SUBALGEBRAS AND HOMOMORPHIC IMAGES OF THE RIEGER-NISHIMURA LATTICE

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1 SUBALGEBRAS AND HOMOMORPHIC IMAGES OF THE RIEGER-NISHIMURA LATTICE Guram Bezhanishvili and Revaz Grigolia Abstract In this note we characterize all subalgebras and homomorphic images of the free cyclic Heyting algebra, also known as the Rieger- Nishimura lattice N. Consequently, we prove that every subalgebra of N is projective, that a finite Heyting algebra is a subalgebra of N iff it is projective, and characterize projective homomorphic images of N. The atoms and co-atoms of the lattice of all subalgebras of N are also characterized, along with the Frattini subalgebra of N. 1 Introduction The free cyclic Heyting algebra was constructed independently by Rieger [7] and Nishimura [6]. Today it is known as the Rieger-Nishimura lattice, and is considered a model of elegance and simplicity. It is constructed from a generator g by means of Nishimura polynomials: P 1 (g) = g P 2 (g) = g P 2n+3 (g) = P 2n+1 (g) P 2n+2 (g) P 2n+4 (g) = P 2n+3 (g) P 2n+1 (g) Here it is assumed that n = 0, 1, 2,.... One can depict the Rieger-Nishimura lattice as shown in Fig.1. 1

2 1 P 9 (g) P 10 (g) P 8 (g) P 7 (g) P 5 (g) P 6 (g) P 4 (g) P 3 (g) g P 2 (g) 0 Fig.1 In spite of the relatively simple structure of N, the structure of the lattice of subalgebras of N is much more involved. The aim of this note is to characterize all subalgebras of N, along with its homomorphic images. The paper is organized as follows: 2 has a preliminary purpose. In it we recall the Priestley type duality for Heyting algebras, as well as define the sum and the ω-sum of Heyting algebras, which are of an utmost importance for the sequel. In 3 we describe the dual space of N, also known as the Rieger- Nishimura ladder, and characterize all homomorphic images of N. 4 is the heart of the paper, and it characterizes all subalgebras of N. In 5 we show that every subalgebra of N is projective, that a finite Heyting algebra is a subalgebra of N iff it is projective, and characterize projective homomorphic images of N. Finally, 6 describes all atoms and co-atoms of the lattice S(N ) of all subalgebras of N, along with the Frattini subalgebra of N. 2 Preliminaries Let us recall that a Heyting algebra (H,,,, 0, 1) is a bounded distributive lattice (H,,, 0, 1) with an additional binary operation : H H H satisfying the following condition for any a, b H: x a b iff a x b. 2

3 The category of all Heyting algebras and all Heyting homomorphisms will subsequently be denoted by HA. 2.1 Dual spaces of Heyting algebras The dual category of HA is constructed by means of ordered topological spaces. Recall that a topological space (X, Ω) is called a Stone space if it is 0-dimensional, compact and Hausdorff. A X is said to be clopen if A is both closed and open. We denote the set of all clopen subsets of X by CP(X). Let (X, R) be a partially ordered set. For any x X and A X, let R(x) = {y X : xry}, R(A) = x A R(x), R 1 (x) = {y X : yrx}, R 1 (A) = x A R 1 (x). A is said to be an R-cone of X if R(A) = A. We call A a down R-cone if R 1 (A) = A. Suppose a triple (X, Ω, R) is given, where (X, Ω) is a Stone space and R is a partial order on X. R is said to be point-closed if R(x) is a closed set for every x X. R is said to be an Esakia relation if R is point-closed and the following condition is satisfied: A CP(X) implies R 1 (A) CP(X). A triple (X, Ω, R) is called an Esakia space, if (X, Ω) is a Stone space and R is an Esakia relation on it. For any two partial orders (X 1, R 1 ) and (X 2, R 2 ), call a function f : X 1 X 2 monotone if xr 1 y implies f(x)r 2 f(y), for any x, y X 1. Call f strongly monotone if in addition f(x)r 2 y implies ( z X 1 )(xr 1 z and f(z) = y), for any x X 1 and y X 2. Now f is said to be an Esakia morphism if f is continuous and strongly monotone. 3

