Assume $\Gamma$ is an infinite set composed of formulas (of finite length), and $A$ is a formula (of finite length). For example, $\Gamma=\{x,y\sqcup z,x_1,x_2,x_3,\cdots\}$, $A=(x\sqcap y)\sqcup(x\sqcap z)$. Obviously, $\mathop⨅\Gamma\le A$ holds in any of the complete distributive lattices $\langle L,\le\rangle$ (i.e. $\mathop⨅\Gamma\le A$ holds for any $x,y,z,x_1,x_2,x_3,\cdots\in L$). Meanwhile, we also can find a finite subset $\{x,y\sqcup z\}\subseteq\Gamma$ such that $\mathop⨅\{x,y\sqcup z\}\le A$ holds in any of the complete distributive lattices $\langle L,\le\rangle$ (i.e.$\mathop⨅\{x,y\sqcup z\}\le A$ holds for any $x,y,z\in L$).
Let $\mathcal{C}$ be a kind of complete lattice (with extra operations). $\mathcal{C}$ is logic-compact iff $\mathcal{C}$ satisfies the following property:
For any infinite set $\Gamma$ composed of formulas (of finite length), for any formula $A$ (of finite length), if $\mathop⨅\Gamma\le A$ holds in any instance $\mathcal{L}$ of $\mathcal{C}$, then we can always find a finite subset $\Gamma_0\subseteq\Gamma$ such that $\mathop⨅\Gamma_0\le A$ holds in any instance $\mathcal{L}$ of $\mathcal{C}$. (The operators in the formulas mentioned above contain both those in the lattice and extra equipped.)
Another example: the infinite inequality $x\sqcap(x\to y)\sqcap x_1\sqcap x_2\sqcap x_3\cdots\le y$ holds in any of the Heyting algebras, which can be attributed to the finite inequality $x\sqcap(x\to y)\le y$ holding in any of the Heyting algebras. I'm wondering whether any infinite inequality holding in any of the algebra can be attributed to a finite inequality holding in any of the algebra.
So, my question is: Does any kind of complete lattice with any operation have logic-compactness? If not, what conditions do the complete lattice and the operations need to satisfy for having the property? (e.g., bounded, distributive, etc.)
