Boolean Algebra : Using identities, prove x(x + y) = x Example: Prove the ABSORPTION LAW:
$$
x(x + y) = x
$$

$ 
Solution: \\
x(x + y) \\
= (x + 0)(x + y) \;\;\;\;\; Identity \;Law \\ 
= x + (0 · y) \;\;\;\;\;\;\;\;\;\;\; Distributive \;Law \\
= x + y · 0 \;\;\;\;\;\;\;\;\;\;\;\;\;\; Commutative \;Law \\
= x + 0 \; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;Domination \;Law \\
= x \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; Identity \;Law \\
$

I was given an example here and shown one method of solving it. I was told there was an alternative solution to it and I was curious about it.

This is one method of proving the problem.

The alternative method suggested solving the same expression in the order of:

DISTRIBUTIVE LAW, IDEMPOTENT LAW, DISTRIBUTIVE LAW, DOMINANCE LAW, and IDENTITY LAW. 

When I try, I get stuck here:

$ 
Solution: \\
x(x + y) \\
= (x · x) + (x · y) \;\;\;\;\; Distributive \;Law \\ 
= x + (x · y) \;\;\;\;\;\;\;\;\;\;\;\;\; Idempotent \;Law \\
= (x + x)(x + y) \;\;\;\;\;\; Distributive \;Law \\
$

I've been stuck on where I can prove it using the dominance law, unless I can extend this with an identity law proof. My first time in a while, please excuse.
Solution:
$$ \begin{align}x(x+y) 
&amp;= (x\cdot x)+(x\cdot y) \quad\text{[Distributive Law]} \\
&amp;= x + (x\cdot y) \quad\text{[Idempotent Law]} \\
&amp;= x \cdot (1+y) \quad\text{[Distributive Law]}\\
&amp;= x \cdot 1 \quad\text{[Property of 1]} \\
&amp;= x \quad\text{[Property of 1]}
\end{align}$$