Symmetric difference equality Something I was thinking about earlier: If $A\triangle B=A \triangle C$, does $B=C$? Where $\triangle$ is symmetric difference. My intuition is telling me no, but I can't seem to think of an example where this is not true. Thanks for the help! 
 A: Symmetric difference is associative.
if $A\triangle B = A\triangle C$ then consider:
$A\triangle (A\triangle B)=A\triangle (A\triangle C)$
$(A\triangle A)\triangle B=(A\triangle A)\triangle C $, and $A\triangle A = \emptyset$ hence,  $\emptyset \triangle B = \emptyset \triangle C$ and $B = C$.
A: This is just a matter of simplification.
I'm used to the not-so-well-known definition
$$
x \in A \triangle B \;\equiv\; x \in A \not\equiv x \in B
$$
(Note that we can leave out the parentheses, because $\;\equiv\;$ and $\;\not\equiv\;$ are mutually associative, just like $\;\equiv\;$ is associative.  This forces us to make an arbitrary choice of where to put the parentheses, which gives us more freedom of manipulation, as you can see around expression $(*)$ below.)
Using this, we can just try to simplify $\;A \triangle B \;=\; A \triangle C\;$ by calculating (I'm taking baby steps here, for clarity):
\begin{align}
& A \triangle B \;=\; A \triangle C \\
\equiv & \qquad \text{"set extensionality"} \\
& \langle \forall x :: x \in A \triangle B \;\equiv\; x \in A \triangle C \rangle \\
\equiv & \qquad \text{"the above definition of $\;\triangle\;$, twice"} \\
& \langle \forall x :: x \in A \not\equiv x \in B \;\equiv\; x \in A \not\equiv x \in C \rangle \\
\equiv & \qquad \text{"logic: negate both sides of $\;\equiv\;$ -- the simplest thing I see"} \\
(*)\quad\phantom\equiv & \langle \forall x :: x \in A \equiv x \in B \;\equiv\; x \in A \equiv x \in C \rangle \\
\equiv & \qquad \text{"logic: $\;\equiv\;$ is symmetric -- to bring identical parts together"} \\
& \langle \forall x :: x \in A \equiv x \in A \;\equiv\; x \in B \equiv x \in C \rangle \\
\equiv & \qquad \text{"logic: $\;\equiv\;$ is reflexive -- simplify"} \\
& \langle \forall x :: \text{true} \;\equiv\; x \in B \equiv x \in C \rangle \\
\equiv & \qquad \text{"logic: $\;\equiv\;$ has identity $\;\text{true}\;$ -- simplify"} \\
& \langle \forall x :: x \in B \equiv x \in C \rangle \\
\equiv & \qquad \text{"set extensionality"} \\
& B = C \\
\end{align}
(Normally I would take the last five steps together into one, but I've spelled them out just in case you are not familiar with these laws of logic.)
So we've proved
$$
A \triangle B \;=\; A \triangle C \;\;\equiv\;\; B = C
$$
A: The definition of symmetric difference is $A\triangle B = (A\setminus B)\cup(B\setminus A)$.
So what if we choose an $x$ such that $x\notin A$ and $x\notin B$, and then set $C = B\cup\{x\}$? Since $x$ is not an element of $A$ or $B$ it is not in $A\triangle B$. Consider $A\triangle C$, $x\notin A$ so $x\in C\setminus A$ so $x\in A\triangle C$.
Edit: Whoops, then the hypothesis is contradicted. At least I proved the inverse right?
A: The answer is that $A\triangle B=A\triangle C\implies B=C$.
It's not hard to see that $A\triangle (A\triangle B)=B$. The uniqueness of $B$ is guaranteed by the symmetric difference operator.
A: The characteristic function of the symmetric difference $A \Delta B$ if $f_A + f_B \pmod 2$. Here, we have $f_A + f_B = f_A + f_C \pmod 2$ and adding $f_A$ to both sides, we get $f_B = f_C$ and thus $B=C$.
