Hereditarily Structurally Complete Intermediate Logics: Citkin’s Theorem Via Duality

A deductive system is said to be structurally complete if its admissible rules are derivable. In addition, it is called hereditarily structurally complete if all its extensions are structurally complete. Citkin (1978) proved that an intermediate logic is hereditarily structurally complete if and only if the variety of Heyting algebras associated with it omits five finite algebras. Despite its importance in the theory of admissible rules, a direct proof of Citkin’s theorem is not widely accessible. In this paper we offer a self-contained proof of Citkin’s theorem, based on Esakia duality and the method of subframe formulas. As a corollary, we obtain a short proof of Citkin’s 2019 characterization of hereditarily structurally complete positive logics.


Introduction
A rule ρ is said to be admissible in a deductive system if the set of tautologies of is closed under the applications of ρ. On the other hand, a rule ρ is called derivable in if ρ belongs to the consequence relation of the system. 1 Clearly, every derivable rule is admissible. While the converse holds for classical propositional calculus CPC, it fails for many non-classical systems, including intuitionistic propositional calculus IPC.
This motivated the study of criteria for admissibility in modal and intermediate logics, undertaken by Rybakov and others [66]. As a consequence, the problem of finding bases for admissible rules was solved for IPC by Iemhoff [40][41][42], building on the work of Ghilardi [35,36] on unification, and independently by Rozière [63]. Later on, similar results have been obtained for modal and Lukasiewicz logics by Jeřábek [44][45][46], see also [54].
While the first step is an easy exercise, the second and the third require a nontrivial argument. Our strategy for them differs substantially from the previous ones and is based almost entirely on Esakia duality. Let be an intermediate logic such that K does not contain any C i . First, using subframe formulas (see, e.g., [17,Ch. 9] and [7]), we prove that extends the Kuznetsov-Gerčiu logic [33,49] of linear sums of one-generated Heyting algebras. When combined with the technique of universal models (see, e.g., [17,Ch. 8] and [10,Sec. 3.2]), this easily implies that is locally tabular. Furthermore, the connection with the Kuznetsov-Gerčiu logic allows to obtain a transparent description of the finite subdirectly irreducible members of K which, in turn, yields that they are weakly projective in K . Citkin's proof of the second and third steps is purely algebraic, while Rybakov's proof requires more complex arguments on universal models and relies on modal companions [26,51] and Fine's completeness theorem.
The paper is organized as follows. In Section 2 we introduce the main definitions of the paper. We also discuss our main proof strategy: The problem of characterizing hereditarily structurally complete intermediate logics is equivalent to that of describing non-trivial primitive varieties of Heyting algebras. In the rest of the paper we focus on the latter problem. In Section 3 we review the main tool of the paper, Esakia's duality for Heyting algebras. Building on Esakia duality, in Section 4 the description of finitely generated free Heyting algebras by means of universal models is recalled. In Section 5 we introduce Citkin's five finite algebras C 1 , . . . , C 5 , and show that these are omitted by any primitive variety of Heyting algebras (Lemma 5.1), thus proving one direction of Citkin's theorem. To prove the other direction, we shift the focus to varieties of Heyting algebras omitting C 1 , . . . , C 5 , which are investigated in Section 6 by means of subframe formulas. In particular, we show that these varieties are locally finite and we describe the structure of their finite subdirectly irreducible members (Theorem 6.13). Section 7 completes the proof of Citkin's theorem (Theorem 6.13 and Corollary 7.8). The obtained results and techniques are employed, in Section 8, to derive a new proof of Citkin's description of hereditarily structurally complete positive logics (Corollary 8.3). We conclude the paper by Section 9 where we review some important properties of hereditarily structurally complete intermediate and positive logics.

Hereditary Structural Completeness
Let F m be the set of formulas in countably many variables of some fixed, but arbitrary, algebraic language. A deductive system is a consequence relation , 3 defined over the set of formulas F m, that is substitution-invariant in the following sense: for every substitution σ and set of formulas Γ ∪ {ϕ} ⊆ F m, if Γ ϕ, then σ[Γ] σ(ϕ).
In addition, all the deductive systems considered in this paper will be assumed to be finitary, in the sense that for every set Γ ∪ {ϕ} ⊆ F m, if Γ ϕ, then there exists a finite set Δ ⊆ Γ such that Δ ϕ.
Let be a deductive system. A deductive system is said to be an extension of if for every set of formulas Γ ∪ {ϕ}, if Γ ϕ, then Γ ϕ.
A rule is an expression of the form Γ ϕ where Γ ∪ {ϕ} is a finite subset of F m. Let be a deductive system. A rule Γ ϕ is said to be admissible in if for all substitutions σ: if ∅ σ(γ) for all γ ∈ Γ, then ∅ σ(ϕ).
Similarly, a rule Γ ϕ is said to be derivable in if Γ ϕ. Accordingly, we say that 1. is structurally complete if every rule that is admissible in is also derivable in .
2. is hereditarily structurally complete if every extension of is structurally complete.
For further variants of structural completeness, we refer the reader to [27,53,55,69]. Under certain assumptions, hereditary structural completeness can be formulated in purely algebraic terms [3,53,60]. To explain how this could be done, it is convenient to recall some basic definitions from universal algebra [4,15]. We denote by I, H, S, P, P u the class operators of closure under isomorphism, homomorphic images, subalgebras, direct products, and ultraproducts, respectively. We assume direct products and ultraproducts of empty families of algebras are trivial algebras. A variety is a class of algebras axiomatized by equations or, equivalently, a class of algebras closed under H, S, and P. A quasi-variety is a class of algebras axiomatized by quasi-equations or, equivalently, a class of algebras closed under I, S, P, and P u . As a consequence, every variety is a quasi-variety, while the converse is not true in general. Given a class of algebras K, we denote by V(K) and Q(K), respectively, the least variety and quasi-variety containing K. It is well known that V(K) = HSP(K) and Q(K) = ISPP u (K). When K is a variety, we say that a class M ⊆ K is a subvariety (resp. subquasi-variety) of K if M is a variety (resp. a quasi-variety). Then a variety K is said to be primitive if every subquasi-variety of K is a variety. When a deductive system is algebraized by a variety K in the sense of [14], the lattice of axiomatic extensions of is dually isomorphic to that of subvarieties of K. In addition, an axiomatic extension of is hereditarily structurally complete if and only if the subvariety of K corresponding to is primitive [60, Thm. 6.12 (2)], see also [3,Prop. 2.4]. Consequently, in this case the task of characterizing hereditarily structurally complete axiomatic extensions of is equivalent to that of characterizing primitive subvarieties of K.
A special instance of this phenomenon is given by intermediate logics, i.e., axiomatic extensions of intuitionistic propositional logic IPC. This is because IPC is algebraized by the variety of Heyting algebras, i.e., algebras of the form A = A; ∧, ∨, →, 0, 1 where A; ∧, ∨, 0, 1 is a bounded lattice with minimum 0 and maximum 1 such that for every a, b, c ∈ A, Thus the task of characterizing hereditarily structurally complete intermediate logics can be rephrased in purely algebraic terms as that of describing primitive varieties of Heyting algebras. This is what we do in the rest of the paper.
To this end, we rely on some basic observation. Let K be a variety. An algebra A ∈ K is said to be weakly projective in K if for every B ∈ K, if A ∈ H(B), then A ∈ IS(B). 4 Moreover, an algebra A is said to be finitely subdirectly irreducible, FSI for short, when the identity relation is meetirreducible in the congruence lattice of A. The  Proof. Consider a finite nontrivial FSI algebra A ∈ K. Then let B ∈ K be such that A ∈ H(B). Since K is primitive, all its subquasi-varieties are varieties, whence A ∈ H(B) ⊆ V(B) = Q(B). Now, it is well known that all FSI members of Q(B) belong to ISP u (B) [23,Lem. 1.5]. Thus A ∈ ISP u (B). Since A is finite and nontrivial, and the type of K is finite, this yields A ∈ IS(B). We conclude that A is weakly projective in K.
A variety is said to be locally finite when its finitely generated members are finite. We also rely on the following observation [38,Prop. 5.1.24], see also [37].

