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Lattice and
Boolean Algebra
Lattice and Boolean Algebra Slide 2
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.
Lattice and Boolean Algebra Slide 3
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
Lattice and Boolean Algebra Slide 4
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.
Lattice and Boolean Algebra Slide 5
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)
Lattice and Boolean Algebra Slide 6
Examplesdistributive non-distributive
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a ⋅(b∨c) ≠ a ⋅b∨a ⋅c
Lattice and Boolean Algebra Slide 7
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
Lattice and Boolean Algebra Slide 8
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)
Lattice and Boolean Algebra Slide 9
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
Lattice and Boolean Algebra Slide 10
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
Lattice and Boolean Algebra Slide 11
Example
• prove the idempotent laws given Huntington’s postulates:a = a 0 = a aa
= (a a) (a a)
= (a a) 1
= a a
Lattice and Boolean Algebra Slide 12
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 Set
Lattice and Boolean Algebra Slide 13
Examples
B3 P({a,b,c}) {n | n|30}
Lattice and Boolean Algebra Slide 14
Isomorphic Boolean Algebra
• Two Boolean algebras A, , , , 0A,1A and B, , , , 0B,1B are isomorphic iff there is a mapping f:AB, such that
• for arbitrary a,b A, f(ab) = f(a)f(b), f(a b) = f(a) f(b), and f(a) = f(a)
• 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.
Lattice and Boolean Algebra Slide 15
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
€
x1∨x2∨K ∨xn = x1 ⋅x2 ⋅K ⋅xn
x1 ⋅x2 ⋅K ⋅xn = x1∨x2∨K ∨xn
Lattice and Boolean Algebra Slide 16
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.
Lattice and Boolean Algebra Slide 17
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• Add parenthesis according to the order of
operations• Interchange with
• Interchange xi with xi
• Interchange 0 with 1
Example
€
x∨(y ⋅z) = x ⋅(y∨z )
Lattice and Boolean Algebra Slide 18
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.
Lattice and Boolean Algebra Slide 19
Dual Boolean Expressions• Let A be a Boolean expression. The dual AD
is defined recursively as follows:1. 0D = 12. 1D = 0
3. if xi is a variable, then xiD = 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 )
Lattice and Boolean Algebra Slide 20
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
a 1
Lattice and Boolean Algebra Slide 21
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 fg fg 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
Lattice and Boolean Algebra Slide 22
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
Lattice and Boolean Algebra Slide 23
Evaluation of logical Expressions
• An assignment mapping :{xi} {0,1} (i = 1, … , n)
• The valuation mapping |F| of a logical expression is obtained:
1. |0| = 0 and |1| = 1
2. If xi is a variable, then | xi | = (xi)
3. If F is a logical expression, then |F| = 1 |F| = 0
4. If F and G are logical expressions, then |F G| = 1 (|F| = 1 or |G| = 1)
5. If F and G are logical expressions, then |F G| = 1 (|F| = 1 and |G| = 1)
Example: F:x y z
(x) = 0, (y) = 0, (z) = 1
Lattice and Boolean Algebra Slide 24
Equivalence of Logic Expressions
• Let F and G be logical expressions. If |F| = |G| hold for every assignment , 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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