# 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.

You should use the commutativity of the $$\textit{or}$$.

$$x \in A \cup B \iff x \in A \lor x \in B \iff x \in B \lor x \in A \iff x \in B \cup A.$$

• Thank you for your advice i highly appreciate it! Apr 18 '20 at 8:49
• You're welcome ;) Apr 18 '20 at 8:50
• 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 .

• Your proofs are very pleasant to my eye! Great work! Apr 18 '20 at 11:08
• @Jstinz. Thanks:)
– user655689
Apr 18 '20 at 11:09