a)The absorption Law:

Give algebraic proof of absorption law of boolean algebra.

(i)

X+XY=X
It can be proved algebraically as

L.H.S. = X+XY
           = X(1 + Y)

Putting 1+ Y = 1   (ref. properties of 0, 1 Theorem 1)
X.1=X = R.H.S. (ref. properties of 0, 1)

Hence proved.

(ii)

X(X + Y) = X

The algebraic proof of this law.

L.H.S. = X(X+Y)= X.X + XY
= X.X + XY
= X + XY  (X.X=X: Indempotence Law)
= X(1 + Y)
= X.1  (using 1+ Y =1, properties of 0, 1)
= X (X.1=X, using property of 0, 1)
= R.H.S

Hence proved.

b)DeMorgan's Law:

Statements :
1. (x+y)'=x'.y'
2..(x.y)'=x'+y'

Proof:
Here we can see that we need to prove that the two propositions are complement to each other.
We know that A+A'=1 and A.A'=0 which are annihilation laws. Thus if we prove these conditions for the above statements of the laws then we shall prove that they are complement of each other.

For statement 1:
We need to prove that:

(x+y)+x'.y'=1  and  (x'.y').(x+y)=0

Case 1.

(x+y)+x'.y'=1

LHS

(x+y)+x'.y'=(x+y).(x+x')+x'.y'

= x.x+x.y+y.x'+x'.y' = x+x.y+y.x'+x'.y'   {Using Distributive Property)

= x+x.y+x'(y+y')

=x+x.y+x'=x+x'+x.y

=1+x.y=1=RHS

Hence proved.

Case 2.

(x'.y').(x+y)=0

LHS:

(x'.y').(x+y)=x.(x'y')+y.(x'.y')

=x.0+o.x'=RHS

Hence proved.

For statement 2:

We need to prove that:

x.y(x'+y')=1  and  x.y.(x'+y')=0

Case 1.

x.y+(x'+y')=1
LHS:

x.y+(x'+y+)=(x+x'+y').(y+x'+y')            {We know that A+BC=(A+B).(A+C)}

=(1+y').(1+x')=1=RHS
 

Hence proved.

Case 2.

x.y.(x'+y')=x.x'.y+x.y.y'

=0=RHS

Hence Proved.

This proves the De-Morgan’s theorems using identities of Boolean Algebra.