Logical proof of an equivalence

Question

I want to prove that: $$x + y.z = (x + y)(x + z)$$ in logic. The right statement equals: $$(x.x) + (x.y) +(x.z) + (y.z)$$ So if I assume that $x=1$, then the whole statement equals 1, no matter the rest of the statement: $$1 + y + z + y.z = 1 + A = 1$$
And if I assume that $x=0$ the whole statement equals $y.z$: $$0 + 0 + 0 + y.z = y.z$$ So it is equivalent to $$x + y.z$$ I have proved the equivalence in this way. But I want to know is there any better way with strong mathematical or logical proof for it. Best Regards

Answer 1

The right statement equals: $$(x.x) + (x.y) +(x.z) + (y.z)$$

Note that $(x.x)=x$ by the idempotence law and $x+(x.y)=x$ by the absorption one.