### SlideShare for iOS

by Linkedin Corporation

FREE - On the App Store

- Total Views
- 245
- Views on SlideShare
- 245
- Embed Views

- Likes
- 2
- Downloads
- 15
- Comments
- 0

No embeds

Uploaded via SlideShare as Microsoft PowerPoint

© All Rights Reserved

- 1. MATHEMATICAL LOGIC By Septi Ratnasari (4101412082) SEMARANG STATE UNIVERSITY 2013
- 2. Statement Statement is sentences which may be true or false, but noth both. True or false refers to the actually of event. In mathematics , statement is usually called proposition. Examples : 1. Jakarta is the capital city of the Republic of Indonesia. (true) 2. Merapi is located in West Java. (false) 3. 2 + 3 = 6 (false) A statement is usually written with a small letter symbol such as p, q, r, s, and so on.
- 3. Truth Value of a Statement A truth value is used to determine whether a statement is true or false The truth value of a statement can be denoted as . = tao, which is a Greek letter that chosen to represent the word truth. Examples 1. p : Human is breathing with lungs. (p) = T ( read : the truth value of p is true ) 2. q : 12 + 3 = 5 (q) = F ( read : the truth value of q is false )
- 4. Open Sentence An open sentence is a sentence that still contain a variable, hence its truth value cannot yet be determined. A variable is a symbol that is used to represent an arbitrary element of a universal set. Examples : 2x – 3 < 9, x∈R x is prime number less than 20 A constant that replaces a variable which converts an open sentence into a true statement is called an open sentence solution. A set that consists of all solutions is called a set of
- 5. Negation of a Statement A new statement that is constructed from a previous statement such that it is true if the previous statement is false, and it is false if the previous statement is true. Symbol of negation is ( ~ ). p ~p ~(~p) T F T F T F
- 6. Truth Value of Compound Statements Compound statement is statements which is obtained by combining two or more statements. Logical conjunctions Symbol Term ... and ... ... or ... ˄ ˄ Conjunction Disjunction If ... then... => Implication ... if and only if ... Bi-implication Single statements that are used to form compound statement are called the components of the resulting compound statement.
- 7. 1. Conjunction A compound statement is called p q p ˄q T T T a conjunction if two statements T F F p and q are combined to form a F T F compound statement with a F F F conjunction “and”, denoted as “p ˄ q”. The truth value of p ˄ q is true Examples : (T)pif both + 3 =components a) : 2 of its 5 ( T ) b) p : 12 is completely areqtrue. 5 is a prime : divided by 3. ( T ) number. ( T ) q : 15 is completely p ˄ q : 2 + 3 = 5 and divided by 2. ( F ) 5 is a prime p ˄ q : 12 is completely
- 8. 2. Disjunction A compound statement is called a disjunction if two statements p and q can be combined by using the logical conjunction “or”, denoted by “p ˄ q” which is read “p or q”. The truth value of p ˄ q is only Examples : true if :at least= 8 ( Tp or q is a)p 5 + 3 either ) b) p true. : 8 is an even q number. ( T ) p˄q :5+3=8 or 8 is an even p T T F F q T F T F pvq T T T F :8>8(F) q :8=8(T) p ˄ q : 8 > 8 or 8 = 8 ( T ), can also be stated as 8 ≥ 8. ( T )
- 9. 3. Implication A compound statement is called an implication if two statements p and q are combined to form a compound statement by using the logical conjunction “if ... then ...” , denoted by “p => q”, which can be read : 1. If p then q, 2. p implies q, 3. q only if p, 4. p is sufficient for q, 5. q is required for p. p is called the antecedent (hypothesis) and q is called the consequant. The truth value of p => q is false (F) if p is true and q is false. And for all the other composition of p => q is