Esakia Duality
The study of Heyting algebras is simplified by their topological representation, known as Esakia duality [29,30], which we will briefly recall here. Given a poset X; ≤ and a set U ⊆ X, the smallest upset and downset containing U are denoted respectively by ↑U and ↓U . In case U = {x}, we will write ↑x and ↓x instead of ↑{x} and ↓{x}, respectively. Then an Esakia space X = X; τ, ≤ comprises a zero-dimensional compact Hausdorff space X; τ and a poset X; ≤ such that (i) ↑x is closed for all x ∈ X, and (ii) ↓U is clopen, for every clopen U ⊆ X.
Observe that the topology of finite Esakia spaces is necessarily discrete (because they are Hausdorff), and that finite posets endowed with the discrete topology are Esakia spaces. We will make a systematic use of this observation, since most Esakia spaces considered in this paper will be finite.
For Esakia spaces X and Y , an Esakia morphism f : X → Y is a continuous order-preserving map f : X → Y such that for all x ∈ X and y ∈ Y , if f (x) ≤ y, then there is z ∈ X such that x ≤ z and f (z) = y.
Esakia duality states that the category ESP of Esakia spaces endowed with Esakia morphisms is dually equivalent to the category HA of Heyting algebras and Heyting algebra homomorphisms [30,Thm. 3.4.4].
The dual equivalence functors are defined as follows. Given a Heyting algebra A, we denote the set of its (non-empty proper) prime filters of A by PrA, and set γ A (a):={F ∈ PrA : a ∈ F } (2) for every a ∈ A. It turns out that the structure A * := PrA; τ, ⊆ is an Esakia space, where τ is the topology on PrA with subbasis {γ A (a) : a ∈ A} ∪ {γ A (a) c : a ∈ A}. Moreover, for every Heyting algebra homomorphism f : A → B, let f * : B * → A * be the Esakia morphism defined by the rule Conversely, let X be an Esakia space. We denote by CupX the set of clopen upsets of X. Then the structure X * := CupX; ∩, ∪, →, ∅, X , where U → V :=X ↓(U V ), is a Heyting algebra. Moreover, for every Esakia morphism f : X → Y , let f * : Y * → X * be the homomorphism of Heyting algebras given by the rule Observe that the dual equivalence functors preserve finiteness.
Let X be an Esakia space. An Esakia subspace (E-subspace for short) of X is a closed upset of X, equipped with the subspace topology and the restriction of the order. For every x ∈ X, the upset ↑x endowed with the subspace topology is easily seen to be an E-subspace of X.
A bisimulation equivalence on X is an equivalence relation R on X such that for every x, y, z ∈ X, (i) if x, y ∈ R and x ≤ z, then there is w ∈ ↑y such that z, w ∈ R, and (ii) if x, y / ∈ R, then there is a clopen U such that x ∈ U and y / ∈ U , which in addition is a union of equivalence classes of R.
In this case, we denote by X/R the Esakia space consisting of the quotient space of X with respect to R, equipped with the partial order ≤ X /R defined as follows for every x, y ∈ X: x/R ≤ X /R y/R ⇐⇒ there are x , y ∈ X such that x, x , y, y ∈ R and x ≤ X y .
The map x → x/R for every x ∈ X is an Esakia morphism from X to X/R, and the kernel of f is a bisimulation equivalence on X for every Esakia morphism f : X → Y . If, moreover, f is surjective, then X/ ker f ∼ = Y .
Remark 3.1. Observe that condition (i) in the definition of a bisimulation equivalence is equivalent to the requirement that for every x, y, z ∈ X with x, y ∈ R, x, z / ∈ R, x = y, and x ≤ z, there is y ≤ w ∈ X such that z, w ∈ R. We rely on this observation without further notice.
The disjoint union X 1 · · · X n of finitely many Esakia spaces X 1 , . . . , X n is their order-disjoint and topologically disjoint union, which is also an Esakia space.

(i) A is non-trivial and FSI if and only if its top element is prime (i.e., if
x ∨ y = 1 then x = 1 or y = 1), or, equivalently, the poset underlying A * is rooted (i.e., it has a least element).
(ii) There is a dual lattice isomorphism σ from the congruence lattice of A to that of E-subspaces of A * , such that (A/θ) * ∼ = σ(θ) for any congruence θ of A, and for any E-subspace Y of A * , we have that (iii) There is a dual lattice isomorphism ρ from the lattice of subalgebras of A to that of bisimulation equivalences on A * , such that if B is a subalgebra of A then B * ∼ = A * /ρ(B), and if R is a bisimulation equivalence on A * then (A * /R) * ∼ = ρ −1 (R).
Remark 3.3. Proofs in this paper would often require the reader to check whether there exists a surjective Esakia morphism between two given finite Esakia spaces. To simplify this task, we will recall a general criterion. Let X be a finite Esakia space and x, y ∈ X.
1. Suppose that y is the only immediate successor of x. Then let R be the least equivalence relation on X such that x, y ∈ R. Observe that R is a bisimulation equivalence on X. The natural map f : X → X/R is called an α-reduction.
2. Suppose that the set of immediate successors of x and y coincide. Then the least equivalence relation R on X such that x, y ∈ R is a a bisimulation equivalence on X, and the natural map f : X → X/R is called a βreduction.
Now, let X and Y be finite Esakia spaces. In [10,Lem. 3.1.7] it is shown that there exists a surjective Esakia morphism f : X → Y if and only if there exists a finite sequence f 1 , . . . , f n of α or β-reductions f i : Z i → Z i+1 such that Z 1 = X and Z n+1 ∼ = Y . In other words, in order to determine whether there exists a surjective Esakia morphism from X to Y , it suffices to check whether X can be "transformed" into Y by means of α and β-reductions.

