Contents
1 Discrete structures
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 1 / 23 Discrete structures Section outline
Lattices (contd.) Boolean lattice 1 Discrete structures Boolean lattice structure Sets Boolean algebra Relations Additional Boolean algebra Lattices properties
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 2 / 23 S = {X|X 6∈ X} S ∈ S? [Russell’s paradox]
Set union: A ∪ B A B
Set intersection: A ∩ B A B U Complement: S A
Set difference: A − B = A ∩ B A B
+ Natural numbers: N = {0, 1, 2, 3,...} or {1, 2, 3,...} = Z Integers: Z = {..., −3, −2, −1, 0, 1, 2, 3,...} Universal set: U Empty set: ∅ = {}
Discrete structures Sets Sets
A set A of elements: A = {a, b, c}
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 3 / 23 S = {X|X 6∈ X} S ∈ S? [Russell’s paradox]
Set union: A ∪ B A B
Set intersection: A ∩ B A B U Complement: S A
Set difference: A − B = A ∩ B A B
Integers: Z = {..., −3, −2, −1, 0, 1, 2, 3,...} Universal set: U Empty set: ∅ = {}
Discrete structures Sets Sets
A set A of elements: A = {a, b, c} + Natural numbers: N = {0, 1, 2, 3,...} or {1, 2, 3,...} = Z
F TECHN O OL E O T G U Y IT K T H S A
N R
I
A
N G
A P
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19 51
yog, km s kOflm^
Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 3 / 23 S = {X|X 6∈ X} S ∈ S? [Russell’s paradox]
Set union: A ∪ B A B
Set intersection: A ∩ B A B U Complement: S A
Set difference: A − B = A ∩ B A B
Universal set: U Empty set: ∅ = {}
Discrete structures Sets Sets
A set A of elements: A = {a, b, c} + Natural numbers: N = {0, 1, 2, 3,...} or {1, 2, 3,...} = Z Integers: Z = {..., −3, −2, −1, 0, 1, 2, 3,...}
F TECHN O OL E O T G U Y IT K T H S A
N R
I
A
N G
A P
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19 51
yog, km s kOflm^
Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 3 / 23 S = {X|X 6∈ X} S ∈ S? [Russell’s paradox]
Set union: A ∪ B A B
Set intersection: A ∩ B A B U Complement: S A
Set difference: A − B = A ∩ B A B
Discrete structures Sets Sets
A set A of elements: A = {a, b, c} + Natural numbers: N = {0, 1, 2, 3,...} or {1, 2, 3,...} = Z Integers: Z = {..., −3, −2, −1, 0, 1, 2, 3,...} Universal set: U Empty set: ∅ = {}
F TECHN O OL E O T G U Y IT K T H S A
N R
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A
N G
A P
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U
D
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19 51
yog, km s kOflm^
Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 3 / 23 S = {X|X 6∈ X} S ∈ S? [Russell’s paradox]
Set union: A ∪ B A B
Set intersection: A ∩ B A B U Complement: S A
Set difference: A − B = A ∩ B A B
Discrete structures Sets Sets
A set A of elements: A = {a, b, c} + Natural numbers: N = {0, 1, 2, 3,...} or {1, 2, 3,...} = Z Integers: Z = {..., −3, −2, −1, 0, 1, 2, 3,...} Universal set: U Empty set: ∅ = {}
F TECHN O OL E O T G U Y IT K T H S A
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A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 3 / 23 Set union: A ∪ B A B
Set intersection: A ∩ B A B U Complement: S A
Set difference: A − B = A ∩ B A B
Discrete structures Sets Sets
A set A of elements: A = {a, b, c} + Natural numbers: N = {0, 1, 2, 3,...} or {1, 2, 3,...} = Z Integers: Z = {..., −3, −2, −1, 0, 1, 2, 3,...} Universal set: U Empty set: ∅ = {} S = {X|X 6∈ X} S ∈ S? [Russell’s paradox]
F TECHN O OL E O T G U Y IT K T H S A
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A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 3 / 23 Set union: A ∪ B A B
Set intersection: A ∩ B A B U Complement: S A
Set difference: A − B = A ∩ B A B
Discrete structures Sets Sets
A set A of elements: A = {a, b, c} + Natural numbers: N = {0, 1, 2, 3,...} or {1, 2, 3,...} = Z Integers: Z = {..., −3, −2, −1, 0, 1, 2, 3,...} Universal set: U Empty set: ∅ = {} S = {X|X 6∈ X} S ∈ S? [Russell’s paradox]
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 3 / 23 Set intersection: A ∩ B A B U Complement: S A
Set difference: A − B = A ∩ B A B
Discrete structures Sets Sets
A set A of elements: A = {a, b, c} + Natural numbers: N = {0, 1, 2, 3,...} or {1, 2, 3,...} = Z Integers: Z = {..., −3, −2, −1, 0, 1, 2, 3,...} Universal set: U Empty set: ∅ = {} S = {X|X 6∈ X} S ∈ S? [Russell’s paradox]
Set union: A ∪ B A B
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 3 / 23 U Complement: S A
Set difference: A − B = A ∩ B A B
Discrete structures Sets Sets
A set A of elements: A = {a, b, c} + Natural numbers: N = {0, 1, 2, 3,...} or {1, 2, 3,...} = Z Integers: Z = {..., −3, −2, −1, 0, 1, 2, 3,...} Universal set: U Empty set: ∅ = {} S = {X|X 6∈ X} S ∈ S? [Russell’s paradox]
Set union: A ∪ B A B
Set intersection: A ∩ B A B
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 3 / 23 Set difference: A − B = A ∩ B A B
Discrete structures Sets Sets
A set A of elements: A = {a, b, c} + Natural numbers: N = {0, 1, 2, 3,...} or {1, 2, 3,...} = Z Integers: Z = {..., −3, −2, −1, 0, 1, 2, 3,...} Universal set: U Empty set: ∅ = {} S = {X|X 6∈ X} S ∈ S? [Russell’s paradox]
Set union: A ∪ B A B
Set intersection: A ∩ B A B U Complement: S A
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 3 / 23 Discrete structures Sets Sets
A set A of elements: A = {a, b, c} + Natural numbers: N = {0, 1, 2, 3,...} or {1, 2, 3,...} = Z Integers: Z = {..., −3, −2, −1, 0, 1, 2, 3,...} Universal set: U Empty set: ∅ = {} S = {X|X 6∈ X} S ∈ S? [Russell’s paradox]
Set union: A ∪ B A B
Set intersection: A ∩ B A B U Complement: S A
F TECHN O OL E O T G U Y IT K T H S A
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Set difference: A − B = A ∩ B A B A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 3 / 23 Complement of intersection (De Morgan): A ∩ B = A ∪ B U A B Power set of A: P(A) P({a, b}) = {∅, {a} , {b} , {a, b}} Non-empty X1,..., Xk is a partition of A if A = X1 ∪ ... ∪ Xk and Xı ∩ X = ∅ |ı6=j A ∩ B, B ∩ A, A ∩ B and A ∪ B constitute a partition of U
Discrete structures Sets Sets (contd.)
Complement of union (De Morgan): A ∪ B = A ∩ B U A B
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 4 / 23 Power set of A: P(A) P({a, b}) = {∅, {a} , {b} , {a, b}} Non-empty X1,..., Xk is a partition of A if A = X1 ∪ ... ∪ Xk and Xı ∩ X = ∅ |ı6=j A ∩ B, B ∩ A, A ∩ B and A ∪ B constitute a partition of U
Discrete structures Sets Sets (contd.)
Complement of union (De Morgan): A ∪ B = A ∩ B U A B Complement of intersection (De Morgan): A ∩ B = A ∪ B U A B
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 4 / 23 P({a, b}) = {∅, {a} , {b} , {a, b}} Non-empty X1,..., Xk is a partition of A if A = X1 ∪ ... ∪ Xk and Xı ∩ X = ∅ |ı6=j A ∩ B, B ∩ A, A ∩ B and A ∪ B constitute a partition of U
Discrete structures Sets Sets (contd.)
Complement of union (De Morgan): A ∪ B = A ∩ B U A B Complement of intersection (De Morgan): A ∩ B = A ∪ B U A B Power set of A: P(A)
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 4 / 23 Non-empty X1,..., Xk is a partition of A if A = X1 ∪ ... ∪ Xk and Xı ∩ X = ∅ |ı6=j A ∩ B, B ∩ A, A ∩ B and A ∪ B constitute a partition of U
Discrete structures Sets Sets (contd.)
Complement of union (De Morgan): A ∪ B = A ∩ B U A B Complement of intersection (De Morgan): A ∩ B = A ∪ B U A B Power set of A: P(A) P({a, b}) = {∅, {a} , {b} , {a, b}}
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 4 / 23 A ∩ B, B ∩ A, A ∩ B and A ∪ B constitute a partition of U
Discrete structures Sets Sets (contd.)
Complement of union (De Morgan): A ∪ B = A ∩ B U A B Complement of intersection (De Morgan): A ∩ B = A ∪ B U A B Power set of A: P(A) P({a, b}) = {∅, {a} , {b} , {a, b}} Non-empty X1,..., Xk is a partition of A if A = X1 ∪ ... ∪ Xk and Xı ∩ X = ∅ |ı6=j
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 4 / 23 Discrete structures Sets Sets (contd.)
