If $A∆B=A∆C$, then $B=C$ If $A\Delta B=A\Delta C$, then $B=C$.
I use the formula  $X∆Y=(X\setminus Y)\cup(Y\setminus X)$ in $A∆B=A∆C$. And further use the formula $X\setminus Y=X\cap Y^c$. I am not  able to reach the conclusion. Please help me.
 A: There are two ways: know or prove that $. \Delta .$ is a self-inverse group operation:
$$B = \emptyset \Delta B = (A \Delta A) \Delta B = A \Delta (A \Delta B) = A \Delta (A \Delta C) = (A \Delta A) \Delta C = \emptyset \Delta C = C$$
Or a direct proof using inclusion and the definitions:
Suppose $A \Delta B ( = (A \setminus B) \cup (B \setminus A) )= A \Delta C$.
Let $x \in B$. If $x \notin A$ then $x \in A \Delta B$ so $x \in A \Delta C$ and as $x \notin A$ we get $x \in C$.
If $x \in A$. the $x \notin A \Delta B$ so $x \notin A \Delta C$, and as $x \in A$ this means that $x \in C$ as well. 
So $B \subseteq C$. The other inclusion is the same, interchanging  the rôles of $B$ and $C$.
A: The key to such problems is to follow a single, arbitrary, element through the equations.
I will prove $B\subseteq C$.  Let $b\in B$ be arbitrary.  We now have two cases.  If $b\in A$, then $b\notin A\Delta B$.  Since $A\Delta B=A\Delta C$, then $b\notin A\Delta C$.  But $b\in A$, so we must have $b\in C$.  The second case is if $b\notin A$.  Now $b\in A\Delta B = A\Delta C$.  Since $b\notin A$, we must have $b\in C$.  So, in both cases, for any $b\in B$, we must have $b\in C$.  This proves $B\subseteq C$.
We now do a similar argument (or use symmetry) to prove that $C\subseteq B$.  Combining with the previous we conclude that $B=C$.
A: Using the indicator functions in $\Bbb{Z}/2\Bbb{Z}$: consider $X := A\cup B\cup C$ so that for $D\subset X$, $\Bbb{1}_D: X \to \Bbb{Z}/2\Bbb{Z}$ is defined as usual. Then $\Bbb{1}_{D\Delta E}= \Bbb{1}_D + \Bbb{1}_E$ for all $D,E\subset X$ (simple calculation).
Thus $A\Delta B = A\Delta C \iff \Bbb{1}_{A\Delta B} = \Bbb{1}_{A\Delta C} \iff \Bbb{1}_A + \Bbb{1}_B = \Bbb{1}_A + \Bbb{1}_C \iff \Bbb{1}_B =\Bbb{1}_C \iff B=C$
A: Since $A\setminus B\cup A\setminus C=A\setminus B\cap C$ and $B\setminus A \cup C\setminus A= B\cup C\setminus A$
$$A\Delta B\cup A\Delta C=(A\cup B\cup C)\setminus (A\cap B\cap C)$$.
But since $A\Delta B\cup A\Delta C=A\Delta B$, $C\setminus A\Delta B=(A\cap B\cap C)$
Similarly since $A\Delta B\cup A\Delta C=A\Delta C$, $B\setminus A\Delta C=(A\cap B\cap C)=B\setminus A\Delta B$.
Then $C\setminus A\Delta B=B\setminus A\Delta B$. Hence $B=C\space\space\space\blacksquare$
A: You can do element chasing.
Or you can manipulate symbols and properties and identities.
But you can also divide and conquer.
Divide the space into $8$ disjoint sets determined by the elements that are are not in $A,B,C$.
They are:
1) $A^c \cap B^c \cap C^c$ (the elements that are in none)
2)  $A^c \cap B^c \cap C$
3)  $A^c \cap B \cap C^c$
4)  $A^c \cap B \cap C$
5)  $A \cap B^c \cap C^c$
6) $A \cap B^c \cap C$
7) $A \cap B \cap C^c$
8) $A \cap B \cap C$
$A\Delta B$ consists of $3$ and $4$ and $5$ and $6$  but none of the others $1,2,7,8$.
$A \Delta C$ consists of $2$ and $4$ and $5$ and $7$ but none of the others $1,3,6,8$.
So we get the seemingly paradoxical result that sets $2,3,6,7$ are all both subsets of $A \Delta B = A \Delta C$, and completely disjoint from $A \Delta B = A \Delta C$.  The only way a set can be both a subset and disjoint from another set, is if the set is empty.  So $2,3,6,7$ are all empty.
$B = 3 \cup 4\cup 7 \cup 8 = 4 \cup 8$.
$C = 2 \cup 4 \cup 6 \cup 8 = 4 \cup 8 = B$.
