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# Lattice and Boolean Algebra - PowerPoint PPT Presentation

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 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  aa

= (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:AB, such that

• for arbitrary a,b  A, f(ab) = 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