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Prove that $P(F_n) = \sum_{k=0}^n (-1)^k \binom n k$
Suppose that $F_n = F \cup F^{ -1}$ where $F = \{f: \mathbb Z_+ \to \mathbb Z_+ | f(x+1) = f(x) \ \forall x \in \mathbb Z_+\}$
Prove that $$P(F_n) = \sum_{k=0}^n (-1)^k \binom n k$$
I tried to show that $P(F_n) = P(F) P(F^{ -1})$ but I don’t know what to do in the next step. How do I prove this?
A:
Let $\mathcal{F}$ be the class of functions $f\colon \mathbb{Z}_+\to \mathbb{Z}_+$ such that $f(x+1)=f(x)$ for every $x\in\mathbb{Z}_+$.
We have, for every $n\in\mathbb{N}$,
$$
\{0,\ldots,n\}=\bigcup_{f\in \mathcal{F}^n}\{0,\ldots,n\}
$$
where
$$
\mathcal{F}^n=\{f\colon \mathbb{Z}_+\to \mathbb{Z}_+\mid f\in\mathcal{F} \text{ for every }x\in\mathbb{Z}_+\}.
$$
It follows that
$$
P(\mathcal{F})=1.
$$
Now let
$$
A=\left\{f\colon \mathbb{Z}_+
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