$A \Delta C = B \Delta C$, then prove that $A = B$ where $\Delta$ is a symmetric difference operation. I suppose that if I can prove that every element that belongs to set $A$ also belongs to set $B$ and vice versa and also any element that doesn't belong to set $A$ doesn't belong to set $B$ either and vice versa, then I can prove that $A = B$.
From Venn diagrams, it is clear that an element that belongs to ($A \Delta C$) then it is either in set $A$ or in set $C$ but not in both. Similarly, if the element is not in ($A \Delta C$), then either it doesn't belong to either or belongs to the common region in the Venn diagram. But I'm unable to see how to proceed from here.
$X \Delta Y = (X-Y) \cup (Y-X)$
Also this problem seems trivial to me. But formally proving seems difficult.
 A: Suppose $x \in A$. Then $x \in A \setminus C$ or $x \in A \cap C$.
Case 1. $x \in A \setminus C$. Then, since $A \Delta C = B \Delta C = (B\setminus C) \cup (C \setminus B)$, $x \in B \setminus C$ as $x \not\in C$. In particular, $x \in B$.
Case 2. $x \in A \cap C$. Again, since $A \Delta C = B \Delta C$, $x \not\in B \Delta C$. That is, $x \in B \cap C$ or $x \not\in B \cup C$. Since $x \in C$, it must be the case that $x \in B \cap C$. In particular, $x \in B$.
Hence, $A \subseteq B$. The other direction is similar.
A: If you’ve already proved that symmetric difference is associative, you can prove the desired result without having to chase elements. Since $A\mathrel{\triangle}C=B\mathrel{\triangle}C$, we have
$$A\mathrel{\triangle}(C\mathrel{\triangle}C)=(A\mathrel{\triangle}C)\mathrel{\triangle}C=(B\mathrel{\triangle}C)\mathrel{\triangle}C=B\mathrel{\triangle}(C\mathrel{\triangle}C)\;.$$
And $C\mathrel{\triangle}C=\varnothing$, so $A\mathrel{\triangle}\varnothing=B\mathrel{\triangle}\varnothing$. Finally, $X\mathrel{\triangle}\varnothing=X$ for any set $X$, so we have $A=B$.
A: *

*Since $A\Delta C=B\Delta C$, intersecting with $C^c$ yields $A\cap C^c=B\cap C^c$.

*Prove that $(X\Delta Y)^c=(X\cap Y)\cup (X^c\cap Y^c)$. Apply this to both sides of the identity $(A\Delta C)^c=(B\Delta C)^c$, then intersect with $C$. This yields $A\cap C=B\cap C$.
