# How does Absorption work in boolean algebra?

So I understand the basic outlines of the property:

$a(a+b)=a$

From that its pretty clear to me that no matter what $b$ is the result will be $a$ regardless.

However I don't understand how that translates to:

$\bar x + \bar y + z + \bar xy\bar z = \bar x + \bar y + z$

I'm assuming you are working in a boolean algebra and are writing $xy$ or $x.y$ for the conjunction of $x$ and $y$ and $x + y$ for their disjunction and $\overline{x}$ for the negation of $x$.. (See https://en.wikipedia.org/wiki/Boolean_algebra for the notation that is more commonly used these days.)
Using distributivity, idempotency and absorption you have $$\overline{x} + \overline{x}.y.\overline{z} = \overline{x}(\overline{x} + y.\overline{z}) = \overline{x}$$
Hence using commutativity and associativity you have $$\overline{x} + \overline{y} + z + \overline{x}.y.\overline{z} = \overline{y} + z + (\overline{x} + \overline{x}.y.\overline{z}) = \overline{y} + z +\overline{x} = \overline{x} + \overline{y} + z$$