- 10. p T T F F q T F T F p => q T F T T Examples : a) p : 5 + 3 = 8 ( T ) q : 8 is an even number. ( T ) p => q : If 5 + 3 = 8, then 8 is an even number. ( T ) b) p : 5 > 3 ( T ) q : 5 is an even number. ( F ) p => q : If 5 > 3, then 5 is an even number.
- 11. 4. Bi-implication A compound statement is called a bi-implication if two statements p and q are combined to form a compound statement with a logical conjunction “... if and only if ...”, denoted by “p q” that means “p if and only if q”, i.e “if p then q and if q then p”. Hence p q ≅ (p => q) ˄(q =>qp). p => q (p => q) ˄ ( => p) ≅ p q p q => p q T T T T T T F F T F F T T F F F F T T T
- 12. Tautology Tautology is a compound statement which is always true for all possibilities from its components. Example of Tautology : p q ~p ~q p => q (p => q) ˄ ~q ((p => q) ˄ ~q) => ~p T T F F T F T T F F T F F T F T T F T F T F F T T T T T
- 13. Contradiction Contradiction (the reverse of tautology) is a compound statement which is false for all possibilities from its components. Example of Contradiction : p q p˄q ~ (p ˄ q) ~ (p ˄ q) ˄ p T T T F F T F T F F F T T F F F F F T F
- 14. Contingency Contingency is a compound statement which is not a tautology and not a contradiction for all possibilities from its components. p q p˄q (p q) p T T T T T F T T F T T F F F F T
- 15. Negation of Compound Statements a. Negation of Conjunction p q ~p ~q p˄q ~(p ˄ q) ~p ˄ ~q T T F F T F F T F F T F T T F T T F F T T F F T T F T T equivalent From the truth table above we see that ~(p ˄q) ≅ ~p ˄~q. Therefore, (~p ˄~q) is the negation of (p ˄q). It can be concluded that : ~(p ˄q) ≅ ~p ˄~q
- 16. b. Negation of Disjunction p q ~p ~q p˄q ~(p ˄ q) ~p ˄ ~q T T F F T F F T F F T T F F F T T F T F F F F T T F T T equivalent From the truth table above we see that ~(p ˄q) ≅ ~p ˄~q. Therefore, (~p ˄~q) is the negation of (p ˄q). It can be concluded that : ~(p ˄q) ≅ ~p ˄~q
- 17. c. Negation of Implication p q ~p ~q p => q ~p ˄ q ~( p => q) ~(~p ˄ q) ≅ (p ˄ ~q) T T F F T T F F T F F T F F T T F T T F T T F F F F T T T T F F equivalent From the table, we can conclude that : ~( p => q) ≅ ~(~p ˄q) ≅ (p ˄~q) equivalent
- 18. d. Negation of Bi-implication p q ≅ (p => q) ˄ ( => p) q Remember that From this equivalence, we can derive : ~( p q) = ~((p => q) ˄ ( => p)) q = = = ~(p => q) ˄ ~( => p) q ~(~p ˄q) ˄~(~q ˄p) (p ˄~q) ˄ ( ˄~p) q p q ~p ~q pq ~( p q) T T F F T F F F F T F F T F T T F T F T T F F T F T T F F T T T F F F F (p ˄ ~q) (q ˄ ~p) equivalent Hence, ~( p q) ≅ (p ˄~q) ˄ ( ˄~p) q (p ˄ ~q) ˄ (q ˄ ~p)
- 19. Converse, Inverse, and Contrapositive Implication = p => q Converse = q => p Inverce = ~p => ~q Contrapositive = ~q => ~p
- 20. The relationships among the implications can be shown by the following diagram. p => q Converse q=>p Inverse Contrapositive Inverse ~p =>~q Converse ~q=>~p
- 21. implication converse inverse contrapositive p q ~p ~q p => q q => p ~p => ~q ~q => ~p T T F F T T T T T F F T F T T F F T T F T F F T F F T T T T T T equivalent equivalent
- 22. Statements with Quantor 1. Universal Quantor
- 23. 2. Exsistential Quantor The solution set from an open sentence p(x) which has at least one member of the whole set S, can be written as follow Symbol is called existential quantor, read as ' There exists' or 'At least one of '.
- 24. Example Given an open sentence 2x + 1 = 7. Expree it by using exsistential quantor, and then determine its truth value, whether the whole set is real number R. Answer :
- 25. Negation of Quantor Statement
- 26.
