Proof Review - Discrete Mathematics - Lecture Slides, Slides of Discrete Mathematics

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Proof Review

Proof methods in this course

• Logical equivalences
  • via truth tables
  • via logical equivalences
• Set equivalences
  • via set identities
  • via membership tables
  • via mutual subset proof
  • via set builder notation and logical equivalences
• Rules of inference
  • for propositions
  • for quantified statements
• Proofs without a category
  • Pigeonhole principle
  • Combinatorial proofs
    • Ten proof methods in chapter 3:
      • Direct proofs
      • Indirect proofs
      • Vacuous proofs
      • Trivial proofs
      • Proof by contradiction
      • Proof by cases
      • Proofs of equivalence
      • Existence proofs
        • Constructive
        • Non-constructive
      • Uniqueness proofs
      • Counterexamples
    • Induction
      • Weak mathematical induction
      • Strong mathematical induction
      • Structural induction

Idempotent Law

Associativity of Or

Using Logical Equivalences

(¬ p ∨ r )∨(¬ q ∨ r )≡(¬ p ∨¬ q )∨ r

¬ p ∨¬ q ∨ r ≡ ¬ p ∨¬ q ∨ r

(¬ p ∨ r )∨(¬ q ∨ r ) ≡ ¬( p ∧ q )∨ r

¬ p ∨¬ q ∨ r ∨ r ≡ ¬ p ∨¬ q ∨ r

Definition of implication

( p → r )∨( q → r ) ≡ ( p ∧ q ) → r

¬ p ∨ r ∨¬ q ∨ r ≡ ¬ p ∨¬ q ∨ r

Re-arranging

Original statement

DeMorgan’s Law

p → q ≡ ¬ p ∨ q

¬( p ∧ q ) ≡ ¬ p ∨¬ q

(¬ p ∨ r )∨(¬ q ∨ r ) ≡ ¬ p ∨ r ∨¬ q ∨ r

r ∨ r ≡ r

Definition of difference

Definition of difference

DeMorgan’s law

Complementation law

Distributive law

Complement law

Identity law

Commutative law

Proof by using basic set identities

• Prove that A∩B=B-(B-A)

A  B = B-(B  A )

= B  (B  A )

= B  (B  A )

= B  (B  A)

= (B  B)  (B  A)

= ∅ (B  A)

= (B  A)

= A  B

Proof by showing each set

is a subset of the other 1

• Assume that x∈B-(B-A)
  • By definition of difference, we know that x∈B and x∉B-A
• Consider x∉B-A
  • If x∈B-A, then (by definition of difference) x∈B and x∉A
  • Since x∉B-A, then only one of the inverses has to be true (DeMorgan’s law): x∉B or x∈A
• So we have that x∈B and (x∉B or x∈A)
  • It cannot be the case where x∈B and x∉B
  • Thus, x∈B and x∈A
  • This is the definition of intersection
• Thus, if x∈B-(B-A) then x∈A∩B

Proof by showing each set

is a subset of the other 2

• Assume that x∈A∩B
  • By definition of intersection, x∈A and x∈B
• Thus, we know that x∉B-A
  • B-A includes all the elements in B that are also not in A not include any of the elements of A (by definition of difference)
• Consider B-(B-A)
  • We know that x∉B-A
  • We also know that if x∈A∩B then x∈B (by definition of intersection)
  • Thus, if x∈B and x∉B-A, we can restate that (using the definition of difference) as x∈B-(B-A)
• Thus, if x∈A∩B then x∈B-(B-A)

Modus Ponens 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

q

p q

p

Modus Tollens

• Assume that we know: ¬q and p → q

• Recall that p → q = ¬q → ¬p

• Thus, we know ¬q and ¬q → ¬p

• We can conclude ¬p

p

p q

q

More rules of inference

• Conjunction: if p and q are true

separately, then p∧q is true

• Elimination: If p∨q is true, and p

is false, then q must be true

• Transitivity: If p→q is true, and

q→r is true, then p→r must be

true

p q

q

p

q

p

p q

p r

q r

p q

Even more rules of inference

• Proof by division into cases: if at

least one of p or q is true, then r

must be true

• Contradiction rule: If ¬p→c is

true, we can conclude p (via the

contra-positive)

• Resolution: If p∨q is true, and

¬p∨r is true, then q∨r must be

true

• Not in the textbook

p q p r q r

r

p c

p

q r

p r

p q

Example of proof

1. ¬p ∧ q 1 st^ hypothesis
2. ¬p Simplification using step 1
3. r → p 2 nd^ hypothesis
4. ¬r Modus tollens using steps 2 & 3
5. ¬r → s 3 rd^ hypothesis
6. s Modus ponens using steps 4 & 5
7. s → t 4 th^ hypothesis
8. t Modus ponens using steps 6 & 7

p

p q

q

p q

p

p

p q

q

Example of proof

• Given the hypotheses:

• “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”?

(¬r ∨ ¬f) →

(s ∧ l)

s → t

¬t

r

Corollary 1: algebraic proof

• Let n be a non-negative integer. Then

• Algebraic proof

=

n 

k

n k

n

0

=

−  

n

k

k n k k

n

0

2 n^ =( 1 + 1 )^ n

=

n

k k

n

0

=

−  

n

j

n (^) xn jyj j

n x y 0

Corollary 1: combinatorial proof

• Let n be a non-negative integer. Then
• Combinatorial proof
  • A set with n elements has 2 n^ subsets
    • By definition of power set
  • Each subset has either 0 or 1 or 2 or … or n elements
    • There are subsets with 0 elements, subsets with 1 element, … and subsets with n elements
    • Thus, the total number of subsets is
  • Thus,

=

n 

k

n k

n

0

n n

k k

n 2 0

^ =

=

 0  n

=

n 

k k

n

0

 1  n

 n  n