1 Basic Proof Methods I Basic Proof Methods II Proofs Involving Quantifiers

 Consider (p  (p→q)) → q 2 Example: Assume you are given the following two statements: “you are in this class” “if you are in this class, you will get a grade” Let p = “you are in this class” Let q = “you will get a grade” By Modus Ponens, you can conclude that you will get a grade

Assume that we know: ¬q and p → q We can conclude ¬p 3 Example: Assume you are given the following two statements: “you will not get a grade” “if you are in this class, you will get a grade” Let p = “you are in this class” Let q = “you will get a grade” By Modus Tollens, you can conclude that you are not in this class

 Addition: If you know that p is true, then p  q will ALWAYS be true  Simplification: If p  q is true, then p will ALWAYS be true 4

5 “It is not sunny this afternoon and it is colder than yesterday” “It is not sunny this afternoon and it is colder than yesterday” “We will go swimming only if it is sunny” “We will go swimming only if it is sunny” “If we do not go swimming, then we will take a canoe trip” “If we do not go swimming, then we will take a canoe trip” “If we take a canoe trip, then we will be home by sunset” “If we take a canoe trip, then we will be home by sunset” Does this imply that “we will be home by sunset”? p q r s t ¬p  q¬p  q r → p ¬r → s s → t t

1. ¬p  q1 st hypothesis 2. ¬pSimplification using step 1 3. r → p2 nd hypothesis 4. ¬rModus tollens using steps 2 & 3 5. ¬r → s3 rd hypothesis 6. sModus ponens using steps 4 & 5 7. s → t 4 th hypothesis 8. tModus ponens using steps 6 & 7 6

 Conjunction: if p and q are true separately, then p  q is true  Disjunctive: If p  q is true, and p is false, then q must be true  Resolution: If p  q is true, and ¬p  r is true, then q  r must be true  Conditional Transitivity: If p→q is true, and q→r is true, then p→r must be true 7

 “If it does not rain or if it is not foggy, then the sailing race will be held and the lifesaving demonstration will go on”  “If the sailing race is held, then the trophy will be awarded”  “The trophy was not awarded” Can you conclude: “It rained”? 8 (¬r  ¬f) → (s  l) s → t ¬t r

1. ¬t3 rd hypothesis 2. s → t2 nd hypothesis 3. ¬sModus tollens using steps 2 & 3 4. (¬r  ¬f)→(s  l)1 st hypothesis 5. ¬(s  l)→¬(¬r  ¬f)Contrapositive of step 4 6. (¬s  ¬l)→(r  f)DeMorgan’s law and double negation law 7. ¬s  ¬lAddition from step 3 8. r  fModus ponens using steps 6 & 7 9. rSimplification using step 8 9

(If p then q): p→q 10 Example: Show that the square of an even number is an even number (i.e. if n is even, then n 2 is even) Proof: Assume n is even Thus, n = 2k, for some k (definition of even numbers) But, n 2 = (2k) 2 = 4k 2 = 2(2k 2 ) As n 2 is 2 times an integer, n 2 is thus even

To show: p→q, instead show the contra positive ¬q→¬p 11 Example: If n 2 is an odd integer then n is an odd integer Proof: by contrapositive, we need to show: If n is an even integer, then n 2 is an even integer Since n is even n=2k for some integer k (definition of even numbers) n 2 = (2k) 2 = 4k 2 = 2(2k 2 ) Since n 2 is 2 times an integer, it is even

When do you use a direct proof versus an indirect proof? - If it’s not clear from the problem, try direct first, then indirect second - If indirect fails, try the other proofs 12

 Problem:  Prove that if n is an integer and n 3 +5 is odd, then n is even  Via direct proof  n 3 +5 = 2k+1 for some integer k (definition of odd numbers)  n 3 = 2k-4   Then what …  So direct proof didn’t work out. Next up: indirect proof 13

 Problem:  Prove that if n is an integer and n 3 +5 is odd, then n is even  Via indirect proof (Contrapositive)  Need to show: If n is odd, then n 3 +5 is even  Assume n is odd, and show that n 3 +5 is even  n=2k+1 for some integer k (definition of odd numbers)  n 3 +5 = (2k+1) 3 +5 = 8k 3 +12k 2 +6k+6 = 2(4k 3 +6k 2 +3k+3)  As 2(4k 3 +6k 2 +3k+3) is 2 times an integer, it is even 14

 Given a statement of the form p→q Just consider the case where p is true and q is false then arrive to a contradiction. 15 Example: Prove that if n is an integer and n 3 +5 is odd, then n is even Rephrased: If n 3 +5 is odd, then n is even Proof: Assume p is true and q is false. Assume n 3 +5 is odd, and n is odd). Since n is odd, n=2k+1 for some integer k (definition of odd numbers) But, n 3 +5 = (2k+1) 3 +5 = 8k 3 +12k 2 +6k+6 = 2(4k 3 +6k 2 +3k+3) As 2(4k 3 +6k 2 +3k+3) is 2 times an integer, it must be even Contradiction!

