I was trying to prove $(A\cup B)-(A\cap B)=(A-B)\cup(B-A)$ and came across issues in translating (pertaining to what I did with $\emptyset$) and got through the proof but was doubting its accuracy so if someone could please affirm or deny its accuracy that would be great. Also, I'm guessing there's much shorter ways of proving this and was thinking of what those ways might be and all I could come up with was set construction. Any ideas on others?

My proof is as follows:

Suppose $x\in(A\cup B)-(A\cap B)$.

Thus $x\in(A\cup B):x\notin(A\cap B)$.

Hence $x\in A$ or $x\in B:x\notin A$ or $x\notin B$.

This presents us with 4 cases:

  1. $x\in A:x\notin A$
  2. $x\in A:x\notin B$
  3. $x\in B:x\notin A$
  4. $x\in B:x\notin B$

Case 1 and 4 are $\emptyset$ however so we really have $x\in\emptyset$, or $x\in A:x\notin B$, or $x\in B:x\notin A$, or $x\in\emptyset$ which is equivalent to $x\in\emptyset$, or $x\in(A-B)$, or $x\in(B-A)$, or $x\in\emptyset$. Which is the same as $x\in\emptyset\cup(A-B)\cup(B-A)\cup\emptyset$ And since the union of the empty set with any set X is the set X itself, we have $x\in(A-B)\cup(B-A)$, which shows that $\forall x[x\in(A\cup B)-(A\cap B)\Rightarrow x\in(A-B)\cup(B-A)]$.

Alternatively, suppose that $x\in(A-B)\cup(B-A)$. Since the union of any set X is the set X itself, we can say that $x\in\emptyset_1\cup(A-B)\cup(B-A)\cup\emptyset_2$ where $\emptyset_1={x\in A:x\notin A}$ and $\emptyset_2={x\in B:x\notin B}$ so $x\in\emptyset_1$, or $x\in(A-B)$, or $x\in(B-A)$, or $x\in\emptyset_2$, thus $x\in [x\in A:x\notin A] $, or $x\in A:x\notin B$, or $x\in B:x\notin A$, or $x\in B:x\notin B$, hence $x\in A:x\notin A$, or $x\in A:x\notin B$, or $x\in B:x\notin A$, or $x\in B:x\notin B$, therefore $x\in A$ or $x\in B$, and $x\notin A$ or $x\notin B$ such that $x\in(A\cup B)$ and $x\notin(A\cap B)$ which means that $x\in(A\cup B)-(A\cap B)$.

The former shows that $\forall x[x\in(A-B)\cup(B-A)\Rightarrow x\in(A\cup B)-(A\cap B)]$ , and thus $(A-B)\cup(B-A)\subset(A\cup B)-(A\cap B)$ by the definition of a subset.

Since, as we showed earlier, $(A\cup B)-(A\cap B)\subset(A-B)\cup(B-A)$ and $(A-B)\cup(B-A)\subset(A\cup B)-(A\cap B)$, by the definition of set equality, $(A\cup B)-(A\cap B)=(A-B)\cup(B-A)$.

Thank you!

  • $\begingroup$ Did you look at how this was formatting before posting? $\endgroup$ Aug 28, 2015 at 2:04
  • $\begingroup$ Yes. I thought I finally had it this time. I'm not a computer guy and am slow to learn how to "write math" on this site. I'm currently trying to fix it though. I apologize. The first quarter or so should be readable now $\endgroup$ Aug 28, 2015 at 2:09
  • $\begingroup$ What does a sentence such as "Thus $x\in(A\cup B):x\notin(A\cap B)$" mean? That is not a standard use of the colon, and it is difficult to guess at a meaning for it that makes sense for all of the colons in your text. $\endgroup$ Aug 28, 2015 at 2:16
  • $\begingroup$ It means "such that". I know it's not used by everyone but I thought it was relatively conventional. My apologies is that isn't the case. $\endgroup$ Aug 28, 2015 at 2:20
  • $\begingroup$ I was trying to edit your post and thus hopefully offer some help, but I surrendered. Your formatting is a disaster. Moreover, you are making many nonsensical statements (which makes the editing even more difficult). $\endgroup$
    – user230734
    Aug 28, 2015 at 2:20

4 Answers 4


It means "such that". I know it's not used by everyone but I thought it was relatively conventional. My apologies is that isn't the case.

