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Proof Methods and Concepts: Theorems, Axioms, and Fallacies

Learn about proof methods, including direct proof, indirect proof, and proof by contradiction. Understand essential concepts such as theorems, axioms, fallacies, and rules of inference.

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Proof Methods and Concepts: Theorems, Axioms, and Fallacies

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  1. CMSC 203Fall 2004 September 9, 2004 Prof. Marie desJardins (for Prof. Matt Gaston)

  2. PROOF METHODS

  3. CONCEPTS / VOCABULARY • Theorems • Axioms / postulates / premises • Hypothesis / conclusion • Lemma, corollary, conjecture • Rules of inference • Modus ponens (P and P→Q imply Q) • Modus tollens (P→Q and ¬Q imply ¬P) • Hypothetical syllogism (P→Q and Q→R imply that P→R) • Disjunctive syllogism (PQ and ¬P imply Q) • Universal instantiation, universal generalization, existential instantiation (skolemization or Everybody Loves Raymond), existential generalization

  4. CONCEPTS / VOCABULARY II • Fallacies • Affirming the conclusion [abductive reasoning] • P→Q and Q do not imply P • Denying the hypothesis • P→Q and ¬P do not imply ¬Q • Begging the question (circular reasoning) • Proof methods • Direct proof (show that P→Q by a series of logical steps from P to Q) • Indirect proof (show that P→Q by proving that ¬Q→¬P) • Proof by contradiction (show P by showing that ¬P implies FALSE) • Trivial proof (show that P→Q by showing that Q is always true) • Proof by cases • Existence proofs (constructive, nonconstructive)

  5. Let’s make a proof • Prove that the premises P1, P2, and P3 imply the conclusion Q: • P1: Students who work hard will do well on the test. • P2: Students who do well on the test and study will get an A. • P3: Bill is a student who gets a B. • Q: Bill either didn’t work hard or didn’t study. • General problem-solving approach to proof construction: • Restate the problem, writing the premise and conclusion in mathematical language. • Decide what type of proof to use. • Apply any relevant definitions, axioms, laws, or theorems to simplify the premise, make it look more like the conclusion, or connect (relate) multiple premises. • Carefully write down and justify each step of the proof, in a sequence of connected steps. • Write a conclusion statement. • Write “Q.E.D.” or □.

  6. Restate the problem • Prove that the premises P1, P2, and P3 imply the conclusion Q. • P1: Students who work hard will do well on the test. • P2: Students who do well on the test and study will get a good grade. • P3: Bill is a student who doesn’t get a good grade. • Q: Bill either didn’t work hard or didn’t study. • P1: x student(x)  work-hard(x) → do-well(x) • P2: x student(x)  do-well(x)  study(x) → good-grade(x) • P3: student(Bill)  ¬good-grade(Bill) • Q: ¬work-hard(Bill)  ¬study(Bill)

  7. Restate the problem • We wish to prove that if x student(x)  work-hard(x) → do-well(x) (P1) and x student(x)  do-well(x)  study(x) → good-grade(x) (P2) and student(Bill)  ¬good-grade(x), (P3) then ¬work-hard(Bill)  ¬study(Bill). (Q)

  8. Select a proof type • Proof by contradiction • Negate the proposition to be proved and derive FALSE • The negation of P1  P2  P3 → Q is P1  P2  P3  ¬Q • Mini-quiz: why? • The negated conclusion ¬Q is¬Q  ¬ (¬work-hard(Bill)  ¬study(Bill))  ¬¬work-hard(Bill)  ¬¬study(Bill) De Morgan’s law  work-hard(Bill)  study(Bill) Double negation • (see Table 5, p. 24) • So if we can prove FALSE from P1  P2  P3  ¬Q, then we can conclude that P1  P2  P3 → Q.

