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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

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    It's not true that $x+yz = (x+y)(x+z)$. Let $x=1, y=1, z=1$. – 2017-02-16
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    It is a logical equivalence. so $1+1=1$. Plus sign means "or" and Dot sign means "and". – 2017-02-16
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    @PatrickStevens By the `logic` tag I assume that's meant to be a [boolean algebra](https://en.wikipedia.org/wiki/Boolean_algebra) question. Granted, the OP could/should have spelled that out more clearly. – 2017-02-16

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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.

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    Thanks a lot. The key of the solution was absorption law. – 2017-02-16