4 Theorem 1. (Esakia [4]) The category of Heyting algebras and Heyting homomorphisms is dual (dually equivalent) to the category of Esakia spaces and Esakia morphisms. For the reader s convenience we recall here the construction of the dual Esakia space (X, Ω, R) of a given Heyting algebra H. X is the set of all prime filters of H, R is the set-theoretical inclusion, and Ω is defined on X by letting the Boolean closure of the set φ(h) to be a basis for the topology. Here φ(a) = {x X : a x} for every a H, and φ(h) = {φ(a)} a H. Then (φ(h),,,, ) constitutes a Heyting algebra, where A B = R 1 (A B) for every A, B φ(h), and it is isomorphic to the initial H. Moreover, φ(h) coincides with the set of all clopen R-cones of X. Hence we obtain the following representation theorem for Heyting algebras: Theorem 2. (Esakia [4]) Every Heyting algebra is represented as the algebra of all clopen R-cones of the corresponding Esakia space. 2.2 Dual characterization of subalgebras and homomorphic images of Heyting algebras Having Theorems 1 and 2 at hand, one can obtain a rather neat dual characterization of subalgebras and homomorphic images of a given Heyting algebra. For this recall that a partition E of an Esakia space X is said to be correct if the following two conditions are satisfied: 1) For all x, y, z X, if xey and yrz, then there exists w X such that xrw and wez; 2) For all x, y X, if it is not the case that xey, then there exists an E-saturated clopen A such that x A and y / A. 1 (Note that 2) implies that every equivalence class of E is closed.) Theorem 3. (Esakia [4]) 1) Subalgebras of a Heyting algebra H correspond to correct partitions of the dual space X of H. 2) Homomorphic images of a Heyting algebra H correspond to closed R- cones of the dual space X of H. 1 Recall that a set A is said to be E-saturated if E(A) = A, where E(A) = w A E(w) and E(w) = {v X : wev}. 4

5 2.3 The sum of Heyting algebras The sum H 1 H 2 of two Heyting algebras H 1 and H 2 is defined by putting H 1 H 2 = H 1 H 2, where H 1 and H 2 are isomorphic copies of H 1 and H 2 respectively with H 1 H 2 = {1 H1 } = {0 H2 }. Further, for any a, b H 1 H 2 put a H1 H 2 b if a H1 b, a H2 b, or a H 1 and b H 2 (a and b are taken from H 1 H 2 up to isomorphism). Figuratively speaking, we put H 2 over H 1. It is easy to prove that H1 H 2 is a lattice order on H 1 H 2, and that H 1 H 2 forms a Heyting algebra with respect to H1 H 2 (see e.g. Troelstra [8]). The dual space X H1 H 2 of H 1 H 2 is constructed from the dual spaces X H1 of H 1 and X H2 of H 2 in the following way. Consider the disjoint union X H1 X H2 of X H1 and X H2. The topology on X H1 X H2 is defined as the standard sum of topological spaces X H1 and X H2, and the order on X H1 X H2 is defined by putting xry if xr H1 y, xr H2 y, or x X H2 and y X H1 (here x and y are taken from X H1 X H2 up to isomorphism). In other words, we put X H2 under X H1. It is a rather easy exercise to check that (X H1 X H2, Ω, R) is an Esakia space, and that X H1 X H2 is isomorphic to the dual X H1 H 2 of H 1 H 2. We call this construction the sum of two Esakia spaces, and denote it by X H1 X H The ω-sum of Heyting algebras The ω-sum, or the countable sum, of Heyting algebras generalizes the ordinary (finite) sum of Heyting algebras described in the previous subsection. It is defined as follows. Suppose a countable family of Heyting algebras {H n } n ω is given. Define n ω H n by putting n ω H n = n ω H n {1}, where H n are isomorphic copies of H n with H n H n+1 = {1 Hn } = {0 Hn+1 }, and 1 / H n for any n. Further, for any a, b n ω H n put a n ω Hn b if there is n ω such that a Hn b, or a H m, b H k and m k. We also let a n ω Hn 1 for any a n ω H n. Figuratively speaking, we form a tower from a countable family of Heyting algebras by putting one algebra over another, and then adjoining a new top element. It is easy to prove that n ω Hn is a lattice order on n ω H n, and that n ω H n forms a Heyting algebra with respect to n ω Hn. 5

