Need Confirmation on one of the Boolean Laws

Thread Starter

KinkyBlok

Joined Oct 29, 2010
1
I just need to know if the boolean law
A+BA' = A+B

can be reversed to

A' + B'A = A'+B'

and if you want to help extra. i derived this equation from a truth table and was wondering if you could tell me if i have simplified it wrong in anyway:

L1=A'B'C+A'BC'+A'BC+AB'C'+AB'C+ABC'

L1= A'(B'C+BC'+BC) + A(B'C'+B'C+BC')

L1=A'(B'C+B(C'+C)) + A(B'(C'+C)+BC')

L1=A'(B'C+B(1)) + A(B'(1)+BC')

L1=A'(B+C) + A(B'+BC')

Any help would definitely be appreciated. Thank You
 

Georacer

Joined Nov 25, 2009
5,182
Of course you can use the formula in the form you described.

You did a fine job in reforming the equation in the second part, but you messed up in the last step.

L1=A'(B'C+B)+A(B'+BC')
L1=A'(B+C)+A(B'+C')

Which, incidentally isn't the same as, A XOR (B + C)
 

zgozvrm

Joined Oct 24, 2009
115
I just need to know if the boolean law
A+BA' = A+B

can be reversed to

A' + B'A = A'+B'
To answer your 1st question: "Yes"

In Boolean algebra, addition is distributive over multiplication and multiplication is distributive over addition.

That is, A(B + C) = AB + AC and A + BC = (A + B)(A + C)

So, A' + B'A = (A' + B')(A' + A) = (A' + B')(1) = A' + B'
 
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