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Boolean algebra

Definition: Let B be a non empty set with two binary operation + and *, a unary operation (complement), and two distinct element 0 and 1. Then B is called a Boolean algebra if the following axioms hold where a,b,c are any elements in B:

[B1]   Commutative laws:

(1a)      a+b=b+a                                             (1b) a*b=b*a

[B2] Distributive laws

(2a)        a+(b*c)=(a+b)*(a+c)                           (2b) a*(b+c)=(a*b)+(a*c)

[B3] Identity laws

(3a)        a+0=a                                                    (3b) a*1=a

[B4] Complement laws

(4a)         a+a/=1                                                     (4b) a*a/=1

 

 

Example: Let B={0,1},the site of bits (binary digit), with the binary operations of + and * and the unary operation / defined by fig- 1. Then B is Boolean algebra.

Screenshot_37

Sub algebras: Suppose C is a non empty subset of Boolean algebra B. We say C is a sub algebra of B if itself is a Boolean algebra .

Isomorphic Boolean algebra: Two Boolean algebras B and B/ are said to be isomorphic if there is a one to one correspondence f:BB/ which preserves the three operations, i.e., such that

f (a+b)= f(a)+ f(b)             f (a*b)= f(a)* f(b)               f (a/)= f (a)/

for any elements a,b in B.

Duality

The dual of any statement in a Boolean algebra B is the statement obtain by interchanging the operations + and *, and interchanging their identity elements 0 and 1in the original statement. The dual of (1+a)*(b+0)=b is (0*a)+(b*1)=b.

DeMorgan’s laws: (a+b)/=a/*b/

Proof: We need to show that (a+b)/+a/*b/=1 and (a+b)*(a/*b/)=0 ; then by uniqueness of complement, (a+b)/=a/*b/.

We have (a+b)+a/*b/=b+a+a/*b/

= b+(a+a/)*(a+b/)

= b+1*(a+b/)

=b+a+b/

= b+ b/+a

= 1+a

= 1

Also,

(a+b)*(a/*b/)= ((a+b)*a/)*b/)

= ((a*a/)+(b*a/))*b/

=(0+b*a/))*b/

= (b*a/)*b/

= (b+b/)*a/

= 0*a/

= 0

Prime Implicants

A fundamental product P is called a prime implicant of a Boolean expression E if                                                                  P+E=E

But no other fundamental product contained in P has this property.

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