Alizadeh-Ardeshir-On the Linear Lindenbaum Algebra of Basic Propositional Logic

Math. Log. Quart. 50, No. 1, 65 70 (2004) / DOI 10.1002/malq.200310077

On the linear Lindenbaum algebra of Basic Propositional LogicMajid Alizadeh1 and Mohammad Ardeshir 21 Institute for Studies in Theoretical Physics and Mathematics, P. O. Box 19395-5746, Tehran, Iran2 Department of Mathematics, Sharif University of Technology, P. O. Box 11365-9415, Tehran, Iran

Received 12 June 2003, revised 2 October 2003, accepted 6 October 2003Published online 1 December 2003

Key words Basic propositional logic, basic algebra, Heyting algebra.MSC (2000) 03G25, 03D20, 03B20We study the linear Lindenbaum algebra of Basic Propositional Calculus, called linear basic algebra.

c 2004 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

1 Introduction and preliminariesMost well-known logical systems have associated with them a natural algebraic structure, namely their Linden-baum algebra. The Lindenbaum algebra of Basic Propositional Calculus BPC, called basic algebra (Ba), isintroduced in [1] (see also [2]).

The lattice of formulas of BPC up to provable equivalence is a Ba . Bas, defined below, are for BPC whatHeyting algebras are for Intuitionistic Propositional Calculus IPC, and Boolean algebras are for Classical Propo-sitional Calculus CPC.

Definition 1.1 A basic algebra B = B,,,, 0, 1 (Ba) is a structure with constants 0 and 1, and binaryfunctions , , and , such that

1. B,,, 0, 1 is a distributive lattice with bottom and top, and2. for we have the additional equations a a = 1, a 1 a, a b c = (a b) (a c),

b c a = (b a) (c a), and (a b) (b c) a c.The relation is expressible in term of equations with or in the standard way, i. e.

a b iff a b = a iff a b = b.

We recall that a Heyting algebra (Ha) is a distributive lattice with constants 0 and 1, and a binary relation ,which satisfies:

a b c iff a b c.

The fourth item of the following Proposition shows that in Bas we have only one direction of the abovebi-conditional:

Proposition 1.2 Let B be a Ba . Then for a, b, c B,1. if a b, then c a c b,2. if a b, then b c a c,3. if a b, then a b = 1,4. if a b c, then a b c,5. a (a b) = a (1 b). e-mail: [email protected] Corresponding author: e-mail: [email protected]


66 M. Alizadeh and M. Ardeshir: On the linear Lindenbaum algebra of Basic Propositional Logic

P r o o f.1. Let a b. Then a b = a, so c a = c a b = (c a) (c b) c b.2. Let a b. Then a b = b, so b c = a b c = (a c) (b c) a c.3. Let a b. By 1., 1 = a a a b. So a b = 1.4. Let a b c. Then by 1., b a b b c, so (b a) (b b) b c. Then b a b c.

Since b 1, by 2., 1 a b a. So a 1 a b a b c.5. a (a b) (1 a) (a b) (1 b). On the other hand, a (1 b) a (a b), by 2.Note that every Ha is a Ba . As a trivial example of a Ba which is not a Ha , let B = {0, 1} with the usual

definition of and . Define a b = 1, for a, b {0, 1}. Then B = B,,,, 0, 1 is a Ba . B is not a Ha ,since 1 1 0, but 1 1 0. So the class of all Has is a proper subclass of all Bas.

The following proposition shows a tight relation between Bas and Has:Proposition 1.3 Let B be a Ba . Then B is a Ha iff for every a B, 1 a = a.P r o o f. It is enough to show that if x a b, then x a b. Suppose x a b. Then

a x a (a b) = a (1 b) = a b.So a x b.

As usual we define:Definition 1.4 A subset F B of a Ba B is called a filter on B if 1 F , a, b F implies a b F , and

a b and b F implies a F .A filter F on a Ba B is called prime, if a b F implies that a F or b F . For two filters F , F on B, let

F F iff a b F and a F implies b F . It is easy to see that the relation is a transitive relationon the set of all filters on B.

Lemma 1.5 Let F be a filter on a Ba B such that a b / F . Then there is a proper prime filter F on Bsuch that F F , a F and b / F .

P r o o f. See [1] or [6].Lemma 1.6 Let B be a Ba and a, b B such that a b. Then there is a prime filter F on B such that a F

and b / F .P r o o f. See, e. g., [5].Corollary 1.7 Let B be a Ba and 1 = a B. Then there is a prime filter F on B such that a / F .P r o o f. Since 1 a, by Lemma 1.6, there is a prime filter F such that a / F .

