Inclusion–exclusion principle [Proof verification]

Hint:

• The proof of the lemma is correct. In fact it states the clear relation, that with $[k]:=\{1,2,\ldots,k\}$ the powerset of $[k+1]$ minus the powerset of $[k]$ is the set of all subsets which contain the element $k+1$. $$ \left. \begin{array}{rl} A&=\mathcal{P}\left([k]\right)\setminus\emptyset\\ B&=\mathcal{P}\left([k+1]\right)\setminus\emptyset\\ \end{array} \right\} \Longrightarrow B\setminus A=\mathcal{P}\left([k+1]\right)\cap\{\{k+1\}\} $$

• The proof of the Inclusion-Exclusion principle seems to be correct. Good work. We might skip the representation $$A_{k+1}=\sum_{J\in\{\{k+1\}\}}(-1)^{|J|-1}\left|\bigcap_{j\in J}A_j\right|$$ as it is not relevant for the proof and stick with $A_{k+1}$ enhancing readability.