Consider an implication: p→q If it can be shown that p is false, then the implication is always true (by definition of an implication) 16 Example: Consider the statement: All criminology majors in Math 3313 are female (i.e. If you are a criminology major and you are in Math 3313, then you are female) Proof: Since there are no criminology majors in this class, the antecedent is false, and the implication is true

Consider an implication: p→q If it can be shown that q is true, then the implication is always true by definition of an implication 17 Example: Consider the statement: If you are tall and are in Math 3313 then you are a student Proof: Since all people in Math 3313 are students, the implication is true regardless

Show a statement is true by showing all possible cases are true You are showing 18

 Prove that  Note that b ≠ 0  Cases:  Case 1: a ≥ 0 and b > 0  Then |a| = a, |b| = b, and  Case 2: a ≥ 0 and b < 0  Then |a| = a, |b| = -b, and  Case 3: a 0  Then |a| = -a, |b| = b, and  Case 4: a < 0 and b < 0  Then |a| = -a, |b| = -b, and 19

 This is showing the definition of a bi- conditional  Given a statement of the form “p if and only if q”  Show it is true by showing (p→q)  (q→p) is true 20

Problem:  Show that m 2 =n 2 if and only if m=n or m=-n (i.e. (m 2 =n 2 ) ↔ [(m=n)  (m=-n)] ) Proof: Need to prove two parts:  [(m=n)  (m=-n)] → (m 2 =n 2 )  Proof by cases!  Case 1: (m=n) → (m 2 =n 2 )  (m) 2 = m 2, and (n) 2 = n 2, so this case is proven  Case 2: (m=-n) → (m 2 =n 2 )  (m) 2 = m 2, and (-n) 2 = n 2, so this case is proven  (m 2 =n 2 ) → [(m=n)  (m=-n)]  Subtract n 2 from both sides to get m 2 -n 2 =0  Factor to get (m+n)(m-n) = 0  Since that equals zero, one of the factors must be zero  Thus, either m+n=0 (which means m=n)  Or m-n=0 (which means m=-n) 21

 A theorem may state that only one such value exists  To prove this, you need to show:  Existence: that such a value does indeed exist  Either via a constructive or non-constructive existence proof  Uniqueness: that there is only one such value 22

 If the real number equation 5x+3=b has a solution then it is unique  Existence  We can manipulate 5x+3=b to yield x=(b-3)/5  Uniqueness  If there are two such numbers, then they would fulfill the following: b = 5x+3 = 5y+3  We can manipulate this to yield that x = y  Thus, the one solution is unique! 23

DISPROVING a UNIVERSAL statement by a counterexample Example: Every positive integer is the square of another integer Proof: The square root of 5 is 2.236, which is not an integer 24

  x P ( x )  P ( o )(substitute any object o )  P ( g )(for g a general element of universe.)  x P ( x )   x P ( x )  P ( c )(substitute a new constant c )  P ( o ) (substitute any extant object o )  x P ( x ) 25

“ Everyone in this foundation math class has taken a course in computer science ” and “ Marla is a student in this class ” imply “ Marla has taken a course in computer science ” F(x): “ x is in foundation math class ” C(x): “ x has taken a course in computer science ”  x (F(x)  C(x)) F(Marla)  C(Marla) 26

StepProved by 1.  x (F(x)  C(x)) Premise #1. 2. F(Marla)  C(Marla)Univ. instantiation. 3. F(Marla)Premise #2. 4. C(Marla)Modus ponens on 2,3. 27

“ A student in this class has not read the book ” and “ Everyone in this class passed the first exam ” imply “ Someone who passed the first exam has not read the book ” C(x): “ x is in this class ” B(x): “ x has read the book ” P(x): “ x passed the first exam ”  x (C(x)   B(x))  x (C(x)  P(x))   x( P(x)   B(x)) 28

StepProved by 1.  x (C(x)   B(x))Premise #1. 2. C(a)   B(a) Exist. instantiation. 3. C(a)Simplification on  x (C(x)  P(x)) Premise #2. 5. C(a)  P(a) Univ. instantiation. 6. P(a)Modus ponens on 3,5 7.  B(a) Simplification on 2 8. P(a)   B(a) Conjunction on 6,7 9.  x( P(x)   B(x))Exist. generalization 29

