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
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)