6 The dual space X of n ω Hn n ω H n is constructed from the dual spaces X Hn of every H n in the following way. Consider the disjoint union n ω X H n with the topology of the disjoint union and take the one-point compactification α( n ω X H n ) of n ω X H n. So, α( X Hn ) = X Hn { }. n ω n ω The order on n ω X H n { } is defined by putting xry if there is n such that xr Hn y, or x X Hk, y X Hm and m k. We also let Rx for any x n ω X H n. In other words, we form a tower from a countable family of Esakia spaces by putting one space under another, and then adjoining a new least element. One can prove that ( n ω X H n { }, Ω, R) is an Esakia space, and that n ω X H n { } is isomorphic to the dual X of n ω Hn n ω H n. We call this construction the ω-sum, or the countable sum of a countable family of Esakia spaces, and denote it by n ω X H n. If each H n is equal to H, then we simply write ω H and ω X H. In 4 we will use the following useful fact, which is easy to prove: if H n is a subalgebra of H n for every n ω, then n ω H n is (isomorphic to) a subalgebra of n ω H n. 3 The dual space of N Having the above described duality at hand, we can easily construct the dual space of the Rieger-Nishimura lattice. To do so, we need to describe all prime filters of N. For this we first describe all proper filters of N. Denote by [a) the principal filter generated by a. Then it is obvious that [P n (g)) are proper filters of N for every n = 1, 2,.... Moreover, it is obvious that the trivial filter {1} is a proper filter of N, and that every proper filter of N is equal either to {1} or to [P n (g)) for some n 1. Furthermore, {1}, [g) and [P 2n (g)), for n = 1, 2,..., are the only prime filters of N. So, one can depict the dual space of the Rieger-Nishimura lattice as shown in Fig.2, where odd numbers correspond to the polynomials g, P 4 (g), P 8 (g), P 12 (g),..., even numbers correspond to the polynomials P 2 (g), P 6 (g), P 10 (g),..., and ω corresponds to {1}. 6

7 ω Fig.2 Subsequently, we will call the dual space of the Rieger-Nishimura lattice the Rieger-Nishimura ladder, and denote it by L. We also let L 0 to denote L {ω}. Note that the topology on L is defined by letting finite subsets of L 0 and their complements (in L) to form a basis. Given this topology, every n L 0 is an isolated point, while ω is closed but not open. So, as a topological space L 0 is discrete, and L is the one-point compactification of L Homomorphic images of N Now we are in a position to characterize homomorphic images of N. As follows from Theorem 3, homomorphic images of N correspond to closed R- cones of L. Since every infinite closed subset of L contains ω, it follows that the only infinite closed R-cone of L is L itself. Hence every proper closed R-cone of L is finite. (It is also obvious that every finite R-cone of L is closed.) It follows that every proper homomorphic image of N is finite. Taking a closer look at proper homomorphic images of N, it is obvious that {1} and {2} are (closed) R-cones of L. Hence the two-element Boolean algebra 2 can be obtained as a homomorphic image of N in two different ways: one is to filter N by the filter [g), and the other is to filter N by [P 2 (g)). Further, {1, 3} is obviously a (closed) R-cone of L. So, the threeelement Heyting algebra 3 is also a homomorphic image of N, which is obtained by filtering N by [P 4 (g)). Furthermore, from the structure of L 7

8 it directly follows that there are four more different configurations for finite (closed) R-cones of L, which are shown in Fig.3. (a) (b) (c) (d) Fig.3 Their corresponding Heyting algebras are shown in Fig.4. Obviously any of these algebras is cyclic. Hence, there exist infinitely many finite cyclic Heyting algebras, and the only infinite cyclic Heyting algebra is N itself. As a direct consequence of these observations, we have the following: Proposition 4. 1) Up to isomorphism the only homomorphic images of N are 2, 3, the algebras shown in Fig.4, and N. 2) Every proper homomorphic image of N is finite. 3) A Heyting algebra H is cyclic iff H is isomorphic to either N, 2, 3, or one of the algebras shown in Fig.4. 4) The only infinite cyclic Heyting algebra is N the free cyclic Heyting algebra. In 5 we will show that among homomorphic images of N only 2, 3, 2 2 2, and N itself are subalgebras of N. Here 2 2 denotes the four-element Boolean algebra; the dual space of is (isomorphic to) {1, 2, 4}, which is a (closed) R-cone of L. So, is the smallest algebra of the configuration (c). 8

