1. 1 Introduction to Abstract Mathematics Chapter 2: The Logic of Quantified Statements. Predicate Calculus Instructor: Hayk Melikya melikyan@nccu.edu 2.3 Arguments with Quantified Statements

2. 2 Introduction to Abstract Mathematics Argument #1 Universal instantiation All men are mortal. Socrates is a man. Therefore, Socrates is mortal (  x  U)P(x) ├ P(a) Universal instantiation is the fundamental tool for deductive reasoning

3. 3 Introduction to Abstract Mathematics Arguments with Quantified Statements Rule of universal instantiation: if some property is true of everything in the domain, then this property is true for any subset in the domain Universal Modus Ponens: u Premises: (  x, if P(x) then Q(x)); (major) – P(a) for some a (minor) u Conclusion: Q(a) Universal Modus Tollens: u Premises: (  x, if P(x) then Q(x)); ~Q(a) for some a u Conclusion: ~P(a) Converse and inverse errors Definition: An argument form is valid if no matter what particular predicates are substituted for predicate symbols in it promises, if resulting promise statements are true, then the conclusion is also true An argument is called valid iff its form is valid

4. 4 Introduction to Abstract Mathematics Validity of Arguments using Diagrams Premises: All human beings are mortal; Zeus is not mortal. Conclusion: Zeus is not a human being Premises: All human beings are mortal; Felix is mortal. Conclusion: Felix is a human being Premises: No polynomial functions have horizontal asymptotes; This function has a horizontal asymptote. Conclusion: This function is not a polynomial

5. 5 Introduction to Abstract Mathematics Diagrams for Validity (p. 104) Diagrams can sometimes be used to: –support a validity of an argument –or, show that an argument is invalid Diagrams are not a formal proof! mortal human SocratesSocrates pneumonia fever patientpatient

6. 6 Introduction to Abstract Mathematics True no matter what the Domain is, or the predicates are.  z [Q(z)  P(z)] → [  x.Q(x)   y.P(y)] Predicate Calculus Validity True no matter what the truth values of A and B are Propositional validity Predicate calculus validity That is, logically correct, independent of the specific content.

7. 7 Introduction to Abstract Mathematics Arguments with Quantified Statements Universal instantiation: Universal modus ponens: Universal modus tollens:

8. 8 Introduction to Abstract Mathematics ├ (  x  U)P(x ) To prove a theorem of the form (  x  U)P(x) which states “for all elements x in a given universe U, the proposition P(x) is true” we select an arbitrary a  U from the universe, and then prove the assertion P(a). Let a be an arbitrary constant from the universe U. If P(a) contains no particular constant from U then P(a) ├ (  x  U)P(x) This is called Universal Generalization

9. 9 Introduction to Abstract Mathematics Example 1 (Universal Direct Proof) Show that all integers divisible by 6 are even. Proof: In the language of predicate logic, we write (  x  Z) ( 6 divides x  x is even) where Z = {0,±1,± 2,...} is the universe of integers. Letting a be an integer, we assume a is divisible by 6, which means there exists an integer y which satisfies a = 6y. Rewriting this as a= 2(3y) we have a = 2k for some integer k = 3y, which proves that a is an even integer. ▌

10. 10 Introduction to Abstract Mathematics Universal Instantiation (UI) (  x  U)P(x) ├ P(a) If a is an arbitrary constant from the universe U then (  x  U)P(x) ├ P(a) This is refered as Universal Instantiation rule. Existential Instantiation (EI) (  x  U)P(x) ├ P(a) where a is a paricular contant from univers Existantial Generalization(EG) P(a) ├ (  x  U)P(x) This rule says that if P(a) true for some constan from the universe U then (  x  U)P(x) is true

11. 11 Introduction to Abstract Mathematics ├ (  x  U)P(x ) To prove a theorem of the form (  x  U)P(x) which states “there exists an element x in a given universe U that satisfies the proposition P(x) ” the strategy is to show one or more elements x  U satisfy the assertion P(x). Example 2 (Proof by Demonstration or construction) Show there exists an even prime integer. Proof: In language of predicate logic, we would write (  x  N)(x is even  x is prime) The proof is simple because 2 is both prime and even

12. 12 Introduction to Abstract Mathematics Proof of (  x)P(x) by Contradiction: To prove the theorem (  x)P(x) which says “for all x, P(x) is true” by contradiction, assume the contrary; i.e. ~(  x)P(x) which is equivalent to (  x) ~P(x), which says “there exists an x such that P(x) is not true”. One then continues the Proof until arriving at a contradiction.

