Prove for sets $A$ and $B$ that $A\cup{B}=B\cup{A}$. Prove for sets $A$ and $B$ that $A\cup{B}=B\cup{A}$. 
Here is my attempt on proving this, by the definition of subset theorem we have $(A\cup{B})\subseteq(B\cup{A})$ and $(B\cup{A})\subseteq(A\cup{B})$ then $x\in({B\cup{A}})$, $x\in({B\cup{A}})$ therefore $A\subseteq{B\cup{A}}\land{B\subseteq{A\cup{B}}}$ this implies $x\in{A}\lor{x\in{B}}\iff{x\in{B}}\lor{x\in{A}}$. I know there is no much work can be done here, I just want to improve my proof's written.
 A: You should use the commutativity of the $\textit{or}$.
$$x \in A \cup B \iff x \in A \lor x \in B \iff x \in B \lor x \in A \iff x \in B \cup A.$$
A: *

*If you want to prove it elementwise, you'll have to apply the fact that the OR logical operator is commutative, since the union operation is defined using he OR-operator. With this method, the proof is immediate. 

*The result  can also be proved at the set level, without resorting to logical operators. 

*We have to prove a set equality, which amounts to a reciprocal inclusion. That is, we have to prove : $A\cup B \subseteq B\cup A$ and $B\cup A \subseteq A\cup B$

*Admitting as a definition that $\color{blue} {S\subseteq T \iff S\cap \overline T = \emptyset}$ , our goal becomes : 
$$(1) (A\cup B)\cap \overline{(B\cup A)} = \emptyset$$ 
and 
$$(2) (B\cup A)\cap \overline{(A\cup B)} = \emptyset$$


*

*This can be shown using DeMorgan's law, Distributive Law, Commutative and Associative law for $\cap$,  and Identity Law for sets. 

*Let me do it for (1) 
$(A\cup B)\cap \overline{(B\cup A)}$
$ = (A\cup B)\cap (\overline B \cap \overline A)$
$ = [ (A\cup B)\cap \overline B] \cap [ (A\cup B)\cap \overline A]$
$ =  [(A\cap \overline B) \cup (B \cap \overline B)] \cap [ (A\cap\overline A) \cup (B \cap \overline A)]$
$ =  [(A\cap \overline B) \cup \emptyset ] \cap [ \emptyset \cup (B \cap \overline A)]$
$ =  [A\cap \overline B] \cap [ B \cap \overline A]$
$ = [A\cap \overline A] \cap [B\cap \overline B]$
$ = \emptyset \cap \emptyset$
$ = \emptyset$. 
Which proves that : $A\cup B \subseteq B\cup A$. 
The reverse inclusion is still to be proved, in order to reach the goal completely . 