9 Fig (a) (b) (c) (d) 4 Subalgebras of N This section is devoted to a characterization of subalgebras of N. We will show that up to isomorphism there are three different types of subalgebras of N. In order to describe them, we will heavily use the sum and the ω-sum of Heyting algebras. Recall that 2 and 2 2 denote the two-element and the four-element Boolean algebras respectively. Consider n ω B n, where each B n is either 2 or 2 2. For the reader s convenience the dual spaces of ω 2 and ω 22 are shown in Fig.5 (a) and (b) respectively. Lemma 5. Every n ω B n is isomorphic to a subalgebra of N. Proof: First let us show that ω 22 is isomorphic to a subalgebra of N. Consider the partition E of L shown in Fig.6. It is an easy exercise to check that E is a correct partition of L and that L/ E is isomorphic to the dual space of ω 22 shown in Fig.5(b). Hence, by Theorem 3, ω 22 is isomorphic to a subalgebra of N. Now, since every n ω B n is isomorphic to a subalgebra of ω 22, every n ω B n is isomorphic to a subalgebra of N as well. 9

10 (a) (b) Fig.5 Corollary 6. Every ( k n=1 B n) 2 is isomorphic to a subalgebra of N. Proof: First note that for every k, ( k 22 ) 2 is isomorphic to a subalgebra of ω 22. The corresponding correct partition is constructed by identifying all the points of X of the depth greater than k. Now since ω 22 every ( k n=1 B n) 2 is isomorphic to a subalgebra of ( k 22 ) 2, every ( k n=1 B n) 2 is isomorphic to a subalgebra of ω 22 either. Hence every ( k n=1 B n) 2 is isomorphic to a subalgebra of N. ω Fig.6 10

11 Lemma 7. Every ( k n=1 B n) N is isomorphic to a subalgebra of N. Proof: First let us show that for every k, ( k 22 ) N is isomorphic to a subalgebra of N. For every k, consider the partition E of N shown in Fig.7, which is analogous to the partition from Lemma 5, but stops identifying points starting from the depth k , , 6 7, ω ω L L/ E Fig.7 It is easy to check that E is a correct partition, and that L/ E is isomorphic to the dual space of ( k 22 ) N. Hence, by Theorem 3, ( k 22 ) N is isomorphic to a subalgebra of N. Now since every k n=1 B n is isomorphic to a subalgebra of k 22, every ( k n=1 B n) N is isomorphic to a subalgebra of ( k 22 ) N, and hence is isomorphic to a subalgebra of N as well. Theorem 8. A Heyting algebra H is a subalgebra of N iff either H = N, or H is isomorphic to n ω B n, ( k n=1 B n) 2 or ( k n=1 B n) N. Proof: It is obvious that N is a subalgebra of itself. Further Lemmas 5 and 7 and Corollary 6 imply that n ω B n, ( k n=1 B n) 2 and ( k n=1 B n) N are isomorphic to subalgebras of N. 11

12 In order to prove the converse, we need some preparation. Claim 9. For any correct partition E of L, every equivalence class C of E is convex. That is, given x, y C and xrzry it follows that z C. Proof: Suppose x, y C and xrzry. Since L is dually well-founded, so is C, and we can assume that y is a R-maximal point of C (otherwise we would exchange y with a R-maximal point u of C such that yru). Since yex and xrz, there exists w such that yrw and zew. wez and zry imply that there exists v such that wrv and vey. Therefore, yrv and yev, and since y is a R-maximal point of C, y = v. Hence y = w, yez, and z C. Now suppose H is a subalgebra of N. Let E be the corresponding correct partition of L. Then either every equivalence class of E is finite (hence H is infinite), or there are only finitely many equivalence classes of E (hence H is finite). In the latter case, all but one of the equivalence classes are finite. Indeed, since every equivalence class of E is closed, if a class contains infinitely many points, it also should contain ω. Hence there can not be two infinite equivalence classes of E. In either case (L/ E, R) is dually well-ordered, where R is defined on L/ E componentwise. v is said to cover w if wrv and whenever wrurv, either w = u or u = v. Call a point w of L/ E degenerate if there exist two R- incomparable points w 1, w 2 L/ E such that w 1 is the only cover of w, and w and w 2 are R-incomparable too. (Note that in L there is just one degenerate point denoted by 3 in Fig.2.) Claim 10. If there is no degenerate point in L/ E, then L/ E is isomorphic to the dual of either n ω B n or ( k n=1 B n) 2. Proof: Since the width (that is the cardinality of a maximal anti-chain of points) of L is 2, the width of L/ E is no more than 2. Further, L/ E is dually well-ordered and contains the least element E(ω). (Note that either E(ω) = {ω} or E(ω) contains the whole lower part of L.) Furthermore, for any w L/ E, if wru and u and v are of the same R-depth, then wrv as well. Hence L/ E is the sum of 1 and 2 element sets, and thus is isomorphic to the dual of either n ω B n or ( k n=1 B n) 2. Claim 11. There may exist only one degenerate point in L/ E, say x E. Moreover, the equivalence class x E consists of just one point, and so do all other equivalence classes which are R-related to x E. 12

