booleanboolean-logicboolean-expressionsimplifydemorgans-law

simplifying a boolean expression using DeMorgans


This is the expression I need to simplify:

[{(AB)'*(BC)'} + (A'C)']'

Right now I have my answer as (AB + BC).(A' + C)

I don't think my answer is right but I'm not sure. IF someone could show steps on how to simplify that would be really appreciated, thanks!


Solution

  • ((AB)'(BC)' + (A'C)')' = ((AB)'(BC)')'(A'C)''           ; DeMorgan
                           = ((AB)'' + (BC)'')(A'C)         ; DeMorgan + double negation
                           = (AB + BC)(A'C)                 ; double negation
                           = ABA'C + BCA'C                  ; distribution
                           = BCA'                           ; AA'=0