It's not quite used this way.   The convention is used in set construction notation to separate the domain of the bound variable and the predicate that constructs the set.   In set construction, we may also use pipes ("|") for this.

For example : $\;2\Bbb N= \{x \in \Bbb N : \exists k\in \Bbb N \; (x=2k)\}\;$, reads as: "the set $2\Bbb N$ is the set of natural numbers, $x$, such that there exists a natural number, $k$ so that $x=2k$."

Thusly, what you appear to be trying to say is:

$$\begin{align} &(A\cup B)-(A\cap B) \; \\[1ex] = & \{ x\in A\cup B: x\notin A\cap B\} \\[1ex] & \vdots \\[1ex] = & \{x\in A:x\notin B\}\cup\{x\in B:x\notin A\} \\[1ex] = & (A-B)\cup (B-A) \end{align}$$

Which is okay.

To use my words:

Take an arbitrary element of the union of $A$ and $B$ that is not in their intersection, that is $\;(A\cup B)-(A \cap B)\;$.  This element is in either $A$ or in $B$ but it is not in both $A$ and $B$.   Considering case by case, (1) if it is in $A$, then it is not in $B$, and (2) if it is in $B$, then it is not in $A$.   So it is either in $A$ but not $B$ or it is in $B$ but not $A$.   Which is to say it is in the symmetric difference of the sets. $$(A\cup B)-(A\cap B)\subseteq (A-B)\cup(B-A)$$

Take any arbitrary element of the union of set differences that is $(A-B)\cup(B-A)$.   This element is either in $A$ but not $B$, or it is in $B$ but not $A$.   Hence it is in either $A$ or $B$ but it is not in both. $$(A\cup B)-(A\cap B)\supseteq (A-B)\cup(B-A)$$

Hence the two are demonstrably equivalent, which is what was to be shown. $$(A\cup B)-(A\cap B) \equiv (A-B)\cup(B-A) \quad \Box$$

  • $\begingroup$ Thank you. So when I see that though it looks exactly like what I wrote. I'm guessing the difference then is that you have what I wrote in {} thus indicating its a set which allows us to use that notation $\endgroup$ Aug 28, 2015 at 3:19
  • $\begingroup$ Thank you! What I'm not sure people are realizing though (perhaps people are just skimming over my proof and not reading it thoroughly due to its verbosity and length or maybe I'm just missing something here) is that I feel like I basically did what you just did EXCEPT I noticed that where you said there were 2 cases, I said there were 4 with $x\in A$ such that $x\notin A$ and $x\in B$ such that $x\notin B$. And that's why my proof got really long; I considered the 2 other two cases which were the empty set. $\endgroup$ Aug 28, 2015 at 5:02
  • $\begingroup$ So, is the vibe I'm getting then just that my proof wasn't wrong but my consideration of the other 2 cases which were the empty set unnecessary but my proof wasn't technically wrong. Or was it wrong DUE to my dealings with the empty set? I'd really like to get to the bottom of this. I made a similar comment on another answer to this question. Please tell me if I'm not being clear $\endgroup$ Aug 28, 2015 at 5:07
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    $\begingroup$ @LiamCooney Indeed, your argument was not technically wrong, but rather just unnecessarily cluttered and in a non-standard format. You have the concept down okay; so you just need to work on the presentation and streamline it a bit. $\endgroup$ Aug 28, 2015 at 5:31
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    $\begingroup$ @LiamCooney Still, there is nothing at all wrong with "displaying even the most obvious steps, for clarity's sake". It can even be helpful! (Most especially in an exam.) Just try not to hide the forest with too many trees. $\endgroup$ Aug 28, 2015 at 6:06

Let $x$ be in $A \cup B$. Then either $x \in A$ and $x \not \in B$, $x \in A$ and $x \in B$, or $x\not \in A$ and $x \in B$.

Therefore, if $x$ is in $A \cup B$ but not in $A \cap B$, then $x$ is in $A - B$ or $B-A$. Vice versa, if $x$ is in $A - B$ or $B-A$, then $x$ is in $A \cup B$ but not in $A \cap B$.

