Set theory identity Prove that $(A\triangle B)\triangle (A\cap B) = A\cup B$ .
I tried and mostly went in circles. The idea is to use that 
$(A\triangle B)=(A\setminus B)\cup (B\setminus A)$ and also 
$A\setminus B = \bar A\cap B$ where $\bar A$ is absolute complement of A together with the De Morgan's laws . These are the few lines:
$(A\triangle B)\triangle (A\cap B)$ = $[(\bar A\cap B)\cup(\bar B\cap A)]\triangle(A\cap B)$ = ...
I would post my solution so far but I suck at LaTeX so it would take me too long.
 A: OK, that's a good start!  Please do know that $A - B = A \cap \bar B$ rather than $\bar A \cap B$
Now:
$$(A\triangle B)\triangle (A\cap B) =$$
$$[(A- B)\cap(B-A)]\triangle (A\cap B) =$$
$$[(A\cap \bar B)\cup(B\cap \bar A)]\triangle(A\cap B)= $$
$$\big( [(A\cap \bar B)\cup(B\cap \bar A)]-(A\cap B)\big) \cup \big( (A\cap B)-[(A\cap \bar B)\cup(B\cap \bar A)]\big)=$$
$$\big( [(A\cap \bar B)\cup(B\cap \bar A)]\cap \overline{ (A\cap B)}\big) \cup \big( (A\cap B)\cap \overline{[(A\cap \bar B)\cup(B\cap \bar A)]}\big) = \text{ (DeMorgan)}$$
$$\big( [(A\cap \bar B)\cup(B\cap \bar A)]\cap (\bar A\cup \bar B)\big) \cup \big( (A\cap B)\cap \overline{(A\cap \bar B)}\cap \overline{(B\cap \bar A)]}\big) = \text{ (Distribution and DeMorgan)}$$
$$\big( [(A\cap \bar B)\cap (\bar A\cup \bar B)]\cup[(B\cap \bar A)\cap (\bar A\cup \bar B)]\big) \cup \big( (A\cap B)\cap (\bar A\cup B)\cap (\bar B\cup A)]\big) = \text{ (Absorption x 4)}$$
$$(A \cap \bar B) \cup(B\cap \bar A) \cup  (A\cap B) = \text{ (Adjacency)}$$
$$(A \cap \bar B) \cup B = \text{ (Reduction)}$$
$$A \cup B$$
Tedious ... but not hard.
To make this a little less tedious, please note that $\triangle$ is associative, and so:
$$(A \triangle B) \triangle (A \cap B) =$$
$$A \triangle (B \triangle (A \cap B)) =$$
$$A \triangle [(B - (A \cap B)) \cup ((A \cap B) - B)]=$$
$$A \triangle [(B \cap (A \cap B)^C) \cup ((A \cap B) \cap B^C)]=$$
$$A \triangle [(B \cap (A^C \cup B^C)) \cup (A \cap (B \cap B^C))]=$$
$$A \triangle [(B \cap A^C) \cup (B \cap B^C) \cup (A \cap \emptyset)]=$$
$$A \triangle [(B \cap A^C) \cup \emptyset \cup \emptyset)]=$$
$$A \triangle (B \cap A^C)=$$
$$(A - (B \cap A^C)) \cup ((B \cap A^C)-A)=$$
$$(A \cap (B \cap A^C)^C) \cup ((B \cap A^C) \cap A^C)= \text{ (Idempotence)}$$
$$(A \cap (B^C \cup A)) \cup (B \cap A^C)= \text{ (Absorption)}$$
$$A \cup (B \cap A^C)=\text{ (Reduction)}$$
$$A \cup B $$
... hmm. still tedious ....
A: Rather than endless algebraic manipulations of expressions involving unions, intersections and complements, it would be far easier to use the definitions of set equality, intersection and union directly: that is, prove that for all $x$, you have
$$x \in (A \triangle B) \triangle (A \cap B) \quad \text{if and only if} \quad x \in A \cup B$$
I'll start you off:

Suppose $x \in (A \triangle B) \triangle (A \cap B)$. Then by definition of $\triangle$, we have $x \in A \triangle B$ or $x \in A \cap B$, but not both.
  
  
*
  
*If $x \in A \cap B$, then in particular $x \in A$, so $x \in A \cup B$ and you're done.
  
*If $x \in A \triangle B$, then $x \in A$ or $x \in B$ (but not both); in any case, $x \in A \cup B$.
  
  
  In both cases, we've proved $x \in A \cup B$.

This proves that $(A \triangle B) \triangle (A \cup B) \subseteq A \cup B$.
You now need to prove the converse.
This method has the additional benefit of making the result more intuitive. What it's saying is that $x \in A \cup B$ if and only if either $x$ is in exactly one of $A$ or $B$, or $x \in A \cap B$, and moreover these latter cases are mutually exclusive.
