.1. Propositions and Logical Operation • Example: Give the negation of the following statements (a) p: 2+3 >1 (b) q: It is snowing. Solution: (a) ~p: 2+3 is not greater than 1, namely, 2+3 <=1 (b) ~q: It is not the case that it is snowing. More simply, ~q: It is not snowing.

8. 2.1. Propositions and Logical Operation • Conjunction If p and q are statements, the conjunction of p and q is the compound statement “p and q” denoted by p∧ q . Truth Table : Note: p∧ q is T if and only if p is T and q is T.

9. 2.1. Propositions and Logical Operation • Example Form the conjunction of p and q p: 1>3 q: It is raining. Solution: p∧ q : 1 > 3 and It is raining. Note: In logic, unlike in everyday English, we may join two totally unrelated statements by logical connectives.

10. 2.1. Propositions and Logical Operation • Disjunction If p and q are statements, the disjunction of p and q is the compound statement “p or q”, denoted by p V q Truth Table : Note: p V q is F is and only if q is F and q is F.

11. 2.1. Propositions and Logical Operation • The connective “or” (a) I left for Spain on Monday or I left for Spain on Friday. (b) I passed math or I failed French” Note: Case (a): Both could not have occurred. “or” is an excusive sense in this case. Case (b): Both could have occurred. “or” is an inclusive sense in this case. In mathematics and computer science. We agree to use the connective “or” in inclusive manner.

12. 2.1. Propositions and Logical Operation • Example: Form the disjunction of p and q p: 2 is a positive integer q : sqrt(2) is a rational number Solution: p V q: 2 is a positive integer or sqrt(2) is a rational number. Since p is true, p V q is true, even through q is false.

13. 2.1. Propositions and Logical Operation Algorithm for making Truth Table: • Step 1: The first n columns of the table are labeled by the component propositional variables. Further columns are included for all intermediate combinations of the variables, culminating in a column for the full statement. • Step 2: Under each of the first n headings, we list the 2n possible n-tuples of truth values for the n component statements. • Step 3: For each of the remaining columns, we compute, in sequence, the remaining truth values.

14. 2.1. Propositions and Logical Operation • Example 5: Make a truth table for the statement (p ∧ q) V (~ p) Truth Table: (1) (2) (3)

15. 2.1. Propositions and Logical Operation • Propositional function (predicate) An element of a set {x | P(x)} is an object t for which the statement P(t) is true. P(x) is called a propositional function (or predicate) , because each choice of x produces a proposition P(x) that is either true or false (well-defined) E.g. Let A={ x | x is an integer less than 8}. Here P(x) is the sentence “x is an integer less than 8” P(1) denotes the statement “1 is an integer less than 8” (true) P(8) denotes the statement “8 is an integer less than 8” (false)

16. 2.1. Propositions and Logical Operation • Universal Quantifiers (∀) The Universal Quantifiers of a predicate P(x) is the statement “for all values of x, P(x) is true” , denoted by ∀ x P(x) Example 8 (a) P(x) : -(-x) = x is a predicate that makes sense for all real number x. then ∀ x P(x) is true statement. Since ∀ x ∈ R, -(-x) = x (b) Q(x): x+1<4. then ∀ x Q(x) is a false statement, since Q(5) is false

17. 2.1. Propositions and Logical Operation • Existential Quantifiers (∃) The Existential Quantifiers of a predicate P(x) is the statement “there exists a value of x for which P(x) is true”, denoted by ∃ x P(x) Example 9 (a) Let Q(x): x+1<4. then the existential quantification of Q(x), ∃ x Q(x), is a true statement, since Q(1) is a true statement (b) The statement ∃ y y+2=y is false since there is no value of y for which the propositional function y+2=y produces a true statement.

18. 2.1. Propositions and Logical Operation • The order of the Quantifiers ∀ &∃ The order does not affect the output for the same quantifiers, while it may produce different results for different quantifiers. E.g. P(x, y) : x + y =1 ∀ x ∃ y P(x) is true, ∃ y ∀ x is false. P(x, y): x *y = 0 ∀ x ∃ y P(x) is true, ∃ y ∀ x is true too.

19. 2.1. Propositions and Logical Operation • The negation of Quantifiers ∀ &∃ (a) let p: ∀x P(x), then ~p: there must be at least one value of x for which P(x) is false, namely, ~ ∀x P(x) = ∃x ~P(x) (b) let p: ∃x P(x), then ~p: for all x, P(x) is false, namely, ~ ∃x P(x)= ∀x ~P(x)

20. 2.1. Propositions and Logical Operation • Homework Ex. 2, Ex. 4, Ex. 16, Ex. 28

21. 2.2. Conditional Statements • Conditional statement If p and q are statements, the compound statement “if p then q”, denoted p=>q, is called a conditional statement or implication. p : antecedent or hypothesis q : consequent or conclusion if … then : => Truth Table : Note: when p is F, then p=>q is T.

