1. Arguments with Quantified Statements Lecture 10 Section 2.4 Thu, Jan 27, 2005

2. Universal Modus Ponens • The universal modus ponens argument form: x  S, P(x)  Q(x) P(a) for a particular a  S Q(a)

3. Example • Let F be the set of all functions from R to R. • f  F, if f is differentiable, then f is continuous. • The function f(x) = x2 + 1 is differentiable. • Therefore, f(x) is continuous.

4. An Argument within in Argument • f  F, if f is differentiable, then f is continuous. • f  F, if f is a polynomial, then f is differentiable. • The function f(x) = x2 + 1 is a polynomial. • Therefore, f(x) is differentiable. • Therefore, f(x) is continuous.

5. Universal Transitivity • The previous example could have been handled differently using the argument form of universal transitivity: x  S, P(x)  Q(x) x  S, Q(x)  R(x)  x  S, P(x)  R(x)

6. Universal Transitivity • Equivalently, P(x)  Q(x) Q(x)  R(x)  P(x)  R(x)

7. Universal Modus Tollens • The universal modus tollens argument form: x  S, P(x)  Q(x) ~Q(a) for a particular a  S  ~P(a)

8. Diagrams • The statement x  S, P(x)  Q(x) means that the truth set of P is a subset of the truth set of Q. • The statement P(a) means that a is in the truth set of P. • Therefore, a must be in the truth set of Q.

9. Truth set of Q Truth set of P a Diagrams • Therefore, we can represent these statements by using Venn diagrams.

10. Truth set of Q Truth set of P a A Diagram for Universal Modus Ponens x  S, P(x)  Q(x) P(a) for a particular a  S Q(a)

11. Continuous functions Example

12. Continuous functions Differentiable functions Example

13. Continuous functions Differentiable functions Polynomial functions Example

14. Example Continuous functions Differentiable functions Polynomial functions f(x) = x2 + 1

15. Diagrams • Recall the example that showed that x  R, x/(x + 2)  3  x  -3.

16. {xR | x -3} {xR | x/(x + 2)  3} Example

17. 0 -4 1 -3 2 -2 3 -1 4 Example • A better representation:

18. Statements with “No” • Rewrite the statement “No HSC student would ever lie” using quantifiers. • ~(x  {HSC students}, x would lie) • x  {HSC students}, ~(x would lie) • x  {HSC students}, x would not lie • Thus, this is a universal statement.

19. Arguments with “No” • Which arguments are valid? • No HSC student would ever lie. Joe is an HSC student. Therefore, Joe would never lie. • No HSC student would ever lie. Buffy is an RMC student. Therefore, Buffy would lie.

20. Joe Arguments with “No” • The diagram shows that Joe cannot be a liar. People HSC Students Liars

21. Statements with “No” • Note that the following two statements are equivalent. • No HSC student is a liar. • No liar is an HSC student.

22. Joe Arguments with “No” • Where would we place the oval for RMC students? People HSC Students Liars

23. Arguments with “No” • Where would we place the oval for RMC students? People RMC Students ? HSC Students Liars

24. Arguments with “No” • Where would we place the oval for RMC students? People RMC Students ? HSC Students Liars

25. Arguments with “No” • Where would we place the oval for RMC students? People HSC Students Liars RMC Students ?

26. Arguments with “No” • Where would we place the oval for RMC students? RMC Students ? People HSC Students Liars

27. Arguments with “No” • Where would we place Buffy? People HSC Students Liars RMC Students

28. Buffy Arguments with “No” • Where would we place Buffy? People HSC Students Liars RMC Students

29. Arguments with “No” • Where would we place Buffy? People HSC Students Liars RMC Students Buffy

30. Arguments with “No” • Where would we place Buffy? People HSC Students Liars RMC Students Buffy

31. Arguments with “No” • Which fallacy is committed in the “Buffy” argument?

32. A Logical Conclusion? • Is the following argument valid? x, y, z, if x is better than y and y is better than z, then x is better than z. A peanut butter sandwich is better than nothing. Nothing is better than sex.  A peanut butter sandwich is better than sex.