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Course notes1

1. 1. PART I: MATHEMATICAL LOGIC“Although mathematics is both a science and an art, special characteristics distinguishmathematics from the humanities and from other sciences. Particularly important is the kind ofreasoning that typifies mathematics. The natural or social scientist generally makes observations ofparticular cases or phenomena and seeks a general theory that describes or explains theobservations. This approach is called inductive reasoning, and it is tested by making furtherobservations. If the results are incompatible with theoretical expectations, the scientist usually mustreject or modify his theory.The mathematician, too, frequently uses inductive reasoning in attempting to describe patternsand relationships among quantities and structures. The characteristic thinking of the mathematician,however, is deductive reasoning, in which one uses logic to draw conclusions based on statementsaccepted as true. The conclusions of a mathematician are proved to be true, provided thatassumptions are true. If a mathematical theory predicts results incompatible with reality, the faultwill lie not in the theory but with the inapplicability of the theory to that portion of reality. Indeed,the mathematician is not restricted to the study of observable phenomena, even though mathematicscan trace its development back to the need to describe spatial relations (Geometry) and motion(Calculus) or to solve numerical problems (Algebra). Using logic, the mathematician can drawconclusions about any mathematical structure imaginable.”The basics of logic and the idea of proof are important in many areas other than mathematics.For example, the thought processes used to construct an algorithm for a computer program aremuch like those used to develop the proof of a theorem.What is Logic?• An instrument used to appraise the correctness of reasoning.• The study of propositions / reasoningWhat is a proposition?• Declarative statement that is either true or false, but not both.Examples:The square root of 2 is irrational.1+1 = 5In the year 2000, more than a thousand Filipinos migrated to Canada.-5 < 75Examples of sentences that are not propositions:What did you say?x = 6.This sentence is false.Notation:• The truth values of propositions are:1 or T for true0 or F for false• Capital letters are used to denote propositions.PROPOSITIONS AND CONNECTIVES:1. Negation of Propositions (  P )• Formed by introducing the word “not” in a proper place in the proposition.• Suppose P is a propositionP is the negation of P
2. 2. • Truth tableP  P1 00 1EXAMPLE: It is not the case that the square root of 10 is greater than 4.10 is not divisible by 2.2. Conjunction ( P ∧ Q )• Truth tableP Q P ∧ Q1 1 11 0 00 1 00 0 0EXAMPLE: 6<7 and 7<8.2*4=16 and a quart is larger than a liter.3. Disjunction ( P ∨ Q )• Truth tableP Q P ∨ Q1 1 11 0 10 1 10 0 0EXAMPLES: Venus is smaller than earth or 1+4 = 5.The square root of 2 is less than the square root of 3 or chicken have lips.4. Implication or Conditional Statement (P⇒Q) read as: If P then Q• P is called the hypothesis or the premise• Q is called the conclusion• Truth tableP Q P ⇒ Q1 1 11 0 00 1 10 0 1EXAMPLES:If triangle ABC is isosceles, then the base angles A and B are equal.Mars has 10 moons implies that 1+1 = 2.1+2=3 implies that 1<0.If the sun shines tomorrow then I will play basketball.If you get 100 in the final exam then you will pass this course.
