# Boolean algebra, ABC'+AB'C'+A'BC' = (A+BC'+C)C', can anyone show it please

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

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Show that both sides are (A+B)C', using the fact that, for every D and E, (D+D')E=E (once in the RHS) and that D+D'E=D+E (once in the LHS and once in the RHS).

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

AC' + BC'C' + CC'

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...

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