# 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) &= (x\cdot x)+(x\cdot y) \quad\text{[Distributive Law]} \\ &= x + (x\cdot y) \quad\text{[Idempotent Law]} \\ &= x \cdot (1+y) \quad\text{[Distributive Law]}\\ &= x \cdot 1 \quad\text{[Property of 1]} \\ &= x \quad\text{[Property of 1]} \end{align}