discrete mathematics - predicates and sets

Discrete MathematicsPredicates and Sets

H. Turgut Uyar Aysegul Gencata Yayımlı Emre Harmancı

2001-2013

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Topics

1 PredicatesIntroductionQuantifiersMultiple Quantifiers

2 SetsIntroductionSubsetSet OperationsInclusion-Exclusion

Topics

Predicate

Definition

predicate (or open statement): a declarative sentence which

contains one or more variables, and

is not a proposition, but

becomes a proposition when the variables in itare replaced by certain allowable choices

Universe of Discourse

Definition

universe of discourse: Uset of allowable choices

examples:

Z: integersN: natural numbersZ+: positive integersQ: rational numbersR: real numbersC: complex numbers

Universe of Discourse

Definition

universe of discourse: Uset of allowable choices

examples:

Z: integersN: natural numbersZ+: positive integersQ: rational numbersR: real numbersC: complex numbers

Predicate Examples

Example

U = Np(x): x + 2 is an even integer

p(5): Fp(8): T

¬p(x): x + 2 is not an even integer

Example

U = Nq(x , y): x + y and x − 2y are even integers

q(11, 3): F , q(14, 4): T

Predicate Examples

Example

p(5): Fp(8): T

Example

q(11, 3): F , q(14, 4): T

Predicate Examples

Example

p(5): Fp(8): T

Example

q(11, 3): F , q(14, 4): T

Topics

Quantifiers

Definition

existential quantifier:predicate is true for some values

symbol: ∃read: there exists

symbol: ∃!read: there exists only one

Definition

universal quantifier:predicate is true for all values

symbol: ∀read: for all

Quantifiers

Definition

Quantifiers

Definition

Quantifiers

existential quantifier

U = {x1, x2, . . . , xn}∃x p(x) ≡ p(x1) ∨ p(x2) ∨ · · · ∨ p(xn)

p(x) is true for some x

universal quantifier

U = {x1, x2, . . . , xn}∀x p(x) ≡ p(x1) ∧ p(x2) ∧ · · · ∧ p(xn)

p(x) is true for all x

Quantifiers

existential quantifier

U = {x1, x2, . . . , xn}∃x p(x) ≡ p(x1) ∨ p(x2) ∨ · · · ∨ p(xn)

p(x) is true for some x

universal quantifier

U = {x1, x2, . . . , xn}∀x p(x) ≡ p(x1) ∧ p(x2) ∧ · · · ∧ p(xn)

p(x) is true for all x

Quantifier Examples

Example

p(x) : x ≥ 0

q(x) : x2 ≥ 0

r(x) : (x − 4)(x + 1) = 0

s(x) : x2 − 3 > 0

are the following expressions true?

∃x [p(x) ∧ r(x)]

∀x [p(x) → q(x)]

∀x [q(x) → s(x)]

∀x [r(x) ∨ s(x)]

∀x [r(x) → p(x)]

Quantifier Examples

Example

p(x) : x ≥ 0

q(x) : x2 ≥ 0

r(x) : (x − 4)(x + 1) = 0

s(x) : x2 − 3 > 0

∃x [p(x) ∧ r(x)]

∀x [p(x) → q(x)]

∀x [q(x) → s(x)]

∀x [r(x) ∨ s(x)]

∀x [r(x) → p(x)]

Quantifier Examples

Example

p(x) : x ≥ 0

q(x) : x2 ≥ 0

r(x) : (x − 4)(x + 1) = 0

s(x) : x2 − 3 > 0

∃x [p(x) ∧ r(x)]

∀x [p(x) → q(x)]

∀x [q(x) → s(x)]

∀x [r(x) ∨ s(x)]

∀x [r(x) → p(x)]

Quantifier Examples

Example

p(x) : x ≥ 0

q(x) : x2 ≥ 0

r(x) : (x − 4)(x + 1) = 0

s(x) : x2 − 3 > 0

∃x [p(x) ∧ r(x)]