- 27. Making Conclusion • One of the important purposes of mathematical logic is to have knowledge regarding testing argument or making conclusion. • In logic, a premise is a claim that is a reason ( or element of a set of reason ) for, or objection against, some other claim. In other words, it is a statement presumed true within the context of an argument toward a conclusion. • In the context of ordinary argumentation, the rational acceptability of a disputed conclusion depends on both the
- 28. 1. Modus Ponens Premise 1 Premise 2 Conclusion : p => q : p : q The argument above can be written in a form of implication as follow [(p => q) ˄p] => q
- 29. 1. Modus Tollens Premise 1 Premise 2 Conclusion : p => q : ~q : ~p The argument above can be written in a form of implication as follow [(p => q) ˄~q ] => ~p
- 30. 3. Syllogism Premise 1 Premise 2 Conclusion : p => q : q => r : p => r The argument above can be written in a form of implication as follow [(p => q) ˄(q => r)] => (p => r)
- 31. Direct and Indirect Proof 1. Direct Proof In order to direcly prove a conditional statement of the form ( p => q ), it is only necessary to consider situations where the statement p is true to conclution q. The common proof rules used are modus ponens, modus tollens, and syllogism.
- 32. Example Show that for all integer n, if n is an odd numbers, then n2 is also an odd number ! Answer : For example p : n is an odd integer q : n2 is an odd integer It will be shown that p => q is true. Because n is an odd number, thus n = 2k + 1, k ∊ C Then, we have n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1 = 2m + 1 Where m = 2k2 + 2k, so n2 is an odd integer. Thus, it is shown that p => q is true.
- 33. 2. a. Indirect proof Indirect Proof by Contradiction To prove statement (p => q) is true, we can proceed by assuming ~q is true. And show that it leads to a logical contradiction. Thus, according to the law of contradiction, ~q must be true, and so, statement (p => q) is true.
- 34. Example Show that ' If n2 is an odd number, then n will be also an odd number ' by using indirect proof by contradiction. Answer : Suppose that n is an even number, that is n = 2k, k ∊ B. Because n = 2k Then n2 = (2k) 2 = 4k2 = 2(2k2 ) = 2m Where m = 2k2 We get n2 is an even number, contradiction with n2 is an odd number. Thus, it is shown that if If n2 is an odd number, then n will be also an odd number.
- 35. b. Indirect Proof by Contraposition To prove statement (p => q) is true, we can proceed by assuming ~q is true. And show that it leads ~p is true. Thus ~q according to the law of contraposition ~p, is true. So statement (~q => ~p) is true.
- 36. Example Show that for all integer n2 is an odd number, then n will be also an odd number. Answer : Use the indirect contraposition to prove the statement above. Suppose that p : n2 is an odd number q : n is an odd number We asssume that ~q is true, it is mean that n is an even number, n = 2k We will get n2 = (2k2) = 4k2 = 2(2k2) = 2m, where m = 2k2 It is mean that n2 is an even number. Thus, ~p : n2 is an even number ~q : n is an even number Because (~q => ~p) is true and p => q ≅ (~q => ~p) Then, p => q is true. Thus, it is shown that if n2 is an odd number, then n will be also an odd
- 37. Exercises Give the negation of the statement “If all leaders put forward the interest of their people, then all people will live prosperously”. 2. Determine the inverse,converse, and contrapositive of the statement form (p ˄q) =>r. 3. Write the valid conclusions of these premises. p1 : Students do not like math or teachers like to teach. p2 : If teachers like to teach, then math grades are good. p3 : Students like math. 1. ∴ ...
- 38. 4. Write the valid conclusions of the following premises.

Full NameComment goes here.M. Wirman Darmawan, Instrument Control Engineer at PT Indah Kiat Pulp & Paper 5 months agomfhusein5 months ago