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.

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
$$[(A \Delta B) \cup (B \Delta C)] \setminus (A \Delta C)=(A \Delta B)\cap (B \Delta C)$$