∀x [p(x) → q(x)]

∀x [q(x) → s(x)]

∀x [r(x) ∨ s(x)]

∀x [r(x) → p(x)]

Quantifier Examples

Example

p(x) : x ≥ 0

q(x) : x2 ≥ 0

r(x) : (x − 4)(x + 1) = 0

s(x) : x2 − 3 > 0

∃x [p(x) ∧ r(x)]

∀x [p(x) → q(x)]

∀x [q(x) → s(x)]

∀x [r(x) ∨ s(x)]

∀x [r(x) → p(x)]

Quantifier Examples

Example

p(x) : x ≥ 0

q(x) : x2 ≥ 0

r(x) : (x − 4)(x + 1) = 0

s(x) : x2 − 3 > 0

∃x [p(x) ∧ r(x)]

∀x [p(x) → q(x)]

∀x [q(x) → s(x)]

∀x [r(x) ∨ s(x)]

∀x [r(x) → p(x)]

Negating Quantifiers

replace ∀ with ∃, and ∃ with ∀negate the predicate

¬∃x p(x) ⇔ ∀x ¬p(x)

¬∃x ¬p(x) ⇔ ∀x p(x)

¬∀x p(x) ⇔ ∃x ¬p(x)

¬∀x ¬p(x) ⇔ ∃x p(x)

replace ∀ with ∃, and ∃ with ∀negate the predicate

¬∃x p(x) ⇔ ∀x ¬p(x)

¬∃x ¬p(x) ⇔ ∀x p(x)

¬∀x p(x) ⇔ ∃x ¬p(x)

¬∀x ¬p(x) ⇔ ∃x p(x)

Theorem

¬∃x p(x) ⇔ ∀x ¬p(x)

Proof.

¬∃x p(x) ≡ ¬[p(x1) ∨ p(x2) ∨ · · · ∨ p(xn)]

⇔ ¬p(x1) ∧ ¬p(x2) ∧ · · · ∧ ¬p(xn)

≡ ∀x ¬p(x)

Theorem

¬∃x p(x) ⇔ ∀x ¬p(x)

Proof.

¬∃x p(x) ≡ ¬[p(x1) ∨ p(x2) ∨ · · · ∨ p(xn)]

⇔ ¬p(x1) ∧ ¬p(x2) ∧ · · · ∧ ¬p(xn)

≡ ∀x ¬p(x)

Theorem

¬∃x p(x) ⇔ ∀x ¬p(x)

Proof.

¬∃x p(x) ≡ ¬[p(x1) ∨ p(x2) ∨ · · · ∨ p(xn)]

⇔ ¬p(x1) ∧ ¬p(x2) ∧ · · · ∧ ¬p(xn)

≡ ∀x ¬p(x)

Theorem

¬∃x p(x) ⇔ ∀x ¬p(x)

Proof.

¬∃x p(x) ≡ ¬[p(x1) ∨ p(x2) ∨ · · · ∨ p(xn)]

⇔ ¬p(x1) ∧ ¬p(x2) ∧ · · · ∧ ¬p(xn)

≡ ∀x ¬p(x)

Predicate Equivalences

Theorem

∃x [p(x) ∨ q(x)] ⇔ ∃x p(x) ∨ ∃x q(x)

Theorem

∀x [p(x) ∧ q(x)] ⇔ ∀x p(x) ∧ ∀x q(x)

Predicate Equivalences

Theorem

∃x [p(x) ∨ q(x)] ⇔ ∃x p(x) ∨ ∃x q(x)

Theorem

∀x [p(x) ∧ q(x)] ⇔ ∀x p(x) ∧ ∀x q(x)

Predicate Implications

Theorem

∀x p(x) ⇒ ∃x p(x)

Theorem

∃x [p(x) ∧ q(x)] ⇒ ∃x p(x) ∧ ∃x q(x)

