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
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
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...
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).