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?

Since you tagged boolean algebra, let's try the following.
First let $x,x_i$ be elements of a boolean algebra, then
$$\begin{aligned}
\prod^n_{i=1}(x+x_i) &= x^n+(x_1+\cdots+x_n)x^{n-1}+(x_1x_2+\cdots)x^{n-2}+\cdots + (x_1\cdots x_n)\\
&=x(1+(x_1+\cdots x_n)+(x_1x_2+\cdots)+\cdots+(x_1\cdots x_n))+(1+x)\prod x_i\\
&=x\prod(1+x_i) +(1+x) \prod x_i\\
\end{aligned}
$$
Add $1$ to both sides and replace $x=1+b_k$, $x_i=b_i$, we have
$$1+\prod(1+(b_k+b_i))=1+(1+b_k)\prod(1+b_i)+b_k\prod b_i$$
Let $b_i$ be the characteristic function of $B_i$, we have
$$\left(\bigcap(B_k\triangle B_i)^c\right)^c = \left(\bigcap B_i^c\right)^c\triangle\bigcap B_i$$
or
$$\bigcup B_k\triangle B_i = \bigcup B_i\triangle \bigcap B_i =\bigcup B_i-\bigcap B_i
$$