Theorem

∀x p(x) ∨ ∀x q(x) ⇒ ∀x [p(x) ∨ q(x)]

Theorem

∀x p(x) ⇒ ∃x p(x)

Theorem

∃x [p(x) ∧ q(x)] ⇒ ∃x p(x) ∧ ∃x q(x)

Theorem

∀x p(x) ∨ ∀x q(x) ⇒ ∀x [p(x) ∨ q(x)]

Theorem

∀x p(x) ⇒ ∃x p(x)

Theorem

∃x [p(x) ∧ q(x)] ⇒ ∃x p(x) ∧ ∃x q(x)

Theorem

∀x p(x) ∨ ∀x q(x) ⇒ ∀x [p(x) ∨ q(x)]

Topics

Multiple Quantifiers

∃x∃y p(x , y)

∀x∃y p(x , y)

∃x∀y p(x , y)

∀x∀y p(x , y)

Multiple Quantifier Examples

Example

U = Zp(x , y) : x + y = 17

∀x∃y p(x , y):for every x there exists a y such that x + y = 17

∃y∀x p(x , y):there exists a y so that for all x , x + y = 17

what if U = N?

Example

U = Zp(x , y) : x + y = 17

what if U = N?

Example

U = Zp(x , y) : x + y = 17

what if U = N?

Example

U = Zp(x , y) : x + y = 17

what if U = N?

Example

Ux = {1, 2} ∧ Uy = {A,B}

∃x∃y p(x , y) ≡ [p(1,A) ∨ p(1,B)] ∨ [p(2,A) ∨ p(2,B)]

∃x∀y p(x , y) ≡ [p(1,A) ∧ p(1,B)] ∨ [p(2,A) ∧ p(2,B)]

∀x∃y p(x , y) ≡ [p(1,A) ∨ p(1,B)] ∧ [p(2,A) ∨ p(2,B)]

∀x∀y p(x , y) ≡ [p(1,A) ∧ p(1,B)] ∧ [p(2,A) ∧ p(2,B)]

Example

Ux = {1, 2} ∧ Uy = {A,B}

∃x∃y p(x , y) ≡ [p(1,A) ∨ p(1,B)] ∨ [p(2,A) ∨ p(2,B)]

∃x∀y p(x , y) ≡ [p(1,A) ∧ p(1,B)] ∨ [p(2,A) ∧ p(2,B)]

∀x∃y p(x , y) ≡ [p(1,A) ∨ p(1,B)] ∧ [p(2,A) ∨ p(2,B)]

∀x∀y p(x , y) ≡ [p(1,A) ∧ p(1,B)] ∧ [p(2,A) ∧ p(2,B)]

Example

Ux = {1, 2} ∧ Uy = {A,B}

∃x∃y p(x , y) ≡ [p(1,A) ∨ p(1,B)] ∨ [p(2,A) ∨ p(2,B)]

∃x∀y p(x , y) ≡ [p(1,A) ∧ p(1,B)] ∨ [p(2,A) ∧ p(2,B)]

∀x∃y p(x , y) ≡ [p(1,A) ∨ p(1,B)] ∧ [p(2,A) ∨ p(2,B)]

∀x∀y p(x , y) ≡ [p(1,A) ∧ p(1,B)] ∧ [p(2,A) ∧ p(2,B)]

Example

Ux = {1, 2} ∧ Uy = {A,B}

∃x∃y p(x , y) ≡ [p(1,A) ∨ p(1,B)] ∨ [p(2,A) ∨ p(2,B)]

∃x∀y p(x , y) ≡ [p(1,A) ∧ p(1,B)] ∨ [p(2,A) ∧ p(2,B)]

∀x∃y p(x , y) ≡ [p(1,A) ∨ p(1,B)] ∧ [p(2,A) ∨ p(2,B)]

∀x∀y p(x , y) ≡ [p(1,A) ∧ p(1,B)] ∧ [p(2,A) ∧ p(2,B)]

