In order to understand two different definitions of the same function $f$ (as stated by Besner, 2022), I am trying to prove that those expressions are equal:
$\Delta(A) = f(A) - \sum_{B \subset A} \Delta(B)$, (formula 1)
$\Delta(A) = \sum_{B \subseteq A} (-1)^{|A|-|B|} \ f(B)$ (formula 2).
I tried doing this by induction on $|A|$. For $|A|=1$, we can clearly see that the statement is true: $f(A) - \sum_{B \subset A} \Delta(B) = f(A)$ and $\sum_{B \subseteq A} (-1)^{|A|-|B|} \ f(B) = (-1)^0 \ f(A) = f(A)$.
Assuming the statement is true for all $|A|<n$, I try to prove this for $|A|=n$:
$\Delta(A) = f(A) - \sum_{B \subset A} \Delta(B)$
$= f(A) - \sum_{B \subset A} \sum_{C \subseteq B} (-1)^{|B|-|C|} \ f(C)$.
This is equivalent to the expression
$f(A) = \sum_{B \subseteq A} \sum_{C \subseteq B} (-1)^{|B|-|C|} \ f(C)$
Unfortunately, this is as far as I got. Earlier suggestions were to use Fubini's theorem or Newton's Binomium, but I could not figure out whether these suggestions were right and how to use it. Hence, the aim here is to interchange or rewrite the summations in order to arrive at formula 2.