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This question already has an answer here:

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

please help with this am stuck with these identities on the set A and B

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

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$A\cap (A\cup B)\subseteq A$:

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.

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$A \cap(A \cup B)=(A \cup \varnothing)\cap (A \cup B)=A \cup(B \cap \varnothing)=A \cup\varnothing=A$

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