No such first-order theory exists. Suppose $T$ axiomatized the Heyting algebras having only finitely many <regular element of a Heyting algebra>[regular elements]. Expand the language by constants $c_n$ and add
$$
\neg\neg c_n=c_n,\qquad c_n\neq c_m\quad(n\neq m).
$$
Every finite subset has a model: take a sufficiently large finite Boolean algebra, in which every element is regular. By the <compactness theorem>, the entire expanded theory has a model. Its reduct is a model of $T$ with infinitely many distinct regular elements, contradicting the proposed axiomatization.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 120 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-120/3/c/solution