Curious identity involving symmetric difference While studying the properties of measurable null sets, I found the following identity:
$\bigcup_i B_k\triangle B_i=\bigcup_i B_i - \bigcap_i B_i $ Or in other words, the value of the expression is independent of the choice of $k$. 
I can prove this using a case separation in first principles, but it doesn't give me much intuition on why it is true.
So I'm asking if the above can be deduced just using basic set algebra?
Edit: I mean is there a sequence of rewritings that reveal the symmetry of the statement?

Let $x \in \bigcup_{i} B_i$. If $x$ is contained in all the $B_i$ than $x$ is not contained in $\bigcup_i B_k \triangle B_i$ and also not contained in $\bigcup_i B_i - \bigcap_i B_i$. Now suppose $x$ is contained in some, but not all, of the $B_i$. Then $x$ is contained in $\bigcup_i B_i - \bigcap_i B_i$. Furthermore, if $x \in B_k$ there is a $i \neq k$ such that $x \not\in B_i$ and if $x \not\in B_k$ there is a $i \neq k$ such that $x \in B_i$. That is, there exists $i \neq k$ with $x \in B_i \triangle B_k$ and $x$ is contained in $\bigcup_i B_k \triangle B_i$. Since this holds for all $x$, the conclusion follows.
