Relative strength and propositional indistinguishability of non-distributive lattices Consider the class of bounded non-distributive lattices $\mathbf{Mn}$ ($n\geqslant 3$).

From left to right: M3, M4, Mn
Now consider a propositional language over $\{\wedge,\vee,\neg\}$ with the following semantics: $\wedge$ and $\vee$ correspond to the meet and join of a lattice while $\neg$ works as follows: $\neg\top=\bot$, $\neg\bot=\top$, $\neg\mathbf{k}=\mathbf{k}$ ($\mathbf{k}\leqslant\mathbf{n}$). A valuation $v$ maps propositional variables to the underlying set of some fixed lattice.
For formulas $\phi$ and $\chi$ in our language say that $\phi\vDash_{\mathbf{Mn}}\chi$ iff $$\forall v:v(\phi)\leqslant_{\mathbf{Mn}}v(\phi)$$
Now we call two lattices $\mathfrak{L}$ and $\mathfrak{L}'$ propositionally indistinguishable iff for any two formulas over $\{\wedge,\vee,\neg\}$ $\phi$ and $\chi$ $$\phi\vDash_{\mathfrak{L}}\chi\Leftrightarrow\phi\vDash_{\mathfrak{L}'}\chi$$
We say that $\mathfrak{L}$ is stronger that $\mathfrak{L}'$ iff for any $\phi$ and $\chi$ $$\phi\nvDash_{\mathfrak{L}'}\chi\Rightarrow\phi\nvDash_{\mathfrak{L}}\chi$$
The question is, thus, as follows.
Is it true that all $\mathbf{Mn}$ lattices are indistinguishable? If yes, how can we prove it (or where it has been proved)?
 A: It is not true. 
Consider the lattices $\mathbf{M}_3,\mathbf{M}_4, \dots$ only with the language $\{\vee,\wedge\}$ instead of $\{\vee,\wedge,\neg\}$. First of all, one can show that $\mathbf{M}_n$ is simple for $n\geq 3$. Second of all, since the class of all lattices is a congruence distibutive variety, we can use Jonsson's lemma to say that the subdirectly irreducibles in $\mathsf{V}(\mathbf{M}_n)$ belong to $\mathsf{HS}(\mathbf{M}_n)$. Since the only sublattices of $\mathbf{M}_n$ are the lattices $\mathbf{M}_k$ for $k=0,1,2,\dots,n$, and each of those (except $\mathbf{M}_1$ and $\mathbf{M}_2$) are simple, we get that the subdirectly irreducibles in $\mathsf{V}(\mathbf{M}_n)$ are exactly $\mathbf{M}_0,\mathbf{M}_3,\dots,\mathbf{M}_n$.
So $\mathsf{V}(\mathbf{M}_m)$ is strictly contained in $\mathsf{V}(\mathbf{M}_n)$ whenever $3\leq m<n$. So, by Birkhoff's theorem, there must be some universally quantified equation $\forall v, \phi(v)=\chi(v)$ satisfied by $\mathbf{M}_m$ that is not satisfied by $\mathbf{M}_n$. So in $\mathbf{M}_n$ $\exists v,\phi(v)\neq\chi(v)$ (if $\phi(v)<\chi(v)$, then switch $\phi$ and $\chi$). Then $\phi\models_{\mathbf{M}_m}\chi$, but $\phi\not\models_{\mathbf{M}_n}\chi$. Since $\phi,\chi$ are formulas in $\{\vee,\wedge\}$, they are also formulas in $\{\vee,\wedge,\neg\}$.
As a specific example, $\mathbf{M}_3$ satisfies the equation
$
x_1\vee((x_2\vee(x_3\wedge x_4))\wedge(x_3\vee(x_1\wedge x_4)))\approx x_1\vee(((x_1\vee x_2)\vee(x_3\wedge x_4))\wedge(x_3\wedge(x_2\vee x_4)))
$
but $\mathbf{M}_4$ does not. Let
$\chi(x_1,x_2,x_3,x_4)=x_1\vee((x_2\vee(x_3\wedge x_4))\wedge(x_3\vee(x_1\wedge x_4)))$
and
$\phi(x_1,x_2,x_3,x_4)=x_1\vee(((x_1\vee x_2)\vee(x_3\wedge x_4))\wedge(x_3\wedge(x_2\vee x_4)))$
Now $\phi\models_{\mathbf{}_3}\chi$ since $\phi(v)$ always returns the same element as $\chi(v)$, but $\phi\not\models_{\mathbf{M}_4}\chi$ since $\chi(1,2,3,4)=1$ and $\phi(1,2,3,4)=\top$.
A: After much time and a bit of effort, it seems to me that I have constructed a family of pairs of formulas $\phi$ and $\chi$ separating $\mathbf{Mn}$ from $\mathbf{Mn+1}$.
Consider the following
$$\mathscr{D}_{n}=\bigwedge\limits_{i=2}^{n+1}(p_1\!\vee\!p_i)\vDash\!p_1\!\vee\!\bigvee\limits_{\begin{matrix}i,j\in\binom{n+1}{2}\end{matrix}}(p_i\!\wedge\!p_j)\label{MnMn+1separation}$$
It is clear that $\mathscr{D}_n$ does not hold in $\mathbf{Mn+1}$ (simply set $v(p_i)=\mathbf{i}$). On the other hand, it holds in $\mathbf{Mn}$ for $n>2$.
The argument should be rather straightforward.
