$A\,\triangle\, B = C$ implies that $A\,\triangle\, C = B$
I understand that the delta is the symmetric difference and that the symmetric difference of $A$ and $B$ is the set of elements that belong to exactly one of $A$ and $B$. How do I prove the above statement?
We need to prove that given $C = A \ \Delta \ B$, then $B = A \ \Delta \ C$
A simple and elementary proof is to just use the definition of the symmetric difference, or $XOR$ as I like to call it, and setup some truth tables:
$\begin{array}{|c|c|c|} \hline A & B & \mathbf{C} \\ \hline 0 & 0 & \mathbf{0} \\ \hline 0 & 1 & \mathbf{1} \\ \hline 1 & 0 & \mathbf{1} \\ \hline 1 & 1 & \mathbf{0} \\ \hline \end{array}$
$\begin{array}{|c|c|c|c|} \hline A & B & \mathbf{C} & \mathbf{A \ \Delta \ C} \\ \hline 0 & 0 & \mathbf{0} & \mathbf{0}\\ \hline 0 & 1 & \mathbf{1} & \mathbf{1}\\ \hline 1 & 0 & \mathbf{1} & \mathbf{0}\\ \hline 1 & 1 & \mathbf{0} & \mathbf{1}\\ \hline \end{array}$
In the last truth table we can clearly see that $B = A \ \Delta \ C$, completing our proof.
If you've proved that $\Delta$ is associative and that $A \Delta A = \Lambda$ (the empty set) just multiply both sides by $A$.
To say that two sets are equal is to say that they contain each other. First, try showing that $A\Delta C\subseteq B$. That is, if $x\in A\Delta C$, then $x\in B$. To say that $x\in A\Delta C$ is to say that either $x\in A\setminus C$ ($x$ is in $A$ but not in $C$) or $x\in C\setminus A$ ($x$ is in $C$ but not in $A$). That gives you two cases. For each case, try a proof by contradiction. First assume $x\in A\setminus C$ and $x\notin B$. Using the fact that $A\Delta B=C$, you should arrive at a contradiction. Then assume $x\in C\setminus A$ and $x\notin B$. Again, using that fact, you should reach a contradiction.
Next, try showing that $B\subseteq A\setminus C$. That is, if $x\in C$, then $x\in A\Delta B$. Remember that saying $x\in A\Delta B$ is to say $x\in A\setminus B\lor x\in B\setminus A$. Assume $x\in C$ and $x\notin A\setminus B$. If you can show that $x\in B\setminus A$, then you are done. Remember to use the fact that $A\Delta B=C$.
While here, it's worth noting that for the characteristic functions we have $$\chi_{A \Delta B} = \chi_A \oplus \chi_B$$ ( modular addition, $\mod 2$).