2 Linear basic algebrasIt is well-known that the set of all formulas which are valid in every linear Heyting algebra is an intermediatelogic between IPC and CPC, called Dummett Logic, DL, and is axiomatizable by IPC + (A B) (B A)(see [3] or [4]). A linear Heyting algebra is a Heyting algebra which is a chain, i. e., such that for every x, y,x y or y x. Since in any Ha , x y = 1 iff x y, the above axiomatization seems natural. Thatwould be more convincing if we remember that DL is also strongly complete with respect to the class of all linearKripke models of IPC (see, e. g., [7]).

Horn [4] called a Ha with the additional axiom (x y) (y x) = 1 to be an Ha (he denoted it asL-algebra) and proved that DL is complete with respect to the class of Has. Note that every chain is an Ha .

The case for Bas is very different from Has. The implication operation in Bas is wild. Recall thatwe have in Bas only if a b c, then a b c, and not the other way around. On the other hand,Visser in [8] proved that BPC + (A B) ((A B) A) is strongly complete with respect to the classof all linear Kripke models of BPC. It is easy to see that although two axioms (x y) (y x) = 1 and(x y) ((x y) x) = 1 are equivalent in any Ha , in Ba we have only one direction:


Math. Log. Quart. 50, No. 1 (2004) / www.interscience.wiley.com 67

Lemma 2.1 In any Ba , for all x, y, if (x y) ((x y) x) = 1, then (x y) (y x) = 1.P r o o f. It is enough to show that in any Ba , (x y) x y x. That is clear by Proposition 1.2, since

y 1 y x y.The following example shows a linear Ba which does not satisfy (x y) ((x y) x) = 1:Example 2.2 Let B = {0, x, y, 1}, where 0 < x < y < 1, and and are defined as usual. Define

a b = y if b < a, and a b = 1, otherwise. Then B = (B,,,, 0, 1) is a linear Ba . But we have(x 0) ((x 0) x) = y.

On the other hand, the example before Proposition 1.3 shows a Ba which is not a Ha and satisfies(x y) ((x y) x) = 1.

Definition 2.3 An Ba is a basic algebra in which for all x, y, (x y) (y x) = 1.Definition 2.4 A chain is a linear Basic algebra, i. e. a Ba such that for all x, y, x y or y x.Clearly every chain is an Ba .Let B be a Ba and F a filter on B. For any a, b B, we shall write a F b iff a b F . The relation

F is a quasi-ordering on B. Write a F b iff (a b) (b a) F . It is easy to check that F is anequivalence relation on B. We shall call F the equivalence relation determined by the filter F , and we writeB/F instead of B/ F . Elements in B/F will be denoted by [a]. We obtain a relation on B/F by [a] [b]iff a b F . This relation is transitive and anti-symmetric, so B/F, is a partially ordered set. For[a], [b] B/F define

[a] [b] = [a b], [a] [b] = [a b], [a] [b] = [a b].It is easy to check that the above relations are well defined. The interesting case is for . We show that for allx, x, y, y B, if x x and y y, then x y x y. Suppose y y. We have y y F , so(x y) (x y) F . Then [x y] [x y]. By the same argument, we have [x y] [x y].So (a) [x y] = [x y]. Now suppose x x. Then x x F . Since (x x) (x y) x y, wehave ((x y) (x y)) F , so [x y] [x y]. By the same argument, we have [x y] [x y].So (b) [x y] = [x y]. Now, by (b) and (a), we have [x y] = [x y] = [x y].

It is also routine to see that [a b] and [a b] are the meet and the join of [a] and [b], respectively, and that satisfies the axioms of a Ba . Thus we have:

Theorem 2.5 Let B be a Ba and F a proper filter on B. Then B/F = B/F,,,, [0], [1] is also a Ba .

Lemma 2.6 If F is a prime filter on an Ba B, then B/F is a chain.P r o o f. Let x, y B. Since (x y) (y x) = 1, we have (x y) F or (y x) F . So [x] [y]

or [y] [x].Now we wish to determine the set of all formulas which are valid in every chain.The language of BPC is L = {,,,,}. The small letters p, q, r, . . . propositional variables and

capital letters A, B, C, . . . denote formulas, where the notion of formula is defined as usual. We use the sequentnotation to axiomatize BPC. A sequent is an expression of the form A B, in which A and B are formulas inL. In the rules of BPC a double horizontal line means that we have two-direction rules.

A A, A , A,A (B C) (A B) (A C), (A B) (B C) A C,(A B) (A C) A B C, (A B) (C B) A C B,A B B C

A C ,A B CA B C ,

A B A CA B C ,

A B C BA C B .