I'll let you finish the proof (it only involves one sentence).

Note: it's normal to struggle with those kind of proofs at first. Draw diagrams and work hard until you really understand what is going on.

  • $\begingroup$ Thank you. I see the logic here and can certainly appreciate its brevity, but I'm just a little hesitant to personally use this because I'm new to proofing and while I follow what you're saying and it makes sense to me, I likely wouldn't be able to recreate it because I don't see the logical "flow" to it (not the proofs problem but mine). For example, I'm not exactly sure what the last sentence you speak of would be. "Vice versa" I'm guessing you say to take care of the proof going in "the other direction" so would it just be that each set is then a subset of the other? $\endgroup$ Aug 28, 2015 at 2:41
  • $\begingroup$ You should not use this. You should try to understand the idea, and then write your own (similar) proof, with the appropriate amount of details. Drawing a « Venn diagram » will also help. $\endgroup$
    – Olivier
    Aug 28, 2015 at 2:43
  • $\begingroup$ I usually start my reading it to myself and verifying it to myself logically without a proof. Then sometimes I also do a Venn Diagram especially if I don't immediately see the truth of it. I think that making the Venn diagrams can be a good tool, but I don't personally see how they aid in proof writing. $\endgroup$ Aug 28, 2015 at 2:47
  • $\begingroup$ I think you understood it right. Saying « vice versa » is often a bit ambiguous. I updated my answer. $\endgroup$
    – Olivier
    Aug 28, 2015 at 2:47
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    $\begingroup$ Did you read it or skim it? I only ask because I just retread it (fixed only 1 or 2 things that I thought could lead to ambiguity) but overall I thought it made sense. What did you think was "nonsensical"?. If you skimmed over it I could see how the stuff I said about $\emptyset$ would appear nonsensical but if you read it on whole I think those things make sense, but if they're not true then my proof is wrong. I don't know if that means the same thing as "nonsensical" though... $\endgroup$ Aug 28, 2015 at 3:15

Here's a different understanding: $$\begin{align*} (A \cup B) - (A \cap B) &= (A \cup B) \cap (A \cap B)^c \\ &= (A \cup B) \cap (A^c \cup B^c) \quad (1)\\ &= (A \cap B^c) \cup (B \cap A^c) \quad (2) \\ &=(A-B) \cup (B-A) \end{align*}$$

If you need help understanding how to get from $(1) \to (2)$, let me know.
Hint: It follows from $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

  • $\begingroup$ No, I understand your translations, thank you. The one I used is just more intuitive to me I guess. I tried doing using the translation you used instead but it got me to the same place. I'd like to point out though that, I basically did the same thing you did here EXCEPT using the other translation AND (although, as I said, I used your translation and got to the same place) I added the 2 other cases that you did not (in your proof, going from line 3-4) which was where I got the empty set stuff. So are you basically telling me that what I did with the empty set was just unnecessary? $\endgroup$ Aug 28, 2015 at 4:56
  • $\begingroup$ Or if it's just plain wrong id appreciate knowing that as well. Please tell me if you don't understand my previous comment. $\endgroup$ Aug 28, 2015 at 5:00

Here is the way to do it by algebra \begin{align} (A\cup B)-(A\cap B) &=(A-A\cap B)\cup(B-A\cap B) \\ &=(A-B)\cup(B-A) \end{align}

  • $\begingroup$ I was trying to let him get the intermediate steps on his own... $\endgroup$ Aug 28, 2015 at 3:25
  • $\begingroup$ I did not see your post when I submit mine. Sorry about it. $\endgroup$
    – hermes
    Aug 28, 2015 at 3:28
  • $\begingroup$ Aha! Thank you. So what I was struggling with was the empty sets. I don't see how this is really any different than what I though except for the fact that I added a lot of detail; yours is obviously more succinct. Question though: is what you wrote a proof? Like can you just show equivalency like that or was this just an outline and you still have to suppose x belongs to some set like I did and all? This is obviously shorter because you didn't do that and you only went in one direction. I'm assuming you can do that because you showed equivalency instead of one set being a subset of another. $\endgroup$ Aug 28, 2015 at 5:11

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