Universal Models
Even if finitely generated free Heyting algebras are not fully understood, major insights in their dual structure were provided by [2,39,64,67], see also [11,24,32,34]. Our presentation is reminiscent of [10] and [17]. Given 1 ≤ n ∈ ω and a poset X; ≤ , an element x ∈ X is said to have depth n if the upset ↑x contains at least one chain of length n, and no chain of length n + 1. Moreover, a finite sequence of zeros and ones is said to be a colour. Given two colours of the same length a = a 1 , . . . , a n and c = c 1 , . . . , c n , we set a ≤ c ⇐⇒ a i ≤ c i for every i = 1, . . . , n, and a < c ⇐⇒ a ≤ c and a i < c i for some i = 1, . . . , n.
Accordingly, when we write a ≤ c or a < c, it should be understood that the colours a and c have the same length.
For every n ∈ ω, we will define a poset U(n) = U (n); ≤ as the union of a chain of posets {D m : 1 ≤ m ∈ ω}. To this end, observe that there are exactly 2 n distinct colours of length n. Then let D 1 be a set of 2 n elements painted with distinct colours of length n, and D 1 = D 1 ; ≤ 1 the poset obtained equipping D 1 with the discrete partial order. Moreover, if D m has already been defined, then let D m+1 be the poset obtained extending D m in accordance to the following rules: (i) For every point x of D m of depth m and of colour a, and every colour c < a, we add to D m a unique point y labeled by c such that (ii) For every antichain Z in D m such that |Z| ≥ 2 containing at least one point of depth m, and every colour c such that c ≤ a for every colour a of some element in Z, we add to D m a unique point y labeled by c such that It is clear that D m is a subposet of D m+1 for every 1 ≤ m ∈ ω, whence it makes sense to define U(n) as the union of the chain {D m : 1 ≤ m ∈ ω}. The importance of the poset U(n) is captured by the following observation:  (ii) If x ∈ F (n) * , then either x has finite depth or for every 1 ≤ n ∈ ω there is an element y ∈ F (n) * of depth n such that x ≤ y.
(iii) For all m ∈ ω, the poset U(n) has only finitely many points of depth ≤ m.
The statements of (i) and (ii) are [10, Thms. 3.2.9 and 3.1.10(4)], which in turn follow from Kuznetsov's theorem [48] (see also [18], [8,Lem  Let n ∈ ω, and let F (n) be the free n-generated Heyting algebra. If X is an infinite E-subspace of F (n) * , then X contains an element of depth m for every 1 ≤ m ∈ ω.
Proof. Consider an infinite E-subspace X of F (n) * and suppose, with a view to contradiction, that X does not contain any element of depth m for some 1 ≤ m ∈ ω. We have two cases: either X contains an element of infinite depth or not. If X contains an element of infinite depth, then we obtain a contradiction because of condition (ii) of Theorem 4.1. Then all elements of X must have finite depth and, therefore, depth < m. As X is infinite, this means that X has infinitely many elements of depth < m. Moreover, since X is an E-subspace of F (n) * , the same holds for F (n) * . But this contradicts conditions (i) and (iii) of Theorem 4.1. Thus we have arrived at a contradiction.
In the rest of the paper we will rely on the following observation, which follows from [20,Lem. 18] or, alternatively, can be deduced from Kuznetsov's theorem [48]. The proof supplied below differs from that of [20], however, as it uses duality and universal models. Theorem 4.3. Let K be a variety of Heyting algebras. Then K is locally finite if and only if K has, up to isomorphism, only finitely many finite n-generated FSI members, for every n ∈ ω.
Proof. The "only if" part is straightforward. To prove the "if" part, we reason by contrapostion: suppose that K is not locally finite. Then there is some n ∈ ω and an n-generated infinite algebra A ∈ K. Clearly A is a homomorphic image of the free n-generated Heyting algebra F (n), whence A * can be identified with an E-subspace of F (n) * in the light of condition (ii) of Lemma 3.2. Moreover, the fact that A is infinite guarantees that so is A * . As a consequence, we can apply Corollary 4.2, obtaining that for every Moreover, observe that the size of the spaces {↑ A * x m : 1 ≤ m ∈ ω} is not bounded by any natural number, as each x m has depth m. As a consequence, also the cardinality of the algebras {A m : 1 ≤ m ∈ ω} cannot be bounded by any natural number. Since the algebras A m are finite, we conclude that there must an infinite subset C ⊆ {A m : 1 ≤ m ∈ ω} of pairwise nonisomorphic algebras. Thus C is an infinite set of pairwise nonisomorphic finite n-generated FSI members of K.

Citkin's Five Algebras
Consider the following FSI Heyting algebras: The following result relates primitive varieties of Heyting algebras with the algebras C 1 , . . . , C 5 . Proof. Suppose, with a view to contradiction, that K is a primitive variety of Heyting algebras containing some algebra in {C 1 , . . . , C 5 }. Consider the following Esakia spaces X 1 , . . . , X 5 endowed with the discrete topology: (3) Moreover, by inspection one sees that for each C i * there is a bisimulation equivalence R i on the disjoint union On the other hand, it is not hard to check that there is no surjective Esakia morphism from X i to C i * . By Lemma 3.2(iii) this implies Now, by assumption there is some i = 1, . . . , 5 such that C i ∈ K. By (4) also X * i ∈ K. Moreover, by (3) and (5) we have C i ∈ H(X * i ) and C i / ∈ IS(X * i ). As a consequence, we conclude that C i is not weakly projective in K. Since C i is a finite nontrivial FSI member of K and K is primitive, this contradicts Lemma 2.1. Hence we reached a contradiction.

