Number of join-irreducibles and rank in a finite lattice

Question

Let $(L, r)$ be a finite, ranked lattice. Is $$r(x) = \#\{j \leq x : j \in L \text{ is join-irreducible}\}$$ for all $x \in L$?

This may be a naive question, though my lattice theory is weak.

Answer 1

I could give a very trivial negative answer just by stating that, if $|L|=n$, and we define a rank function such that $r(0)=n+1$, then the formula can't hold (notice that, at least in the Wikipedia article, it's not necessary that $r(0)=0$).

That said, it's reasonable to think that what you have in mind is a rank function $r$ such that $r(0)=0$.
Even in that case, the result doesn't hold, as the following example shows:

This is a ranked lattice with ranks $0,1,2,3$, but for the top element, $1$, which satisfies $r(1)=3$, we have $4$ join-irreducibles below it.
All other elements satisfy your formula.

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Update

As I noted in a comment below, the $\mathbf M_3$ lattice (see here) is an even simpler counter-example.