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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.

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

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

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    $\begingroup$ Thank you for your advice i highly appreciate it! $\endgroup$
    – EquDox
    Apr 18 '20 at 8:49
  • $\begingroup$ You're welcome ;) $\endgroup$
    – Riccardo
    Apr 18 '20 at 8:50
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  • 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 .

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

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