# Feedback on my proof that $(A\setminus B)\cup(B\setminus A)=(A\cup B)\setminus (A\cap B)$

I would like to prove that $(A\setminus B)\cup(B\setminus A)=(A\cup B)\setminus (A\cap B)$. Please could you offer some feedback?

Firstly, I will show that:

$(A\setminus B)\cup(B\setminus A)\subseteq(A\cup B)\setminus (A\cap B).\tag{1}$

Assume $x \in (A\setminus B)\cup(B\setminus A)$. Then, by the definition of union, $x \in A\setminus B$, or $x \in B\setminus A$, or both.

(i) but if $x \in A\setminus B$ - then $x \in A$, $x \notin B$ and $x \notin A \cap B$.

(ii) or if $x \in B\setminus A$, then $x \in B$ and $x \notin A$ and $x \notin B \cap A$.

Putting the (i) and (ii) together, either $x \in A$ or $x \in B$ but $x\notin A\cap B$. This proves (1).

$(A\cup B)\setminus (A\cap B)\subseteq(A\setminus B)\cup(B\setminus A)\tag{2}$

Assume $x \in (A \cup B) \setminus (A \cap B)$.Then $x \in A$ or $x \in B$ or both by the definition of union. However we know that $x \notin A \cap B$. Therefore it must be the case that either $x \in A$ and $x \notin B$ or $x \in B$ and $x \notin A$. This proves (2).

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Looks correct. One comment: In mathematics "or" almost always means "or both" already. – Julian Kuelshammer Jan 3 '13 at 22:45
cheers @JulianKuelshammer – bosra Jan 3 '13 at 22:46
+1 Looks good to me. – Derek Allums Jan 3 '13 at 23:14
Thanks @unit3000-21 – bosra Jan 3 '13 at 23:22

I think you did quite well, and it's the most appropriate approach for you at this point in time:

Just expanding on your work (filling in some gaps which were easy enough to assume):

Assume $x \in (A\setminus B)\cup(B\setminus A)$. Then, by the definition of union,

(a) $x \in A\setminus B$, or $x \in B\setminus A$.

(a.i) $x \in A\setminus B$, then $x \in A$ and $x \notin B$, so $x \notin A \cap B$.

or

(a. ii) $x \in B\setminus A$, then $x \in B$ and $x \notin A$, so $x \notin B \cap A$.

From (a), and (a.i) and (a.ii), we have ($x \in A$ or $x \in B$) and $x\notin (A\cap B)$. Hence, $x \in (A\cup B)$ and $x \notin (A\cap B)$. That is, $x \in (A\cup B)\setminus(A\cap B)$.

Therefore, $(A\setminus B)\cup(B\setminus A)\subseteq(A\cup B)\setminus (A\cap B).\tag{1}$

Now, Assume $x \in [(A \cup B) \setminus (A \cap B)]$.

Then $x \in (A\cup B)$ and $x\notin (A \cap B)$, by the definition of setminus. Then $x \in A$ or $x \in B$ by the definition of union. However we also know that $x \notin (A \cap B)$. Therefore it must be the case that either $(x \in A$ and $x \notin B)$ or ($x \in B$ and $x \notin A$). That is, either $x \in (A \setminus B)$ or $x \in (B \setminus A)$. So $x \in (A\setminus B) \cup (B \setminus A)$.

Therefore, $(A\cup B)\setminus (A\cap B)\subseteq(A\setminus B)\cup(B\setminus A)\tag{2}$

Very nicely done, by the way.

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+1 @amWhy thanks for the feedback.. – bosra Jan 4 '13 at 8:12
I should point out that a post by @Brian M. Scott helped me a lot. Other interested readers should see the following math.stackexchange.com/questions/239875/element-chasing-proof – bosra Jan 4 '13 at 9:33
Yes, that's indeed a good post. – amWhy Jan 4 '13 at 18:17

This is not a feedback, but just a better way to do it. I will use the distributivity of $\cup$ over $\cap$. $$(A/ B)\cup(B/A)=$$ $$(A\cap B^c)\cup(A^c\cap B)=$$ $$(A\cup A^c)\cap(A\cup B)\cap(B^c\cup A^c)\cap(B^c\cup B)=$$ $$(A\cup B)\cap(B^c\cup A^c)=$$ $$(A\cup B)\cap(A\cap B)^c=$$ $$(A\cup B)/(A\cap B)$$

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(1) You are using / where \ is needed (\setminus); (2) I don't know about better. Many times element chasing is a much better exercise pedagogically. At least on the very basic level. In this aspect your solution obfuscates this sort of reasoning. – Asaf Karagila Jan 3 '13 at 23:42
I don't know about better either. The better route, when first learning, is element chasing. – amWhy Jan 3 '13 at 23:44
Yes may be both of you are right. I just felt its neater to do it this way. – Amr Jan 3 '13 at 23:45
More rigorus than @borsa proof. I like rigor. – PyRulez Jan 6 '13 at 2:15