# Prove by double inclusion the set identity $A\cap(A\cup B) = A$ [duplicate]

Prove by double inclusion the set identity $A\cap(A\cup B) = A$

## marked as duplicate by Shailesh, Martin Sleziak, JonMark Perry, Joey Zou, Daniel W. FarlowAug 6 '16 at 14:43

$A\cap (A\cup B)\subseteq A$:
Let $x\in A\cap (A\cup B)$. Then $x\in A$ (and $x\in A \cup B$), by definition of intersection.
$A\subseteq A\cap (A\cup B)$:
Let $x\in A$. Then $x\in A\cup B$ (by definition of union). Thus $x\in A\cap (A\cup B)$, by the definition of intersection.
$A \cap(A \cup B)=(A \cup \varnothing)\cap (A \cup B)=A \cup(B \cap \varnothing)=A \cup\varnothing=A$