Proof of addition rule of probabilities with 4 events I need to show that, given a event space and four events. If the only non-empty intersections between them are $A\cap B$, $B\cap C$, $C\cap D$, $D\cap A$, then:

$P(A\cup B\cup C\cup D)=P(A)+P(B)+P(C)+P(D)-P(A\cap B)-P(B\cap C)-P(C\cap D)-P(A\cap D)$

I think that i need to start with an equivalent form of $A\cup B\cup C\cup D$.
For example, i know that: $A\cup B=(A-B)\cup B$. But i dont know how to do this with the four sets.
 A: Note that
\begin{align*}
\mathbb P(A)-\mathbb P(A\cap B)=&\,\mathbb P(A\cap B^c)\\
\mathbb P(B)-\mathbb P(B\cap C)=&\,\mathbb P(B\cap C^c),\\
\mathbb P(C)-\mathbb P(C\cap D)=&\,\mathbb P(C\cap D^c),\\
\mathbb P(D)-\mathbb P(D\cap A)=&\,\mathbb P(D\cap A^c).
\end{align*}
Then, the right-hand side of the formula to be shown can be rewritten as
\begin{align*}
\mathbb P(A\cap B^c)+\mathbb P(B\cap C^c)+\mathbb P(C\cap D^c)+\mathbb P(D\cap A^c). \tag{$\spadesuit$}
\end{align*}
Given that $A\cap C=B\cap D=\varnothing$, these four events are pairwise disjoint, so $(\spadesuit)$ further equals:
$$\mathbb P\left[(A\cap B^c)\cup(B\cap C^c)\cup(C\cap D^c)\cup(D\cap A^c)\right].$$
If we can show that the event between the brackets is the same as $A\cup B\cup C\cup D$, the proof will be complete.
$\textbf{Claim:}\phantom{---}(A\cap B^c)\cup(B\cap C^c)\cup(C\cap D^c)\cup(D\cap A^c)=A\cup B\cup C\cup D$.
$\textit{Proof:}\phantom{---}$Since $A\cap B^c\subseteq A$, $B\cap C^c\subseteq B$, $C\cap D^c\subseteq C$, and $D\cap A^c\subseteq D$, it is clear that
$$(A\cap B^c)\cup(B\cap C^c)\cup(C\cap D^c)\cup(D\cap A^c)\subseteq A\cup B\cup C\cup D.$$ To see the other direction, suppose that $x\in A\cup B\cup C\cup D$. We need to show that $x$ is contained in at least one of $(A\cap B^c)$, $(B\cap C^c)$, $(C\cap D^c)$, or $(D\cap A^c)$.
$\textit{Case 1:}\phantom{---}$$x\in A$. If $x\in B^c$, then $x\in (A\cap B^c)$. If $x\in B$, then, since $x\in A\subseteq C^c$ (given that $A\cap C=\varnothing$), it follows that $x\in (B\cap C^c)$.
The other three cases ($x\in B$, $x\in C$, or $x\in D$) are completely analogous. $\blacksquare$
