Meet-irreducible element of lattice I have a question concerning the exercise 5.7 of the Davey & Priestley's book. Here are the questions:
Let $L$ be a finite distributive lattice. Prove by the steps below that $\mathcal{J}(L)\cong\mathcal{M}(L)$.
(i) Let $x\in\mathcal{J}(L)$. Show that there exists $\hat{x}\in L$ such that $\downarrow \hat{x}=L\backslash\uparrow x$. [Hint. Let $\hat{x}:=\bigvee (L\backslash\uparrow x)$ and then use Lemma 5.11 to show that $\hat{x}\not\geq x$].


I have done it and it's ok.


(ii) Show that for all $x\in\mathcal{J}(L)$ the element $\hat{x}$ defined in (i) is meet-irreducible.


Here, I would need some help. We have to show that if $\bigvee (L\backslash\uparrow x)<a$ and $\bigvee (L\backslash\uparrow x)<b$ then $\bigvee (L\backslash\uparrow x)<a\wedge b$. 
    Can we show first that: 
    $\forall y\in L\backslash\uparrow x$ if $y<a$ and $y<b$ then $y<a\wedge b$


Thank you. 

Relating to your question 
"Can we show first that...": no, that's not true.
As an example, take $L$ to be the power-set of $\{a,b,c\}$, which is distributive.
Singletons are join-irreducible. For example, take $x=\{a\}$; $y=\{b\} \in L\setminus\uparrow x$.
Then $y=\{a,b\}\cap\{b,c\}$, so it is not meet-irreducible.
Now, for the main problem: prove that $\hat{x}$ is meet-irreducible.
Suppose, for a contradiction, that it isn't, that is, there exist $a,b\in L$ such that $\hat{x}=a \wedge b$, but $x\neq a,b$ (and so $x<a$ and $x<b$).
So $a,b \notin \downarrow\hat{x}=L\setminus\uparrow x$, and thus $a,b \in \uparrow x$, that is, $x\leq a$ and $x \leq b$.
It follows that $x\leq a\wedge b =\hat{x}$, a contradiction to what you have proved in (i) (that $\hat{x}\ngeq x$).
