Prove that $(A\setminus B)\cup (B \setminus A)\subseteq (A\cup B)\setminus(A\cap B)$

I am asked to prove that

$$(A\setminus B)\cup (B\setminus A)\subseteq (A\cup B)\setminus(A\cap B)$$ where $$A$$ and $$B$$ are sets. Could someone please check my solution and see if it is correct?

I suppose that sets $$A$$ and $$B$$ are subsets of a global set $$X$$. Fix $$x\in (A\setminus B)\cup (B\setminus A)$$. Then

\begin{align} x\in (A\setminus B)\cup (B\setminus A)&\Rightarrow x\in A\setminus B\text{ or }x\in B\setminus A\\ &\Rightarrow (x\in A\text{ and }x\not\in B)\text{ or }(x\in B \text{ and } x\not\in A)\\ &\Rightarrow [(x\in A\text{ and }x\not\in B)\text{ or }x\in B]\text{ and } [(x\in A\text{ and }x\not\in B)\text{ or }x\not\in A]\\ &\Rightarrow [(x\in A\text{ or }x\in B) \text{ and } (x\in B\text{ or }x\not\in B)]\text{ and }\\ &\qquad [(x\in A\text{ or }x\not\in A)\text{ and } (x\not\in A\text{ or }x\not\in B)]\\ &\Rightarrow [(x\in A\cup B \text{ and } (x\in B\text{ or }x\in X\setminus B)]\text{ and }\\ &\qquad [(x\in A\text{ or }x\in X\setminus A)\text{ and } \neg(x\in A\text{ and }x\in B)]\\ &\Rightarrow [x\in A\cup B \text{ and } x\in B\cup(X\setminus B)] \text{ and }\\ &\qquad [x\in A\cup(X\setminus A)\text{ and }\neg(x\in A\cap B)]\\ &\Rightarrow [x\in A\cup B \text{ and }x\in X]\text{ and } [x\in X\text{ and }x\not\in A\cap B]\\ &\Rightarrow x\in A\cup B\text{ and } x\not\in A\cap B\\ &\Rightarrow x\in(A\cup B)\setminus(A\cap B) \end{align}

• Please use $A \setminus B$ (' A\setminus B') instead of $A /B$. – Kavi Rama Murthy Apr 12 at 9:04
• @KaviRamaMurthy Fixed it. – user663414 Apr 12 at 9:09
• @ShubhamJohri where? – user663414 Apr 12 at 9:10
• @ShubhamJohri I do not see why that is illegal. – user663414 Apr 12 at 9:15
• Nevermind, I thought you used the distributive property. The proof is correct, although I see a lot of unnecessary details; for example, $x\in B\text{ or }x\notin B$ is a true statement, and you seemed to have overcomplicated reaching that inference. – Shubham Johri Apr 12 at 9:31