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Is correct this proof ?

$A\ \Delta \ B=(A\ ∪\ B)-(B\ ∩\ A)$

\begin{split} x \in (A \ \cup B)- (B\ ∩\ A) & \Rightarrow x\in A \ \cup B\ \wedge x \notin A\ \cap \ B\ \\ & \Rightarrow (x \in A\ \vee x\in B) \wedge\ x\notin A\ \cap \ B\ \\ & \Rightarrow (x \in A \ \wedge x\notin A\cap B ) \ \vee \ (x \in B \ \wedge \ x\notin A\ \cap B\ )\\ &\Rightarrow (x \in A \ \wedge x\notin B ) \ \vee \ (x \in B \ \wedge \ x\notin A) \\&\Rightarrow\ x\in \boldsymbol{(A-B)\ \cup \ (B-A)} \end{split}


Second part

$ x \in (A-B) \cup \ (B-A) \Rightarrow x\in (A - B)\ \vee x \in (B- A)$

\begin{split} x\in (A - B) & \Rightarrow x \in A\ \wedge x\notin B\\ & \Rightarrow x \in A \cup B \ \wedge x\notin B \\ &\Rightarrow x \in A \cup B \ \wedge x\notin A \cap B \ \\&\Rightarrow\ x\in \boldsymbol{(A\cup B)\ - \ (A\cap B)} \end{split}

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    $\begingroup$ No, it is not correct. You have only proved half. Next you need to do the other direction. Or check that all those $\Rightarrow$ can be made into $\Leftrightarrow$. $\endgroup$ – GEdgar Sep 18 '17 at 20:45
  • $\begingroup$ Now it is correct? $\endgroup$ – B. David Sep 18 '17 at 21:54
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Yes, it's correct, but (was) incomplete; you need(ed) to prove the reverse inclusion. Just replace all your implications (in your first half) with biconditionals.

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  • $\begingroup$ OP's work is correct, but the proof is incomplete, namely the reverse inclusion has not yet been shown. $\endgroup$ – Kevin Long Sep 18 '17 at 21:04
  • $\begingroup$ Well spotted, @KevinLong. I've edited it now. $\endgroup$ – Shaun Sep 18 '17 at 21:09
  • $\begingroup$ Now it is correct? $\endgroup$ – B. David Sep 18 '17 at 21:55
  • $\begingroup$ Yes, @Daniel. ${}$ $\endgroup$ – Shaun Sep 18 '17 at 21:56
  • $\begingroup$ Thank you @Shaun and KevingLong :D Can you give me a hint to prove if $A \Delta B=\emptyset$ then $A \subset B$ or $B\subset A$ please ? $\endgroup$ – B. David Sep 18 '17 at 22:06

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