Kindly help in Simplifying Y = BCD + BC'D.
I have been trying to simplify the expression for sometime now, using the the 10 rules but cannot simplify fully.
Kindly help in Simplifying Y = BCD + BC'D.
I have been trying to simplify the expression for sometime now, using the the 10 rules but cannot simplify fully.
Using the distributive property \begin{align} Y = BCD + BC'D = BD(C + C') \end{align} Since $C + C'$ is always true \begin{align} Y = BD(C + C') = BD(1) = BD \end{align} Therefore $Y = BD$.