Complement of union (De Morgan): A ∪ B = A ∩ B U A B Complement of intersection (De Morgan): A ∩ B = A ∪ B U A B Power set of A: P(A) P({a, b}) = {∅, {a} , {b} , {a, b}} Non-empty X1,..., Xk is a partition of A if A = X1 ∪ ... ∪ Xk and Xı ∩ X = ∅ |ı6=j A ∩ B, B ∩ A, A ∩ B and A ∪ B constitute a partition of U
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 4 / 23 Discrete structures Sets Set algebra
Idempotence A ∪ A = A A ∩ A = A Associativity (A ∪ B) ∪ C = A ∪ (B ∪ C) (A ∩ B) ∪ C = A ∩ (B ∪ C) Commutativity A ∪ B = B ∪ A A ∩ B = B ∩ A Distributivity A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) Identity A ∪ {} = A, A ∪ U = U A ∩ {} = {} , A ∪ U = A Involution A¯ = A Complements U¯ = {} , A ∪ A¯ = U {}¯ = U, A ∩ A¯ = {} DeMorgan A ∪ B = A¯ ∩ B¯ A ∩ B = A¯ ∪ B¯
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 5 / 23 Cartesian product: A × B = {ha, bi | a ∈ A, b ∈ B} {a, b, c} × {α, β} = {ha, αi , hb, αi , hc, αi , ha, βi , hb, βi , hc, βi} N × N = {hı, i | ı, ≥ 1} Binary relation R on sets A and B: R ⊆ A × B 1 if ha, bi ∈ R Characteristic function of R: χ (a, b) = R 0 if ha, bi 6∈ R R ⊆ A × A is reflexive if ∀x ∈ A. xRx R ⊆ A × A is symmetric if ∀x, y ∈ A. xRy ⇒ yRx R ⊆ A × A is transitive if ∀x, y, z ∈ A. xRy ∧ yRz ⇒ xRz R ⊆ A × A is antisymmetric if ∀x, y ∈ A. xRy ∧ yRx ⇒ x = y Equivalence relation: R is reflexive, symmetric and transitive An equivalence relation induces a partition and vice versa Partial order: R is reflexive, antisymmetric and transitive
Discrete structures Relations Relations
Tuple: ha, bi , h4, b, αi
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 6 / 23 {a, b, c} × {α, β} = {ha, αi , hb, αi , hc, αi , ha, βi , hb, βi , hc, βi} N × N = {hı, i | ı, ≥ 1} Binary relation R on sets A and B: R ⊆ A × B 1 if ha, bi ∈ R Characteristic function of R: χ (a, b) = R 0 if ha, bi 6∈ R R ⊆ A × A is reflexive if ∀x ∈ A. xRx R ⊆ A × A is symmetric if ∀x, y ∈ A. xRy ⇒ yRx R ⊆ A × A is transitive if ∀x, y, z ∈ A. xRy ∧ yRz ⇒ xRz R ⊆ A × A is antisymmetric if ∀x, y ∈ A. xRy ∧ yRx ⇒ x = y Equivalence relation: R is reflexive, symmetric and transitive An equivalence relation induces a partition and vice versa Partial order: R is reflexive, antisymmetric and transitive
Discrete structures Relations Relations
Tuple: ha, bi , h4, b, αi Cartesian product: A × B = {ha, bi | a ∈ A, b ∈ B}
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 6 / 23 N × N = {hı, i | ı, ≥ 1} Binary relation R on sets A and B: R ⊆ A × B 1 if ha, bi ∈ R Characteristic function of R: χ (a, b) = R 0 if ha, bi 6∈ R R ⊆ A × A is reflexive if ∀x ∈ A. xRx R ⊆ A × A is symmetric if ∀x, y ∈ A. xRy ⇒ yRx R ⊆ A × A is transitive if ∀x, y, z ∈ A. xRy ∧ yRz ⇒ xRz R ⊆ A × A is antisymmetric if ∀x, y ∈ A. xRy ∧ yRx ⇒ x = y Equivalence relation: R is reflexive, symmetric and transitive An equivalence relation induces a partition and vice versa Partial order: R is reflexive, antisymmetric and transitive
Discrete structures Relations Relations
Tuple: ha, bi , h4, b, αi Cartesian product: A × B = {ha, bi | a ∈ A, b ∈ B} {a, b, c} × {α, β} = {ha, αi , hb, αi , hc, αi , ha, βi , hb, βi , hc, βi}
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 6 / 23 Binary relation R on sets A and B: R ⊆ A × B 1 if ha, bi ∈ R Characteristic function of R: χ (a, b) = R 0 if ha, bi 6∈ R R ⊆ A × A is reflexive if ∀x ∈ A. xRx R ⊆ A × A is symmetric if ∀x, y ∈ A. xRy ⇒ yRx R ⊆ A × A is transitive if ∀x, y, z ∈ A. xRy ∧ yRz ⇒ xRz R ⊆ A × A is antisymmetric if ∀x, y ∈ A. xRy ∧ yRx ⇒ x = y Equivalence relation: R is reflexive, symmetric and transitive An equivalence relation induces a partition and vice versa Partial order: R is reflexive, antisymmetric and transitive
Discrete structures Relations Relations
Tuple: ha, bi , h4, b, αi Cartesian product: A × B = {ha, bi | a ∈ A, b ∈ B} {a, b, c} × {α, β} = {ha, αi , hb, αi , hc, αi , ha, βi , hb, βi , hc, βi} N × N = {hı, i | ı, ≥ 1}
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 6 / 23 1 if ha, bi ∈ R Characteristic function of R: χ (a, b) = R 0 if ha, bi 6∈ R R ⊆ A × A is reflexive if ∀x ∈ A. xRx R ⊆ A × A is symmetric if ∀x, y ∈ A. xRy ⇒ yRx R ⊆ A × A is transitive if ∀x, y, z ∈ A. xRy ∧ yRz ⇒ xRz R ⊆ A × A is antisymmetric if ∀x, y ∈ A. xRy ∧ yRx ⇒ x = y Equivalence relation: R is reflexive, symmetric and transitive An equivalence relation induces a partition and vice versa Partial order: R is reflexive, antisymmetric and transitive
Discrete structures Relations Relations
Tuple: ha, bi , h4, b, αi Cartesian product: A × B = {ha, bi | a ∈ A, b ∈ B} {a, b, c} × {α, β} = {ha, αi , hb, αi , hc, αi , ha, βi , hb, βi , hc, βi} N × N = {hı, i | ı, ≥ 1} Binary relation R on sets A and B: R ⊆ A × B
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 6 / 23 R ⊆ A × A is reflexive if ∀x ∈ A. xRx R ⊆ A × A is symmetric if ∀x, y ∈ A. xRy ⇒ yRx R ⊆ A × A is transitive if ∀x, y, z ∈ A. xRy ∧ yRz ⇒ xRz R ⊆ A × A is antisymmetric if ∀x, y ∈ A. xRy ∧ yRx ⇒ x = y Equivalence relation: R is reflexive, symmetric and transitive An equivalence relation induces a partition and vice versa Partial order: R is reflexive, antisymmetric and transitive
Discrete structures Relations Relations
Tuple: ha, bi , h4, b, αi Cartesian product: A × B = {ha, bi | a ∈ A, b ∈ B} {a, b, c} × {α, β} = {ha, αi , hb, αi , hc, αi , ha, βi , hb, βi , hc, βi} N × N = {hı, i | ı, ≥ 1} Binary relation R on sets A and B: R ⊆ A × B 1 if ha, bi ∈ R Characteristic function of R: χ (a, b) = R 0 if ha, bi 6∈ R
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 6 / 23 R ⊆ A × A is symmetric if ∀x, y ∈ A. xRy ⇒ yRx R ⊆ A × A is transitive if ∀x, y, z ∈ A. xRy ∧ yRz ⇒ xRz R ⊆ A × A is antisymmetric if ∀x, y ∈ A. xRy ∧ yRx ⇒ x = y Equivalence relation: R is reflexive, symmetric and transitive An equivalence relation induces a partition and vice versa Partial order: R is reflexive, antisymmetric and transitive
Discrete structures Relations Relations
Tuple: ha, bi , h4, b, αi Cartesian product: A × B = {ha, bi | a ∈ A, b ∈ B} {a, b, c} × {α, β} = {ha, αi , hb, αi , hc, αi , ha, βi , hb, βi , hc, βi} N × N = {hı, i | ı, ≥ 1} Binary relation R on sets A and B: R ⊆ A × B 1 if ha, bi ∈ R Characteristic function of R: χ (a, b) = R 0 if ha, bi 6∈ R R ⊆ A × A is reflexive if ∀x ∈ A. xRx
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 6 / 23 R ⊆ A × A is transitive if ∀x, y, z ∈ A. xRy ∧ yRz ⇒ xRz R ⊆ A × A is antisymmetric if ∀x, y ∈ A. xRy ∧ yRx ⇒ x = y Equivalence relation: R is reflexive, symmetric and transitive An equivalence relation induces a partition and vice versa Partial order: R is reflexive, antisymmetric and transitive
Discrete structures Relations Relations
