Proving $ C = D $ from $ A \triangle C = A \triangle D$ I have been working on one part of the proof where my aim is to show that $C =  D$. I have been able to prove that $ A \triangle C = A \triangle D$.
Is it reasonable to conclude $ C = D $ from $ A \triangle C = A \triangle D$ ?
 A: I'll assume that you define
$$
X\mathbin{\triangle}Y=(X\cup Y)\setminus (X\cap Y).
$$
Then it can be shown that, for any three sets $X,Y,Z$,
$$
X\mathbin{\triangle}(Y\mathbin{\triangle}Z)=
(X\mathbin{\triangle}Y)\mathbin{\triangle}Z
$$
This is the property you can try and prove.
It's easier (almost obvious, actually) proving that
$$
X\mathbin{\triangle}X=\emptyset,\quad
X\mathbin{\triangle}\emptyset=X,\quad
\emptyset\mathbin{\triangle}X=X.
$$
Thus, from $A\mathbin{\triangle}C=A\mathbin{\triangle}D$ we get
$$
A\mathbin{\triangle}(A\mathbin{\triangle}C)=
A\mathbin{\triangle}(A\mathbin{\triangle}D)
$$
so, by associativity,
$$
(A\mathbin{\triangle}A)\mathbin{\triangle}C=
(A\mathbin{\triangle}A)\mathbin{\triangle}D
$$
and so
$$
\emptyset\mathbin{\triangle}C=\emptyset\mathbin{\triangle}D
$$
and, finally,
$$
C=D.
$$
A: Suppose $x \in A$. Then $$x \in C \iff x \not\in A \triangle C \iff x \not\in A \triangle D \iff x \in D.$$
Suppose $x \not\in A$. Then
$$ x \in C \iff x \in A \triangle C \iff x \in A \triangle D \iff x \in D.$$
So for any $x$ we have $x \in C \iff x \in D$, hence $C=D$.
A: Yes, for a set $X$ with power set $P(X)$, the symmetric difference gives you an abelian group $(P(X),Δ,∅))$ on it, so you can cancel $A$.
You can show associativity of $Δ$ by writing down a truth table and any set is obviously self-inverse with respect to $Δ$.
