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For any three sets $A, B$ and $C$, show that $$A\Delta B = C \iff A = B\Delta C.$$

I am a student and wish some more information on the above. Kindly help.

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Do you know that $\Delta$ is associative and $A\Delta A=\emptyset$? – azarel Aug 28 '12 at 21:29
What is $\Delta$ here? Symmetric difference? – Henning Makholm Aug 28 '12 at 21:37
@HenningMakholm: Yes, that is the customary meaning of the symbol. – Harald Hanche-Olsen Aug 28 '12 at 21:56
Not so, @HaraldHanche-Olsen...according to WA ( ), one must dispose of that symbol as it is already used in other contexts. – DonAntonio Aug 29 '12 at 2:23
@DonAntonio: right. I will bow to the authority and will immediately stop following a custom that goes back to the beginnings of measure theory and set theory just because W|A (of all people) says so... FWIW: I've never seen any of the other symbols (except maybe +) that are suggested on that page in any measure theory or topology book. Deprecated. That's ridiculous. – t.b. Aug 29 '12 at 13:23

Here is a fact which is easy to check:

The set $C=A\triangle B$ is uniquely defined by the identity $\mathbf 1_C=\mathbf 1_A+\mathbf 1_B\pmod{2}$.


$C=A\triangle B\iff\mathbf 1_{C}=\mathbf 1_A+\mathbf 1_B\pmod{2}\iff\mathbf 1_A=\mathbf 1_C-\mathbf 1_B=\mathbf 1_C+\mathbf 1_B\pmod{2}$

$\iff A=C\triangle B$

Edit: @Harald Hanche-Olsen's answer is based on an equivalent formulation of the easy fact above, equally worthy of notice and perhaps more interesting because it is more symmetrical, namely:

The set $C=A\triangle B$ is uniquely defined by the identity $\mathbf 1_A+\mathbf 1_B+\mathbf 1_C=0\pmod{2}$.

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Here is one way: Notice that $A\bigtriangleup B=C$ is equivalent to the statement that every $x$ belongs to an even number (i.e., none or two) of the sets $A$, $B$, $C$. From that the claim follows by symmetry.

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MathJax's \bigtriangleup seems a bit too big. Oh well. – Harald Hanche-Olsen Aug 28 '12 at 22:00
There is \Delta $\Delta$ v. $\bigtriangleup$ – Henry Aug 28 '12 at 22:50
@Harold, if you have the STIX fonts installed, that might be the culprit, as the big triangle in that font is a bit larger than TeX's version. There is not a lot of choice in the triangles within a font, I'm afraid. – Davide Cervone Aug 29 '12 at 0:18
@Henry: Yes, I know, but $\Delta$ is an ordinary math symbol. Though I could of course have fixed that by typing \mathbin\Delta to get $A\mathbin\Delta B=C$. – Harald Hanche-Olsen Aug 29 '12 at 7:15

Using the associativity and commutativity of the symmetric difference operator, you get:

$$A \Delta B = (B \Delta C) \Delta B = C \Delta (B \Delta B) = C \Delta \emptyset = C$$

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And how do you prove that this operator is associative and commutative? – Godot Aug 28 '12 at 21:55
Your assumptions and conclusions don't seem very clear to me. It would be simpler, I think, if you phrased your argument as "if [...] then [...]". – Ben Millwood Aug 28 '12 at 21:55
@ Godot: That is a different question I guess. The commutativity is absolutely trivial. The associativity follows rather easily. However, I didn't feel that it was important when I wrote my answer. I just wanted to make Michael Kessy aware of that what he is asking follows easily from these two properties – DoubleTrouble Aug 28 '12 at 22:15
@ Ben Millwood: You are absolutely right, it was sloppy. I just wanted to demonstrate that using these two properties is the way to go. Michael Kessy will have to do the formalities himself when he applies this to his problem, or whatever. – DoubleTrouble Aug 28 '12 at 22:18

Here's the long messy way. We want to show $A\triangle B = C \iff A = B\triangle C$

$\Rightarrow$: Assume $C = A\triangle B$

Recall $A\triangle B = (A\cup B) - (A\cap B) = (B-A)\cup(A-B)$

$$\begin{align} (B\cup C) - (B\cap C) &= (B\cup [(A - B) \cup (B - A)]) - (B\cap [(A\cup B) - (A\cap B)]) \\ &= ([B\cup (A- B)] \cup (B - A)) - ([B\cap (A\cup B)] - [B\cap (A\cap B)]) \\ &= ((A\cup B) \cup (B - A)) - (B - (A\cap B))\\ &= ((A\cup B) \cup (B - A)) - (B - A)\\ &= (A\cup B) - (B - A)\\ &= A \end{align}$$

$\Leftarrow$ is, dare is say, symmetric.

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The algebraic structure of classical sets under union, intersection and complement (note that I've included complement, so we do have a universal set here), comes as the same as the algebra of classical logic under disjunction, conjunction, and negation. So, you can first rewrite the symmetric difference formulas in terms of union and complement. Then you can check out the corresponding truth tables in terms of disjunction and negation. After that you can invoke the sameness of algebraic structure and the result follows (unless it's not correct, I haven't checked myself).

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