A Structure Theorem
In this section we give a description of the structure of varieties of Heyting algebras omitting C 1 , . . . , C 5 (Theorem 6.13). To this end, recall that the Rieger-Nishimura lattice RN (depicted below) is the free one-generated Heyting algebra [57,61,62]. As a consequence, H(RN ) is the class of all one-generated Heyting algebras.
Let A and B be Heyting algebras. The sum A+B is the Heyting algebra obtained by pasting B below A and gluing the top element of B to the bottom element of A. As + is clearly associative, there is no ambiguity in writing A 1 + · · · + A n for the descending chain of finitely many Heyting algebras A 1 , . . . , A n , each glued to the previous one.
Then the Kuznetsov-Gerčiu variety is defined as follows: The variety KG was introduced in the study of finite axiomatizability, and of the finite model property in varieties of Heyting algebras [33,49] (see also [6,10,56]). We will see that varieties of Heyting algebras omitting C 1 , . . . , C 5 are subvarieties of KG (Theorem 6.13).
To this end, it is convenient to recall some basic concept. In [70], every finite rooted Esakia space Z is associated with a formula β(Z) in the language of Heyting algebras, called the subframe formula of Z (see also [7,12,17]). For the present purpose, the way in which subframe formulas are concretely defined is immaterial and, to explain their importance, it is sufficient to recall the following definition.
and for every clopen U of Y , the downset generated by U with respect to ≤ X is clopen in X. The following result clarifies the role of subframe formulas [12, Thm. 3.13]: Theorem 6.1. Let X and Z be Esakia spaces such that Z is finite and rooted. Then X * |= β(Z) ≈ 1 if and only if Z is not the image of an Esakia morphism, whose domain is a subspace of X.
Remark 6.2. Recall that finite Esakia spaces coincide with finite posets endowed with the discrete topology. Thus if X is a finite Esakia space, then the above theorem specializes as follows: X * |= β(Z) ≈ 1 if and only if Z is not the image of an Esakia morphism, whose domain is a subposet of X.
For the present purpose, the interest in subframe formulas is that they provide a convenient axiomatization of KG. To explain how this is obtained, consider the discrete rooted Esakia spaces P 1 , P 2 , and P 3 whose underlying posets are depicted below: The proof of the following result can be found in [10,Thm. 4.3.4] (see also [6,47]): KG is the variety of Heyting algebras axiomatized by the equations Given a positive integer n, a poset X; ≤ has width ≤ n if there is no x ∈ X such that ↑x contains an antichain of n + 1 elements. Accordingly, a Heyting algebra A is said to have width ≤ n when so does the poset underlying A * . Lemma 6.4. Let K be a variety of Heyting algebras omitting C 1 , . . . , C 5 . Then every finite member of K has width ≤ 2 and satisfies β(P 1 ) ≈ 1.
Proof. Suppose, with a view to contradiction, that there is a finite A ∈ K of width > 2. Then A * contains a subposet isomorphic to P 1 . We label its elements as follows: As H(K) ⊆ K and A * is finite, by Lemma 3.2(ii) we may assume without loss of generality that the following holds: Fact 6.5.
(i) ⊥ is the minimum of A * and the unique common lower bound of x, y, z.
(ii) {x, y, z} in A * is the unique three-element antichain in ↓{x, y, z}.
Then consider the following relation on A * : Bearing in mind that A * is finite and, therefore, endowed with the discrete topology, it is easy to see that R is a bisimulation equivalence on A * . Accordingly, we consider the Esakia space A * /R. In the light of Lemma 3.2(iii), we obtain (A * /R) * ∈ IS(A) ⊆ K. Therefore, we may assume without loss of generality that R is the identity relation (otherwise, we replace A by (A * /R) * in the proof). Observe that R identifies everything in A * ↓{x, y, z}. Thus the assumption that R is the identity on A * means that A * contains at most one element not in ↓{x, y, z}. Denoting by Y the subposet ↓{x, y, z} of A * , we obtain the following: Fact 6.6. There is an Esakia space X such that X * ∈ K and one of the following holds: Proof. Suppose that conditions (i) and (ii) fail. Then, in particular, A * = ↓{x, y, z}, otherwise A * would satisfy condition (i). Consequently, We will see that is comparable with some element among x, y, and z. Suppose the contrary, with a view to contradiction. Then the least equivalence relation S on A * that identifies with x is easily seen to be a bisimulation equivalence on A * . Moreover, the poset underlying A * /S is isomorphic to Y . As by Lemma 3.2(iii), (A * /S) * ∈ IS(A) ⊆ K, taking X:=A * /S we would obtain that condition (i) holds, which is false. Thus we conclude that is comparable with some element among x, y, z, as desired. We may assume without loss of generality that this element is x. Since / ∈ Y , this implies x < .
An argument analogous to the one described above shows that the assumption that y and z leads to a contradiction. Then we can assume without loss of generality that y ≤ and, therefore, y < (as / ∈ Y ). Finally, if z , then condition (ii) holds, contradicting the assumption. Then we conclude that z ≤ , whence is the maximum of A * . Thus taking X:=A * , we obtain that condition (iii) holds, as desired.
Fact 6.7. The following relation is a bisimulation equivalence on X: Proof. First observe that S is an equivalence relation. Then it only remains to show that S satisfies conditions (i) and (ii) in the definition of a bisimulation equivalence. Since X is finite, its topology is discrete, whence condition (ii) is obviously satisfied. To prove condition (i), consider three elements t, u, v ∈ X such that t, u ∈ S, t, v / ∈ S, t = u, and t ≤ v. We need to find some u ≤ w ∈ X such that v, w ∈ S. Clearly First consider the case in which {x, y, z}∩↑v = ∅. From Fact 6.6 it follows If condition (iii) of Fact 6.6 holds, then, by taking w:= , we are done. Now suppose that condition (iii) of Fact 6.6 fails. Together with the fact that X = { } ∪ Y and / ∈ Y , this implies that condition (ii) of Fact 6.6 holds. Thus we may assume without loss of generality that x, y < and z . Since t = v = , clearly t ∈ Y . Now, if t ∈ ↓{x, y}, then also u ∈ ↓{x, y} (as t, u ∈ S). Consequently, u ≤ = v and, by taking w:=v, we are done. Next we consider the case where t / ∈ ↓{x, y}. We will see that this case leads to a contradiction. To this end, observe that in this case t ≤ z, as t ∈ Y = ↓{x, y, z} and t / ∈ ↓{x, y}. Moreover, since t ≤ v = and z , we obtain t < z. But the fact that t x, y and t < z implies that {x, y, t} is a three-element antichain in Y different from {x, y, z}, contradicting Fact 6.5(ii).
If {x, y, z} ∩ ↑v = {x}, then v, x ∈ S. Moreover, from t ≤ v ≤ x and t, u ∈ S it follows that u ≤ x. Thus, by setting w:=x, we are done.
A similar argument works if v ∩ ↑{x, y, z} is {y} or {z} (take respectively w:=y or w:=z).
By (7) (8), this guarantees that t ≤ x, y, z, whence also u ≤ x, y, z as u, t ∈ S. By Fact 6.5(i), we have that ⊥ is the unique common lower bound of x, y, z, whence t = ⊥ = u, contradicting the fact that t = u. Thus we conclude that S is a bisimulation equivalence on X.
Recall that X * ∈ K and that S is a bisimulation equivalence on X by Facts 6.6 and 6.7. Thus by Lemma 3.2(iii), we have that (X/S) * ∈ IS(X * ) ⊆ K. Accordingly, we may assume without loss of generality that S is the identity relation on X.
Bearing this in mind, if case (i) of Fact 6.6 holds, then the poset underlying X is one of the rooted posets depicted below (in which the elements other than ⊥, x, y, and z are marked with squares): Observe that Z 1 ∼ = C 3 * and Z 3 ∼ = C 5 * . Moreover, there are bisimulation equivalences T and T , respectively on Z 2 and Z 4 , such that Z 2 /T ∼ = C 1 * and Z 4 /T ∼ = C 4 * . By Lemma 3.2(iii), this implies that IS( contradicting the assumption that K omits C 1 , C 3 , C 4 , C 5 . Thus we conclude that case (i) of Fact 6.6 cannot hold. Now, suppose that case (ii) of Fact 6.6 holds. Since S is the identity, the poset underlying X is one of the rooted posets depicted below (in which the elements other than ⊥, x, y, z, and are marked with squares): E E E y y y y y y y • y y y y y y y • y y y y y y y • y y y y y y y For every i = 1, . . . , 6 there is a bisimulation equivalence T i on Z i such that contradicting the assumption that K omits C 1 , C 2 , C 4 . Thus we conclude that also case (ii) of Fact 6.6 cannot hold. Thus condition (iii) of Fact 6.6 holds necessarily. Since S is the identity, the poset underlying X is one of the rooted posets depicted below (in which the elements other than ⊥, x, y, z, and are marked with squares): Observe that Z 1 ∼ = C 4 * . Moreover, for every i = 2, 3, 4 there is a bisimulation equivalence T i on Z i such that Z 2 /T 2 ∼ = Z 3 /T 3 ∼ = C 2 * and Z 4 /T 4 ∼ = C 4 * . As in the previous cases, this implies K ∩ {C 2 , C 4 } = ∅, contradicting the assumption that K omits C 2 and C 4 . Thus we reached the desired contradiction. As a consequence, A has width ≤ 2. This immediately implies that finite members of K validate β(P 1 ) ≈ 1. Proof. Suppose, with a view to contradiction, that there is a finite algebra A ∈ K in which the equation β(P 2 ) ≈ 1 fails. By Theorem 6.1 there is a subframe X of A * and a surjective Esakia morphism from X to the space obtained endowing P 2 with the discrete topology. Because of the definition of an Esakia morphism, this implies that there is a subposet of A * isomorphic to P 2 . We label the elements of this a copy of P 2 inside A * as follows: Moreover, by Lemma 3.2(ii) and H(K) ⊆ K, we may assume without loss of generality that ⊥ is the minimum of A * and the unique common lower bound of x and y . In addition, as in the proof of Lemma 6.4, we may assume without loss of generality that A * contains at most one element not in ↓{x, y}. By Lemma 6.4 we know that A * has depth ≤ 2, whence, provided that exists, it must be comparable either with x or y. As x, y, this implies that either does not exist or x < or y < . Consequently, we deduce: Fact 6.9. One of the following conditions holds: (ii) A * has a maximum and A * = { } ∪ ↓{x, y}; or (iii) A * has a maximal element strictly above exactly one between x and y, and A * = { } ∪ ↓{x, y}.
Observe that in case (iii) if x < (resp. y < ), then y (resp. x ), and x and y are incomparable.
Our aim is to show that conditions (i), (ii), and (iii) lead to a contradiction. First suppose that condition (i) holds. We will see that for all z ∈ A * , {x, x , y, y } ∩ ↑z ∈ {{x}, {y}, {x, x }, {y, y }, {x, x , y}, {y, y , x}, {x, x , y, y }}. (9) To prove this, consider z ∈ A * . Clearly {x, x , y, y } ∩ ↑z is an upset of the copy of P 2 in A * given by {⊥, x, x , y, y }. Moreover, this upset must be non-empty by assumption (i), since z ∈ ↓{x, y}. Thus, in order to establish the above display, it suffices to show that {x, x , y, y }∩↑z = {x, y}. Suppose the contrary with a view to contradiction. Then observe that ⊥ ≤ z, x , y and recall that x and y are incomparable. Since A * has width ≤ 2 by Lemma 6.4, this implies that z is comparable either with x or with y . We may assume without loss of generality that z is comparable with x . Since {x, x , y, y } ∩ ↑z = {x, y}, this implies x < z and, therefore, x ≤ z ≤ y . But this contradicts the fact that x y, whence establishing (9).
