Boolean algebra Simplification of "xy + x'z + yz" I'd like to simplify the following expression "xy + x'z + yz":

         xy + x'z + yz = xy + z(x' +y)
                       = (xy + z)(xy + x' + y)
                       = (xy + z)(y(x + 1) + x') 
                       = (xy + z) ( y + x')

What do I do after the last step?

By either drawing a Karnaugh map or recognizing that this is just the Consensus Theorem, observe that:
\begin{align*}
xy + x'z + yz
&= xy + x'z + (1)yz \\
&= xy + x'z + (x + x')yz \\
&= xy + x'z + (xyz + x'yz) \\
&= (xy + xyz) + (x'z + x'yz) \\
&= xy(1 + z) + x'z(1 + y) \\
&= xy(1) + x'z(1) \\
&= xy + x'z
\end{align*}