Tuple: ha, bi , h4, b, αi Cartesian product: A × B = {ha, bi | a ∈ A, b ∈ B} {a, b, c} × {α, β} = {ha, αi , hb, αi , hc, αi , ha, βi , hb, βi , hc, βi} N × N = {hı, i | ı, ≥ 1} Binary relation R on sets A and B: R ⊆ A × B 1 if ha, bi ∈ R Characteristic function of R: χ (a, b) = R 0 if ha, bi 6∈ R R ⊆ A × A is reflexive if ∀x ∈ A. xRx R ⊆ A × A is symmetric if ∀x, y ∈ A. xRy ⇒ yRx
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 6 / 23 R ⊆ A × A is antisymmetric if ∀x, y ∈ A. xRy ∧ yRx ⇒ x = y Equivalence relation: R is reflexive, symmetric and transitive An equivalence relation induces a partition and vice versa Partial order: R is reflexive, antisymmetric and transitive
Discrete structures Relations Relations
Tuple: ha, bi , h4, b, αi Cartesian product: A × B = {ha, bi | a ∈ A, b ∈ B} {a, b, c} × {α, β} = {ha, αi , hb, αi , hc, αi , ha, βi , hb, βi , hc, βi} N × N = {hı, i | ı, ≥ 1} Binary relation R on sets A and B: R ⊆ A × B 1 if ha, bi ∈ R Characteristic function of R: χ (a, b) = R 0 if ha, bi 6∈ R R ⊆ A × A is reflexive if ∀x ∈ A. xRx R ⊆ A × A is symmetric if ∀x, y ∈ A. xRy ⇒ yRx R ⊆ A × A is transitive if ∀x, y, z ∈ A. xRy ∧ yRz ⇒ xRz
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 6 / 23 Equivalence relation: R is reflexive, symmetric and transitive An equivalence relation induces a partition and vice versa Partial order: R is reflexive, antisymmetric and transitive
Discrete structures Relations Relations
Tuple: ha, bi , h4, b, αi Cartesian product: A × B = {ha, bi | a ∈ A, b ∈ B} {a, b, c} × {α, β} = {ha, αi , hb, αi , hc, αi , ha, βi , hb, βi , hc, βi} N × N = {hı, i | ı, ≥ 1} Binary relation R on sets A and B: R ⊆ A × B 1 if ha, bi ∈ R Characteristic function of R: χ (a, b) = R 0 if ha, bi 6∈ R R ⊆ A × A is reflexive if ∀x ∈ A. xRx R ⊆ A × A is symmetric if ∀x, y ∈ A. xRy ⇒ yRx R ⊆ A × A is transitive if ∀x, y, z ∈ A. xRy ∧ yRz ⇒ xRz R ⊆ A × A is antisymmetric if ∀x, y ∈ A. xRy ∧ yRx ⇒ x = y
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 6 / 23 An equivalence relation induces a partition and vice versa Partial order: R is reflexive, antisymmetric and transitive
Discrete structures Relations Relations
Tuple: ha, bi , h4, b, αi Cartesian product: A × B = {ha, bi | a ∈ A, b ∈ B} {a, b, c} × {α, β} = {ha, αi , hb, αi , hc, αi , ha, βi , hb, βi , hc, βi} N × N = {hı, i | ı, ≥ 1} Binary relation R on sets A and B: R ⊆ A × B 1 if ha, bi ∈ R Characteristic function of R: χ (a, b) = R 0 if ha, bi 6∈ R R ⊆ A × A is reflexive if ∀x ∈ A. xRx R ⊆ A × A is symmetric if ∀x, y ∈ A. xRy ⇒ yRx R ⊆ A × A is transitive if ∀x, y, z ∈ A. xRy ∧ yRz ⇒ xRz R ⊆ A × A is antisymmetric if ∀x, y ∈ A. xRy ∧ yRx ⇒ x = y Equivalence relation: R is reflexive, symmetric and transitive
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 6 / 23 Partial order: R is reflexive, antisymmetric and transitive
Discrete structures Relations Relations
Tuple: ha, bi , h4, b, αi Cartesian product: A × B = {ha, bi | a ∈ A, b ∈ B} {a, b, c} × {α, β} = {ha, αi , hb, αi , hc, αi , ha, βi , hb, βi , hc, βi} N × N = {hı, i | ı, ≥ 1} Binary relation R on sets A and B: R ⊆ A × B 1 if ha, bi ∈ R Characteristic function of R: χ (a, b) = R 0 if ha, bi 6∈ R R ⊆ A × A is reflexive if ∀x ∈ A. xRx R ⊆ A × A is symmetric if ∀x, y ∈ A. xRy ⇒ yRx R ⊆ A × A is transitive if ∀x, y, z ∈ A. xRy ∧ yRz ⇒ xRz R ⊆ A × A is antisymmetric if ∀x, y ∈ A. xRy ∧ yRx ⇒ x = y Equivalence relation: R is reflexive, symmetric and transitive An equivalence relation induces a partition and vice versa F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 6 / 23 Discrete structures Relations Relations
Tuple: ha, bi , h4, b, αi Cartesian product: A × B = {ha, bi | a ∈ A, b ∈ B} {a, b, c} × {α, β} = {ha, αi , hb, αi , hc, αi , ha, βi , hb, βi , hc, βi} N × N = {hı, i | ı, ≥ 1} Binary relation R on sets A and B: R ⊆ A × B 1 if ha, bi ∈ R Characteristic function of R: χ (a, b) = R 0 if ha, bi 6∈ R R ⊆ A × A is reflexive if ∀x ∈ A. xRx R ⊆ A × A is symmetric if ∀x, y ∈ A. xRy ⇒ yRx R ⊆ A × A is transitive if ∀x, y, z ∈ A. xRy ∧ yRz ⇒ xRz R ⊆ A × A is antisymmetric if ∀x, y ∈ A. xRy ∧ yRx ⇒ x = y Equivalence relation: R is reflexive, symmetric and transitive An equivalence relation induces a partition and vice versa F TECHN O OL E O T G U Y IT K T H S A
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19 51 Partial order: R is reflexive, antisymmetric and transitive yog, km s kOflm^
Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 6 / 23 Total order: Connected partial order (eg on R) Irreflexive relation: ∀x ∈ A, hx, xi 6∈ R Asymmetric relation: hx, yi ∈ R ⇒ hy, xi 6∈ R Strict order: R is irreflexive and transitive (∴ asymmetric) If is a PO on A, then <: x < y ≡ x y ∧ x 6= y is a SO on A If < is a SO on A, then : x y ≡ x < y ∨ x = y is a PO on A {0, 1, 2} Suppose hA, i is a poset, M ∈ A (m ∈ A), S ⊆ A
{0, 1} {0, 2} {1, 2} M (m) is a maximal (minimal) element of S iff M ∈ S (m ∈ S) and 6 ∃x ∈ S st M < x (x < m) {0} {1} {2} M (m) is a maximum (minimum) of S iff M ∈ S (m ∈ S) and
{} ∀x ∈ S, x M (m x) A PO can be embedded in a total order
Discrete structures Relations Relations (contd.) Connected relation: ∀x, y ∈ A, either xRy or yRx
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 7 / 23 Irreflexive relation: ∀x ∈ A, hx, xi 6∈ R Asymmetric relation: hx, yi ∈ R ⇒ hy, xi 6∈ R Strict order: R is irreflexive and transitive (∴ asymmetric) If is a PO on A, then <: x < y ≡ x y ∧ x 6= y is a SO on A If < is a SO on A, then : x y ≡ x < y ∨ x = y is a PO on A {0, 1, 2} Suppose hA, i is a poset, M ∈ A (m ∈ A), S ⊆ A
{0, 1} {0, 2} {1, 2} M (m) is a maximal (minimal) element of S iff M ∈ S (m ∈ S) and 6 ∃x ∈ S st M < x (x < m) {0} {1} {2} M (m) is a maximum (minimum) of S iff M ∈ S (m ∈ S) and
{} ∀x ∈ S, x M (m x) A PO can be embedded in a total order
Discrete structures Relations Relations (contd.) Connected relation: ∀x, y ∈ A, either xRy or yRx Total order: Connected partial order (eg on R)
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 7 / 23 Asymmetric relation: hx, yi ∈ R ⇒ hy, xi 6∈ R Strict order: R is irreflexive and transitive (∴ asymmetric) If is a PO on A, then <: x < y ≡ x y ∧ x 6= y is a SO on A If < is a SO on A, then : x y ≡ x < y ∨ x = y is a PO on A {0, 1, 2} Suppose hA, i is a poset, M ∈ A (m ∈ A), S ⊆ A
{0, 1} {0, 2} {1, 2} M (m) is a maximal (minimal) element of S iff M ∈ S (m ∈ S) and 6 ∃x ∈ S st M < x (x < m) {0} {1} {2} M (m) is a maximum (minimum) of S iff M ∈ S (m ∈ S) and
{} ∀x ∈ S, x M (m x) A PO can be embedded in a total order
Discrete structures Relations Relations (contd.) Connected relation: ∀x, y ∈ A, either xRy or yRx Total order: Connected partial order (eg on R) Irreflexive relation: ∀x ∈ A, hx, xi 6∈ R