Then we will see that the following relation is a bisimulation equivalence on A * : As before, it suffices to show that condition (i) in the definition of a bisimulation equivalence holds. To this end, consider t, u, v ∈ A * such that t, u ∈ S, t = u, t < v, and t, v / ∈ S. We need to find an element w ≥ u such that v, w ∈ S. By (9) {x, x , y, y }∩↑v ∈ {{x}, {y}, {x, x }, {y, y }, {x, x , y},{y, y , x},{x, x , y,  But an argument analogous to the one detailed in the last paragraph of the proof of Fact 6.7 shows that this case leads to a contradiction. Hence we conclude the S is a bisimulation equivalence on A * .
In particular, by Lemma 3.2(iii) this implies (A * /S) * ∈ K. Consequently, we may assume without loss of generality that S is the identity relation on A * . Together with (9) and the fact that {⊥, x, x , y, y } forms a subposet of A * isomorphic to P 2 , this implies that A * is isomorphic to one of the following rooted posets (in which the elements other than ⊥, x, x , y, y are marked with squares): Observe that C 1 * is isomorphic to an E-subspace of Z 2 and Z 3 . Moreover, there is a bisimulation equivalence T on Z 1 such that Z 1 /T ∼ = C 1 * . By Lemma 3.2(ii, iii) this implies C 1 ∈ H(A) ∪ IS(A) ⊆ K, contradicting the assumption that C 1 / ∈ K. Thus condition (i) cannot hold.
Next we consider the case where condition (ii) holds. An argument analogous to the one detailed for case (i) shows that for every z ∈ A * , We will see that the following relation is a bisimulation equivalence on A * : To this end, consider t, u, v ∈ A * such that t, u ∈ S, t = u, t < v, and t, v / ∈ S. We need to find an element w ≥ u such that v, w ∈ S. First we consider the case where v = . In this case, v ∈ ↓{x, y} by assumption (ii). As t ≤ v, we also get t ∈ ↓{x, y}. In turn, this guarantees that u ∈ ↓{x, y}, since t, u ∈ S. Consequently, t, u, v ∈ ↓{x, y}. This allows us to repeat the argument detailed in the case of condition (i), obtaining the desired element w. Then it only remains to consider the case where v = . But assumption (ii) guarantees that u ≤ = v. Thus, by setting w:= , we are done. This establishes that S is a bisimulation equivalence on A * .
Consequently, we may assume without loss of generality that S is the identity relation on A * . Together with (10) and the fact that {⊥, x, x , y, y , } forms a subposet of A * isomorphic to P 2 plus a new top element, this implies that A * is isomorphic to one of the following rooted posets (in which the elements other than ⊥, x, x , y, y , are marked with squares): Observe that C 2 * is isomorphic to an E-subspace of Z 2 and Z 3 . Moreover, there is a bisimulation equivalence T on Z 1 such that Z 1 /T ∼ = C 2 * . By Lemma 3.2(ii, iii) this implies C 2 ∈ H(A) ∪ IS(A) ⊆ K, contradicting the assumption that C 2 / ∈ K. Thus also condition (ii) cannot hold.
Consequently, by Fact 6.9 condition (iii) holds. We may assume without loss of generality that > x and y. Observe that for every z ∈ A * , This is an immediate consequence of the fact that {x, x , y } ∩ ↑z must be an upset of the subposet of A * with universe {x, x , y }.
We will see that the following relation is a bisimulation equivalence on A * : To prove this, consider t, u, v ∈ A * such that t, u ∈ S, t = u, t < v, and t, v / ∈ S. As usual, we need to find an element w ≥ u such that v, w ∈ S. First we consider the case where {x, x , y } ∩ ↑v = ∅. Observe that {x, x , y } ∩ ↑t = ∅, since t, v / ∈ S. Thus either t ≤ x or t ≤ y . As t, u ∈ S, this implies that either u ≤ x or u ≤ y . Consequently, either But an argument analogous to the one detailed in the last paragraph of the proof of Fact 6.7 shows that this case leads to a contradiction. Hence we conclude the S is a bisimulation equivalence on A * .
Consequently, we may assume without loss of generality that S is the identity relation on A * . Observe that the subposet of A * with universe {⊥, x, x , y , } is isomorphic to one of the following rooted posets: For every i = 1, . . . , 4 there is a bisimulation equivalence T i on Z i such that But this contradicts the fact that K omits C 1 and C 2 . Hence we reached the desired contradiction. Proof. Suppose, with a view to contradiction, that there is a finite algebra A ∈ K in which the equation β(P 3 ) ≈ 1 fails. By Theorem 6.1 there is a subframe X of A * and a surjective Esakia morphism from X to the space obtained endowing P 3 with the discrete topology. Because of the definition of an Esakia morphism, this implies that there is a subposet of A * isomorphic to P 3 . We label the elements of this copy of P 3 inside A * as follows: Moreover, by Lemma 3.2(ii) and H(K) ⊆ K, we may assume without loss of generality that ⊥ is the minimum of A * and the unique common lower bound of x 3 and y. First observe that for every z ∈ A * , This is an immediate consequence of the fact that {x 2 , x 3 , y}∩ ↑z is an upset of the subposet of A * with universe {x 2 , x 3 , y}. Now, we will see that the following relation is a bisimulation equivalence on A * : To this end, consider t, u, v ∈ A * such that t, u ∈ S, t = u, t < v, and t, v / ∈ S. As usual, we need to find an element w ≥ u such that v, w ∈ S. First we consider the case where {x 2 , x 3 , y} ∩ ↑v = ∅. If t ≤ x 2 , then also u ≤ x 2 ≤ x 1 (as t, u ∈ S). Consequently, by setting w:=x 1 , we are done. Then suppose that t x 2 . Since t, v / ∈ S and {x 2 , x 3 , y} ∩ ↑v = ∅, we get {x 2 , x 3 , y} ∩ ↑t = ∅. This implies that either t ≤ x 2 or t ≤ y. As t x 2 , we conclude t ≤ y. First consider the case where t = y. Then y = t < v. As t, u , we have u ≤ y ≤ v. Thus, by taking w:=v, we are done. We will see that the case where t < y never happens. To this end, suppose the contrary, with a view to contradiction. We will see that As t x 2 , clearly t = ⊥, whence ⊥ < t. Consequently, ⊥ < t < y. Now, as t x 2 , we have t x 2 , x 3 . Moreover, since t ≤ y and x 2 , x 3 y, clearly x 2 , x 3 t. Thus t is incomparable with x 2 and x 3 . This establishes (13). Then {⊥, x 2 , x 3 , t, y} is the universe of a subposet of A * isomorphic to P 2 . Thus A does not satisfy β(P 2 ). But this contradicts Lemma 6.8. Hence we reached a contradiction, as desired. This concludes the analysis of the case where {x 2 , x 3 , y} ∩ ↑v = ∅.
If {x 2 , x 3 , y} ∩ ↑v is equal to {x 2 }, {x 2 , x 3 }, or {y}, then, by taking respectively w:=x 2 , w:=x 3 or w:=y, we are done. In the light of (12), the only case that remains to be considered is the one where But an argument analogous to the one detailed in the last paragraph of the proof of Fact 6.7 shows that this case leads to a contradiction. Hence we conclude the S is a bisimulation equivalence on A * .
Consequently, we may assume without loss of generality that S is the identity relation on A * . Now, either y is a maximal element of A * or it is not. If y is a maximal element of A * , then the fact that S is the identity relation and condition (12) imply that A * is isomorphic to one of the following rooted posets: ? ? •y In both cases, C 1 ∈ IS(A) ⊆ K by Lemma 3.2(iii). But this contradicts the assumption that K omits C 1 .
We conclude that y is not a maximal element of A * . Together with the fact that S is the identity relation and condition (12), this implies that A * is isomorphic to one of the following rooted posets: ? ? •y In both cases, C 2 ∈ IS(A) ⊆ K by Lemma 3.2(iii). But this contradicts the assumption that K omits C 2 . Hence we reached a contradiction.
Corollary 6.11. Let K be a variety of Heyting algebras omitting C 1 , . . . , C 5 . Then the finite members of K belong to KG.
We rely on the following observation, which specializes 5 [10, Cor. 4.3.10]: Lemma 6.12. If A ∈ KG is a nontrivial finite FSI algebra, then A = B 1 + · · · + B n for some Heyting algebras B 1 , . . . , B n ∈ H(RN ) such that B 1 is the two-element Boolean algebra.
Let 2 and 4 be, respectively, the two and four-element Boolean algebras. Moreover, let D be the Heyting algebra depicted below: (ii) Every nontrivial finite FSI member of K has the form B 1 + · · · + B n for some Heyting algebras B i such that B 1 ∼ = 2, and if n > 1, then B n ∈ I{2, 4, D} and B j ∈ I{2, 4} for all 1 < j < n.
Moreover, the above conditions hold for every primitive variety K of Heyting algebras.
Proof. (ii): Let A be a finite nontrivial FSI member of K. By Corollary 6.11 and Lemma 6.12 we obtain that A = B 1 +· · ·+B n for some finite nontrivial B 1 , . . . , B n ∈ H(RN ) such that B 1 = 2. We may assume without loss of generality that no B i can be written as a sum with at least two nontrivial components.
Suppose with a view to contradiction that n > 1 and B n / ∈ I{2, 4, D}. Then observe Now, recall that B n is a finite nontrivial member of H(RN ). To visualize B n , it is convenient to observe that the order-type of finite homomorphic images of RN is that of principal nontotal downsets of RN . Since B n cannot be written as a sum with at least two nontrivial components, this implies that the order type of B n is that of ↓ RN a for some a ∈ RN {0} which is not prime (i.e., it is not the join of two strictly smaller elements).
If ↓ RN a has at least 8 elements, the assumption that a is not prime guarantees that C 1 ∈ H(B n ), whence C 1 ∈ H(B n ) ⊆ H(2 + B n ) ⊆ K, a contradiction. Then we consider the case where ↓ RN a has less than 8 elements. Since a is not prime and B n is nontrivial, this implies that B n ∈ I{2, 4, D}, which is also a contradiction. Hence, we conclude that if n > 1, then B n ∈ I{2, 4, D}.
As above, the order type of B j is that of ↓ RN a for some a ∈ RN {0} which is not prime. Together with the fact that B j is nonisomorphic to 2, 4 and inspecting the structure of RN , this yields C 2 ∈ S(2 + B j + 2), whence C 2 ∈ K which is false. Hence we conclude that if 1 < j < n, then B j ∈ I{2, 4}.
(i): Using the layer-structure given by condition (ii) and the fact that an n-generated Heyting algebra cannot be a sum of more than 2n + 1 nontrivial algebras, it is not hard to see that for every n ∈ ω there are, up to isomorphism, only finitely many n-generated finite FSI algebras in K. By Theorem 4.3 we conclude that K is locally finite. Now, Corollary 6.11 guarantees that the finite members of K belong to KG. As K is locally finite and, therefore, generated by its finite members, this implies that K ⊆ KG.
Finally, the fact that conditions (i) and (ii) hold for all primitive varieties of Heyting algebras is a consequence of Lemma 5.1.
Remark 6.14. Theorem 6.13(i) is also a consequence of Corollary 6.11 and a general criteria of local finiteness in subarieties of KG [10, Thm. 4.6.5] stating that a subvariety of KG is locally finite if and only if it omits RN + 2. We chose to provide a full proof of this result, in order to keep the paper self contained.