3. 3. 5. Logical Equivalence or the Biconditional Statement (P⇔Q) read as: P if and only if Q• P ⇔ Q is equal to (( P ⇒ Q ) ∧ (Q ⇒ P))• Truth tableP Q P ⇒ Q Q ⇒ P P ⇔ Qor( P ⇒ Q ) ∧ ( Q⇒P)1 1 1 1 11 0 0 1 00 1 1 0 00 0 1 1 1EXAMPLES : A rectangle is a square if and only if the rectangle’s diagonals are perpendicular.5+6=6 if and only if 7+1=10.Types of Propositional FormsTautology• A proposition that is always true under all possible combination of truth values for allcomponent propositions.example: (P∧Q) ⇒ PP Q P∧Q (P∧Q) ⇒ P1 1 1 11 0 0 10 1 0 10 0 0 1Contradiction• A proposition that is always false under all possible combination of truth values of thecomponent propositions.example: P∧¬PP ¬P P∧¬P0 1 01 0 0Contingency• The combination of tautology and Contradictiona proposition which is neither a tautology nor a contradictionex.(P v Q ) ⇒ PP Q P v Q ( P v Q) ⇒ P1 1 1 11 0 1 10 1 1 00 0 0 1
4. 4. Rules of Replacement( equivalent propositions)Any logically equivalent expressions can replace each other wherever they occur.1. IdempotenceP ⇔ ( P ∨ P )P ⇔ ( P ∧ P )2. Commutativity( P ∨ Q ) ⇔ ( Q ∨ P )( P ∧ Q ) ⇔ ( Q ∧ P )3. Associativity( P ∨ Q ) ∨ R ⇔ P ∨ ( Q ∨ R )( P ∧ Q ) ∧ R ⇔ P ∧ ( Q ∧ R )4. De Morgan’s Laws ( P ∨ Q ) ⇔  P ∧  Q ( P ∧ Q ) ⇔  P ∨  Q5. Distributivity of ∧ over ∨P ∧ ( Q ∨ R ) ⇔ ( P ∧ Q ) ∨ ( P ∧ R )6. Distributivity of ∨ over ∧P ∨ ( Q ∧ R ) ⇔ ( P ∨ Q ) ∧ ( P ∨ R )7. Double Negation  P ⇔ P8. Material Implication( P ⇒ Q ) ⇔ (  P ∨ Q )9. Material Equivalence( P ⇔ Q ) ⇔ ( P ⇒ Q ) ∧ ( Q⇒P )10. Exportation[ ( P ∧ Q ) ⇒ R ] ⇔ [ P ⇒ ( Q ⇒ R ) ]11. Absurdity[ ( P ⇒ Q ) ∧ ( P ⇒  Q )] ⇔  P12. Contrapositive( P ⇒ Q ) ⇔ (  Q ⇒  P )13. IdentityP ∨ 1 ⇔ 1 P ∧ 1 ⇔ PP ∨ 0 ⇔ P P ∧ 0 ⇔ 0P ∨  P ⇔ 1 P ∧  P ⇔ 0 0 ⇔ 1  1 ⇔ 013. AbsorptionP ∧ ( P ∨ R ) ⇔ PP ∨ ( P ∧ R ) ⇔ PEXAMPLES:1. SHOW THAT [P<=>Q] <=>[(P∧Q) ∨ (¬P ∧ ¬Q)]P Q P<=>QP∧Q ¬P∧¬Q (P∧Q)∨(¬P∧¬Q)1 1 1 1 0 11 0 0 0 0 0
5. 5. 0 1 0 0 0 00 0 1 0 1 1Corresponding TRUTH VALUES OF THE PROPOSITIONS ARE EQUAL.Therefore, [P<=>Q] <=>[(P∧Q) ∨ (¬P ∧ ¬Q)] is a tautology. Hence, P<=>Q is equivalent to [(P∧Q)∨ (¬P ∧ ¬Q)].2. SIMPLIFY [(P⇒Q) v (P⇒R)] ⇒( Q v R )[(¬P v Q )v(¬P v R)] ⇒ (Q v R) MI[(¬P v¬P)v(Q v R)] ⇒( Q v R ) Assoc.[¬Pv(QvR) ] ⇒ ( Q v R ) Idempotence¬[¬P v( Q v R ) v ( Q v R ) MI[P∧¬(QvR)]v(QvR) Dem, DN[Pv(QvR)]∧[¬(QvR)v(QvR)] Distributivity[Pv (QvR)]∧1 IdentityPv(QvR) Identity3. On a rare TV game show, a young couple was shown three keys labeled A, B and C and was toldthat they could win a beautiful new home simply by picking the key which opened its front door. Tohelp the couple make their decision, the EMCEE of the show gave them the following pieces ofinformation:“ Either key A will open the door, or key B will not and key C will. ““ If key B will not open the door then key C will. ““ It is impossible that either both keys A and B will open the door or both keys A and C will.”After a few minute-reflection the couple made the correct choice. WHICH KEY DID THECOUPLE PICK? Show a solution that uses a TRUTH TABLE.By truth table, show that every condition is true if key chosen is A or B or C.Solution:A B C A ∨ ( B ∧ C)  B ⇒ C  [(A ∧ B) ∨ (A ∧ C)]1 0 0 1 0 10 1 0 0 1 10 0 1 1 1 1Key C is the one chosen.EXERCISE. Prove each of the rules of replacement using truth tablesArgument• An argument is a collection of propositions where it is claimed that one of the propositions ( theconclusion ), follows from the other propositions (the premises).Inductive Argument• One where it is claimed that within a certain probability of error the conclusion follows from thepremisesDeductive Argument• One where it is claimed that the conclusion absolutely follows from the premises.valid Argument
6. 6. • A deductive argument is said to be valid if whenever the premises are all true and the conclusionis also true.Rules of Inference ( Basic Valid Arguments )1.AdditionP______∴P v Q2.ConjunctionPQ______∴P∧Q3.SimplificationP∧Q_______∴P4.Modus PonensP⇒ QP______∴ Q5. Modus TolensP⇒ Q¬Q______∴¬P6.Disjunctive SyllogismP v Q¬P______∴Q7. Hypothetical SyllogismP⇒QQ⇒R______∴P⇒ R8.Constructive Dilemma(P⇒Q)∧(R⇒S)P v R
