Let A,B,C be sets. If A△B=A△C, does this imply that B=C? I believe that a counterexample exists, but I have struggled to find one since I need to satisfy (A\B) U (B\A) = (A\C) U (C\A), which is a very complex statement. Also I used △ to represent symmetric difference.
 A: Suppose $x\in B$ and $x\notin C$, and the equality of symmetric differences holds.
There are two cases:


*

*$x\in A$: here you easily see it is not in $(A\setminus B)\cup (B\setminus A)$ but it is in $(A\setminus C)\cup (C\setminus A)$, a contradiction

*$x\notin A$: then it is not in $(A\setminus C)\cup (C\setminus A)$ but it is in $(A\setminus B)\cup (B\setminus A)$, another contradiction.


Therefore no such $x$ exists, and $B\subseteq C$.  By the symmetric argument, $C\subseteq B$.

Another algebraic way to do it is if you know that (assuming $A,B,C$ are subsets of $X$) $\triangle$ is an abelian group operation on the subsets of $X$, one in which all elements have order $2$, and the emptyset is the identity.
Suppose we then write $A\triangle B=A+B$. Then the equation $A+B=A+C$ can be simplified by adding $-A=A$ to both sides, whereupon you immediately get $C=B$.
A: For me, manipulations with $\triangle$ are easier if one uses the characteristic function
noting that $1_{A \triangle B} = 1_A \oplus 1_B$,
where $\oplus$ is 'exclusive or' (mild abuse of notation).
If $x,y$ have values in $\{0,1\}$, then $x \oplus x = 0$, $0 \oplus x = x$.
(Equivalently, $A \triangle A = \emptyset$ and $\emptyset \triangle A = A$.)
Using characteristic functions, we have
$1_A \oplus 1_B = 1_A \oplus 1_C$, so $1_A \oplus 1_A \oplus 1_B = 1_{B}$ and similarly 
$1_A \oplus 1_A \oplus 1_C = 1_{C}$. Hence $1_{B} = 1_{C}$ and so $B=C$.
A: For an elementary approach, you can split into cases - it's most convenient to make cases based on whether $x\in A$. We want to show that, given that $A\triangle B=A\triangle C$, we have that $x\in B$ if and only if $x\in C$. 
So, first let's handle the case where $x\in A$. Note that $x\in A\triangle B$ if and only if $x\not\in B$ and $x\in A\triangle C$ if and only if $x\not\in C$. However, $x\in A\triangle B$ if and only if $x\in A\triangle C$ as these sets are equal, so we find that, if $x\in A$, then $x\not\in B$ if and only if $x\not\in C$. The case where $x\not\in A$ is handled similarly - though if you were trying to write this out very formally, you might split into further cases to avoid confusions from using statement like "if this, then this if and only if that".
A more elegant way to show this would be to show the following identity:
$$A\triangle (A\triangle S) = S.$$
Then, starting from $A\triangle B=A\triangle C$, you can just apply $A\triangle \cdot$ to both sides to get
$$A\triangle (A\triangle B)=A\triangle (A\triangle C)$$
And then use the identity on both sides to get
$$B=C$$
A: I'd like to add one more proof -- by contradiction. Suppose there exist such $x \in B$ that $x \notin C$. Consider two cases
$$
x \in A 
\xrightarrow{x \notin C} x \in A \triangle C
\xrightarrow{A \triangle C = A \triangle B} x \in A \triangle B
\xrightarrow{x \in A} x \notin B
~~~~~~\text{(contradiction)}
$$
$$
x \notin A 
\xrightarrow{x \in B} x \in A \triangle B
\xrightarrow{A \triangle C = A \triangle B} x \in A \triangle C
\xrightarrow{x \notin A} x \in C
~~~~~~\text{(contradiction)}
$$
Swapping letters 'B' and 'C' we can prove that there is also no such $x$ that $x \notin B$ and $x \in C$.
This completes the proof.
A: $\Delta$ is associative, hence $A\operatorname{\Delta}B=A\operatorname{\Delta}C$ implies
$$ B=(A\operatorname{\Delta}A)\operatorname{\Delta}B=A\operatorname{\Delta}(A\operatorname{\Delta}B)=A\operatorname{\Delta}(A\operatorname{\Delta}C)=(A\operatorname{\Delta}A)\operatorname{\Delta}C=C.$$
