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If we define symmetric difference as $A\triangle B=(A\setminus B)\cup(B\setminus A)$, can we also say that $(A\cup B)\triangle C=(A\triangle C)\cup(B\triangle C)$?

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Have you tried just expanding both sides? One way to study this is that instead of writing $A \backslash B$, write $A \cap B^c$, where the $^c$ stands for complement. In this manner you can use the fact that in a set $X$, there is a boolean ring $(P(X), \cap, \cup)$ which is a ring of sets, and under this operation there is distributivity of the two operations. Perhaps you probably knew but I was just making sure. –  Patrick Da Silva Nov 16 '11 at 5:35
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If $x$ is in $B$ and $C$ but not in $A$, then it's not in the first set but it is in the second. –  Gerry Myerson Nov 16 '11 at 5:37
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Following Gerry's comment, here is an explicit example of how the equality can fail.

Let $A = \emptyset$ and $B = C = \{x\}$. We have $$ \begin{align*} (A \cup B) \Delta C &= \{x\} \Delta \{x\}\\ &= \emptyset, \end{align*} $$ while $$ \begin{align*} (A \Delta C) \cup (B \Delta C) &= \{x\} \cup \emptyset\\ &= \{x\}. \end{align*} $$

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