Statements with symmetric difference - equality proof.

Question

The symmetric difference is defined as follows: $A \triangle B = (A \setminus B) \cup (B \setminus A)$.

I need to prove that the both sides of the following equation are equal: $$ (A \triangle B) \setminus C = (A \setminus C) \triangle (B \setminus C). $$

I tried to do it in this way:

$ RHS = (A \setminus C) \triangle (B \setminus C) = ((A \setminus C) \setminus (B \setminus C)) \cup ((B \setminus C) \setminus (A \setminus C)) $

Let's assume for simplicity that $A$ means $x \in A$

$ RHS = ((A \land \lnot C) \land \lnot (B \land \lnot C)) \lor ((B \land \lnot C) \land \lnot(A \land \lnot C)) = ((A \land \lnot C) \land (\lnot B \lor C)) \lor ((B \land \lnot C) \land (\lnot A \lor C)). $

Now I assumed that we can continue in this way: $ = ((A \land \lnot C) \land \lnot C) \lor ((B \land \lnot A) \land \lnot C) $. The rest of the proof is trivial but my question is if it was correct to make that step. In my opinion, both statements are equal because if the statement is $True$, $C$ is always $False$, so we can omit that. I'm not sure though if this is 'formal' enough and correct.

Answer 1

$\begin{align}(A\mathop\triangle B)\smallsetminus C~&=~[(A\smallsetminus B)\cup (B\smallsetminus A)]\cap C^c\\[1ex]&=~((A\smallsetminus B)\smallsetminus C)\cup ((B\smallsetminus A)\smallsetminus C)\\[1ex]&=~((A\smallsetminus C)\smallsetminus (B\smallsetminus C))\cup ((B\smallsetminus C)\smallsetminus (A\smallsetminus C))\\[1ex]&=~(A\smallsetminus C)\mathop{\triangle} (B\smallsetminus C)\end{align}$

Answer 2

Yes, you can use distribution, annihilation, and identity.

$\begin{align}(A\smallsetminus C)\mathop\Delta(B\smallsetminus C) &= ((A\cap C^\complement)\cap(B\cap C^\complement)^\complement)\cup((B\cap C^\complement)\cap(A\cap C^\complement)^\complement)\\[1ex]&=(A\cap C^\complement\cap(B^\complement\cup C))\cup(B\cap C^\complement\cap(A\cup C))\\[2ex] &=((A\cap C^\complement\cap B^\complement)\cup(A\cap C^\complement\cap C))\cup((B\cap C^\complement\cap A^\complement)\cup(B\cap C^\complement\cap C))\\[1ex]&=((A\cap C^\complement\cap B^\complement)\cup(C^\complement\cap C))\cup((B\cap C^\complement\cap A^\complement)\cup(C^\complement\cap C))\\[2ex]&=(A\cap C^\complement\cap B^\complement)\cup(B\cap C^\complement\cap A^\complement)\\[1ex] &=((A\cap B^\complement)\cup(B\cap A^\complement))\cap C^\complement\\[1ex]&=(A\mathop\Delta B)\smallsetminus C\end{align}$

Often those middle steps are skipped, and we immediately equate $C^\complement\cap (X\cup C)=C^\complement\cap X$ by rule of absorption.

$\begin{align}(A\smallsetminus C)\mathop\Delta(B\smallsetminus C) &= ((A\cap C^\complement)\cap(B\cap C^\complement)^\complement)\cup((B\cap C^\complement)\cap(A\cap C^\complement)^\complement)\\[1ex]&=(A\cap C^\complement\cap(B^\complement\cup C))\cup(B\cap C^\complement\cap(A\cup C))\\[1ex] &=(A\cap C^\complement\cap B^\complement)\cup(B\cap C^\complement\cap A^\complement)\\[1ex] &=((A\cap B^\complement)\cup(B\cap A^\complement))\cap C^\complement\\[1ex]&=(A\mathop\Delta B)\smallsetminus C\end{align}$