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Lattice and Boolean Algebra Slide 3
Algebra
• An algebraic system is defined by the tuple ⟨A,o1, …, ok; R1, …, Rm; c1, … ck⟩, where, A is a non-empty set, oi is a function Api →A, pi is a positive integer, Rj is a relation on A, and ci is an element of A.
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Lattice and Boolean Algebra Slide 4
Lattice
• The lattice is an algebraic system ⟨A, ∨, ⋅⟩, given a,b,c in A, the following axioms are satisfied:1. Idempotent laws: a ∨ a = a, a ⋅ a = a;
2. Commutative laws: a ∨ b = b ∨ a, a ⋅ b = b ⋅ a3. Associative laws: a ∨ (b ∨ c) = (a ∨ b) ∨ c,
a ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c
4. Absorption laws: a ∨ (a ⋅ b) = a, a ⋅ (a ∨ b) = a
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Lattice and Boolean Algebra Slide 5
Lattice - Example
• Let A={1,2,3,6}.• Let a ∨ b be the least common multiple• Let a ∧ b be the greatest common divisor • Then, the algebraic system ⟨A, ∨, ∧⟩ satisfies
the axioms of the lattice.
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Lattice and Boolean Algebra Slide 6
Distributive Lattice
• The lattice ⟨A, ∨, ⋅⟩ satisfying the following axiom is a distributive lattice5. Distributive laws: a ∨ (b ⋅ c) = (a ∨ b) ⋅ (a ∨ c),
a ⋅ (b ∨ c) = (a ⋅ b) ∨ (a ⋅ c)
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Lattice and Boolean Algebra Slide 7
Examplesdistributive non-distributive
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a⋅(b∨c)≠a⋅b∨a⋅c
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Lattice and Boolean Algebra Slide 8
Complemented Lattice
• Let a lattice ⟨A, ∨, ⋅⟩ have a maximum element 1 and a minimum element 0. For any element a in A, if there exists an element xa such that a ∨ xa = 1 and a ⋅ xa = 0, then the lattice is a complemented lattice.
• Find complements in the previous example
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Lattice and Boolean Algebra Slide 9
Boolean Algebra• Let B be a set with at least two elements 0 and 1.
Let two binary operations ∨ and ⋅, and a unary operation are defined on B. The algebraic system
⟨B, ∨, ⋅ , , 0,1⟩ is a Boolean algebra, if the following postulates are satisfied:
1. Idempotent laws: a ∨ a = a, a ⋅ a = a;
2. Commutative laws: a ∨ b = b ∨ a, a ⋅ b = b ⋅ a3. Associative laws: a ∨ (b ∨ c) = (a ∨ b) ∨ c,
a ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c
4. Absorption laws: a ∨ (a ⋅ b) = a, a ⋅ (a ∨ b) = a
5. Distributive laws: a ∨ (b ⋅ c) = (a ∨ b) ⋅ (a ∨ c), a ⋅ (b ∨ c) = (a ⋅ b) ∨ (a ⋅ c)
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Lattice and Boolean Algebra Slide 10
Boolean Algebra
6. Involution:
7. Complements: a ∨ a = 1, a ⋅ a = 0;
8. Identities: a ∨ 0 = a, a ⋅ 1 = a;
9. a ∨ 1 = 1, a ⋅ 0 = 0;
10.De Morgan’s laws:€
a=a
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a∨b=a⋅b
a⋅b=a∨b
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Lattice and Boolean Algebra Slide 11
Huntington’s Postulates
• To verify whether a given algebra is a Boolean algebra we only need to check 4 postulates:1. Identities
2. Commutative laws
3. Distributive laws
4. Complements
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Lattice and Boolean Algebra Slide 12
Example
• prove the idempotent laws given Huntington’s postulates:a = a ∨ 0
= a ∨ a⋅a
= (a ∨ a) ⋅ (a ∨ a)
= (a ∨ a) ⋅ 1 = a ∨ a
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Lattice and Boolean Algebra Slide 13
Models of Boolean Algebra• Boolean Algebra over {0,1}
B={0,1}. ⟨B, ∨, ⋅ , , 0,1⟩
• Boolean Algebra over Boolean VectorsBn = {(a1, a2, … , an) | ai ∈ {0,1}}
Let a=(a1, a2, … , an) and b = (b1, b2, … , bn) ∈ Bn
define
a ∨ b = (a1 ∨ b1, a2 ∨ b2, … , an ∨ bn)
a ⋅ b = (a1 ⋅ b1, a2 ⋅ b2, … , an ⋅ bn)
a=(a1, a2, … , an)
then ⟨Bn, ∨, ⋅ , , 0,1⟩ is a Boolean algebra, where, 0 = (0,0, …, 0) and 1 = (1,1, …, 1)
• Boolean Algebra over Power Setschool.edhole.com
Lattice and Boolean Algebra Slide 14
Examples
B3 P({a,b,c}) {n ∈ Ν| n|30}
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Lattice and Boolean Algebra Slide 15
Isomorphic Boolean Algebra
• Two Boolean algebras ⟨A, ∨, ⋅ , , 0A,1A⟩ and
⟨B, ∨, ⋅ , , 0B,1B⟩ are isomorphic iff there is a mapping f:A→B, such that
1. for arbitrary a,b ∈ A, f(a∨b) = f(a)∨f(b), f(a ⋅ b) = f(a) ⋅ f(b), and f(a) = f(a)
2. f(0A ) = 0B and f(1A ) = 1B
An arbitrary finite Boolean algebra is isomorphic to the Boolean algebra ⟨Bn, ∨, ⋅ , , 0,1⟩
Question: define the mappings for the previous slide.school.edhole.com
Lattice and Boolean Algebra Slide 16
De Morgan’s Theorem
• De Morgan’s Laws hold
• These equations can be generalized
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a∨b=a⋅b
a⋅b=a∨b
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x1∨x2∨K∨xn=x1⋅x2⋅K⋅xn
x1⋅x2⋅K⋅xn=x1∨x2∨K∨xn
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Lattice and Boolean Algebra Slide 17
Definition
• Let ⟨Bn, ∨, ⋅ , , 0,1⟩ be a Boolean algebra. The variable that takes arbitrary values in the set B is a Boolean variable. The expression that is obtained from the Boolean variables and constants by combining with the operators ∨, ⋅ , and parenthesis is a Boolean expression. If a mapping f:Bn →B is represented by a Boolean expression, then f is a Boolean function. However, not all mappings f:Bn →B are Boolean functions.