Primitive Varieties of Heyting Algebras
For the present purpose, it is convenient to describe Esakia spaces dual to sums of Heyting algebras. Let X = X; ≤ X and Y = Y ; ≤ Y be two posets (with disjoint universes). Their sum X + Y is the poset with universe X ∪ Y and whose order relation ≤ is defined as follows for every x, y ∈ X ∪ Y : x ≤ y ⇐⇒ either (x, y ∈ X and x ≤ X y) or (x, y ∈ Y and x ≤ Y y) or (x ∈ X and y ∈ Y ).
So, X + Y is the poset obtained by placing Y above X. Now, let X and Y be two Esakia spaces (with disjoint universes). The sum X + Y is the Esakia space, whose underlying poset is X; ≤ X + Y ; ≤ Y , endowed with the topology consisting of the sets U ⊆ X ∪ Y such that U ∩ X and U ∩ Y are open respectively in X and Y . The following result is Moreover, we will recall a basic concept from universal algebra. Let K be a variety. A nontrivial algebra A ∈ K is said to be a splitting algebra in K [52,68] if there exists the largest subvariety V of K omitting A. In this case, V is always axiomatized relative to K by a single equation, sometimes called the splitting equation. In the realm of Heyting algebras, this phenomenon was first discovered by Jankov [43] 6 , who associated a special formula χ(A)-nowadays known as the Jankov formula-with every finite nontrivial 6 Jankov's approach was subsequently generalized to arbitrary varieties with EDPC in [13,Cor. 3.2] (see also [25,Cor. 3.8] for a similar result). FSI (equiv. finite subdirectly irreducible) Heyting algebra A, validating the following result: Let A and B be Heyting algebras such that A is finite, nontrivial, and FSI.
Bearing this in mind, let Citk be the largest variety of Heyting algebras omitting C 1 , . . . , C 5 , i.e., the variety of Heyting algebras axiomatized by the equations Citikin's Theorem [19] can be phrased in purely algebraic terms as follows: Theorem 7.3. The following conditions are equivalent for a variety K of Heyting algebras: (i) K is primitive; (ii) K is a subvariety of Citk; (iii) K omits the algebras C 1 , . . . , C 5 ; (iv) Every nontrivial finite FSI member of K has the form B 1 + · · · + B n for some Heyting algebras B i such that B 1 ∼ = 2, and B j ∈ I{2, 4} for all 1 < j < n, and, if n > 1, then B n ∈ I{2, 4, D}.
Consequently, Citk is the largest primitive variety of Heyting algebras.
(iv)⇒(i): Observe that the Esakia spaces dual to the algebras 2, 4, and D are respectively the following posets endowed with the discrete topology: We will rely on the following observations. Claim 7.4. Let X be a finite Esakia space such that X * ∈ K. Then for every x ∈ X there are a positive integer n and Esakia spaces X 1 , . . . , X n such that ↑x is isomorphic to X 1 + · · · + X n , where X 1 = 2 * , X j ∈ {2 * , 4 * } for all 1 < j < n, and, if n > 1, then X n ∈ {2 * , 4 * , D * }.
Proof of the Claim. In view of Lemma 3.2(i, ii), the Heyting algebra dual to ↑x is an FSI member of K. Together with the assumption (i.e., condition (iv) of Theorem 7.3) and Lemma 7.1, this yields the desired conclusion.
Claim 7.5. Let X be a finite Esakia space such that X * ∈ K and x ∈ X. If X ↑x = ∅, then there exists a surjective noninjective Esakia morphism f : X → Y that restricts to an order isomorphism from ↑x to an upset U of Y .
Proof of the Claim. In view of Claim 7.4, there are a positive integer n and Esakia spaces X 1 , . . . , X n such that ↑x is isomorphic to X 1 + · · · + X n , where X 1 = 2 * , X j ∈ {2 * , 4 * } for all 1 < j < n, and, if n > 1, then X n ∈ {2 * , 4 * , D * }. Let us label the elements of ↑x (equiv. of X 1 + · · ·+ X n ). Consider one of the X i 's. If X i = 2 * (resp. X i = 4 * ), then we denote the unique element (resp. the two elements) of X i by a i (resp. a i and b i ). Then suppose that X i = D * . In this case, we have necessarily i = n and we denote the elements of X i as follows: c d a n b n Since, by assumption, X ↑x is finite and nonempty, it has a maximal element y. Let then m(y) be the set of minimal elements in ↑x ∩ ↑y. The maximality of y in X ↑x guarantees that m(y) is the set of immediate successors of y.
If m(y) = ∅, then y is a maximal element of X. Then consider a maximal element z ∈ ↑x. Since y / ∈ ↑x, we have that y and z are two distinct maximal elements of X. Consequently, we can identify them through a β-reduction, thus producing the desired Esakia morphism. On the other hand, if m(y) is a singleton, say {z}, then z is the unique immediate successor of y. Consequently, we can identity y and z with an α-reduction, thus producing the desired Esakia morphism.
Therefore, it only remains to consider the case where m(y) has at least two elements. We will prove that By assumption, m(y) has at least two distinct elements z and v. Suppose, with a view to contradiction, that m(y) = {z, v}. Then there is an element w ∈ m(y) {z, v}. As z, v, and w are minimal in ↑x ∩ ↑y, they must be incomparable in ↑x. But this contradicts the fact that ↑x is a rooted poset of width ≤ 2. Hence we conclude that m(y) = {z, v}. Bearing in mind that ↑x = X 1 + · · · + X n and that z and v are incomparable, this easily implies that m(y) is one of the sets {c, d}, {b n , c}, and {a i , b i } for some i ≤ n. Therefore, to conclude the proof of condition (14), it only remains to show that m(y) cannot be {b n , c}.
Suppose the contrary, with a view to contradiction. Therefore, X n = D * and ↑y contains a n , b n , and c. By applying Claim 7.4 to ↑y, we obtain that Together with the fact that ↑y contains two distinct comparable elements, namely, c and a n , both of which are incomparable with an element b n ≥ y, this implies that Y m = D * . Furthermore, as a n and b n are maximal, they must be the maximal elements of Y m . Bearing in mind that Y m = D * , this guarantees the existence of an element z ∈ Y m such that z ≤ a n , b n and z c. Since y ≤ z and y is maximal in X ↑x, we obtain that either y = z or z ∈ ↑x. Now, from c ∈ m(y) it follows that y ≤ c. Together with z c, this implies that y = z and, therefore, that z ∈ ↑x. As m(y) = {b n , c} is the set of minimal elements of ↑x ∩ ↑y, this guarantees that b n ≤ z or c ≤ z. But, since z ≤ a n , b n , this would yield that either b n ≤ a n or c ≤ b n , a contradiction. Hence, we conclude that condition (14) holds.
In view of condition (14) It is easy to see that the set of immediate successors of z is precisely m(y).
Since m(y) is also the set of immediate successors of y, we can identify y and z with a β-reduction, thus obtaining the desired Esakia morphism.
With a series of applications of Claim 7.5, we obtain the following: Claim 7.6. Let X be a finite Esakia space such that X * ∈ K and x ∈ X. Then there exists a surjective Esakia morphism f : X → ↑x, whose restriction to ↑x is the identity function. Now, we turn to the proof of the main statement. The proof of Theorem 6.13(ii) shows that K is locally finite. Hence to establish that K is primitive it suffices, by Theorem 2.2, to show that the finite nontrivial FSI members of K are weakly projective in K.
Consider a finite nontrivial FSI algebra A ∈ K. Let also E ∈ K be such that A ∈ H(E). Clearly, there is a surjective homomorphism f : E → A. Since A is finite, there is a finitely generated subalgebra B ≤ E such that A ∈ H(B). Therefore, to conclude that A is weakly projective in K, it will be enough to show that A ∈ IS(B) and, therefore, A ∈ IS(E). Instead of proving directly that A ∈ IS(B), we will establish the existence of a surjective Esakia morphism g : B * → A * (see Lemma 3.2(iii)).
To this end, observe that B is finite, since K is locally finite and B is finitely generated. Thus the Esakia space B * is also finite and such that (B * ) * ∈ K. Furthermore, in view of Lemma 3.2(ii) and A ∈ H(B), we can identify A * with a principal upset of B * . Therefore, we can apply Claim 7.6 obtaining that there is a surjective Esakia morphism g : B * → A * , as desired.
Remark 7.7. The proof of the implication (iv)⇒(i) in Theorem 7.3 shows that A is projective in K in the classical sense. This is because Claim 7.6 guarantees that the restriction of g to A * is the identity map. As a consequence, the embedding g * : A → B can be viewed as an embedding g * : A → E such that f • g * is the identity map on A and, therefore, A is a retract of E. It follows that the nontrivial finite FSI members of a primitive variety K of Heyting algebras are projective in K.