22. 2.2. Conditional Statements • Example Form the implication p=>q for each the following (a) p: I am hungry. q: I will eat. (b) p: 2+2=5 q: I am the king of England. Solution (a) If I am hungry, then I will eat (b) If 2+2=5, then I am the king of English Note: There is no cause-and effect relationship between p and q in case (b). And (b) is true, since 2+2=5 is false.

23. 2.2. Conditional Statements • Converse and Contrapositive If p=>q is an implication, then its converse is the implication q => p and its contrapositive is the implication ~ q => ~p E.g. Give the converse and the contrapositive of the implication “If it is raining, then I get wet” Converse: If I get wet, then It is raining. Contrapositive: If I do not get wet, then It is not raining.

24. 2.2. Conditional Statements • Equivalence (biconditional) If p and q are statements, the compound statement p if and only if q, denoted by p  q, is called an equivalence or biconditional. Truth Table : Note: p <=> q is T when p and q are both T or both F.

25. 2.2. Conditional Statements • Example 3: Is the following equivalence a true statement? 3>2 if and only if 0< 3 – 2 Solution: Let p: 3>2 and q : 0< 3 – 2, since p and q are both true, we then conclude that p  q is true statement.

26. 2.2. Conditional Statements • Example 4. Compute the truth table of the statement (p=>q)  (~q => ~p) Truth Table :

27. 2.2. Conditional Statements • Tautology A statement that is true for all possible values of its propositional variables called Tautology. (e.g. Example 4) • Contradiction (Absurdity) A statement that is false for all possible values of its propositional variables called Contradiction or Absurdity. (e.g. p ∧ ~p) • Contingency A statement that can be either true or false, depending on the truth values of its propositional variables, is called a contingency. (e.g. p => q)

28. 2.2. Conditional Statements • Logically Equivalent If two statements p and q are always either both true or both false, for any values of the propositional variables, namely p  q is a tautology Then we call p and q are logically equivalent. Denoted by p ≡ q

29. 2.2. Conditional Statements • Example 6 Show that p V q and q V p are logically equivalent The truth table of (p V q )  (q V p ) are shown as follows Truth Table :

30. 2.2. Conditional Statements • Two Structures with similar properties (Theorem 1) Structure 1 (logic operations): (propositions, ∧, V, ~) Structure 2 (sets operations) (sets, U, ∩ , - )

31. 2.2. Conditional Statements • Theorem 1 Commutative properties p ∨ q ≡ q ∨p, p∧ q ≡ q ∧ p Associative Properties p ∨ (q ∨ r) ≡ (p ∨ q) ∨r , p∧ (q ∧ r) ≡ (p ∧ q) ∧ r Distributive Properties p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r) p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)

32. 2.2. Conditional Statements Idempotent Properties p ∨ p ≡ p , p∧ p ≡ p Properties of Negation ~(~p) ≡ p ~(p ∨ q) ≡ (~p) ∧ (~q) ~(p ∧ q) ≡ (~p) ∨ (~q) De Morgan’s laws

33. 2.2. Conditional Statements • Theorem 2 (p ⇒ q) ≡ ((~p) ∨ q) (p ⇒ q) ≡ (~q ⇒ ~p) (p  q) ≡ ((p ⇒ q) ∧ (q ⇒ p)) ~(p ⇒ q) ≡ (p ∧ ~q) ~(p  q) ≡ ((p ∧ ~q) ∨ (q ∧ ~p))

34. 2.2. Conditional Statements • Theorem 3 ~(∀xP(x)) ≡ ∃x~P(x) ~(∃xP(x)) ≡ ∀x(~P(x)) ∃x(P(x) ⇒ Q(x)) ≡ ∀xP(x) ⇒ ∃xQ(x) ∃x(P(x) ∨ Q(x)) ≡ ∃xP(x) ∨ ∃xQ(x) ∀x(P(x) ∧ Q(x)) ≡ ∀xP(x) ∧ ∀xQ(x) ∀xP(x) ∨ ∀xQ(x) ⇒ ∀x(P(x) ∨ Q(x)) is a tautology ∃x(P(x) ∧ Q(x)) ⇒ ∃xP(x) ∧ ∃xQ(x) is a tautology

35. 2.2. Conditional Statements • Theorem 4 : Each of the following is a tautology (p ∧ q) ⇒ p , (p ∧ q) ⇒ q p ⇒ (p ∨ q) , q ⇒ (p ∨ q) ~p ⇒ (p ⇒ q) , ~(p ⇒ q) ⇒ p (p ∧ (p ⇒ q)) ⇒ q , (~p ∧ (p ∨ q)) ⇒ q (~q ∧ (p ⇒ q)) ⇒ ~p ((p ⇒ q) ∧ (q ⇒ r)) ⇒ (p ⇒ r)

