[Math] Boolean algebra, ABC’+AB’C’+A’BC’ = (A+BC’+C)C’, can anyone show it please

boolean-algebra

I was asked to prove the next formula –
(A+BC'+C)C' = ABC'+AB'C'+A'BC'

i need to show all the stages of the simplification, i have all of the rules/identities.
i have tried many times and i did not make it, i always get something else
thank you

Best Answer

opening the brackets is done as with "regular" algebra, which yields: 

AC' + BC'C' + CC'

from your identites: 

C'C' = C'  
CC' = 0  
X + 0 = X

which imply that the above expression can be simplified to: 

AC' + BC'

from here you should look to expand the expression by the following identities: 

X*1 = X  
X + X' = 1

which yields: 

AC'(B + B') + BC'(A + A')

expand the brackets as usual, change the order of multiplication with the identity XY = YX, cancel common terms based on X + X = X, and youre done...

someone should add a h/w tag to this question. i dont have any points...

Related Question