Example

Ux = {1, 2} ∧ Uy = {A,B}

∃x∃y p(x , y) ≡ [p(1,A) ∨ p(1,B)] ∨ [p(2,A) ∨ p(2,B)]

∃x∀y p(x , y) ≡ [p(1,A) ∧ p(1,B)] ∨ [p(2,A) ∧ p(2,B)]

∀x∃y p(x , y) ≡ [p(1,A) ∨ p(1,B)] ∧ [p(2,A) ∨ p(2,B)]

∀x∀y p(x , y) ≡ [p(1,A) ∧ p(1,B)] ∧ [p(2,A) ∧ p(2,B)]

References

Required Reading: Grimaldi

Chapter 2: Fundamentals of Logic

2.4. The Use of Quantifiers

Supplementary Reading: O’Donnell, Hall, Page

Chapter 7: Predicate Logic

Topics

Definition

set: a collection of elements that are

distinct

unordered

non-repeating

Set Representation

explicit representationelements are listed within braces: {a1, a2, . . . , an}

implicit representationelements that validate a predicate: {x |x ∈ G , p(x)}

∅: empty set

let S be a set, and a be an element

a ∈ S : a is an element of set Sa /∈ S : a is not an element of set S

|S |: number of elements (cardinality)

Set Representation

∅: empty set

Set Representation

∅: empty set

Set Representation

∅: empty set

Set Representation

∅: empty set

Explicit Representation Example

Example

{3, 8, 2, 11, 5}11 ∈ {3, 8, 2, 11, 5}|{3, 8, 2, 11, 5}| = 5

Implicit Representation Examples

Example

{x |x ∈ Z+, 20 < x3 < 100} ≡ {3, 4}{2x − 1|x ∈ Z+, 20 < x3 < 100} ≡ {5, 7}

Example

A = {x |x ∈ R, 1 ≤ x ≤ 5}

Example

E = {n|n ∈ N,∃k ∈ N [n = 2k]}A = {x |x ∈ E , 1 ≤ x ≤ 5}

Example

{x |x ∈ Z+, 20 < x3 < 100} ≡ {3, 4}{2x − 1|x ∈ Z+, 20 < x3 < 100} ≡ {5, 7}

Example

A = {x |x ∈ R, 1 ≤ x ≤ 5}

Example

E = {n|n ∈ N,∃k ∈ N [n = 2k]}A = {x |x ∈ E , 1 ≤ x ≤ 5}

Example

{x |x ∈ Z+, 20 < x3 < 100} ≡ {3, 4}{2x − 1|x ∈ Z+, 20 < x3 < 100} ≡ {5, 7}

Example

A = {x |x ∈ R, 1 ≤ x ≤ 5}

Example

E = {n|n ∈ N,∃k ∈ N [n = 2k]}A = {x |x ∈ E , 1 ≤ x ≤ 5}

Set Dilemma

There is a barber who lives in a small town.He shaves all those men who don’t shave themselves.He doesn’t shave those men who shave themselves.

Does the barber shave himself?

yes → but he doesn’t shave men who shave themselves→ no

no → but he shaves all men who don’t shave themselves→ yes

Set Dilemma

let S be the set of sets that are not an element of themselvesS = {A|A /∈ A}

Is S an element of itself?

yes → but the predicate is not valid → no

no → but the predicate is valid → yes

Set Dilemma

Topics

Subset

Definition

A ⊆ B ⇔ ∀x [x ∈ A → x ∈ B]

set equality:A = B ⇔ (A ⊆ B) ∧ (B ⊆ A)

proper subset:A ⊂ B ⇔ (A ⊆ B) ∧ (A 6= B)

∀S [∅ ⊆ S ]

Subset

Definition

A ⊆ B ⇔ ∀x [x ∈ A → x ∈ B]

∀S [∅ ⊆ S ]

Subset

Definition

A ⊆ B ⇔ ∀x [x ∈ A → x ∈ B]