36. 2.2. Conditional Statements • Properties of Quantifiers ∃ and ∀ (a) ∃ x (P(x) V Q(x)) ≡ ∃x P(x) V ∃x (Q(x)) (b) ∀ x (P(x) ∧ Q(x)) ≡ ∀x P(x) ∀x Q(x)

37. 2.2. Conditional Statements • Homework ex.2, ex.7, ex.12, ex.15, ex.21 ex.34

38. 2.3. Method of Proof • Logically Follow If an implication p ⇒ q is a tautology, where p and q may be compound statements involving any number of proposition variables, we say that q logically follows from p. Suppose that an implication of the form (p1 ∧ p2 ∧… ∧ pn) ⇒ q is a tautology. We say that q logically follows from p1, p2, …, pn, denoted by

39. 2.3. Method of Proof (p1 ∧ p2 ∧… ∧ pn) ⇒ q The pi’s are called the hypotheses or premises, and q is called the conclusion. Note: we are not trying to show that q is true, but only that q will be true if all the pi are true. ∴ denotes “therefore”

40. 2.3. Method of Proof • Rules of inference Arguments based on tautologies represent universally correct methods of reasoning. Their validity depends only on the form of the statements involved and not on the truth values of the variables they contain. Example 1 ((p ⇒ q) ∧ (q ⇒ r)) ⇒ (p ⇒ r) is tautology, then the argument is universally valid, and so is a rule of inference.

41. 2.3. Method of Proof • Example 2: Is the following argument valid? If you invest in the stock market, then you will get rich. If you get rich, then you will be happy. ∴ If you invest in the stock market, then you will be happy. let p: you invest in the stock market, q: you will get rich r: you will be happy The above argument is of the form given in Example 1, hence the argument is valid!

42. 2.3. Method of Proof • Example 3 The tautology (p  q) ((p ⇒ q) ∧ (q ⇒ p)) means that the following two arguments are valid p ⇒ q p  q q ⇒ p ∴ (p ⇒ q) ∧ (q ⇒ p) ∴ p  q

43. 2.3. Method of Proof • Equivalence (p  q) They are usually stated “p if and only if q”. We need to prove both p=>q and q=>p by the tautology mentioned in example 3. Algorithm : Step one: Assuming p is true, show q must be true. Step two: Assuming q is true, show p must be true.

44. 2.3. Method of Proof • Modus Ponens p is true, and p=>q is true, so q is true (Theorem 4(g) in Section 2.2.) p p=>q ∴ q

45. 2.3. Method of Proof • Example 4: Is the following argument valid? Smoking is healthy. If smoking is healthy, then cigarettes are prescribed by physicians. ∴ Cigarettes are prescribed by physicians p: Smoking is healthy. q: cigarettes are prescribed by physicians The argument is valid since it is of form modus ponens.

46. 2.3. Method of Proof • Example 5 Is the following argument valid? If taxes are lowered, then income rises Income rises ∴ taxes are lowered Solution: Let p: taxes are lowed q: income rise p=>q q ∴ p Then argument is not valid , since p=>q and q can be both true with p being false. (or show the truth table of (p=>q) ∧ q => p , and determine whether or not it is a tautology)

47. 2.3. Method of Proof • Indirect Method of Proof The tautology (p=>q )  (~q) => (~p ) (An implication is equivalent to its contrapositive) Note: To proof p=>q indirectly, we assume q is false (~q) and show that p is then false (~p)

48. 2.3. Method of Proof • Example 6 Let n be an integer. Prove that if n2 is odd, then n is odd. Solution: Let p: n2 is odd , q: n is odd. To prove that (p=>q) We try to prove its contrapositive ~q=>~p Assuming that n is even (~q), let n=2k, k is an integer, then we have n2 = (2k) 2 =4k2 is even (~p). Hence, the given statement has been proved.

49. 2.3. Method of Proof • The tautology ((p ⇒ q) ∧ (~q)) ⇒ ~p If a statement p implies a false statement q, then p must be false. Proof by contradiction To prove (p1 ∧ p2 ∧… ∧ pn) ⇒ q , We add (~q) into hypothesis p1 ∧ p2 ∧… ∧ pn, if the enlarged hypothesis p1 ∧ p2 ∧… ∧ pn ∧ (~q) implies a contradiction, then we can conclude that q follows from p1 ,p2 , … and pn .

50. 2.3. Method of Proof • Example 7 Prove there is no rational number a/b whose square is 2, namely, sqrt(2) is irrational. Solution: Let p: a, b are integers and no common factors, and b is not 0 q: (a/b)2 is not 2 In order to prove p => q , We try to find the contradiction from p ∧ ~q Refer to Example 7 for more details.