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For two sets we define $A \Delta B = (A \cup B) \setminus (A \cap B)$ (symmetric difference).

Problem: Proof that \begin{align*} A \Delta C \subset (A \Delta B) \cup (B \Delta C). \end{align*}

Attempt at proof: Let $x \in A \Delta C$. Then $x \in (A \cup C)$ and $x \notin (A \cap C)$. Thus $x \in A$ or $x \in C$. When $x \in A$, then also $x \in A \cup B$. When $x \in C$, then also $x \in B \cup C$.

Now, I don't know how to proceed. I'm not sure how to use the fact that $x \notin (A \cap C)$ in my proof.

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  • $\begingroup$ Do you mean $\subseteq$? $\endgroup$ – simonzack Aug 6 '15 at 9:30
  • $\begingroup$ No, I mean a proper subset. $\endgroup$ – Kamil Aug 6 '15 at 9:32
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    $\begingroup$ If $B=\varnothing$, and $A\cap C=\varnothing$, then you get equality. $\endgroup$ – Asaf Karagila Aug 6 '15 at 9:35
  • $\begingroup$ What would be an example such that the equality does not hold? What would I have to choose $A, B$ and $C$ as? $\endgroup$ – Kamil Aug 6 '15 at 9:56
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Suppose that $x\in A$, what would cause it not to be in the RHS set? If $x\notin A\mathbin{\triangle} B$ and $x\notin B\mathbin{\triangle} C$.

Since we already assumed that $x\in A$, then for $x\notin A\mathbin{\triangle}B$, it would have to be the case that $x\in B$ as well. But then in order for $x$ not to be an element of $B\mathbin{\triangle}C$, it means that $x\in C$ as well. So $x\in A\cap C$. But since $x\notin A\cap C$, this cannot be the case.


An alternative method, would be to simply note that:

$$A\mathbin{\triangle}C=(A\mathbin{\triangle}B)\mathbin{\triangle}(B\mathbin{\triangle}C)=\bigl((A\mathbin{\triangle}B)\cup (B\mathbin{\triangle}C)\bigr)\setminus\bigl((A\mathbin{\triangle}B)\cap (B\mathbin{\triangle} C)\bigr)\subseteq\ldots$$

But this only works after you've proved that $\mathbin{\triangle}$ is associative, of course.

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  • $\begingroup$ I understand that $x \in B$ would have to be the case. But then what you say after that, i.e. that in order for $x$ not to be an element of $B \Delta C$, why do we need $x \in C$ for that? $\endgroup$ – Kamil Aug 6 '15 at 9:33
  • $\begingroup$ Because it's already an element of $B$. So it is in $B\cup C$, but it is not an element of $B\mathbin{triangle}C$, so it is necessarily the case that $x\in C$ as well. $\endgroup$ – Asaf Karagila Aug 6 '15 at 9:34
  • $\begingroup$ I see. So what is your proof technique here? Did you derive a contradiction? What is the conclusion? $\endgroup$ – Kamil Aug 6 '15 at 9:38
  • $\begingroup$ The conclusion is that if $x\in A$, and $x$ is not in the RHS, then $x\in C$. Therefore $x\notin A\mathbin{\triangle} C$. So it's a contrapositive proof. $\endgroup$ – Asaf Karagila Aug 6 '15 at 9:42

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