Proof of De-Morgan’s laws in boolean algebra

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+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.

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.