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Proof of De-Morgan’s laws in boolean algebra

  • Difficulty Level : Easy
  • Last Updated : 14 May, 2020

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.

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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+0.x'=0=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')=0
LHS: (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.




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