6. Use the Boolean identities to prove the following:
a) The absorption laws
b) DeMorgan’s laws
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.