Let A,B,C be sets. If A△B=A△C, does this imply that B=C?
$\Delta$ is associative, hence $A\operatorname{\Delta}B=A\operatorname{\Delta}C$ implies $$ B=(A\operatorname{\Delta}A)\operatorname{\Delta}B=A\operatorname{\Delta}(A\operatorname{\Delta}B)=A\operatorname{\Delta}(A\operatorname{\Delta}C)=(A\operatorname{\Delta}A)\operatorname{\Delta}C=C.$$
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 $C=B$.
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 $B=C$.