  9. Apply relevant knowledge / Justify • P1: x student(x)  work-hard(x) → do-well(x) • (1) student(Bill)  work-hard(Bill) → do-well(Bill) by P1 and univ. instantiation [Table 2, p. 62] • P2: x student(x)  do-well(x)  study(x) → good-grade(x) • (2) student(Bill)  do-well(Bill)  study(Bill) → good-grade(Bill) by P2 and univ. instantiation • P3: student(Bill)  ¬good-grade(Bill) • (3) student(Bill) by P3 and simplification [Table 1, p. 58] • (4) ¬good-grade(Bill) by P3 and simplification • ¬Q: work-hard(Bill)  study(Bill) • (5) work-hard(Bill) by ¬Q and simplification • (6) study(Bill) by ¬Q and simplification • (7) student(Bill)  work-hard(Bill) by (3), (5) and conjunction • (8) do-well(Bill) by (1), (7) and modus ponens • (9) student(Bill)  do-well(Bill) study(Bill) by (3), (8), (6) and conjunction • (10) good-grade(Bill) by (2), (9) and modus ponens • (11) good-grade(Bill)  ¬good-grade(Bill) by (4), (10) and conjunction • FALSE by (11) and the negation law [Table 2, p. 24]

  10. Write a conclusion statement • Therefore, P1  P2  P3 → Q.

  11. Write “Q.E.D.” • Q.E.D. or□

  12. The Proof • Theorem. If students who work hard will do well on the test (P1), students who do well on the test and study will get an A (P2), and Bill is a student who gets a B (P3), then Bill either didn’t work hard or didn’t study (Q). • Proof. By contradiction. Suppose that the premises hold and the conclusion Q is false (i.e., Bill didwork hard and didstudy). Then • [insert the sequence of steps on slide 13] Therefore, P1  P2  P3 → Q. □

  13. Perfect numbers • A perfect number is an integer that is equal to the sum of its proper divisors. • 6 is a perfect number, since 1 + 2 + 3 = 6 • 28 is a perfect number, since 1 + 2 + 4 + 7 + 14 = 28 • Show that x=2p-1(2p-1) is a perfect number when 2p-1 is prime.

  14. Restate the problem • Show that x=2p-1(2p-1) is a perfect number when 2p-1 is prime. • Use the definition of a perfect number to restate more precisely/mathematically/explicitly. • Show that the sum of the proper divisors of x=2p-1(2p-1) is equal to x when 2p-1 is prime.

  15. Select a proof type • It seems like we should be able to work forward, so we’ll try a direct proof. • Common error: Proof by cases, applied to a few cases. Not valid, because the cases must be exhaustive, i.e., cover all possible situations: • odd numbers and even numbers • negative numbers, positive numbers, and zero • integers that are 0, 1, or 2 mod 3

  16. Apply relevant knowledge • Often it’s useful to verify the proposition for some specific instances. • x=2p-1(2p-1) is a perfect number when 2p-1 is prime. • Let p=2. Then 2p-1=3, which is prime, and x=6, which we already showed to be a perfect number. • Let p=3. Then 2p-1=7, which is prime, and x=28, which is also a perfect number. • Let p=5. Then 2p-1=31, which is prime, and x=16*31=406. Is that a perfect number? How will we list its divisors?

  17. Apply relevant knowledge, cont. • Premises: • x=2p-1(2p-1) • 2p-1 is prime • What we know about divisors: • The divisors of x are 2p-1, 2p-1, their divisors, and the products of their divisors. • The divisors of 2p-1 are 1, 2, 22, …, 2p-1. • Since 2p-1 is prime, its only divisors are 1 and 2p-1. • The proper divisors of x are the unique combinations of these divisors (except x itself) • One more step: figure out all the divisors and their sum

  18. The Proof • Theorem. x=2p-1(2p-1) is a perfect number when 2p-1 is prime. • Proof. For x to be a perfect number, it must be equal to the sum of its proper divisors. The divisors of 2p-1 are 1, 2, 22, ..., 2p-1. Since 2p-1 is prime, its only divisors are 1 and itself. Therefore, the proper divisors of x are 1, 2, 22, ..., 2p-1, 2(2p-1), 22(2p-1 ), …, 2p-2(2p-1). The sum of these divisors is i=0p-1 2i + (2p-1) i=0p-2 2i = 2p-1 + (2p-1) (2p-1 – 1) = (2p-1) (1 + 2p-1 – 1) = 2p-1 (2p-1) • Therefore, x=2p-1 (2p-1) is a perfect number. • Q.E.D.

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