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

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

  • $\begingroup$ Please use $A \setminus B$ (' A\setminus B') instead of $A /B$. $\endgroup$ – Kavi Rama Murthy Apr 12 at 9:04
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    $\begingroup$ @KaviRamaMurthy Fixed it. $\endgroup$ – user663414 Apr 12 at 9:09
  • $\begingroup$ @ShubhamJohri where? $\endgroup$ – user663414 Apr 12 at 9:10
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    $\begingroup$ @ShubhamJohri I do not see why that is illegal. $\endgroup$ – user663414 Apr 12 at 9:15
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    $\begingroup$ 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. $\endgroup$ – Shubham Johri Apr 12 at 9:31

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