Prove $(x+yz)(y'+x)(y'+z')=x(y'+z')$ in Boolean algebra

Question

How can we prove $(x+yz)(y'+x)(y'+z')=x(y'+z')$ in a Boolean algebra $B$?

Answer 1

The first two factors simplified to $$(x+yz)(y'+x)$$

$$=(x+yz)(x+y')$$ $$=x+\{(yz).y'\}$$ $$=x+0$$ $$=x$$

Answer 2

One way is proof by exhaustion:

\begin{matrix} x & y & z & (x+yz)(y'+x)(y'+z') & x(y'+z')\\ \hline False & False & False & False & False \\ False & False & True & False & False \\ False & True & False & False & False \\ False & True & True & False & False \\ True & False & False & True & True \\ True & False & True & True & True \\ True & True & False & True & True \\ True & True & True & False & False \end{matrix} Since the values match for any assignment of $(x,y,z)$, they are equal.