13 Proof: Suppose x E is a degenerate point of L/ E. Then there exist w 1, w 2 L/ E such that w 1 and w 2 are R-incomparable, w 1 is the only cover of x E, and x E and w 2 are also R-incomparable. Suppose also x E consists of at least two points. From the structure of L it directly follows that one of those points is R-related to a point from w 2, and hence x E Rw 2, a contradiction. Hence x E consists of just one point x. Now suppose x is of depth n, and show that the equivalence class containing another point of depth n also consists only of that point. Indeed, denote another point of depth n by y. Suppose yez and z y. Then zrx. Hence there should exist u such that yru and uex, which is impossible. Thus the equivalence class containing y consists only of y. Now consider a point z of depth n+1 which is not related to y and show that the equivalence class containing z also consists only of z. Suppose zeu and u z. Then ury, and from zeu and ury it follows that there exists v such that zrv and vey, which is again a contradiction. Hence the equivalence class containing z also consists only of z. In the same way one can show that the equivalence class containing another point of depth n + 1 also consists only of that point. Continuing this process we get that every equivalence class R-related to x E consists of just one point. It follows that there may exist only one degenerate point in L/ E. Indeed, since L/ E is dually well-founded, there exists the first degenerate point x E = {x}. Now all the equivalence classes R-related to x E contain only a single point, and L/ E repeats the lower part of L starting from x. Hence, there is just one degenerate point in L/ E. It directly follows from Claim 11 that if L/ E contains a degenerate point, then either L/ E = L, or the upper part of L/ E is the sum of either 1 or 2 element sets, and the lower part of L/ E is isomorphic to L. Hence, either L/ E = L, or L/ E is isomorphic to the dual of ( k n=1 B n) N. Thus, if H is a subalgebra of N, then either H = N, or H is isomorphic to n ω B n, ( k n=1 B n) 2 or H = ( k n=1 B n) N. 5 Connection with projective algebras In this section we will show that every subalgebra of N is projective, that a finite Heyting algebra is a subalgebra of N iff it is projective, and characterize projective homomorphic images of N. Recall that an algebra A from a variety V is said to be projective if it is a retract of a free V-algebra. 13

14 Proposition 12. 1) A finite Heyting algebra is a subalgebra of N iff H is (isomorphic to) ( k n=1 B n) 2. 2) (See also Citkin [2]) A finite Heyting algebra is (isomorphic to) a subalgebra of N iff H is projective. Proof: 1) directly follows from Theorem 8. 2) It follows from Balbes and Horn [1] that a finite Heyting algebra H is projective iff H is (isomorphic to) ( k n=1 B n) 2. Now apply 1). Proposition 13. If H is a subalgebra of N, then H is projective. Proof: It is obvious that N itself is projective. Further, if H is (isomorphic to) ( k n=1 B n) 2, then as follows from Proposition 12, H is projective. Furthermore, it is shown in Grigolia [5] that if either H is (isomorphic to) n ω B n or ( k n=1 B n) N, then H is a retract of F (ω) the free countably generated Heyting algebra. Hence, H is projective. (Though n ω B n and ( k n=1 B n) N are not retracts of N ). From Propositions 12 and 13 it directly follows that among homomorphic images of N only 2, 3, 2 2 2, and N are subalgebras of N. Hence, we arrive at the following useful fact (a different proof can be found in de Jongh [3]): 1) A Heyting algebra is a retract of N iff it is isomorphic to either 2, 3, 2 2 2, or N. 2) A homomorphic image of N is a projective algebra iff it is isomorphic to either 2, 3, 2 2 2, or N. 6 The lattice structure of S(N ) In this final section we investigate the lattice structure of S(N ). It is obvious that 2 and N are a least element and a greatest element of S(N ) respectively. We prove that S(N ) contains countably many atoms, which are all isomorphic to the three element Heyting algebra 3, but are incomparable inside S(N ). We also show that S(N ) contains two co-atoms F 1 and F 2, and that F 1 is isomorphic to a subalgebra of F 2, but is incomparable with F 2 inside S(N ). And finally we describe the Frattini subalgebra of N. As a corollary, we obtain that there exists an isomorphism (outside of S(N )) between the Frattini subalgebra of N and F 1, and that N does not contain a proper subalgebra isomorphic to itself. 14

