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

PROOF METHODS

CONCEPTS / VOCABULARY Theorems Rules of inference 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

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)

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 □.

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)

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)

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.

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]

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

Write “Q.E.D.” Q.E.D. or □

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 did work hard and did study). Then [insert the sequence of steps on slide 13] Therefore, P1  P2  P3 → Q. □

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.

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.

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

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?

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

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.