Question about Distributive Laws of Logic and proof of the Symmetric Difference Here is the definition of the Symmetric Difference:
$$
A \Delta B = (A \backslash B) \cup (B \backslash A),
$$
Our goal is to prove:
$$
A \Delta B = (A \cup B) \backslash(A \cap B).
$$
I want to prove this and at the same time to check if my understanding of the Distributive Laws of Logic is correct.
My main goal is about Distributive Laws of Logic. I want to know if I can always break the parenthesis with this method.
I know that there are similar proofs in the website but I feel that I need help with the Distributive Laws.
Please check if this is correct:
We start with 
$(x \in A$ $\land$ $x \notin  B)$ $\vee$ $(x \notin A$ $\land$ $x \in  B)$
This is the part that I want to know if I am correct, and I want to know if I can do this for any parenthesis regarding of what is inside of the parenthesis of the right part:
$(x \in A$ $\land$ $x \notin  B)$ $\vee$ x $\notin$ A ) $\land$
$(x \in A$ $\land$ $x \notin  B)$ $\vee$ x $\in$ B )
Now we can use the classic version of the Distributive Law:
$(x \in A$ $\vee$ x $\notin$ $A$ ) $\land$
$(x \notin B$ $\vee$ x $\notin$ $A$ ) $\land$
$(x \in A$ $\vee$ x $\in$ $B$ ) $\land$
$(x \notin B$ $\vee$ x $\in$ $B$ )
Now we can simplify:
******Universe****** $\land$
$(x \notin B$ $\vee$ x $\notin$ $A$ ) $\land$
$(x \in A$ $\vee$ x $\in$ $B$ ) $\land$
******Universe******
Finally we have,
$(x \in A$ $\vee$ x $\in$ $B$ ) $\land$
$(x \notin B$ $\vee$ x $\notin$ $A$ )
The last step is to use the Morgan Laws
$(x \in A$ $\vee$ x $\in$ $B$ ) $\land$
$\neg$ $(x \in A$ $\land$ $x$ $\in$ $B$ )
That is our goal:

Q.E.D.
 A: Yes, that is a permissible application of distribution. 
It is basically using the proof for the quadratic distribution: $$(s\wedge t)\vee (u\wedge v)~{= (s\vee (u\wedge v))\wedge(t\vee (u\wedge v))\\ =(s\vee u)\wedge(s\vee v)\wedge(t\vee u)\wedge (t\vee v)}$$

Take an arbitrary $x$ such that $x\in (S\cap T)\cup(U\cap V)$.   By definition of union and disjunction, this is equivalent to $(x\in S\wedge x\in T)\vee(x\in U\wedge x\in V)$.   By distribution, that is equivalent to $(x\in S\vee(x\in U\wedge x\in V))\wedge (x\in T\vee(x\in U\wedge x\in V))$, and distributing again that is $(x\in S\vee x\in U)\wedge (x\in S\vee x\in V)\wedge (x\in T\vee x\in U)\wedge (x\in T\vee x\in V)$. Thus it is equivalent to $x\in(S\cup U)\cap (S\cup V)\cap (T\cup U)\cap (T\cup V)$.

More succinctly, using the set algebra:
$$(S\cap T)\cup(U\cap V)~{=(S\cup(U\cap V))\cap(T\cup(U\cap V))\\=(S\cup U)\cap (S\cup V)\cap (T\cup U)\cap (T\cup V)}$$
Just substitute the relevant sets.

If we presume complementation relative to some superset $\mathcal U$ (the "universe") of both $A,B$, then your proof is essentially:
$$\def\sm{\smallsetminus} {\begin{align}
&\quad (A\Delta B)
\\ &= (A\sm B)\cup(B\sm A) &&\text{definition of symmetric difference}
\\& = (A\cap B^\complement)\cup(B\cap A^\complement) &&\text{definition of set minus}
\\& = (A\cup (B\cap A^\complement))\cap(B^\complement\cup (B\cap A^\complement)) && \text{distribution}
\\& = (A\cup B)\cap(A\cup A^\complement)\cap (B^\complement\cup B)\cap (B^\complement\cup A^\complement)&&\text{distribution}
\\& = (A\cup B)\cap\mathcal U\cap\mathcal U\cap (B^\complement\cup A^\complement)&&\text{complementation}
\\&=  (A\cup B)\cap(B^\complement\cup A^\complement) && \text{identity}
\\&=  (A\cup B)\cap(A^\complement\cup B^\complement) && \text{commutivity}
\\&= (A\cup B)\cap(A\cap B)^\complement & & \text{deMorgan's Rule}
\\&= (A\cup B)\sm(A\cap B) && \text{definition of set minus}\end{align}}$$