∀S [∅ ⊆ S ]

Subset

Definition

A ⊆ B ⇔ ∀x [x ∈ A → x ∈ B]

∀S [∅ ⊆ S ]

Subset

not a subset

A * B ⇔ ¬∀x [x ∈ A → x ∈ B]

⇔ ∃x ¬[x ∈ A → x ∈ B]

⇔ ∃x ¬[¬(x ∈ A) ∨ (x ∈ B)]

⇔ ∃x [(x ∈ A) ∧ ¬(x ∈ B)]

⇔ ∃x [(x ∈ A) ∧ (x /∈ B)]

Subset

not a subset

A * B ⇔ ¬∀x [x ∈ A → x ∈ B]

⇔ ∃x ¬[x ∈ A → x ∈ B]

⇔ ∃x ¬[¬(x ∈ A) ∨ (x ∈ B)]

⇔ ∃x [(x ∈ A) ∧ ¬(x ∈ B)]

⇔ ∃x [(x ∈ A) ∧ (x /∈ B)]

Subset

not a subset

A * B ⇔ ¬∀x [x ∈ A → x ∈ B]

⇔ ∃x ¬[x ∈ A → x ∈ B]

⇔ ∃x ¬[¬(x ∈ A) ∨ (x ∈ B)]

⇔ ∃x [(x ∈ A) ∧ ¬(x ∈ B)]

⇔ ∃x [(x ∈ A) ∧ (x /∈ B)]

Subset

not a subset

A * B ⇔ ¬∀x [x ∈ A → x ∈ B]

⇔ ∃x ¬[x ∈ A → x ∈ B]

⇔ ∃x ¬[¬(x ∈ A) ∨ (x ∈ B)]

⇔ ∃x [(x ∈ A) ∧ ¬(x ∈ B)]

⇔ ∃x [(x ∈ A) ∧ (x /∈ B)]

Subset

not a subset

A * B ⇔ ¬∀x [x ∈ A → x ∈ B]

⇔ ∃x ¬[x ∈ A → x ∈ B]

⇔ ∃x ¬[¬(x ∈ A) ∨ (x ∈ B)]

⇔ ∃x [(x ∈ A) ∧ ¬(x ∈ B)]

⇔ ∃x [(x ∈ A) ∧ (x /∈ B)]

Power Set

Definition

power set: P(S)the set of all subsets of a set, including the empty setand the set itself

if a set has n elements, its power set has 2n elements

Power Set

Definition

power set: P(S)the set of all subsets of a set, including the empty setand the set itself