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 7 / 23 Strict order: R is irreflexive and transitive (∴ asymmetric) If is a PO on A, then <: x < y ≡ x y ∧ x 6= y is a SO on A If < is a SO on A, then : x y ≡ x < y ∨ x = y is a PO on A {0, 1, 2} Suppose hA, i is a poset, M ∈ A (m ∈ A), S ⊆ A
{0, 1} {0, 2} {1, 2} M (m) is a maximal (minimal) element of S iff M ∈ S (m ∈ S) and 6 ∃x ∈ S st M < x (x < m) {0} {1} {2} M (m) is a maximum (minimum) of S iff M ∈ S (m ∈ S) and
{} ∀x ∈ S, x M (m x) A PO can be embedded in a total order
Discrete structures Relations Relations (contd.) Connected relation: ∀x, y ∈ A, either xRy or yRx Total order: Connected partial order (eg on R) Irreflexive relation: ∀x ∈ A, hx, xi 6∈ R Asymmetric relation: hx, yi ∈ R ⇒ hy, xi 6∈ R
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 7 / 23 If is a PO on A, then <: x < y ≡ x y ∧ x 6= y is a SO on A If < is a SO on A, then : x y ≡ x < y ∨ x = y is a PO on A {0, 1, 2} Suppose hA, i is a poset, M ∈ A (m ∈ A), S ⊆ A
{0, 1} {0, 2} {1, 2} M (m) is a maximal (minimal) element of S iff M ∈ S (m ∈ S) and 6 ∃x ∈ S st M < x (x < m) {0} {1} {2} M (m) is a maximum (minimum) of S iff M ∈ S (m ∈ S) and
{} ∀x ∈ S, x M (m x) A PO can be embedded in a total order
Discrete structures Relations Relations (contd.) Connected relation: ∀x, y ∈ A, either xRy or yRx Total order: Connected partial order (eg on R) Irreflexive relation: ∀x ∈ A, hx, xi 6∈ R Asymmetric relation: hx, yi ∈ R ⇒ hy, xi 6∈ R Strict order: R is irreflexive and transitive (∴ asymmetric)
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 7 / 23 If < is a SO on A, then : x y ≡ x < y ∨ x = y is a PO on A {0, 1, 2} Suppose hA, i is a poset, M ∈ A (m ∈ A), S ⊆ A
{0, 1} {0, 2} {1, 2} M (m) is a maximal (minimal) element of S iff M ∈ S (m ∈ S) and 6 ∃x ∈ S st M < x (x < m) {0} {1} {2} M (m) is a maximum (minimum) of S iff M ∈ S (m ∈ S) and
{} ∀x ∈ S, x M (m x) A PO can be embedded in a total order
Discrete structures Relations Relations (contd.) Connected relation: ∀x, y ∈ A, either xRy or yRx Total order: Connected partial order (eg on R) Irreflexive relation: ∀x ∈ A, hx, xi 6∈ R Asymmetric relation: hx, yi ∈ R ⇒ hy, xi 6∈ R Strict order: R is irreflexive and transitive (∴ asymmetric) If is a PO on A, then <: x < y ≡ x y ∧ x 6= y is a SO on A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 7 / 23 {0, 1, 2} Suppose hA, i is a poset, M ∈ A (m ∈ A), S ⊆ A
{0, 1} {0, 2} {1, 2} M (m) is a maximal (minimal) element of S iff M ∈ S (m ∈ S) and 6 ∃x ∈ S st M < x (x < m) {0} {1} {2} M (m) is a maximum (minimum) of S iff M ∈ S (m ∈ S) and
{} ∀x ∈ S, x M (m x) A PO can be embedded in a total order
Discrete structures Relations Relations (contd.) Connected relation: ∀x, y ∈ A, either xRy or yRx Total order: Connected partial order (eg on R) Irreflexive relation: ∀x ∈ A, hx, xi 6∈ R Asymmetric relation: hx, yi ∈ R ⇒ hy, xi 6∈ R Strict order: R is irreflexive and transitive (∴ asymmetric) If is a PO on A, then <: x < y ≡ x y ∧ x 6= y is a SO on A If < is a SO on A, then : x y ≡ x < y ∨ x = y is a PO on A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 7 / 23 Discrete structures Relations Relations (contd.) Connected relation: ∀x, y ∈ A, either xRy or yRx Total order: Connected partial order (eg on R) Irreflexive relation: ∀x ∈ A, hx, xi 6∈ R Asymmetric relation: hx, yi ∈ R ⇒ hy, xi 6∈ R Strict order: R is irreflexive and transitive (∴ asymmetric) If is a PO on A, then <: x < y ≡ x y ∧ x 6= y is a SO on A If < is a SO on A, then : x y ≡ x < y ∨ x = y is a PO on A {0, 1, 2} Suppose hA, i is a poset, 36 M ∈ A (m ∈ A), S ⊆ A
{0, 1} {0, 2} {1, 2} M (m) is a maximal (minimal) 12 18 element of S iff M ∈ S (m ∈ S) 4 6 9 and 6 ∃x ∈ S st M < x (x < m) {0} {1} {2} 2 3 M (m) is a maximum (minimum) of S iff M ∈ S (m ∈ S) and 1 F TECHN O OL E O T G U Y IT K T H S A
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∀ ∈ , N x S x M (m x) I {} Divisors of 36 (60?)
19 51 A PO can be embedded in a total order yog, km s kOflm^ Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 7 / 23 Discrete structures Relations Relations (contd.) Connected relation: ∀x, y ∈ A, either xRy or yRx Total order: Connected partial order (eg on R) Irreflexive relation: ∀x ∈ A, hx, xi 6∈ R Asymmetric relation: hx, yi ∈ R ⇒ hy, xi 6∈ R Strict order: R is irreflexive and transitive (∴ asymmetric) If is a PO on A, then <: x < y ≡ x y ∧ x 6= y is a SO on A If < is a SO on A, then : x y ≡ x < y ∨ x = y is a PO on A {0, 1, 2} Suppose hA, i is a poset, 60 M ∈ A (m ∈ A), S ⊆ A
{0, 1} {0, 2} {1, 2} 20 12 30 M (m) is a maximal (minimal) element of S iff M ∈ S (m ∈ S) 4 10 6 15 and 6 ∃x ∈ S st M < x (x < m) {0} {1} {2} 2 5 3 M (m) is a maximum (minimum) of S iff M ∈ S (m ∈ S) and 1 F TECHN O OL E O T G U Y IT K T H S A
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∀ ∈ , N x S x M (m x) I {} Divisors of 60 19 51 A PO can be embedded in a total order yog, km s kOflm^ Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 7 / 23 ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦
Discrete structures Lattices Lattices
Let hA, i be a poset, let x, y ∈ A The meet of x and y (x ∧ y), is the maximum of all lower bounds for x and y: x ∧ y = max {w ∈ A : w x, w y}, glb for x and y The join of x and y (x ∨ y), is the minimum of all upper bounds for x and y; x ∨ y = min {z ∈ A : x z, y z}, lub for x and y A poset hA, i is a lattice iff every pair of elements in A have both a meet and a join
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 8 / 23 Discrete structures Lattices Lattices
Let hA, i be a poset, let x, y ∈ A The meet of x and y (x ∧ y), is the maximum of all lower bounds for x and y: x ∧ y = max {w ∈ A : w x, w y}, glb for x and y The join of x and y (x ∨ y), is the minimum of all upper bounds for x and y; x ∨ y = min {z ∈ A : x z, y z}, lub for x and y A poset hA, i is a lattice iff every pair of elements in A have both a meet and a join ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 8 / 23 Discrete structures Lattices Lattices (contd.) Basic order properties of meet and join x ∧ y {x, y} x ∨ y x y iff x ∧ y = x x y iff x ∨ y = y If x y, then x ∧ z y ∧ z and x ∨ z y ∨ z If x y and z w, then x ∧ z y ∧ w and x ∨ z y ∨ w Theorem If x y, then x ∧ z y ∧ z and x ∨ z y ∨ z Proof. y ◦ Let v = x ∧ z and u = y ∧ z u ◦z ◦ By transitivity, v is a lb for y and z x ◦ By definition of ∧, v u (as v is the maximum ◦ v among all lbs)
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 9 / 23 (x ∧ y) y [x ∧ y {x, y}] (x ∧ y) ∧ z y ∧ z [If x y, then x ∧ z y] Thus (x ∧ y) ∧ z is a lb of both x and y ∧ z ∴ (x ∧ y) ∧ z x ∧ (y ∧ z) [glb of x and y ∧ z] Also, x ∧ (y ∧ z) (x ∧ y) ∧ z [on similar lines] ∴ (x ∧ y) ∧ z = x ∧ (y ∧ z) [if a b and b a then a = b]
