(Discrete Math) Prove or disprove the equality y(x+~z) + ~yx + (x + ~x)~y~z= x + ~z.?

Proof: x + ~x = 1, so the third term on the left is just ~y~z. Then the combination of the last two terms on the left side is ~yx + ~y~z = ~y(x+~z). Finally the combination of all three terms on the left is y(x+~z) + ~y(x+~z) = (x+~z)(y+~y) = (x+~z)(1)=(x+~z) which is equivalent to the right side.