Post on 05-Dec-2014
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Discrete MathematicsPredicates and Sets
H. Turgut Uyar Aysegul Gencata Yayımlı Emre Harmancı
2001-2013
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c©2001-2013 T. Uyar, A. Yayımlı, E. Harmancı
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Topics
1 PredicatesIntroductionQuantifiersMultiple Quantifiers
2 SetsIntroductionSubsetSet OperationsInclusion-Exclusion
Topics
1 PredicatesIntroductionQuantifiersMultiple Quantifiers
2 SetsIntroductionSubsetSet OperationsInclusion-Exclusion
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
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
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
Topics
1 PredicatesIntroductionQuantifiersMultiple Quantifiers
2 SetsIntroductionSubsetSet OperationsInclusion-Exclusion
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
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
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
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
U = R
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
U = R
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
U = R
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
U = R
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
U = R
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
U = R
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)]
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)
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)
Negating Quantifiers
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)
Negating Quantifiers
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)
Negating Quantifiers
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)
Negating Quantifiers
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)]
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)]
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)]
Topics
1 PredicatesIntroductionQuantifiersMultiple Quantifiers
2 SetsIntroductionSubsetSet OperationsInclusion-Exclusion
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?
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?
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?
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?
Multiple Quantifiers
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)]
Multiple Quantifiers
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)]
Multiple Quantifiers
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)]
Multiple Quantifiers
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)]
Multiple Quantifiers
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
1 PredicatesIntroductionQuantifiersMultiple Quantifiers
2 SetsIntroductionSubsetSet OperationsInclusion-Exclusion
Set
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
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
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
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
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)
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}
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}
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}
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
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
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
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
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
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
Topics
1 PredicatesIntroductionQuantifiersMultiple Quantifiers
2 SetsIntroductionSubsetSet OperationsInclusion-Exclusion
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]
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]
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]
set equality:A = B ⇔ (A ⊆ B) ∧ (B ⊆ A)
proper subset:A ⊂ B ⇔ (A ⊆ B) ∧ (A 6= 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
1 PredicatesIntroductionQuantifiersMultiple Quantifiers
2 SetsIntroductionSubsetSet OperationsInclusion-Exclusion
Set Operations
complement
A = {x |x /∈ A}
intersection
A ∩ B = {x |(x ∈ A) ∧ (x ∈ B)}
if A ∩ B = ∅ then A and B are disjoint
union
A ∪ B = {x |(x ∈ A) ∨ (x ∈ B)}
Set Operations
complement
A = {x |x /∈ A}
intersection
A ∩ B = {x |(x ∈ A) ∧ (x ∈ B)}
if A ∩ B = ∅ then A and B are disjoint
union
A ∪ B = {x |(x ∈ A) ∨ (x ∈ B)}
Set Operations
complement
A = {x |x /∈ A}
intersection
A ∩ B = {x |(x ∈ A) ∧ (x ∈ B)}
if A ∩ B = ∅ then A and B are disjoint
union
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
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
symmetric difference:A4 B = {x |(x ∈ A ∪ B) ∧ (x /∈ 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
A = A
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
A = A
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
A = A
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
A = A
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
A = A
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
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
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
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
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
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
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.
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.
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.
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.
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.
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.
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.
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
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
1 PredicatesIntroductionQuantifiersMultiple Quantifiers
2 SetsIntroductionSubsetSet OperationsInclusion-Exclusion
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| =∑
i
|Ai | −∑i ,j
|Ai ∩ Aj |
+∑i ,j ,k
|Ai ∩ Aj ∩ Ak |
· · ·+−1n−1|Ai ∩ Aj ∩ · · · ∩ An|
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| =∑
i
|Ai | −∑i ,j
|Ai ∩ Aj |
+∑i ,j ,k
|Ai ∩ Aj ∩ Ak |
· · ·+−1n−1|Ai ∩ Aj ∩ · · · ∩ An|
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| =∑
i
|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
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
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
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
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
Inclusion-Exclusion Example
Example (sieve of Eratosthenes)
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|
Inclusion-Exclusion Example
Example (sieve of Eratosthenes)
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|
Inclusion-Exclusion Example
Example (sieve of Eratosthenes)
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|
Inclusion-Exclusion Example
Example (sieve of Eratosthenes)
|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
Inclusion-Exclusion Example
Example (sieve of Eratosthenes)
|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
Inclusion-Exclusion Example
Example (sieve of Eratosthenes)
|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
Inclusion-Exclusion Example
Example (sieve of Eratosthenes)
|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
Inclusion-Exclusion Example
Example (sieve of Eratosthenes)
|A2 ∪ A3 ∪ A5 ∪ A7| = (50 + 33 + 20 + 14)
− (16 + 10 + 7 + 6 + 4 + 2)
+ (3 + 2 + 1 + 0)
− (0)
= 78
number of primes: (100− 78) + 4− 1 = 25
Inclusion-Exclusion Example
Example (sieve of Eratosthenes)
|A2 ∪ A3 ∪ A5 ∪ A7| = (50 + 33 + 20 + 14)
− (16 + 10 + 7 + 6 + 4 + 2)
+ (3 + 2 + 1 + 0)
− (0)
= 78
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