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Lattice and Boolean Algebra Slide 18
Theorem
• Let F(x1, x2, …, xn) be a Boolean expression. Then the complement of the complement of the Boolean expression F(x1, x2, …, xn) is obtained from F as follows1. Add parenthesis according to the order of
operations
2. Interchange ∨ with ⋅ 3. Interchange xi with xi
4. Interchange 0 with 1
Example
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x∨(y ⋅z)=x ⋅(y∨z )school.edhole.com
Lattice and Boolean Algebra Slide 19
Principle of Duality
• In the axioms of Boolean algebra, in an equation that contains ∨, ⋅, 0, or 1, if we interchange ∨ with ⋅ , and/or 0 with 1, then the other equation holds.
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Lattice and Boolean Algebra Slide 20
Dual Boolean Expressions• Let A be a Boolean expression. The dual AD
is defined recursively as follows:1. 0D = 12. 1D = 03. if xi is a variable, then xi
D = xi
4. if A, B, and C are Boolean expressions, and A = B ∨ C, then AD = BD ⋅ CD
5. if A, B, and C are Boolean expressions, and A = B ⋅ C, then AD = BD ∨ CD
6. if A and B are Boolean expressions, and A = B, then
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AD=(BD)
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Lattice and Boolean Algebra Slide 21
Examples
1. Given xy ∨ yz = xy ∨ yz ∨ xz
the dual (x ∨ y)(y ∨ z) = (x ∨ y)(y ∨ z)(x ∨ z)
2. Consider the Boolean algebra B={0,1,a,a} check if f is a Boolean function.
f(x) = xf(0) ∨ xf(1)
f(x) = x ⋅ a ∨ x ⋅ 1f(a) = a ⋅ a ∨ a ⋅ 1 = a
x f(x)
0 a
1 1
a a
a 1school.edhole.com
Lattice and Boolean Algebra Slide 22
Logic Functions• Let B = {0,1}. A mapping Bn →B is always represented
by a Boolean expression–a two-valued logic function.
f ∨ g = h ⇔ f(x1,x2,…,xn) ∨ g(x1,x2,…,xn) = h(x1,x2,…,xn)
f = g ⇔ f(x1,x2,…,xn) = g(x1,x2,…,xn)
x y f g f∨g f⋅g f g
0 0 0 0 0 0 1 1
0 1 1 0 1 0 0 1
1 0 1 0 1 0 0 1
1 1 0 1 1 0 1 0
Example
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Lattice and Boolean Algebra Slide 23
Logical Expressions
1. Constants 0 and 1 are logical expressions
2. Variables x1,x2,…,xn are logical expressions
3. If E is a logical expression, then E is one
4. If E1 and E2 are logical expressions, then (E1 ∨ E2) and (E1 ⋅ E2) are also logical expressions
5. The logical expressions are obtained by finite application of 1 - 4
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Lattice and Boolean Algebra Slide 24
Evaluation of logical Expressions
• An assignment mapping α:{xi} →{0,1} (i = 1, … , n)
• The valuation mapping |F|α of a logical expression is obtained:
1. |0|α = 0 and |1|α = 1
2. If xi is a variable, then | xi |α = α(xi)
3. If F is a logical expression, then |F|α = 1⇔ |F|α = 0
4. If F and G are logical expressions, then |F ∨ G|α = 1⇔ (|F|α = 1 or |G|α = 1)
5. If F and G are logical expressions, then |F ⋅ G|α = 1⇔ (|F|α = 1 and |G|α = 1)
Example: F:x ∨ y ⋅ z α(x) = 0, α(y) = 0, α(z) = 1
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Lattice and Boolean Algebra Slide 25
Equivalence of Logic Expressions
• Let F and G be logical expressions. If |F|α = |G|α hold for every assignment α, then F and G are equivalent ==> F ≡ G
• Logical expressions can be classified into 22n
equivalence classes by the equivalence relation (≡)
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