Search code examples
boolean-logicboolean-expressionsimplifyboolean-operations

Simplifying boolean expressions


I am having trouble simplifying these two boolean expressions and was hoping for a solution. I have put my work so far under each one.

 1)~B(A + A * B)
 My work:
 ~BA + ~BAB
 ~BA + 0
 ~BA
 I wasn't sure if you could distribute ~B into A*B to equal ~BAB like that

 2)~B + ~(A*B) + ~C
 Totally lost on this one. 
 My thinking:
 ~B + (~A + ~B) + ~C (demorgans)
 1 + ~A + ~C
 1 + ~C
 1

Are these correct or am I on the right path? Thanks


Solution

    1. Yes, you distributed the ~B and simplified properly.
    2. No. Good use of DeMorgan's Theorem, but you lost ~B after that. ~B + ~B simplifies not to 1 but to ~B.

      ~B + (~A + ~B) + ~C

      ~B + ~A + ~C

      ~A + ~B + ~C

      Or, if you want to apply DeMorgan's Theorem again:

      ~(ABC)