if a set has n elements, its power set has 2n elements

Example of Power Set

Example

P({1, 2, 3}) = {∅,{1}, {2}, {3},{1, 2}, {1, 3}, {2, 3},{1, 2, 3}

Topics

Set Operations

complement

A = {x |x /∈ A}

intersection

A ∩ B = {x |(x ∈ A) ∧ (x ∈ B)}

if A ∩ B = ∅ then A and B are disjoint

A ∪ B = {x |(x ∈ A) ∨ (x ∈ B)}

Set Operations

complement

A = {x |x /∈ A}

intersection

A ∩ B = {x |(x ∈ A) ∧ (x ∈ B)}

A ∪ B = {x |(x ∈ A) ∨ (x ∈ B)}

Set Operations

complement

A = {x |x /∈ A}

intersection

A ∩ B = {x |(x ∈ A) ∧ (x ∈ B)}

A ∪ B = {x |(x ∈ A) ∨ (x ∈ B)}

Set Operations

difference

A− B = {x |(x ∈ A) ∧ (x /∈ B)}

A− B = A ∩ B

symmetric difference:A4 B = {x |(x ∈ A ∪ B) ∧ (x /∈ A ∩ B)}

Set Operations

difference

A− B = {x |(x ∈ A) ∧ (x /∈ B)}

A− B = A ∩ B

Set Operations

difference

A− B = {x |(x ∈ A) ∧ (x /∈ B)}

A− B = A ∩ B

Cartesian Product

Definition

Cartesian product:A× B = {(a, b)|a ∈ A, b ∈ B}

A× B × C × · · · × N = {(a, b, . . . , n)|a ∈ A, b ∈ B, . . . , n ∈ N}

|A× B × C × · · · × N| = |A| · |B| · |C | · · · |N|

Cartesian Product

Definition

Cartesian product:A× B = {(a, b)|a ∈ A, b ∈ B}

A× B × C × · · · × N = {(a, b, . . . , n)|a ∈ A, b ∈ B, . . . , n ∈ N}

|A× B × C × · · · × N| = |A| · |B| · |C | · · · |N|

Cartesian Product Example

Example

A = {a1.a2, a3, a4}B = {b1, b2, b3}

A× B = {(a1, b1), (a1, b2), (a1, b3),

(a2, b1), (a2, b2), (a2, b3),

(a3, b1), (a3, b2), (a3, b3),

(a4, b1), (a4, b2), (a4, b3)

Equivalences

Double Complement

CommutativityA ∩ B = B ∩ A A ∪ B = B ∪ A

Associativity(A ∩ B) ∩ C = A ∩ (B ∩ C ) (A ∪ B) ∪ C = A ∪ (B ∪ C )

IdempotenceA ∩ A = A A ∪ A = A

InverseA ∩ A = ∅ A ∪ A = U

Equivalences

Double Complement

Equivalences

Double Complement

Equivalences

Double Complement

Equivalences

Double Complement

Equivalences

IdentityA ∩ U = A A ∪ ∅ = A

DominationA ∩ ∅ = ∅ A ∪ U = U

DistributivityA ∩ (B ∪ C ) = (A ∩ B) ∪ (A ∩ C ) A ∪ (B ∩ C ) = (A ∪ B) ∩ (A ∪ C )

AbsorptionA ∩ (A ∪ B) = A A ∪ (A ∩ B) = A

DeMorgan’s LawsA ∩ B = A ∪ B A ∪ B = A ∩ B

Equivalences

DeMorgan’s Laws

Proof.

A ∩ B = {x |x /∈ (A ∩ B)}= {x |¬(x ∈ (A ∩ B))}= {x |¬((x ∈ A) ∧ (x ∈ B))}= {x |¬(x ∈ A) ∨ ¬(x ∈ B)}= {x |(x /∈ A) ∨ (x /∈ B)}= {x |(x ∈ A) ∨ (x ∈ B)}= {x |x ∈ A ∪ B}= A ∪ B

DeMorgan’s Laws

Proof.

DeMorgan’s Laws

Proof.

DeMorgan’s Laws

Proof.

DeMorgan’s Laws

Proof.

DeMorgan’s Laws

Proof.

DeMorgan’s Laws

Proof.

DeMorgan’s Laws

Proof.

Example of Equivalence

Theorem

A ∩ (B − C ) = (A ∩ B)− (A ∩ C )

Equivalence Example

Proof.

(A ∩ B)− (A ∩ C ) = (A ∩ B) ∩ (A ∩ C )

= (A ∩ B) ∩ (A ∪ C )

= ((A ∩ B) ∩ A) ∪ ((A ∩ B) ∩ C ))

= ∅ ∪ ((A ∩ B) ∩ C ))

= (A ∩ B) ∩ C

= A ∩ (B ∩ C )

= A ∩ (B − C )

Equivalence Example

Proof.

(A ∩ B)− (A ∩ C ) = (A ∩ B) ∩ (A ∩ C )

= (A ∩ B) ∩ (A ∪ C )

= ((A ∩ B) ∩ A) ∪ ((A ∩ B) ∩ C ))

= ∅ ∪ ((A ∩ B) ∩ C ))

= (A ∩ B) ∩ C

= A ∩ (B ∩ C )

= A ∩ (B − C )

Equivalence Example

Proof.

