This document discusses lattices and Boolean algebra. It defines lattices as algebraic systems that satisfy certain axioms involving binary operations of join (∨) and meet (⋅). Boolean algebras are lattices that also have a unary complement operation and satisfy additional axioms such as distributivity and the existence of complements. Examples of Boolean algebras include the power set of a set under set operations and vectors of 0s and 1s under component-wise operations. Boolean functions and logical expressions are then introduced and their evaluation is discussed.

1. Top school in India
By:
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2. Lattice
and
Boolean Algebra
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3. 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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4. 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 ⋅ a
3. 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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5. 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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6. Lattice and Boolean Algebra Slide 6
Distributive Lattice
• The lattice 〈A, ∨, ⋅〉 satisfying the following
axiom is a distributive lattice
5. Distributive laws: a ∨ (b ⋅ c) = (a ∨ b) ⋅ (a ∨ c),
a ⋅ (b ∨ c) = (a ⋅ b) ∨ (a ⋅ c)
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7. Lattice and Boolean Algebra Slide 7
Examples
distributive
non-distributive
a⋅(b∨c)≠a⋅b∨a⋅c
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8. 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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9. 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 ⋅ a
3. 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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10. 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
a∨b=a⋅b
a⋅b=a∨b
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11. 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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12. 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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13. 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 Vectors
Bn
= {(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 Set
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14. Lattice and Boolean Algebra Slide 14
Examples
B3 P({a,b,c}) {n ∈ Ν| n|30}
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15. 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
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16. Lattice and Boolean Algebra Slide 16
De Morgan’s Theorem
• De Morgan’s Laws hold
• These equations can be generalized
a∨b=a⋅b
a⋅b=a∨b
x1∨x2∨K∨xn=x1⋅x2⋅K⋅xn
x1⋅x2⋅K⋅xn=x1∨x2∨K∨xn
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17. 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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18. 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 follows
1. Add parenthesis according to the order of
operations
2. Interchange ∨ with ⋅
3. Interchange xi with xi
4. Interchange 0 with 1
Example
x∨(y⋅z)=x⋅(y∨z)
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19. 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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20. 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
= 1
2. 1D
= 0
3. 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 AD
=(BD
)
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21. 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 ⋅ 1
f(a) = a ⋅ a ∨ a ⋅ 1 = a
x f(x)
0 a
1 1
a a
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22. 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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23. 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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24. 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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25. 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 22
n
equivalence classes by the equivalence
relation (≡)
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