# Math proof of absorption law

I have little experience with math proofs and I would like to prove $A \cup (A\cap B)=A$, by showing that the left hand side is a subset of right hand side and vice versa.

Since $A \subseteq A \space and \space A\cap B \subseteq A$

$\therefore \space lhs \subseteq rhs$

Given that $A \cup (A\cap B) \equiv A \cap (A \cup B)$

$A \subseteq A \space and \space A \subseteq (A\cup B)$

$\therefore \space rhs \subseteq lhs$

Does the above proposed solution suffice as a proof?

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Yes, provided that you’ve already proved that $A\cup(A\cap B)=A\cap(A\cup B)$. –  Brian M. Scott Nov 17 '12 at 21:18
It suffices. I would somewhat prefer chasing elements. Kind of depends on how "algebraic" you want to be. –  André Nicolas Nov 17 '12 at 21:18

It also depends on what have you proved before (e.g. $X\subseteq Y$ and $Z\subseteq Y$ implies that $X\cup Z\subseteq Y$).
Note that the for $\text{RHS}\subseteq\text{LHS}$ you only need to note that $A\subseteq A$ and therefore $A\subseteq A\cup X$ for every $X$, in particular $X=A\cap B$.
You may also wish to write an element chasing proof, e.g. to show $A\subseteq A\cup(A\cap B)$:
Let $a\in A$ be an arbitrary element, we will show that $a\in A\cup(A\cap B)$. Since $a\in A$ we have that $a\in A$ or $a\in A\cap B$. Therefore $a\in A\cup(A\cap B)$, and therefore $A\subseteq A\cup(A\cap B)$.
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