1. set theory operation

We want to prove that for each $\displaystyle A,B \in \mathcal{P}(X)$, there exists a unique set $\displaystyle C = A \Delta B$ such that $\displaystyle A \Delta C = B$.

So $\displaystyle A \Delta C = B \Rightarrow (A-C) \cup (C-A) = B \Rightarrow A-C \subseteq B$ and $\displaystyle C-A \subseteq B \Rightarrow$ $\displaystyle C = (A-B) \cup (B-A) \Rightarrow C = A \Delta B$?

Obviously $\displaystyle C = A \Delta B$ because $\displaystyle A \Delta (A \Delta B) = (A \Delta A) \Delta B = \emptyset \Delta B = B$ (by associative law). But that only proves existence and not uniqueness.

Thanks

2. You should know that the symmetric difference operation is associative:
$\displaystyle \left( {K\Delta L} \right)\Delta M = K\Delta \left( {L\Delta M} \right)$.

If $\displaystyle C = A\Delta B$ then consider this:
$\displaystyle A\Delta C = A\Delta \left( {A\Delta B} \right) = \left( {A\Delta A} \right)\Delta B = \emptyset \Delta B = B$. We have shown the property.

3. Yes thats what I did in my first post.

But does that suffice to show that $\displaystyle C = A \Delta B$?

4. Uniqueness
If $\displaystyle A\Delta L = B$ then $\displaystyle \left( {A\Delta L} \right)\Delta B = \emptyset$.
But $\displaystyle \left( {A\Delta L} \right)\Delta B = \left( {A\Delta B} \right)\Delta L = \emptyset \quad \Rightarrow \quad L = A\Delta B$