Discrete structures Lattices Lattices (contd.) Commutativity x ∧ y = y ∧ x, x ∨ y = y ∨ x Associativity (x ∧ y) ∧ z = x ∧ (y ∧ z), (x ∨ y) ∨ z = x ∨ (y ∨ z) Absorption x ∧ (x ∨ y) = x, x ∨ (x ∧ y) = x Idempotence x ∧ x = x, x ∨ x = x
Associativity of meet. (x ∧ y) ∧ z x ∧ y x [x ∧ y {x, y} applied twice] (x ∧ y) ∧ z x [transitivity of ]
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 10 / 23 Thus (x ∧ y) ∧ z is a lb of both x and y ∧ z ∴ (x ∧ y) ∧ z x ∧ (y ∧ z) [glb of x and y ∧ z] Also, x ∧ (y ∧ z) (x ∧ y) ∧ z [on similar lines] ∴ (x ∧ y) ∧ z = x ∧ (y ∧ z) [if a b and b a then a = b]
Discrete structures Lattices Lattices (contd.) Commutativity x ∧ y = y ∧ x, x ∨ y = y ∨ x Associativity (x ∧ y) ∧ z = x ∧ (y ∧ z), (x ∨ y) ∨ z = x ∨ (y ∨ z) Absorption x ∧ (x ∨ y) = x, x ∨ (x ∧ y) = x Idempotence x ∧ x = x, x ∨ x = x
Associativity of meet. (x ∧ y) ∧ z x ∧ y x [x ∧ y {x, y} applied twice] (x ∧ y) ∧ z x [transitivity of ] (x ∧ y) y [x ∧ y {x, y}] (x ∧ y) ∧ z y ∧ z [If x y, then x ∧ z y]
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 10 / 23 Also, x ∧ (y ∧ z) (x ∧ y) ∧ z [on similar lines] ∴ (x ∧ y) ∧ z = x ∧ (y ∧ z) [if a b and b a then a = b]
Discrete structures Lattices Lattices (contd.) Commutativity x ∧ y = y ∧ x, x ∨ y = y ∨ x Associativity (x ∧ y) ∧ z = x ∧ (y ∧ z), (x ∨ y) ∨ z = x ∨ (y ∨ z) Absorption x ∧ (x ∨ y) = x, x ∨ (x ∧ y) = x Idempotence x ∧ x = x, x ∨ x = x
Associativity of meet. (x ∧ y) ∧ z x ∧ y x [x ∧ y {x, y} applied twice] (x ∧ y) ∧ z x [transitivity of ] (x ∧ y) y [x ∧ y {x, y}] (x ∧ y) ∧ z y ∧ z [If x y, then x ∧ z y] Thus (x ∧ y) ∧ z is a lb of both x and y ∧ z ∴ (x ∧ y) ∧ z x ∧ (y ∧ z) [glb of x and y ∧ z]
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 10 / 23 ∴ (x ∧ y) ∧ z = x ∧ (y ∧ z) [if a b and b a then a = b]
Discrete structures Lattices Lattices (contd.) Commutativity x ∧ y = y ∧ x, x ∨ y = y ∨ x Associativity (x ∧ y) ∧ z = x ∧ (y ∧ z), (x ∨ y) ∨ z = x ∨ (y ∨ z) Absorption x ∧ (x ∨ y) = x, x ∨ (x ∧ y) = x Idempotence x ∧ x = x, x ∨ x = x
Associativity of meet. (x ∧ y) ∧ z x ∧ y x [x ∧ y {x, y} applied twice] (x ∧ y) ∧ z x [transitivity of ] (x ∧ y) y [x ∧ y {x, y}] (x ∧ y) ∧ z y ∧ z [If x y, then x ∧ z y] Thus (x ∧ y) ∧ z is a lb of both x and y ∧ z ∴ (x ∧ y) ∧ z x ∧ (y ∧ z) [glb of x and y ∧ z] Also, x ∧ (y ∧ z) (x ∧ y) ∧ z [on similar lines]
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 10 / 23 Discrete structures Lattices Lattices (contd.) Commutativity x ∧ y = y ∧ x, x ∨ y = y ∨ x Associativity (x ∧ y) ∧ z = x ∧ (y ∧ z), (x ∨ y) ∨ z = x ∨ (y ∨ z) Absorption x ∧ (x ∨ y) = x, x ∨ (x ∧ y) = x Idempotence x ∧ x = x, x ∨ x = x
Associativity of meet. (x ∧ y) ∧ z x ∧ y x [x ∧ y {x, y} applied twice] (x ∧ y) ∧ z x [transitivity of ] (x ∧ y) y [x ∧ y {x, y}] (x ∧ y) ∧ z y ∧ z [If x y, then x ∧ z y] Thus (x ∧ y) ∧ z is a lb of both x and y ∧ z ∴ (x ∧ y) ∧ z x ∧ (y ∧ z) [glb of x and y ∧ z] Also, x ∧ (y ∧ z) (x ∧ y) ∧ z [on similar lines]
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19 51 ∴ (x ∧ y) ∧ z = x ∧ (y ∧ z) [if a b and b a then a = b] yog, km s kOflm^
Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 10 / 23 Idempotence. x ∧ x = x ∧ (x ∨ (x ∧ y)) = x [Absorbtion, applied twice]
Principle of Duality The dual of any theorem in a lattice is also a theorem.
Discrete structures Lattices Lattices (contd.)
Absorbtion. x x ∨ y [{x, y} x ∨ y] ∴ x ∧ (x ∨ y) = x [x y iff x ∧ y = x]
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 11 / 23 Discrete structures Lattices Lattices (contd.)
Absorbtion. x x ∨ y [{x, y} x ∨ y] ∴ x ∧ (x ∨ y) = x [x y iff x ∧ y = x]
Idempotence. x ∧ x = x ∧ (x ∨ (x ∧ y)) = x [Absorbtion, applied twice]
Principle of Duality The dual of any theorem in a lattice is also a theorem.
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 11 / 23 1 1 a a b c c b 0 a ∧ (b ∨ c) = a 0 (a ∧ b) ∨ (a ∧ c) = 0 a ∧ (b ∨ c) = a (a ∧ b) ∨ (a ∧ c) = b 1
a Distributive lattice: If ∀x, y, z ∈ A, d e c x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z) and b x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z) 0 Are these lattices distributive? a ∧ (b ∨ c) = a (a ∧ b) ∨ (a ∧ c) = b Is P(A) for set A distributive? ? x ∩ (y ∪ z) = (x ∩ y) ∪ (x ∩ z) and ? x ∪ (y ∩ z) = (x ∪ y) ∩ (x ∪ z)
Discrete structures Lattices (contd.) Lattices (contd.) Bounded lattice: It has a maximum element (1) and a minimum element (0), in which case identity properties hold: 0 ∨ x = x = x ∨ 0, 1 ∧ x = x = x ∧ 1 0 ∧ x = 0 = x ∧ 0, 1 ∨ x = 1 = x ∨ 1 Every finite lattice is bounded
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 12 / 23 Is P(A) for set A distributive? ? x ∩ (y ∪ z) = (x ∩ y) ∪ (x ∩ z) and ? x ∪ (y ∩ z) = (x ∪ y) ∩ (x ∪ z)
Discrete structures Lattices (contd.) Lattices (contd.) Bounded lattice: It has a maximum 1 1 element (1) and a minimum element (0), in a which case identity properties hold: a b c c b 0 ∨ x = x = x ∨ 0, 1 ∧ x = x = x ∧ 1 0 a ∧ (b ∨ c) = a 0 0 ∧ x = 0 = x ∧ 0, 1 ∨ x = 1 = x ∨ 1 (a ∧ b) ∨ (a ∧ c) = 0 a ∧ (b ∨ c) = a (a ∧ b) ∨ (a ∧ c) = b Every finite lattice is bounded 1 a Distributive lattice: If ∀x, y, z ∈ A, d e c x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z) and b x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z) 0 Are these lattices distributive? a ∧ (b ∨ c) = a (a ∧ b) ∨ (a ∧ c) = b
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 12 / 23 Discrete structures Lattices (contd.) Lattices (contd.) Bounded lattice: It has a maximum 1 1 element (1) and a minimum element (0), in a which case identity properties hold: a b c c b 0 ∨ x = x = x ∨ 0, 1 ∧ x = x = x ∧ 1 0 a ∧ (b ∨ c) = a 0 0 ∧ x = 0 = x ∧ 0, 1 ∨ x = 1 = x ∨ 1 (a ∧ b) ∨ (a ∧ c) = 0 a ∧ (b ∨ c) = a (a ∧ b) ∨ (a ∧ c) = b Every finite lattice is bounded 1 a Distributive lattice: If ∀x, y, z ∈ A, d e c x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z) and b x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z) 0 Are these lattices distributive? a ∧ (b ∨ c) = a (a ∧ b) ∨ (a ∧ c) = b Is P(A) for set A distributive?
F TECHN O OL E O ∩ ( ∪ ) = ( ∩ ) ∪ ( ∩ ) T G x y z x y x z and U Y IT K ? T H S A
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19 51 ? x ∪ (y ∩ z) = (x ∪ y) ∩ (x ∪ z) yog, km s kOflm^
Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 12 / 23 1 1
a x¯ = x¯ ∧ 1 = x¯ ∧ (x ∨ z) = 1 a ∨ b a ∨ c b ∨ c ¯ ¯ a (x ∧ x) ∨ (x ∧ z) = c 0 ∨ (x¯ ∧ z) = c (x ∧ z) ∨ (x¯ ∧ z) = b b a b c 0 (x ∨ x¯) ∧ z = 1 ∧ z = z 0 0
Discrete structures Lattices (contd.) Complemented lattice Complement in a bounded lattice: z is the complement of x iff x ∧ z = 0 and x ∨ z = 1 Bounded complemented lattice: every element has a complement In a bounded distributive lattice with minimum 0 and maximum 1, the complements of elements are unique, provided they exist – let x¯ and z be complements of x ...