Primitive Varieties of Brouwerian Algebras
It is well known that the ∧, ∨, → -fragment of IPC, here denoted by IPC + , is algebraized by the variety of Brouwerian algebras, i.e., ∧, ∨, → -subreducts of Heyting algebras. As a consequence of the algebraization phenomenon, the lattice of varieties of Brouwerian algebras is dually isomorphic to that of positive logics, i.e., axiomatic extensions of IPC + . Moreover, a positive logic is hereditarily structurally complete if and only if the variety of Brouwerian algebras associated with it is primitive.
As structural completeness and its variants are very sensitive to (even small) changes of signature, it was natural to wonder whether Citkin's description of hereditarily structurally complete intermediate logics could be extended to positive logics. Recently, a positive solution to this question was supplied by Citkin himself in [22]. However, as we will show below, the results and techniques of the previous sections of this paper yield a very short alternative proof of this result.
Given a Brouwerian algebra A, we denote by A ⊥ the unique Heyting algebra obtained by adding a new bottom element ⊥ to A. As the characterization of FSI algebras given in Lemma 3.2(i) holds for Brouwerian algebras as well, A is FSI if and only if so is A ⊥ . Given a class of Brouwerian algebras K, define Observe that for every class K of Brouwerian algebras, where the class operators H and S are computed in the language of Heyting algebras for H(K ⊥ ) and S(K ⊥ ), and in the language of Brouwerian algebras for H(K) and S(K). Finally, given a Heyting algebra A, we denote by A + its ∧, ∨, → -reduct.
As a consequence, C i has the form B ⊥ for some Brouwerian algebra B such that B ∈ SH(A). (16) As C i = B ⊥ , the bottom element of C i is meet-irreducible. By inspecting C 1 , . . . , C 5 , this guarantees that C i ∈ {C 2 , C 4 }. Together with B ⊥ = C i , this implies B ∈ {C + 1 , C + 3 }. By (16) we conclude that either C + 1 ∈ SH(A) ⊆ K or C + 3 ∈ SH(A) ⊆ K. But this contradicts the fact that K omits C + 1 and C + 3 .
As shown by Jankov [43], Theorem 7.2 generalizes to the case of Brouwerian algebras. More precisely, every finite nontrivial FSI (equiv. finite subdirectly irreducible) Brouwerian algebra A can be associated with a formula χ(A) + such that the largest variety of Brouwerian algebras omitting A exists and is axiomatized by χ(A) + ≈ 1. Bearing this in mind, let Citk + be the largest variety of Brouwerian algebras omitting C + 1 and C + 3 , i.e., the variety of Brouwerian algebras axiomatized by the equations χ(C + 1 ) + ≈ 1 and χ(C + 3 ) + ≈ 1. Citkin's description of hereditarily structurally complete positive logics can be phrased algebraically as follows: Theorem 8.2. The following conditions are equivalent for a variety K of Brouwerian algebras: (i) K is primitive; (ii) K is a subvariety of Citk + ; (iii) K omits the algebras C + 1 and C + 3 ; (iv) Every nontrivial finite FSI member of K has the form B 1 + · · ·+ B n for some Brouwerian algebras B i such that B 1 ∼ = 2 + , and B j ∈ I{2 + , 4 + } for all j > 1.
Consequently, Citk + is the largest primitive variety of Brouwerian algebras.
Proof. Observe that conditions (ii) and (iii) are equivalent by definition of Citk + . Moreover, the proof of (i)⇒(ii) is analogous to that of Lemma 5.1.
(ii)⇒(iv): Let A be a nontrivial finite FSI member of K. Then A ⊥ is a finite nontrivial FSI member of V(K ⊥ ). From Lemma 8.1 and Theorem 7.3 it follows that A ⊥ = B 1 + · · · + B n for some Heyting algebras B i such that B 1 ∼ = 2, and B j ∈ I{2, 4} for all 1 < j < n, and, if n > 1, then B n ∈ I{2, 4, D}. By construction of A ⊥ , its bottom element is meetirreducible. Consequently, necessarily B n ∼ = 2. Also, as A is nontrivial, A ⊥ has at least three elements, whence n > 1. Thus A ⊥ ∼ = 2 + B 2 + · · · + B n−1 + 2.

