Simplify: B + (BC)’
1.0
2.1
3.C + B’
4.B + C’
Simplify: B + (BC)’
1.0
2.1
3.C + B’
4.B + C’
Following is the solution.
B + (BC)'
B + B' + C'
(B + B') + C'
1 + C'
{∵ X + X’ = 1}
1
{∵ 1 + X = 1}
use distrubutive law like a+bc = (a+b)(A+C)