(A ∩ B)− (A ∩ C ) = (A ∩ B) ∩ (A ∩ C )

= (A ∩ B) ∩ (A ∪ C )

= ((A ∩ B) ∩ A) ∪ ((A ∩ B) ∩ C ))

= ∅ ∪ ((A ∩ B) ∩ C ))

= (A ∩ B) ∩ C

= A ∩ (B ∩ C )

= A ∩ (B − C )

Equivalence Example

Proof.

(A ∩ B)− (A ∩ C ) = (A ∩ B) ∩ (A ∩ C )

= (A ∩ B) ∩ (A ∪ C )

= ((A ∩ B) ∩ A) ∪ ((A ∩ B) ∩ C ))

= ∅ ∪ ((A ∩ B) ∩ C ))

= (A ∩ B) ∩ C

= A ∩ (B ∩ C )

= A ∩ (B − C )

Equivalence Example

Proof.

(A ∩ B)− (A ∩ C ) = (A ∩ B) ∩ (A ∩ C )

= (A ∩ B) ∩ (A ∪ C )

= ((A ∩ B) ∩ A) ∪ ((A ∩ B) ∩ C ))

= ∅ ∪ ((A ∩ B) ∩ C ))

= (A ∩ B) ∩ C

= A ∩ (B ∩ C )

= A ∩ (B − C )

Equivalence Example

Proof.

(A ∩ B)− (A ∩ C ) = (A ∩ B) ∩ (A ∩ C )

= (A ∩ B) ∩ (A ∪ C )

= ((A ∩ B) ∩ A) ∪ ((A ∩ B) ∩ C ))

= ∅ ∪ ((A ∩ B) ∩ C ))

= (A ∩ B) ∩ C

= A ∩ (B ∩ C )

= A ∩ (B − C )

Equivalence Example

Proof.

(A ∩ B)− (A ∩ C ) = (A ∩ B) ∩ (A ∩ C )

= (A ∩ B) ∩ (A ∪ C )

= ((A ∩ B) ∩ A) ∪ ((A ∩ B) ∩ C ))

= ∅ ∪ ((A ∩ B) ∩ C ))

= (A ∩ B) ∩ C

= A ∩ (B ∩ C )

= A ∩ (B − C )

Topics

Principle of Inclusion-Exclusion

|A ∪ B| = |A|+ |B| − |A ∩ B||A ∪ B ∪ C | =|A|+ |B|+ |C | − (|A ∩ B|+ |A ∩ C |+ |B ∩ C |) + |A ∩ B ∩ C |

Theorem

|A1 ∪ A2 ∪ · · · ∪ An| =∑

|Ai | −∑i ,j

|Ai ∩ Aj |

+∑i ,j ,k

|Ai ∩ Aj ∩ Ak |

· · ·+−1n−1|Ai ∩ Aj ∩ · · · ∩ An|

Theorem

|A1 ∪ A2 ∪ · · · ∪ An| =∑

|Ai | −∑i ,j

|Ai ∩ Aj |

+∑i ,j ,k

|Ai ∩ Aj ∩ Ak |

· · ·+−1n−1|Ai ∩ Aj ∩ · · · ∩ An|

Theorem

|A1 ∪ A2 ∪ · · · ∪ An| =∑

|Ai | −∑i ,j

|Ai ∩ Aj |

+∑i ,j ,k

|Ai ∩ Aj ∩ Ak |

· · ·+−1n−1|Ai ∩ Aj ∩ · · · ∩ An|

Inclusion-Exclusion Example

Example (sieve of Eratosthenes)

a method to identify prime numbers

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1718 19 20 21 22 23 24 25 26 27 28 29 30

2 3 5 7 9 11 13 15 1719 21 23 25 27 29

2 3 5 7 11 13 1719 23 25 29

2 3 5 7 11 13 1719 23 29

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1718 19 20 21 22 23 24 25 26 27 28 29 30