As a consequence,
where each B + i is isomorphic either to 2 + or to 4 + . (iv)⇒(i): First observe that K omits C + 1 and C + 3 . Therefore, Lemma 8.1 and Theorem 6.13(i) imply that V(K ⊥ ) is locally finite. This, in turn, guarantees that K is also locally finite. By Theorem 2.2 we conclude that, in order to prove that K is primitive, it suffices to show that its finite nontrivial FSI members are weakly projective in K.
Consider a finite nontrivial FSI member A of K. Then let B ∈ K and f : B → A be a surjective homomorphism. Observe that the unique map f ⊥ : B ⊥ → A ⊥ which extends f by f ⊥ (⊥):=⊥ is a homomorphism of Heyting algebras. By assumption, A is a finite linear sum of copies of 2 and 4, whence the same holds for A ⊥ . By [1,Thm. 4.10] this implies that A is projective in the standard sense. Therefore, as f ⊥ is surjective, there is an embedding g : A ⊥ → B ⊥ . Observe that g restricts to an embedding g : A → B of Brouwerian algebras. Consequently, A ∈ IS(B). Hence we conclude that A is weakly projective in K.

Properties of Primitive Varieties
Primitive varieties of Heyting and Brouwerian algebras have a number of interesting properties. Recall that a variety is said to be finitely based if it can be axiomatized by finitely many equations.
We conclude the paper by sketching a proof of the above result. 7 Proof sketch. We consider the case of Heyting algebras only, as that of Brouwerian algebras is analogous. First observe that primitive varieties of Heyting algebras are locally finite by Theorem 7.3 and the last part of Theorem 6.13. This establishes condition (i). Moreover, condition (iii) is an immediate consequence of (ii). Thus, to conclude the proof, it suffices to establish (ii). We will provide a proof sketch only. To this end, recall from Theorem 7.3 that Citk is the largest primitive variety of Heyting algebras. Therefore, to conclude the proof, it only remains to show that all subvarieties of Citk are finitely based. Observe that Citk is finitely based by definition. Moreover, it is locally finite by condition (i). Thus, by general arguments related to Jankov formulas, e.g., [10,Thm. 3.4.14] and [17,Ch. 9], one can reduce the problem of proving that all subvarieties of Citk are finitely based to that of showing that the poset Ord(Citk) of finite nontrivial FSI members of Citk ordered under the relation

A B ⇐⇒ A ∈ HS(B)
has no infinite antichain. Recall from Theorem 2.2 that all nontrivial FSI members of Citk are weakly projective in Citk. As a consequence for every A, B ∈ Ord(Citk),

A B ⇐⇒
Thus, to conclude the proof, it suffices to show that there is no infinite antichain in the poset of finite nontrivial FSI members of Citk ordered under the relation

A B ⇐⇒ A ∈ S(B).
This can be shown by a combinatorial argument similar to the one detailed in [22,Sec. 7] for the case of Brouwerian algebras, using the description of finite nontrivial FSI members of Citk given in Theorem 7.3.
Thus we arrive at the following corollary.