Free Heyting algebra endomorphisms: Ruitenburg’s Theorem and beyond

Representation Theorem for Heyting Lattices

The articles [11], [7], [13], [1], [14], [5], [6], [4], [9], [10], [15], [16], [12], [2], [3], and [8] provide the notation and terminology for this paper. Let us note that every lower bound lattice which is Heyting is also implicative and every lattice which is implicative is also upper-bounded. In the sequel T denotes a topological space and A, B denote subsets of the carrier of T . The follo...

The universal modality, the center of a Heyting algebra, and the Blok-Esakia theorem

We introduce the bimodal logic S4.Grzu, which is the extension of Bennett’s bimodal logic S4u by Grzegorczyk’s axiom ( (p→ p)→ p)→ p and show that the lattice of normal extensions of the intuitionistic modal logic WS5 is isomorphic to the lattice of normal extensions of S4.Grzu, thus generalizing the Blok–Esakia theorem. We also introduce the intuitionistic modal logicWS5.C, which is the extens...

Compact Endomorphisms of Certain Subalgebras of the Disk Algebra

Free Heyting Algebras: Revisited

We use coalgebraic methods to describe finitely generated free Heyting algebras. Heyting algebras are axiomatized by rank 0-1 axioms. In the process of constructing free Heyting algebras we first apply existing methods to weak Heyting algebras—the rank 1 reducts of Heyting algebras—and then adjust them to the mixed rank 0-1 axioms. On the negative side, our work shows that one cannot use arbitr...

Lattice of Substitutions is a Heyting Algebra

(6) For every finite element a of V→̇C holds {a} ∈ SubstitutionSet(V,C). (7) If A a B = A, then for every set a such that a ∈ A there exists a set b such that b ∈ B and b⊆ a. (8) If μ(A a B) = A, then for every set a such that a ∈ A there exists a set b such that b ∈ B and b⊆ a. (9) If for every set a such that a ∈ A there exists a set b such that b ∈ B and b ⊆ a, then μ(A a B) = A. Let V be a s...