Simplifying the boolean expression xy + (!x)z + yz with boolean algebra? So using K-maps I was able to simplify xy + (!x)z + yz to xy + (!x)z, and I double-checked that the truth tables are the same.
I'm having trouble understanding how I would have used boolean algebra to get this result. I tried my usual tricks of adding zero, letting addition distribute over multiplication, etc., without avail.
Any thoughts appreciated.
Thanks.
$$
xy + \bar{x} z + yz =
xy + \bar{x} z + (x+\bar{x})yz =
xy + \bar{x} z + xyz + \bar{x}yz =
xy(1 + z) + \bar{x} z(1 + y) = xy + \bar{x}z.
$$