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 13 / 23 Discrete structures Lattices (contd.) Complemented lattice Complement in a bounded lattice: z is the complement of x iff x ∧ z = 0 and x ∨ z = 1 Bounded complemented lattice: every element has a complement In a bounded distributive lattice with minimum 0 and maximum 1, the complements of elements are unique, provided they exist – let x¯ and z be complements of x ... 1 1
a x¯ = x¯ ∧ 1 = x¯ ∧ (x ∨ z) = 1 a ∨ b a ∨ c b ∨ c ¯ ¯ a (x ∧ x) ∨ (x ∧ z) = c 0 ∨ (x¯ ∧ z) = c (x ∧ z) ∨ (x¯ ∧ z) = b b a b c 0 (x ∨ x¯) ∧ z = 1 ∧ z = z 0
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 13 / 23 De Morgan’s laws in a Boolean lattice hA, , , 0, 1i Join of complements is 1 1 x ∧ y = x ∨ y (x ∧ y) ∨ (x ∨ y) = ((x ∧ y) ∨ x) ∨ y 2 x ∨ y = x ∧ y = ((x ∨ x) ∧ (y ∨ x)) ∨ y Meet of complements is 0 = (1 ∧ (y ∨ x)) ∨ y (x ∧ y) ∧ (x ∨ y) = = (y ∨ x) ∨ y (x ∧ y ∧ x) ∨ (x ∧ y ∧ y) = x ∨ (y ∨ y) = 0 ∨ 0 = 0 = x ∨ 1 = 1
Discrete structures Boolean lattice Boolean lattice Boolean lattice: Bounded complemented distributive lattice Extreme elements: Max: 1, Min: 0 Distributivity holds Every element has unique complement De Morgan’s laws apply
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 14 / 23 Join of complements is 1 (x ∧ y) ∨ (x ∨ y) = ((x ∧ y) ∨ x) ∨ y = ((x ∨ x) ∧ (y ∨ x)) ∨ y Meet of complements is 0 = (1 ∧ (y ∨ x)) ∨ y (x ∧ y) ∧ (x ∨ y) = = (y ∨ x) ∨ y (x ∧ y ∧ x) ∨ (x ∧ y ∧ y) = x ∨ (y ∨ y) = 0 ∨ 0 = 0 = x ∨ 1 = 1
Discrete structures Boolean lattice Boolean lattice Boolean lattice: Bounded complemented distributive lattice Extreme elements: Max: 1, Min: 0 Distributivity holds Every element has unique complement De Morgan’s laws apply
De Morgan’s laws in a Boolean lattice hA, , , 0, 1i
1 x ∧ y = x ∨ y 2 x ∨ y = x ∧ y
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 14 / 23 Join of complements is 1 (x ∧ y) ∨ (x ∨ y) = ((x ∧ y) ∨ x) ∨ y = ((x ∨ x) ∧ (y ∨ x)) ∨ y = (1 ∧ (y ∨ x)) ∨ y = (y ∨ x) ∨ y = x ∨ (y ∨ y) = x ∨ 1 = 1
Discrete structures Boolean lattice Boolean lattice Boolean lattice: Bounded complemented distributive lattice Extreme elements: Max: 1, Min: 0 Distributivity holds Every element has unique complement De Morgan’s laws apply
De Morgan’s laws in a Boolean lattice hA, , , 0, 1i
1 x ∧ y = x ∨ y 2 x ∨ y = x ∧ y
Meet of complements is 0 (x ∧ y) ∧ (x ∨ y) = (x ∧ y ∧ x) ∨ (x ∧ y ∧ y)
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 14 / 23 Discrete structures Boolean lattice Boolean lattice Boolean lattice: Bounded complemented distributive lattice Extreme elements: Max: 1, Min: 0 Distributivity holds Every element has unique complement De Morgan’s laws apply
De Morgan’s laws in a Boolean lattice hA, , , 0, 1i Join of complements is 1 1 x ∧ y = x ∨ y (x ∧ y) ∨ (x ∨ y) = ((x ∧ y) ∨ x) ∨ y 2 x ∨ y = x ∧ y = ((x ∨ x) ∧ (y ∨ x)) ∨ y Meet of complements is 0 = (1 ∧ (y ∨ x)) ∨ y (x ∧ y) ∧ (x ∨ y) = = (y ∨ x) ∨ y (x ∧ y ∧ x) ∨ (x ∧ y ∧ y) = x ∨ (y ∨ y)
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19 51 = 0 ∨ 0 = 0 = x ∨ 1 = 1 yog, km s kOflm^
Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 14 / 23 a 6= 0 is join irreducible if and only if a has a unique immediate predecessor Elements immediately succeeding 0 are atoms (join irreducible) Any element a can be expressed as the join of a set of atoms Not unique for non-distributive lattice (diamond lattice) For finite lattice a = d1 ∨ d2 ∨ ... ∨ dn, dj are join irreducible dj = di ∨ dj for di dj Any di dj can be dropped to make the join irredundant Unique (up to permutation) for distributive lattice
Discrete structures Boolean lattice structure Boolean lattice structure Let A be a lattice with min 0 a ∈ A is join irreducible if a a a
a 6= x ∨ y for x, y a, alternatively b1 b2 c
a = x ∨ y implies a = x or a = y 0 0 b1 b2 0 is join irreducible
If b1 c and b2 c (immediate preds) of c then c = b1 ∨ b2
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 15 / 23 Elements immediately succeeding 0 are atoms (join irreducible) Any element a can be expressed as the join of a set of atoms Not unique for non-distributive lattice (diamond lattice) For finite lattice a = d1 ∨ d2 ∨ ... ∨ dn, dj are join irreducible dj = di ∨ dj for di dj Any di dj can be dropped to make the join irredundant Unique (up to permutation) for distributive lattice
Discrete structures Boolean lattice structure Boolean lattice structure Let A be a lattice with min 0 a ∈ A is join irreducible if a a a
a 6= x ∨ y for x, y a, alternatively b1 b2 c
a = x ∨ y implies a = x or a = y 0 0 b1 b2 0 is join irreducible
If b1 c and b2 c (immediate preds) of c then c = b1 ∨ b2 a 6= 0 is join irreducible if and only if a has a unique immediate predecessor
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 15 / 23 Any element a can be expressed as the join of a set of atoms Not unique for non-distributive lattice (diamond lattice) For finite lattice a = d1 ∨ d2 ∨ ... ∨ dn, dj are join irreducible dj = di ∨ dj for di dj Any di dj can be dropped to make the join irredundant Unique (up to permutation) for distributive lattice
Discrete structures Boolean lattice structure Boolean lattice structure Let A be a lattice with min 0 a ∈ A is join irreducible if a a a
a 6= x ∨ y for x, y a, alternatively b1 b2 c
a = x ∨ y implies a = x or a = y 0 0 b1 b2 0 is join irreducible
If b1 c and b2 c (immediate preds) of c then c = b1 ∨ b2 a 6= 0 is join irreducible if and only if a has a unique immediate predecessor Elements immediately succeeding 0 are atoms (join irreducible)
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 15 / 23 Not unique for non-distributive lattice (diamond lattice) For finite lattice a = d1 ∨ d2 ∨ ... ∨ dn, dj are join irreducible dj = di ∨ dj for di dj Any di dj can be dropped to make the join irredundant Unique (up to permutation) for distributive lattice
Discrete structures Boolean lattice structure Boolean lattice structure Let A be a lattice with min 0 a ∈ A is join irreducible if a a a
a 6= x ∨ y for x, y a, alternatively b1 b2 c
a = x ∨ y implies a = x or a = y 0 0 b1 b2 0 is join irreducible
If b1 c and b2 c (immediate preds) of c then c = b1 ∨ b2 a 6= 0 is join irreducible if and only if a has a unique immediate predecessor Elements immediately succeeding 0 are atoms (join irreducible) Any element a can be expressed as the join of a set of atoms
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 15 / 23 For finite lattice a = d1 ∨ d2 ∨ ... ∨ dn, dj are join irreducible dj = di ∨ dj for di dj Any di dj can be dropped to make the join irredundant Unique (up to permutation) for distributive lattice
Discrete structures Boolean lattice structure Boolean lattice structure Let A be a lattice with min 0 a ∈ A is join irreducible if a a a
a 6= x ∨ y for x, y a, alternatively b1 b2 c
a = x ∨ y implies a = x or a = y 0 0 b1 b2 0 is join irreducible
If b1 c and b2 c (immediate preds) of c then c = b1 ∨ b2 a 6= 0 is join irreducible if and only if a has a unique immediate predecessor Elements immediately succeeding 0 are atoms (join irreducible) Any element a can be expressed as the join of a set of atoms Not unique for non-distributive lattice (diamond lattice)
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 15 / 23 dj = di ∨ dj for di dj Any di dj can be dropped to make the join irredundant Unique (up to permutation) for distributive lattice
Discrete structures Boolean lattice structure Boolean lattice structure Let A be a lattice with min 0 a ∈ A is join irreducible if a a a
a 6= x ∨ y for x, y a, alternatively b1 b2 c
a = x ∨ y implies a = x or a = y 0 0 b1 b2 0 is join irreducible
If b1 c and b2 c (immediate preds) of c then c = b1 ∨ b2 a 6= 0 is join irreducible if and only if a has a unique immediate predecessor Elements immediately succeeding 0 are atoms (join irreducible) Any element a can be expressed as the join of a set of atoms Not unique for non-distributive lattice (diamond lattice) For finite lattice a = d1 ∨ d2 ∨ ... ∨ dn, dj are join irreducible
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 15 / 23 Any di dj can be dropped to make the join irredundant Unique (up to permutation) for distributive lattice