 Proving p  q  Direct proof: Assume p is true, and prove q.  Indirect proof: Assume  q, and prove  p.  Trivial proof: Prove q true.  Vacuous proof: Prove  p is true.  Proving p  Proof by contradiction: Prove  p  (r   r) ( r   r is a contradiction); therefore  p must be false.  Prove ( a  b )  p  Proof by cases: prove ( a  p ) and ( b  p ).  More … 30

 A proof of a statement of the form  x P ( x ) is called an existence proof.  If the proof demonstrates how to actually find or construct a specific element a such that P ( a ) is true, then it is called a constructive proof.  Otherwise, it is called a non-constructive proof. 31

 Theorem: There exists a positive integer n that is the sum of two perfect cubes in two different ways:  equal to j 3 + k 3 and l 3 + m 3 where j, k, l, m are positive integers, and { j, k } ≠ { l, m }  Proof: Consider n = 1729, j = 9, k = 10, l = 1, m = 12. Now just check that the equalities hold. 32

 Theorem: “There are infinitely many prime numbers.”  Any finite set of numbers must contain a maximal element, so we can prove the theorem if we can just show that there is no largest prime number.  i.e., show that for any prime number, there is a larger number that is also prime.  More generally: For any number,  a larger prime.  Formally: Show  n  p>n : p is prime. 33

 Some very simple statements of number theory haven’t been proved or disproved!  E.g. Goldbach’s conjecture : Every integer n ≥2 is exactly the average of some two primes.   n≥ 2  primes p, q : n =( p + q )/2.  There are true statements of number theory (or any sufficiently powerful system) that can never be proved (or disproved) (Gödel). 34

 Assume that we know that  x P(x) is true  Then we can conclude that P(c) is true  Here c stands for some specific constant  This is called “universal instantiation”  Assume that we know that P(c) is true for any value of c  Then we can conclude that  x P(x) is true  This is called “universal generalization” 35

 Given a statement:  x P(x)  We only have to show that a P(c) exists for some value of c  Two types:  Constructive: Find a specific value of c for which P(c) exists  Nonconstructive: Show that such a c exists, but don’t actually find it  Assume it does not exist, and show a contradiction 36

 Show that a square exists that is the sum of two other squares  Proof: = 5 2  Show that a cube exists that is the sum of three other cubes  Proof: =

 Problem: Prove that either 2* or 2* is not a perfect square  A perfect square is a square of an integer  Rephrased: Show that a non-perfect square exists in the set {2* , 2* }  Proof: The only two perfect squares that differ by 1 are 0 and 1  Thus, any other numbers that differ by 1 cannot both be perfect squares  Thus, a non-perfect square must exist in any set that contains two numbers that differ by 1  Note that we didn’t specify which one it was! 38

 Assume that we know that  x P(x) is true  Then we can conclude that P(c) is true for some value of c  This is called “existential instantiation”  Assume that we know that P(c) is true for some value of c  Then we can conclude that  x P(x) is true  This is called “existential generalization” 39

 Given the hypotheses:  “Linda, a student in this class, owns a red convertible.”  “Everybody who owns a red convertible has gotten at least one speeding ticket”  Can you conclude: “Somebody in this class has gotten a speeding ticket”? 40 C(Linda) R(Linda)  x (R(x)→T(x))  x (C(x)  T(x))

1.  x (R(x)→T(x))3 rd hypothesis 2. R(Linda) → T(Linda)Universal instantiation using step 1 3. R(Linda)2 nd hypothesis 4. T(Linda)Modes ponens using steps 2 & 3 5. C(Linda)1 st hypothesis 6. C(Linda)  T(Linda)Conjunction using steps 4 & 5 7.  x (C(x)  T(x))Existential generalization using step 6 41 Thus, we have shown that “Somebody in this class has gotten a speeding ticket”

 Given the hypotheses:  “There is someone in this class who has been to France”  “Everyone who goes to France visits the Louvre”  Can you conclude: “Someone in this class has visited the Louvre”? 42  x (C(x)  F(x))  x (F(x)→L(x))  x (C(x)  L(x))

1.  x (C(x)  F(x))1 st hypothesis 2. C(y)  F(y)Existential instantiation using step 1 3. F(y)Simplification using step 2 4. C(y)Simplification using step 2 5.  x (F(x)→L(x))2 nd hypothesis 6. F(y) → L(y)Universal instantiation using step 5 7. L(y)Modus ponens using steps 3 & 6 8. C(y)  L(y)Conjunction using steps 4 & 7 9.  x (C(x)  L(x))Existential generalization using step 8 43 Thus, we have shown that “Someone in this class has visited the Louvre”

 Experience!  In general, use quantifiers with statements like “for all” or “there exists” 44