**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:

(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.

**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.