Discrete structures Boolean lattice structure Boolean lattice structure Let A be a lattice with min 0 a ∈ A is join irreducible if a a a
a 6= x ∨ y for x, y a, alternatively b1 b2 c
a = x ∨ y implies a = x or a = y 0 0 b1 b2 0 is join irreducible
If b1 c and b2 c (immediate preds) of c then c = b1 ∨ b2 a 6= 0 is join irreducible if and only if a has a unique immediate predecessor Elements immediately succeeding 0 are atoms (join irreducible) Any element a can be expressed as the join of a set of atoms Not unique for non-distributive lattice (diamond lattice) For finite lattice a = d1 ∨ d2 ∨ ... ∨ dn, dj are join irreducible dj = di ∨ dj for di dj
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 15 / 23 Unique (up to permutation) for distributive lattice
Discrete structures Boolean lattice structure Boolean lattice structure Let A be a lattice with min 0 a ∈ A is join irreducible if a a a
a 6= x ∨ y for x, y a, alternatively b1 b2 c
a = x ∨ y implies a = x or a = y 0 0 b1 b2 0 is join irreducible
If b1 c and b2 c (immediate preds) of c then c = b1 ∨ b2 a 6= 0 is join irreducible if and only if a has a unique immediate predecessor Elements immediately succeeding 0 are atoms (join irreducible) Any element a can be expressed as the join of a set of atoms Not unique for non-distributive lattice (diamond lattice) For finite lattice a = d1 ∨ d2 ∨ ... ∨ dn, dj are join irreducible dj = di ∨ dj for di dj Any d d can be dropped to make the join irredundant TECH i j OF NO L E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 15 / 23 Discrete structures Boolean lattice structure Boolean lattice structure Let A be a lattice with min 0 a ∈ A is join irreducible if a a a
a 6= x ∨ y for x, y a, alternatively b1 b2 c
a = x ∨ y implies a = x or a = y 0 0 b1 b2 0 is join irreducible
If b1 c and b2 c (immediate preds) of c then c = b1 ∨ b2 a 6= 0 is join irreducible if and only if a has a unique immediate predecessor Elements immediately succeeding 0 are atoms (join irreducible) Any element a can be expressed as the join of a set of atoms Not unique for non-distributive lattice (diamond lattice) For finite lattice a = d1 ∨ d2 ∨ ... ∨ dn, dj are join irreducible dj = di ∨ dj for di dj Any d d can be dropped to make the join irredundant TECH i j OF NO L E O T G U Y IT K T H S A
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19 51 Unique (up to permutation) for distributive lattice yog, km s kOflm^
Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 15 / 23 Now, ci , dj c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn ∴ ci = ci ∧ (d1 ∨ d2 ∨ ... ∨ dn) = (ci ∧ d1) ∨ (ci ∧ d2) ∨ ... ∨ (ci ∧ dn) Since ci is join irreducible, ∃dj |ci = ci ∧ dj , so that ci dj
But similar working, dj ck , so that ci dj ck
This requires ci = ck , since these are irredundant
Thus, ci dj and dj ci , ci = dj ,
This way, all the ci s may to paired off with the dj s, making the representation unique (up to permutation)
Discrete structures Boolean lattice structure Boolean lattice representation (contd.)
Unique irredundant irreducible sum representation
Let a = c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 16 / 23 ∴ ci = ci ∧ (d1 ∨ d2 ∨ ... ∨ dn) = (ci ∧ d1) ∨ (ci ∧ d2) ∨ ... ∨ (ci ∧ dn) Since ci is join irreducible, ∃dj |ci = ci ∧ dj , so that ci dj
But similar working, dj ck , so that ci dj ck
This requires ci = ck , since these are irredundant
Thus, ci dj and dj ci , ci = dj ,
This way, all the ci s may to paired off with the dj s, making the representation unique (up to permutation)
Discrete structures Boolean lattice structure Boolean lattice representation (contd.)
Unique irredundant irreducible sum representation
Let a = c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn
Now, ci , dj c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 16 / 23 Since ci is join irreducible, ∃dj |ci = ci ∧ dj , so that ci dj
But similar working, dj ck , so that ci dj ck
This requires ci = ck , since these are irredundant
Thus, ci dj and dj ci , ci = dj ,
This way, all the ci s may to paired off with the dj s, making the representation unique (up to permutation)
Discrete structures Boolean lattice structure Boolean lattice representation (contd.)
Unique irredundant irreducible sum representation
Let a = c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn
Now, ci , dj c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn ∴ ci = ci ∧ (d1 ∨ d2 ∨ ... ∨ dn) = (ci ∧ d1) ∨ (ci ∧ d2) ∨ ... ∨ (ci ∧ dn)
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 16 / 23 But similar working, dj ck , so that ci dj ck
This requires ci = ck , since these are irredundant
Thus, ci dj and dj ci , ci = dj ,
This way, all the ci s may to paired off with the dj s, making the representation unique (up to permutation)
Discrete structures Boolean lattice structure Boolean lattice representation (contd.)
Unique irredundant irreducible sum representation
Let a = c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn
Now, ci , dj c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn ∴ ci = ci ∧ (d1 ∨ d2 ∨ ... ∨ dn) = (ci ∧ d1) ∨ (ci ∧ d2) ∨ ... ∨ (ci ∧ dn) Since ci is join irreducible, ∃dj |ci = ci ∧ dj , so that ci dj
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 16 / 23 This requires ci = ck , since these are irredundant
Thus, ci dj and dj ci , ci = dj ,
This way, all the ci s may to paired off with the dj s, making the representation unique (up to permutation)
Discrete structures Boolean lattice structure Boolean lattice representation (contd.)
Unique irredundant irreducible sum representation
Let a = c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn
Now, ci , dj c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn ∴ ci = ci ∧ (d1 ∨ d2 ∨ ... ∨ dn) = (ci ∧ d1) ∨ (ci ∧ d2) ∨ ... ∨ (ci ∧ dn) Since ci is join irreducible, ∃dj |ci = ci ∧ dj , so that ci dj
But similar working, dj ck , so that ci dj ck
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 16 / 23 Thus, ci dj and dj ci , ci = dj ,
This way, all the ci s may to paired off with the dj s, making the representation unique (up to permutation)
Discrete structures Boolean lattice structure Boolean lattice representation (contd.)
Unique irredundant irreducible sum representation
Let a = c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn
Now, ci , dj c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn ∴ ci = ci ∧ (d1 ∨ d2 ∨ ... ∨ dn) = (ci ∧ d1) ∨ (ci ∧ d2) ∨ ... ∨ (ci ∧ dn) Since ci is join irreducible, ∃dj |ci = ci ∧ dj , so that ci dj
But similar working, dj ck , so that ci dj ck
This requires ci = ck , since these are irredundant
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 16 / 23 This way, all the ci s may to paired off with the dj s, making the representation unique (up to permutation)
Discrete structures Boolean lattice structure Boolean lattice representation (contd.)
Unique irredundant irreducible sum representation
Let a = c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn
Now, ci , dj c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn ∴ ci = ci ∧ (d1 ∨ d2 ∨ ... ∨ dn) = (ci ∧ d1) ∨ (ci ∧ d2) ∨ ... ∨ (ci ∧ dn) Since ci is join irreducible, ∃dj |ci = ci ∧ dj , so that ci dj
But similar working, dj ck , so that ci dj ck
This requires ci = ck , since these are irredundant
Thus, ci dj and dj ci , ci = dj ,
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 16 / 23 making the representation unique (up to permutation)
Discrete structures Boolean lattice structure Boolean lattice representation (contd.)
Unique irredundant irreducible sum representation
Let a = c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn
Now, ci , dj c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn ∴ ci = ci ∧ (d1 ∨ d2 ∨ ... ∨ dn) = (ci ∧ d1) ∨ (ci ∧ d2) ∨ ... ∨ (ci ∧ dn) Since ci is join irreducible, ∃dj |ci = ci ∧ dj , so that ci dj
But similar working, dj ck , so that ci dj ck
This requires ci = ck , since these are irredundant
Thus, ci dj and dj ci , ci = dj ,
This way, all the ci s may to paired off with the dj s,
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 16 / 23 Discrete structures Boolean lattice structure Boolean lattice representation (contd.)
