equality between symmetric differences I'm trying to show this property of the symmetric difference.
I want to show in which case this equality holds:
$$
\mathbb{P}(A\Delta C) = \mathbb{P}(A\Delta B)+\mathbb{P}(B\Delta C)
$$
What I know is that $$
\mathbb{P}(A\Delta C)\leq \mathbb{P}(A\Delta B)+\mathbb{P}(B\Delta C)
$$ always holds and the $\subseteq$ comes from the monotonicity.
To show the equality i was thinking about using "$[(A \bigtriangleup B) \bigcup (B \bigtriangleup C)] \setminus (A \bigtriangleup C) = (A \bigtriangleup B) \bigcap (B \bigtriangleup C)$" but I even have to show the latter and i don't know how to do it.
 A: Let $X=\mathbf{1}_A$, $Y=\mathbf{1}_B$ and $Z=\mathbf{1}_C$. Since for sets $E$ and $F$, the indicator function of $E\Delta F$ is $\left\lvert \mathbb{1}_E-\mathbb{1}_F\right\rvert$, the equality $$\tag{*}\mathbb{P}(A\Delta C) = \mathbb{P}(A\Delta B)+\mathbb{P}(B\Delta C)$$ is equivalent to
$$
\mathbb E\left\lvert X-Z\right\rvert=\mathbb E\left\lvert X-Y\right\rvert+\mathbb E\left\lvert Y-Z\right\rvert
$$
and the triangle inequality is an equality if and only if
$$
\mathbb E\left[\left(X-Y\right)\left(Y-Z\right)\right]=0.
$$
Splitting this expectation as
$$ 
\mathbb E\left[\left(X-Y\right)\left(Y-Z\right)Y\right]+\mathbb E\left[\left(X-Y\right)\left(Y-Z\right)(1-Y)\right]
$$
we can see that (*) holds if and only if
$$
\mathbb P\left(A\cap B^c\cap C\right)= \mathbb P\left(A^c\cap B\cap C^c\right)=0,
$$
or equivalently, as found by HackR,
$\mathbb P((A\Delta B)\cap (B\Delta C))=0$
A: For brevity, we write $X=A\Delta B$ and $Y=B\Delta C$. Then we note that $$X\Delta Y=(A\Delta B)\Delta(B\Delta C)=A\Delta(B\Delta B)\Delta C=(A\Delta \varnothing)\Delta C=A\Delta C$$
where we used the associative property of symmetric difference multiple times.
We also have $$X\Delta Y=(X\cap Y^c)\cup(X^c\cap Y)$$
Now, for later use, we note down the following $$X=(X\cap Y^c)\cup(X\cap Y)$$ $$Y=(Y\cap X^c)\cup(X\cap Y)$$
All the three set unions above are disjoint (check it!).
Then, we have
$$\mathbb{P}(A\Delta C) = \mathbb{P}(A\Delta B)+\mathbb{P}(B\Delta C)\iff \mathbb P(X\Delta Y)=\mathbb P(X)+\mathbb P(Y)$$
Plugging in the above set equalities, we get
$$\mathbb P(X\cap Y^c)+\mathbb P(X^c\cap Y)=\mathbb P(X\cap Y^c)+\mathbb P(Y\cap X^c)+2\mathbb P(X\cap Y)$$
or equivalently $$\mathbb P(X\cap Y)=0$$
Plugging back the values for $X$ and $Y$, we are left with
$$\mathbb{P}(A\Delta C) = \mathbb{P}(A\Delta B)+\mathbb{P}(B\Delta C)\iff \mathbb P((A\Delta B)\cap (B\Delta C))=0$$

For the sake of completion, I do want to mention that the set equality mentioned in the question is also true.
For any two sets $M$ and $N$, it is true that $$(M\cup N)\setminus (M\Delta N)=M\cap N$$
Plugging in $M=X,\ N=Y$ and the facts from the previous part of the answer, we get
$$[(A \Delta B) \cup (B \Delta C)] \setminus (A \Delta C)=(A \Delta B)\cap (B \Delta C)$$
