Question

can anyone prove demorgan's laws of boolean algebra?? I need only proof. Please don't give shabby answers. Reply in the event if you know well.

(a+b)' = a'+b'
(a.b)' = a' + b'


Please give the correct answer soon. 'Cause I'm having my exam tomorrow. Please...

Answer 1

Dear mate,
Kindly mark the answer as brainliest if you find it useful.

Please find the answer in the image. I have done the proof for one and the proof for other can be done in an exactly similar fashion.

Hope this clears your doubt.✌

Answer 2

Answer:

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.

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