15 Let us start by describing the atoms of S(N ). For any natural n, consider the three element subset {0, a n, 1} of L (Fig.8). 1 a 3 a 2 a 1 0 Fig.8 It is routine to check that {0, a n, 1} is a subalgebra of N for any n ω, that {0, a n, 1} is incomporable with {0, a m, 1} for n m, and that every {0, a n, 1} is isomorphic to 3. Dually {0, a n, 1} corresponds to the correct partition E n of L defined by partiting L into two classes, one class containing all the points of L of depth n, and another containing all the points of L of depth > n. Moreover, for any correct partition E of L, it is obvious that there is an n such that E E n. Algebraically this means that for any subalgebra H of N, there is an n such that {0, a n, 1} is a subalgebra of H. Hence we arrive at the following Proposition 14. S(N ) contains countably many atoms. They are {0, a n, 1} for every n ω. Now let us investigate the co-atoms of N. Let E 1 be the correct partition of L identifying 1 and 2, and E 2 be the correct partition of L identifying 1 and 3. L/ E1 and L/ E2 are shown in Fig.9. Let F 1 and F 2 denote the subalgebras of N corresponding to E 1 and E 2 respectively. It is obvious that F 1 is isomorphic to 2 N, and that F 2 is isomorphic to 2 2 N. Hence F 1 15

16 , 2 ω Fig.9 1, ω can be isomorphically embedded into F 2. However, since E 1 and E 2 are incomparable correct partitions of L, F 1 and F 2 are incomparable as subalgebras of N. Furthermore, suppose H is a proper subalgebra of N. Let E be the corresponding partition of L. Then there are at least two E-equivalent points in L. This means that at least two back-to-back points 2 are E-equivalent, which in turn means that either 1, 2 or 1, 3 are E-equivalent as well. In the former case E 1 E, and in the latter case E 2 E. Algebraically this means that either H is a subalgebra of F 1, or H is a sublagebra of F 2. Hence we arrrive at the following Proposition 15. F 1 and F 2 are the only co-atoms of S(N ). Finally, recall that the Frattini subalgebra of a universal algebra is the intersection of all maximal proper subalgebras of A. So, the Frattini subalgebra of N is F 1 F 2. The dual space of the Frattini subalgebra of N is shown in Fig.10. Interestingly enough, it follows that F 1 F 2 is isomorphic to 2 N, and hence is isomorphic to F 1. Though inside S(N ), F 1 F 2 is a proper subalgebra of F 1. Another interesting corollary is the following feature of N. 2 By back-to-back points we mean two points which either have the same depth or one of them covers the other. 16

17 1, 2, ω Fig.10 Proposition 16. N contains no proper subalgebra which is isomorphic to N. References [1] R. Balbes and A. Horn, Injective and projective Heyting algebras, Trans. Amer. Math. Soc. 148(1970), pp [2] A. Citkin, On admissible rules of intuitionistic propositional calculus, Math. USSR Sbornik, 31 (1977), pp [3] D. de Jongh, Formulas of one propositional varable in intuitionistic arithmetic, The L.E.J. Brouwer Centenary Symposium, A. Troelstra and D. Van Dalen (editors), North Holland Publishing Company, 1982, pp [4] L. Esakia, Topological Kripke models [in Russian], Dokl. Akad. Nauk SSSR 214(1974), pp [5] R. Grigolia, Projective Heyting algebras, preprint,

18 [6] I. Nishimura, On formulas of one variable in intuitionistic propositional calculus, Journal of Symbolic Logic 25(1960), pp [7] L. Rieger, On the lattice theory of Brouwerian propositional logic, Acta fac. rerum nat. Univ. Car., 189 (1949), pp [8] A. Troelstra, On intermediate propositional logics, Indagationes Mathematicae 27(1965), pp Guram Bezhanishvili Revaz Grigolia Department of Math. Sciences Department of Math. Logic New Mexico State University Institute of Cybernetics Las Cruces, NM, Georgian Academy of Sciences USA Euli 5, Tbilisi 86 GEORGIA 18