A intersection (A union B)

Solution 1:

The statements $A \cap (A \cup B) = A$ and $A \cup (A \cap B) = A$ are known as absorption laws.

When the proof is not clear using the equivalence "if and only if", one can try proving the direct first and later the converse, or viceversa.

Proof. $(\Longrightarrow)$ Suppose $x \in A \cap (A \cup B)$. Then we have simultaneously $x \in A$ and $x \in A \cup B$, by Definition 2. Thus $x \in A$ as desired.

$(\Longleftarrow)$ Suppose $x \in A$. By Definition 1, we have $x \in A \cup B$. Thus, we have $x \in A$ (by hypothesis) "and" $x \in A \cup B$. Hence, by Definition 2, $x \in A \cap (A \cup B)$ as desired.

Definition 1 (Union). $x \in A \cup B \;\iff\; (x \in A \;\;\text{or}\;\; x \in B)$.

Definition 2 (Intersection). $x \in A \cap B \;\iff\; (x \in A \;\;\text{and}\;\; x \in B)$.

If you want to use the if and only if clause, you only need "to assembly" the two parts of the proof.

Remember:

• if $P$ is true, then $P \;\;\text{or}\;\; anything$ is true.
• if $P \;\;\text{and}\;\; Q$ is true, then $P$ is true.
• if $P \;\;\text{and}\;\; Q$ is true, then $Q$ is true.

Solution 2:

From $x \in A$ and ($x \in A$ or $x \in B$), you can conclude that $x \in A$, or the first clause is false and the whole statement is false. You are already there.

Solution 3:

$A\subseteq A\cup B\,$ so $A\subseteq A\cap(A\cup B)\;$, and $\;A\cap(A\cup B)\subseteq A\;$ so $\;A=A\cap(A\cup B)$

Solution 4:

Use the fact that $A$ is always a subset of $A\cup B$. And a subset's intersection with the set containing it always results in the subset only.