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**barb** - On 16-09-2012
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Lattice and Boolean Algebra. Algebra. An algebraic system is defined by the tuple A,o 1 , …, o k ; R 1 , …, R m ; c 1 , … c k , where, A is a non-empty set, o i is a function A p i A, p i is a positive integer, R j is a relation on A, and c i is an element of A. Lattice.

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Lattice and Boolean Algebra

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Lattice andBoolean 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

- The lattice is an algebraic system A, , , given a,b,c in A, the following axioms are satisfied:
- Idempotent laws: a a = a, a a = a;
- Commutative laws: a b = b a, a b = b a
- Associative laws: a (b c) = (a b) c, a (b c) = (a b) c
- Absorption laws: a (a b) = a, a (a b) = a

Lattice and Boolean Algebra

- 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

- 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)

Lattice and Boolean Algebra

distributive

non-distributive

Lattice and Boolean Algebra

- 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

- 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:
- Idempotent laws: a a = a, a a = a;
- Commutative laws: a b = b a, a b = b a
- Associative laws: a (b c) = (a b) c, a (b c) = (a b) c
- Absorption laws: a (a b) = a, a (a b) = a
- Distributive laws: a (b c) = (a b) (a c), a (b c) = (a b) (a c)

Lattice and Boolean Algebra

- Involution:
- Complements: a a = 1, a a = 0;
- Identities: a 0 = a, a 1 = a;
- a 1 = 1, a 0 = 0;
- De Morgan’s laws:

Lattice and Boolean Algebra

- To verify whether a given algebra is a Boolean algebra we only need to check 4 postulates:
- Identities
- Commutative laws
- Distributive laws
- Complements

Lattice and Boolean Algebra

- 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

- 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

Lattice and Boolean Algebra

P({a,b,c})

{n | n|30}

B3

Lattice and 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

- De Morgan’s Laws hold
- These equations can be generalized

Lattice and Boolean Algebra

- 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

- 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

Lattice and Boolean Algebra

- 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

- Let A be a Boolean expression. The dual AD is defined recursively as follows:
- 0D = 1
- 1D = 0
- if xi is a variable, then xiD = xi
- if A, B, and C are Boolean expressions, and A = B C, then AD= BD CD
- if A, B, and C are Boolean expressions, and A = B C, then AD= BD CD
- if A and B are Boolean expressions, and A = B, then

Lattice and Boolean Algebra

- Given xy yz = xy yz xz
the dual (x y)(y z) = (x y)(y z)(x z)

- 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

Lattice and Boolean Algebra

- 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)

Example

Lattice and Boolean Algebra

- Constants 0 and 1 are logical expressions
- Variables x1,x2,…,xn are logical expressions
- If E is a logical expression, then E is one
- If E1 and E2 are logical expressions, then (E1 E2) and (E1 E2) are also logical expressions
- The logical expressions are obtained by finite application of 1 - 4

Lattice and Boolean Algebra

- An assignment mapping :{xi} {0,1} (i = 1, … , n)
- The valuation mapping |F| of a logical expression is obtained:
- |0| = 0 and |1| = 1
- If xi is a variable, then | xi | = (xi)
- If F is a logical expression, then |F| = 1 |F| = 0
- If F and G are logical expressions, then |F G| = 1 (|F| = 1 or |G| = 1)
- 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

- 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 ()

Lattice and Boolean Algebra