2 3 5 7 9 11 13 15 1719 21 23 25 27 29

2 3 5 7 11 13 1719 23 25 29

2 3 5 7 11 13 1719 23 29

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1718 19 20 21 22 23 24 25 26 27 28 29 30

2 3 5 7 9 11 13 15 1719 21 23 25 27 29

2 3 5 7 11 13 1719 23 25 29

2 3 5 7 11 13 1719 23 29

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1718 19 20 21 22 23 24 25 26 27 28 29 30

2 3 5 7 9 11 13 15 1719 21 23 25 27 29

2 3 5 7 11 13 1719 23 25 29

2 3 5 7 11 13 1719 23 29

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1718 19 20 21 22 23 24 25 26 27 28 29 30

2 3 5 7 9 11 13 15 1719 21 23 25 27 29

2 3 5 7 11 13 1719 23 25 29

2 3 5 7 11 13 1719 23 29

number of primes between 1 and 100

numbers that are not divisible by 2, 3, 5 and 7

A2: set of numbers divisible by 2A3: set of numbers divisible by 3A5: set of numbers divisible by 5A7: set of numbers divisible by 7

|A2 ∪ A3 ∪ A5 ∪ A7|

|A2| = b100/2c = 50

|A3| = b100/3c = 33

|A5| = b100/5c = 20

|A7| = b100/7c = 14

|A2 ∩ A3| = b100/6c = 16

|A2 ∩ A5| = b100/10c = 10

|A2 ∩ A7| = b100/14c = 7

|A3 ∩ A5| = b100/15c = 6

|A3 ∩ A7| = b100/21c = 4

|A5 ∩ A7| = b100/35c = 2

|A2| = b100/2c = 50

|A3| = b100/3c = 33

|A5| = b100/5c = 20

|A7| = b100/7c = 14

|A2 ∩ A3| = b100/6c = 16

|A2 ∩ A5| = b100/10c = 10

|A2 ∩ A7| = b100/14c = 7

|A3 ∩ A5| = b100/15c = 6

|A3 ∩ A7| = b100/21c = 4

|A5 ∩ A7| = b100/35c = 2

|A2 ∩ A3 ∩ A5| = b100/30c = 3

|A2 ∩ A3 ∩ A7| = b100/42c = 2

|A2 ∩ A5 ∩ A7| = b100/70c = 1

|A3 ∩ A5 ∩ A7| = b100/105c = 0

|A2 ∩ A3 ∩ A5 ∩ A7| = b100/210c = 0

|A2 ∩ A3 ∩ A5| = b100/30c = 3

|A2 ∩ A3 ∩ A7| = b100/42c = 2

|A2 ∩ A5 ∩ A7| = b100/70c = 1

|A3 ∩ A5 ∩ A7| = b100/105c = 0

|A2 ∩ A3 ∩ A5 ∩ A7| = b100/210c = 0

|A2 ∪ A3 ∪ A5 ∪ A7| = (50 + 33 + 20 + 14)

− (16 + 10 + 7 + 6 + 4 + 2)

+ (3 + 2 + 1 + 0)

− (0)

number of primes: (100− 78) + 4− 1 = 25

|A2 ∪ A3 ∪ A5 ∪ A7| = (50 + 33 + 20 + 14)

− (16 + 10 + 7 + 6 + 4 + 2)

+ (3 + 2 + 1 + 0)

− (0)

number of primes: (100− 78) + 4− 1 = 25

References

Required Reading: Grimaldi

Chapter 3: Set Theory

3.1. Sets and Subsets3.2. Set Operations and the Laws of Set Theory

Chapter 8: The Principle of Inclusion and Exclusion

8.1. The Principle of Inclusion and Exclusion

Supplementary Reading: O’Donnell, Hall, Page

Chapter 8: Set Theory

discrete mathematics - predicates and sets

Education