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Call a partial order $\mathcal{F}=(F, \leq)$ rooted if there is an element $a \in F$ such that for any $b \in F$, $a\leq b$.

Let $\mathcal{F}_0$ and $\mathcal{F}_1$ be two different finite rooted partial orders and then they could be regarded as two Kripke frames. $\textbf{F}_0$ and $\textbf{F}_1$ denote the intermediate logics of $\mathcal{F}_0$ and $\mathcal{F}_1$.

It is obvious that $\textbf{Int}\in \textbf{F}_0$ and $\textbf{Int}\in \textbf{F}_1$, my question is

Is it possible to have $\textbf{Int}=\textbf{F}_0\cap \textbf{F}_1$ for some $\mathcal{F}_0$ and $\mathcal{F}_1$.

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    $\begingroup$ No. Int is not complete w.r.t. a finite set of finite frames (aka tabular). $\endgroup$
    – Emil Jeřábek
    Commented Jan 8 at 21:19
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    $\begingroup$ Even better, the disjunction property of Int implies that it is not the intersection of finitely many intermediate logics strictly containing Int (irrespective of whether they are tabular). $\endgroup$
    – Emil Jeřábek
    Commented Jan 8 at 21:53

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