Unique irredundant irreducible sum representation
Let a = c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn
Now, ci , dj c1 ∨ c2 ∨ ... ∨ cm = d1 ∨ d2 ∨ ... ∨ dn ∴ ci = ci ∧ (d1 ∨ d2 ∨ ... ∨ dn) = (ci ∧ d1) ∨ (ci ∧ d2) ∨ ... ∨ (ci ∧ dn) Since ci is join irreducible, ∃dj |ci = ci ∧ dj , so that ci dj
But similar working, dj ck , so that ci dj ck
This requires ci = ck , since these are irredundant
Thus, ci dj and dj ci , ci = dj ,
This way, all the ci s may to paired off with the dj s, making the representation unique (up to permutation)
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 16 / 23 Join irreducible elements in a Boolean lattice
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A lattice with having a non-zero join irreducible a ∨ b a ∨ c b ∨ c element will not have unique complements
In a Boolean lattice all non-zero join irreducible a b c elements are atoms
0
Discrete structures Boolean lattice structure Boolean lattice structure (contd.) Let z be the complement of a in a lattice as shown Suppose a has b as a unique predecessor a So, a ∨ z = 1 and a ∧ z = 0 z 6= 1
Now, b ∨ z = a ∨ z = 1 and b ∧ z = a ∧ z = 0 as b is the b 6= 0 immediate predecessor of a So, b is also a complement of z
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 17 / 23 Discrete structures Boolean lattice structure Boolean lattice structure (contd.) Let z be the complement of a in a lattice as shown Suppose a has b as a unique predecessor a So, a ∨ z = 1 and a ∧ z = 0 z 6= 1
Now, b ∨ z = a ∨ z = 1 and b ∧ z = a ∧ z = 0 as b is the b 6= 0 immediate predecessor of a So, b is also a complement of z Join irreducible elements in a Boolean lattice
1
A lattice with having a non-zero join irreducible a ∨ b a ∨ c b ∨ c element will not have unique complements
In a Boolean lattice all non-zero join irreducible a b c elements are atoms
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 17 / 23 Non-trivial atomic elements are present for |A| > 1 directly above level 0, let those be S = {a1,..., an}, akin to {a1} , {a2} ,... {an} Join of pairs of elements Y1, Y2 at level i (n > i > 1) st |Y1 − Y2| = |Y2 − Y1| = 1 at level ı + 1 is Y = Y1 ∪ Y2 Meet of pairs of elements X1, X2 at level i (n > i > 1) st |X1 − X2| = |X2 − X1| = 1 at level ı − 1 is Y = Y1 ∩ Y2 n Pn n n There will be i such sets in level i, totaling to 0 i = 2
Discrete structures Boolean lattice structure Stone representation of Boolean lattices
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Atom of a Boolean lattice: Non-trivial a ∨ c a ∨ b b ∨ c minimal element of A \{0} |A| = 2n for some n for a Boolean lattice a b c Its structure is that of the power set of the atomic elements
0
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 18 / 23 Join of pairs of elements Y1, Y2 at level i (n > i > 1) st |Y1 − Y2| = |Y2 − Y1| = 1 at level ı + 1 is Y = Y1 ∪ Y2 Meet of pairs of elements X1, X2 at level i (n > i > 1) st |X1 − X2| = |X2 − X1| = 1 at level ı − 1 is Y = Y1 ∩ Y2 n Pn n n There will be i such sets in level i, totaling to 0 i = 2
Discrete structures Boolean lattice structure Stone representation of Boolean lattices
1
Atom of a Boolean lattice: Non-trivial a ∨ c a ∨ b b ∨ c minimal element of A \{0} |A| = 2n for some n for a Boolean lattice a b c Its structure is that of the power set of the atomic elements
0 Non-trivial atomic elements are present for |A| > 1 directly above level 0, let those be S = {a1,..., an}, akin to {a1} , {a2} ,... {an}
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 18 / 23 Meet of pairs of elements X1, X2 at level i (n > i > 1) st |X1 − X2| = |X2 − X1| = 1 at level ı − 1 is Y = Y1 ∩ Y2 n Pn n n There will be i such sets in level i, totaling to 0 i = 2
Discrete structures Boolean lattice structure Stone representation of Boolean lattices
1
Atom of a Boolean lattice: Non-trivial a ∨ c a ∨ b b ∨ c minimal element of A \{0} |A| = 2n for some n for a Boolean lattice a b c Its structure is that of the power set of the atomic elements
0 Non-trivial atomic elements are present for |A| > 1 directly above level 0, let those be S = {a1,..., an}, akin to {a1} , {a2} ,... {an} Join of pairs of elements Y1, Y2 at level i (n > i > 1) st |Y1 − Y2| = |Y2 − Y1| = 1 at level ı + 1 is Y = Y1 ∪ Y2
F TECHN O OL E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 18 / 23 n Pn n n There will be i such sets in level i, totaling to 0 i = 2
Discrete structures Boolean lattice structure Stone representation of Boolean lattices
1
Atom of a Boolean lattice: Non-trivial a ∨ c a ∨ b b ∨ c minimal element of A \{0} |A| = 2n for some n for a Boolean lattice a b c Its structure is that of the power set of the atomic elements
0 Non-trivial atomic elements are present for |A| > 1 directly above level 0, let those be S = {a1,..., an}, akin to {a1} , {a2} ,... {an} Join of pairs of elements Y1, Y2 at level i (n > i > 1) st |Y1 − Y2| = |Y2 − Y1| = 1 at level ı + 1 is Y = Y1 ∪ Y2 Meet of pairs of elements X1, X2 at level i (n > i > 1) st |X − X | = |X − X | = 1 at level ı − 1 is Y = Y ∩ Y F TECHN O OL 1 2 2 1 1 2 E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 18 / 23 Discrete structures Boolean lattice structure Stone representation of Boolean lattices
1
Atom of a Boolean lattice: Non-trivial a ∨ c a ∨ b b ∨ c minimal element of A \{0} |A| = 2n for some n for a Boolean lattice a b c Its structure is that of the power set of the atomic elements
0 Non-trivial atomic elements are present for |A| > 1 directly above level 0, let those be S = {a1,..., an}, akin to {a1} , {a2} ,... {an} Join of pairs of elements Y1, Y2 at level i (n > i > 1) st |Y1 − Y2| = |Y2 − Y1| = 1 at level ı + 1 is Y = Y1 ∪ Y2 Meet of pairs of elements X1, X2 at level i (n > i > 1) st |X − X | = |X − X | = 1 at level ı − 1 is Y = Y ∩ Y F TECHN O OL 1 2 2 1 1 2 E O T G U Y IT K T H S A
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 18 / 23 Discrete structures Boolean algebra Boolean algebra from Boolean lattice For the Boolean lattice hA, , 0, 1i consider the algebraic system hA, +, ·, , 0, 1i where ∨ 7→ +, ∧ 7→ · and ∀x ∈ A, x 7→ z|x + z = 1, x · z = 0 This system satisfies the Huntington’s postulates for a Boolean algebra B1: Commutative Laws 1 x + y = y + x 2 x · y = y · x B2: Distributive Laws 1 x · (y + z) = x · y + x · z 2 x + (y · z) = (x + y) · (x + z) B3: Identity Laws 1 x + 0 = x = 0 + x 2 x · 1 = x = 1 · x B4: Complementation Laws
F TECHN O OL E O 1 T G U Y IT K ¯ ¯ T H S + = = + A x x 1 x x N R
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19 51 2 x · x¯ = 0 = x¯ · x yog, km s kOflm^
Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 19 / 23 Discrete structures Additional BA properties Additional Boolean algebra properties
These properties carry over from the Boolean lattice May be proven independently from the Huntington’s postulates Idempotence: Axiomatic proof 1 x + x = x x + x = (x + x) · 1 2 x · x = x = (x + x) · (x + x¯) = x + (x · x¯) = x + 0 = x Absorption: Axiomatic proof 1 x + xy = x x + xy = (x · 1) + xy 2 x · (x + y) = x = x(1 + y) = x(y + 1) 3 x + xy¯ = x + y = x · 1 = x
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 20 / 23 Discrete structures Additional BA properties Boolean algebra (contd.)
Boundedness/annihilation: Axiomatic proof 1 x + 1 = 1 x + 1 = 1 · (x + 1) 2 x · 0 = 0 = (x + x¯) · (x + 1) = x + (x¯ · 1) = x + x¯ = 1 x y x x · y x + y 0 0 1 0 0 Truth table for Boolean AND, OR, NOT: 0 1 1 0 1 1 0 0 0 1 1 1 0 1 1 Associativity:
1 F TECHN O OL E O (x + y) + z = x + (y + z) T G U Y IT K T H S A
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19 51 2 (x · y) · z = x · (y · z) yog, km s kOflm^
Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 21 / 23 Discrete structures Additional BA properties Boolean algebra (contd.)
Axiomatic proof of associativity of Boolean + Let x = a + (b + c) and y = (a + b) + c ax = aa + a(b + c) = a + a(b + c) = a bx = ba + b(b + c) = ba + (bb + bc) = ba + (b + bc) = ba + b = b Similarly, cx = c and ay = a, by = b and cy = c yx = ((a+b)+c)x = (a+b)x +cx = (ax +bx)+cx = (a+b)+c = y xy = (a+(b+c))y = ay +(b+c)y = ay +(by +cy) = a+(b+c) = x Thus, x = xy = yx = y
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 22 / 23 Discrete structures Additional BA properties Additional Boolean algebra properties (contd.)
Uniqueness of Complement: If (a + x) = 1 and (a · x) = 0, then x = a . Involution: (a) = a Complements of extreme elements: Axiomatic proof 0 = 1 1 + 0 = 1 [identity] 1 = 0 1 · 0 = 0 [boundedness] ∴ 0 is the complement of 1 DeMorgan’s laws: (x + y) = x · y (x · y) = x + y Let a b if a · b = a or a + b = b then · 7→ ∧ and + 7→ ∨
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Properties from axiomatic proofs allow Boolean algebras to be G
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Chittaranjan Mandal (IIT Kharagpur) SCLD January 21, 2020 23 / 23