# If $A\ \Delta \ B=(A\ ∪\ B)-(B\ ∩\ A)$ prove $A\ \Delta \ B = (A-B)\ \cup \ (B-A)$

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}

• 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 $\Leftrightarrow$. – GEdgar Sep 18 '17 at 20:45
• Now it is correct? – B. David Sep 18 '17 at 21:54

• Yes, @Daniel. ${}$ – Shaun Sep 18 '17 at 21:56
• 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 ? – B. David Sep 18 '17 at 22:06