7. 7. ______∴Q v S9.Destructive Dilemma(P⇒Q)∧(R⇒S)¬Q v ¬S______∴¬P v ¬RProving Invalidity• To show that an argument is invalid , we make an argument of truth values to the componentpropositions that will make the premises true and the conclusion false.example:J ⇒ ( K ⇒ L ) 1 J = 1K ⇒ (  L ⇒ M ) 1 K = 0( L ∨ M ) ⇒ N 1 L = 0M = 0∴ J ⇒ N 0 N = 0InvalidR=>P P = 1P v Q Q = 0______∴P=>Q R = ORule of conditional Proof• To prove P ⇒ Q, include P among the premises and prove Q instead.Ex.1. A=>(B=>C)2. B=>(C=>D)_____∴ A=>(B=>D)3. A Rule of conditional proof∴B=>D4.B Rule of conditional proof∴D5. C=>D MP 2,46. B =>C MP 1,37. C MP 4,68. D MP 5,7Rule of Indirect Proof• To Prove Q, include  Q among the premises and derive a contradiction (prove 0).Ex.1.P⇒Q
8. 8. 2.Q⇒R3.S⇒¬R4.¬S⇒¬T5.¬T⇒¬Q-----------∴ ¬P6.¬¬P Rule of indirect proof______∴ 07.P DN 68.Q MP 1,79.R MP 2,810.¬S MT 3,911.¬T MP 4,1012. ¬Q MP 5,1113. Q ∧ ¬Q Conj. 8,1214. 0 identity 13Examples:1. Prove the following argumentsa. (A v G )=> S(A∧T)______∴ SProof:(1) (A v G )⇒ S(2) (A∧T)______∴ S(3) A Simp. 2(4) A v G Add. 3(5) S MP 1,4b. (1) (A∧B)⇒[A ⇒(D∧E)](2) (A∧B)∧C_____∴D v E(3) A∧B Simp. 2(4) A ⇒(D∧E) MP 1,3(5) A Simp. 3(6) D∧E MP 4, 5(7) D Simp. 6(8) D v E Add. 72.“ If we are a rich nation then we can help the poor. If we can help the poor then surely we should.On the other hand, if the poor can help themselves then we shouldn’t help them. But if the poorcan’t help themselves then there are not sufficient jobs. If there are not sufficient jobs then we can’thelp the poor. But either the poor can help themselves or they cannot. Hence, either way, it seemsthat we are not a rich nation! ”
9. 9. Let P = we are a rich nation, Q = we can help the poor, R = we should help the poor, S = the poorcan help themselves,T = there are sufficient jobsProof:1. P ⇒ Q2. Q ⇒ R3. S ⇒  R4.  S ⇒  T5.  T ⇒  Q6. S ∨  S∴  P7. ¬S ⇒ ¬Q HS (4,5)8. Q ⇒ S Contrapositive (7)9. Q ⇒ ¬R HS ( 8,3)10. ¬R ⇒ ¬Q Contrapositive (2)11 Q ⇒ ¬Q HS (9,10)12. ¬Q ∨ ¬Q Material Implication (11)13. ¬Q Idempotence (12)14. ¬P MT (1,13)3.“ If the test subject asks for food then he is hungry and if he is deprived of food for twenty-four hours,he is hungry. Analysis of the test subject’s blood will indicate an absence of significant amount ofglucose only if he is deprived of food or asks for food. Hence, if the analysis of the subjects bloodshows an absence of significant amount of glucose, he is hungry.”Let P = subject asks for food, Q = he is hungry, R = he is deprived of food, S = Analysis shows anabsence of glucoseProof:1. ( P ⇒ Q ) ∧ ( R ⇒ Q )2. S ⇒ ( R ∨ P )∴ S ⇒ Q3. S Conditional Proof∴ Q4. R ∨ P MP (2,3)5. P ∨ R Comm (4)6. Q ∨ Q CD (1,5)7. Q Idempotence4. If universities charge no tuition, everyone will enter. If universities charge tuition, there is no equalityof opportunity unless deserving students get scholarships. But universities must either not charge orcharge tuition. Moreover, it is false that everyone will enter universities. Accordingly, if there isequality of opportunity, then deserving students get scholarship.LetP- universities charge tuitionQ-everyone will enterR-there is equality in opportunity
10. 10. S-deserving students get scholarships1. ¬P ⇒ Q2. p ⇒ ¬RvS3. P ∨ ¬P4. ¬Q______∴R⇒S5.P6.P∨Q7.¬R∨S8.R⇒S5. If the world is chaos, then it cannot be reformed unless a sage appears. But no sage can appear if theworld is chaos. The world surely is chaos. Hence, it cannot be reformed.Let P= The world is chaos, Q= The world can be reformedR-sage appears1.P⇒(¬Q∨R)2.P⇒¬R3.P______∴¬Q1.P⇒(¬Q∨R)2.P⇒¬R3.P______∴¬Q4.¬Q∨R5.¬R6.¬QUSEFUL TRANSLATIONsP ∧ QP and Q P despite QP but Q Despite P, QP although Q P inspite of QAlthough P, Q P whereas QNot only P, but also Q P yet QWhile P, Q P while QP ∨ QP or Q P and/or QP or Q or both P unless QEither P or QP ⇒ QIf P then Q Q if PIf P, Q Q provided that PP only if Q Q incase of P
11. 11. P implies QP ⇔ QP if and only if Q If p and only then QIf and only if P, Q P just incase QIf P, and only if Q