I need to show that a variety generated by finitely many finite lattices yields only finitely many subvarieties. I think Jónsson's lemma is the key but I am not sure how to use it. For those that do not know here is the lemma:
If $V$ is a congruence distributive variety and $A_i \in V, i \in I$. Then if $B \le \prod_{i \in I}A_i$ and $\theta \in Con(B)$ is such that $B/\theta$ is a nontrivial subdirectly-irreducible algebra. Then, there is an ultrafilter $U$ over $I$ such that $\theta_{U\restriction_B} \subseteq \theta$. Where $\theta_U := (a,b) \in \theta_U \iff [[a=b]] = \{i \in I : a_i = b_i \} \in U $.
There is also a corollary from Jónsson that says for a congruence distributive variety made up of a finite class $K$ of finite algebras, then every SI algebra in the variety $V(K)$ is in $HS(K)$.
It seems that if $K$ is a finite class of finite lattices then $V(K)$ is congruence distributive as the lattice varieties always are congruence distributive. Further, has $K$ is finite, it there are only finitely many SI-algebras or SI-subalgebras within it. So it makes sense that there are only finitely many subvarieties as varieties are generated by the SI-algebras within the class.
Is this the correct idea? I am not quite sure how to go about deriving this from the lemma and maybe the corollary as well.
Claim. If $V$ is a variety, then an algebra $A$ lies in $V$ iff all SI quotients of $A$ lie in $V$.
Reason. $V$ is closed under the formation of quotients, so if $A\in V$, then any SI quotient of $A$ lies in $V$. Conversely, if all SI quotients of $A$ lie in $V$, then since $A$ is a subdirect product of its SI quotients it follows that $A$ is in $V$. \\\
Corollary 1. Every variety is generated by its SI members.
Corollary 2. A variety with finitely many SI's has finitely many subvarieties.