If you have proven that $A \Delta B = (A \cup B) \setminus (A \cap B)$, then yes, you can rewrite the question like you did. The "proof" would be to reference your previous proof that it is true.
Exactly the same way. Use the definition of the symmetric difference.
$\Rightarrow$: Let $x \in B$.
Case 1: $x \in A \Delta B$. Since $x \in B$, it must be that $x \notin A$. But, we also have $A\Delta B = A\Delta C$, so $x \in A$ or $x\in C$, but not $x \in A\cap C$. Since $x \notin A$, we know $x \in C$.
Case 2: $x \notin A \Delta B$. Since $x \in B$, it must be that $x \in A$. Then, since $A\Delta B = A \Delta C$, we know that $x \notin A \Delta C$. But, we know $x \in A$, so it must be that $x \in C$.
$\Leftarrow$: Similar to above.
Thanks , I will try the next direction and I would be happy if you can correct me :
⇐B=C ;
Let x∈B and from The given information x∈C.
Case 1(We want to show that AΔB ⊆ AΔC): x∈AΔB and we want to show that x∈AΔC, if x∈AΔB then x∉A , we know that x∈C from given info then x∈AΔC.
Case 2(We want to show that AΔC ⊆ AΔB): x∈AΔC , we know that x∈C that mean it must be x∉A . we now that x∈B from info given on top and x∉A then x∈AΔB .
proved .
Is that Okay ?
Thanks for all your help.
Close, but for $\Leftarrow$, you are assuming $B=C$ to prove that $A\Delta B = A\Delta C$. So, instead of $x \in B$, you would start with $x \in A\Delta B$ and prove that $x \in A \Delta C$.
Also, I did not complete the proof for $\Rightarrow$. I assumed $A\Delta B = A \Delta C$ and proved that $B \subseteq C$. I did not prove that $C\subseteq B$, which is needed for $B=C$. That can be done with a single line saying "The proof that $C \subseteq B$ follows exactly the same as the proof for $B\subseteq C$."
So, here is the proof:
$\Leftarrow:$ Given $B=C$, we want to show $A\Delta B = A \Delta C$ for any set $A$. First, we will show that $A\Delta B \subseteq A \Delta C$. Let $x \in A\Delta B$. We want to show that $x \in A\Delta C$.
Case 1: $x \in A$. Then, by the definition of symmetric difference, $x \notin B$. Since $B=C$, we know $x \notin C$. Therefore, since $x \in A, x\notin C$, we know $x \in A\Delta C$ again by the definition of symmetric difference.
Case 2: $x \notin A$. Then, since $x\in A\Delta B$, it must be that $x \in B$, and since $B=C$, we know $x \in C$. Since $x \notin A, x \in C$, we have $x \in A \Delta C$, again by the definition of symmetric difference.
Since this is true for any arbitrary $x \in A\Delta B$, we have $A\Delta B \subseteq A\Delta C$ for any set $A$.
The proof that $A\Delta C \subseteq A\Delta B$ follows the exact same proof (just swapping $B$ and $C$). There is no reason to rewrite it.