<s>[INST] 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.
 [/INST] 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.
$B = 3 \cup 4\cup 7 \cup 8 = 4 \cup 8$.
$C = 2 \cup 4 \cup 6 \cup 8 = 4 \cup 8 = B$.
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