$L$ finite and distributive lattice, then $\mathcal{J}(L)$ (join-irreducible's) is isomorphic, as poset, to $\mathcal{M}(L)$ (meet-irreducible's) Show that for any finite distributive lattice $L$, $\mathcal{J}(L)$, that is the associated poset of join-irreducible elements of $L$, is isomorphic to $\mathcal{M}(L)$, the associated poset of meet-irreducible elements of $L$. Obviously $\mathcal{J}(L)$ and $\mathcal{M}(L)$ are poset with the order $\le$ of $L$.
To show that $\mathcal{J}(L) \cong \mathcal{M}(L)$ I need to find an order isomorphism between the two poset.
Now, if $L$ is a finite (not necessarily distributive), it is easy to see that $\forall \,\, x \in L$ we have $x = \bigvee\{a\in\mathcal{J}(L)\,\,| \,\,a \le x\}$, in fact if $H :=\{a \in \mathcal{J}(L)\,\,|\,\,a\le x\}$, then $x \in H^u$ by definition of $H$. Moreover if $y \in H^u$ with $x \nleq y$ then $\exists \,\, a\in \mathcal{J}(L)$ s.t. $a \le x $ and $a \nleq y$, but $a \in H$ and $y \in H^u \Rightarrow a \le y$, a contradiction. By duality we have also that $x = \bigwedge\{b \in\mathcal{M}(L)\,\,| \,\,b \ge x\}$, that is, $\forall x \in L$: $$x = \bigvee\{a\in\mathcal{J}(L)\,\,| \,\,a \le x\} = \bigwedge\{b \in\mathcal{M}(L)\,\,| \,\,b \ge x\}$$
When $L$ is a finite and also a distributive lattice then, for the Birkhoff's representation theorem, we have that there is a distributive lattices isomorphism between $L$ and $\,\,\mathcal{O}(\mathcal{J}(L))$ (the lattice of down-sets of $\mathcal{J}(L)$ ordered by inclusion), by putting $x \to \{a\in\mathcal{J}(L)\,\,| \,\,a \le x\}$. Dued to the duality defining $x \to \{b \in\mathcal{M}(L)\,\,| \,\,b \ge x\}$ gives a distributive lattice isomorphism between $L$ and $\,\,\mathcal{O}(\mathcal{M}(L))$.
How to deduce that $\mathcal{J}(L) \cong \mathcal{M}(L)$ as poset, by all of that?
 A: You have a lot of information there and perhaps you can savage some of it to give a proof.
I'll take a different approach.
Start by proving the following lemma:

Lemma. Let $P$ be a finite set. Then

*

*The map $\Phi:P \to \mathcal J(\mathcal O(P))$ given by $x \mapsto {\downarrow}x$ is an order-isomorphism.

*The map $\Psi : P \to \mathcal M(\mathcal O(P))$ given by $x \mapsto P \setminus {\downarrow}x$ is an order-isomorphism.


Hence
\begin{equation}
\label{equation}
P \cong \mathcal J(\mathcal O(P)) \cong \mathcal M(\mathcal O(P)).
\tag{$\dagger$}
\end{equation}
Now, by Birkhoff's Theorem, for a finite distributive lattice $L$ we have that $L \cong \mathcal O(\mathcal J(L))$.
As $\mathcal O(P)$ is distributive for any $P$ (and obviously finite if $P$ is),
\begin{align}
\mathcal J(L)
&\cong \mathcal J(\mathcal O(\mathcal J(L))) \tag{Theorem of Birkhoff}\\
&\cong \mathcal M(\mathcal O(\mathcal J(L))) \tag{by \eqref{equation}}\\
&\cong \mathcal M(L). \tag{Theorem of Birkhoff}
\end{align}