13. 13 Introduction to Abstract Mathematics Example 4: (Proof by Contradiction ) Show if m, n are integers, then 5m+ 20n  1. Proof: In the language of predicate logic this theorem becomes (  m  Z)(  n  Z) (5m+ 20n  1). Assuming the contrary, we have (  m  Z)(  n  Z) (5m+ 20n = 1). But the equation 5m+ 20n = 1 cannot hold since 5 divides the left side of the equation and not the right. Hence, the denial is false so the theorem is true. ▌

14. 14 Introduction to Abstract Mathematics Proof of (  x)P(x) by Contradiction : To prove “there exists an x such that P(x) is true” assume the theorem is not true: i.e. ~(  x) P(x)  (  x) ~P(x) which states“for all x the assertion P(x) is not true. You then continue the proof until you arrive at a contradiction of some kind.

15. 15 Introduction to Abstract Mathematics Proving Unique Existential Theorems: To prove a theorem of the form (  !x  U )P(x) which states “there exists a unique element x such that P(x) is true” the strategy is to show first that some element x satisfies P(x), then show that if two elements y, z  U satisfy the assertion, then in fact they are the same; i.e. y = z. In predicate logic language, we must show (  x  U )P(x)  (  y  U )(  z  U )[P( y)  P(z)  y = x]

16. 16 Introduction to Abstract Mathematics Important Relations in Predicate Logic a) (  x)(  y)P(x, y)  (  y)(  x)P(x, y) b) (  x)(  y)P(x, y)  (  y)(  x)P(x, y) c) (  x)[P(x)  Q(x)]  [ (  x)P(x)  (  x)Q(x) ] d) (  x)[P(x)  Q(x)]  [(  x)P(x)  (  x)Q(x)] e) [(  x)P(x)  (  x)Q(x)]  (  x)[P(x)  Q(x)] f) (  x)(  y)P(x, y)  (  y)(  x)P(x, y)

17. 17 Introduction to Abstract Mathematics Proof: Give countermodel, where  z [Q(z)  P(z)] is true, but  x.Q(x)   y.P(y) is false. In this example, let domain be integers, Q(z) be true if z is an even number, i.e. Q(z)=even(z) P(z) be true if z is an odd number, i.e. P(z)=odd(z)  z [Q(z)  P(z)] → [  x.Q(x)   y.P(y)] Not Valid Find a domain, and a predicate. Validity  z [Q(z)  P(z)] → [  x.Q(x)   y.P(y)] Proof strategy: We assume  z [Q(z)  P(z)] and prove  x.Q(x)   y.P(y)

18. 18 Introduction to Abstract Mathematics Proof: Assume  z [Q(z)  P(z)]. So Q(z)  P(z) holds for all z in the domain. Now let c be some domain element. So Q(c)  P(c) holds, and therefore Q(c) by itself holds. But c could have been any element of the domain. So we conclude  x.Q(x). We conclude  y.P(y) similarly. Therefore,  x.Q(x)   y.P(y) QED. (by UG)  z [Q(z)  P(z)] → [  x.Q(x)   y.P(y)] Validity

19. 19 Introduction to Abstract Mathematics Proof and Logic We prove mathematical statement by using logic. not valid To prove something is true, we need to assume some axioms! This is invented by Euclid in 300 BC, who begins with 5 assumptions about geometry, and derive many theorems as logical consequences.

20. 20 Introduction to Abstract Mathematics Validity of Arguments Hence validity of arguments is defined in the same way The difference is: –in predicate logic it is not always possible to go through all interpretations to prove that P logically implies Q Why? –The number of interpretations can be infinite Thus, proving arguments with inference rules becomes the method of choice We can also derive new inference rules for our toolbox

21. 21 Introduction to Abstract Mathematics Thm: For any “reasonable” theory that proves basic arithemetic truth, an arithmetic statement that is true, but not provable in the theory, can be constructed. Gödel's Incompleteness Theorem for Arithmetic Power and Limits of Logic No hope to find a complete and consistent set of axioms! Thm : Given a set of axioms, there is no procedure that decide whether quantified assertions are valid. (unlike propositional formulas) That is, starting from a few propositional & simple predicate validities, every valid assertion can be proved using just universal generalization and modus ponens repeatedly! Good news: Gödel's Completeness Theorem

22. 22 Introduction to Abstract Mathematics Practice problems 1. Study the Sections 3.3 and 3.4 from your textbook. 2. Be sure that you understand all the examples discussed in class and in textbook. 3. Do the following problems from the textbook: Exercise 3.3 # 1, 11, 16, 19, 21, 24, 30, 41, 55, 57. Exercise 3.4 # 1, 4, 11, 14, 22, 26, 32, 34.