This question is taken from the book Introduction to Lattices and Order by Davey and Priestley. Question 14 of Chapter 1.
Definition : If $P$ is and ordered set and $Q\subseteq P$, then $Q$ is a down-set if whenever, $x\in Q$, $y\in P$ and $y\leq x$, we have $y\in Q$.
Question : Let $P$ be a finite ordered set. Let $\mathcal{O}(P)$ be the set of all down-sets of $P$. Show that for all $x\in P$, $|\mathcal{O}(P)|=|\mathcal{O}(P\backslash \{x\})|+|\mathcal{O}(P\backslash (\uparrow x \cup \downarrow x))|$.
I suppose the hint given by the question, which says that there exists a one-to-one correspondence between elements in $\mathcal{O}(P)$ and antichains of $P$, is essential here.
If I am to denote $\mathcal{A}(P)$ as the set of all anti-chains of $P$, all I've got is $|\mathcal{O}(P)|=|\mathcal{A}(P)|$. I tried to check if it's the case that $|\mathcal{A}(P)|=|\mathcal{A}(P\backslash \{x\})|+|\mathcal{A}(P\backslash (\uparrow x \cup \downarrow x))|$. However, I really could not see any link between them.
Thus, I would like to seek your input to help me see the links here. Thank you.
HINT: Suppose that $A$ is an antichain in $P$. If $x\notin A$, then $A$ is an antichain in $P\setminus\{x\}$. If $x\in A$, then $A\setminus\{x\}$ is an antichain in $P\setminus(\uparrow\!\!x\cup\downarrow\!\!x)$.