This ends the axiomatization of BPC. For notations, basic definitions and properties of BPC, see [2]. We use Afor A, and A B is an abbreviation for A B and B A.

Let us define BPC to be BPC plus the axiom schema (A B) (B A). As usual the elements ofthe Lindenbaum algebra of BPC are the equivalence classes |A| of formulas in the language of BPC, where|A| = {B : BPC A B}. The algebraic operations of the structure are just those inherited from the logicaloperations of BPC:

|A| |B| = |A B|, |A| |B| = |A B|, |A| |B| = |A B|, || = 1, || = 0.

By the Substitution property (see [2, Proposition 2.4]), these operations are well defined.Let U be the Lindenbaum algebra of BPC. Then U is an Ba . It is easy to see that for every sequent A B,

|A| |B| iff BPC A B, so for any formula A, |A| = 1 iff BPC A.Definition 2.7 An algebraic model of BPC consists of a pair B = B, I with B an Ba and I a map from

the set of all propositional variables of the language of BPC to B. The map I can be uniquely extended to allformulas of the language of BPC by the following definition:

I() = 1, I() = 0, I(A B) = I(A) I(B), I(A B) = I(A) I(B), I(A B) = I(A) I(B).

A sequent A B is satisfied in model B (B A B) if I(A) I(B). A sequent A B is valid inan Ba B if it is satisfied in B, I for all interpretations I .

Now let U = U , I be the term model for BPC with the canonical interpretation function I , i. e., I(p) = |p|for every propositional variable p. Then for every formula A, I(A) = |A|. So BPC A iff I(A) = 1, and forevery sequent A B, BPC A B iff I(A) I(B).

Theorem 2.8 (Soundness and Completenes) Let A |A| be a map, as described above, from the languageof BPC to a Ba B. Then for all sequents A B, if BPC A B, then |A| |B| in B. Conversely, letU = U , I be the term model of BPC with I : A |A|. Then |A| |B| if and only if BPC A B.

P r o o f. By induction on the length of proofs. We consider only the Implication Introduction case. SupposeA B C follows from A B C. By induction |A| |B| |C|. Then |B| |A| |B| |B| |C|and |A| 1 |A| |B| |A| = |B| |A| |B|, by Proposition 1.2. So |A| |B| |C|. The converse isobserved above.

Before we continue this topic, we prove a useful property of BPC, which we need in the sequel. In thefollowing definition means deduction in BPC.

Definition 2.9 A set of sequents is called faithful if (A1 B1) (An Bn) A B implies {(A1 B1), . . . , (An Bn)} (A B).

All extensions of IPC are faithful, since in these extensions C D is equivalent to C D for all formulasC, D. In particular CPC is faithful. Let E1 be the theory axiomatized by . E1 is not faithful, otherwise,it would be inconsistent. We want to show that BPC is faithful. Our proof is Kripke model theoretic. A Kripkemodel for BPC is like the one for IPC, expect that the underlying set of nodes need not be reflexive. So for aKripke modelK = (K,,), the forcing relation for implication is

for k K , k A B iff for all k k, if k A, then k B.

Given a Kripke model K with root k, let K be the model formed from K by adding a new root node k0 kwhich is reflexive exactly when k is, and such that k0 p exactly when k p.

Proposition 2.10 Let be a set of sequents such that its class of Kripke models is closed under the followingtransformation: If K is a rooted Kripke model of with irreflexive root, then so is K. Then is faithful.

P r o o f. See [2].


Math. Log. Quart. 50, No. 1 (2004) / www.interscience.wiley.com 69

Theorem 2.11 BPC is faithful.P r o o f. Let K be a rooted Kripke model of BPC with root k and let k0 be the root of K. Suppose

k0 (A B) (B A). Then k0 A B and k0 B A. So there are k1, k2 k0 such thatk1 A, k1 B, k2 B and k2 A. Since K is a model of BPC, it is not the case that both k1, k2 k. Onthe other hand, if one of k1 or k2, say k1, is k, then k1 k2. This contradicts with k1 A and k2 A. So K isa model of BPC.

Now we go back to Ba .Corollary 2.12 Let F be the unit filter on U , i. e., F = {1}. Then U/F U .P r o o f. We show that for every formula A, [|A|] = {|A|}. Let |B| [|A|], then || = |A B| and

|| = |B A|. By faithfulness of BPC, BPC A B. Thus |A| = |B|, and so [|A|] = {|A|}. ThusU/F U .

Theorem 2.13 The basic algebra U is a subalgebra of a direct product of chains.P r o o f. For every prime filter F , let F be the natural homomorphism from U to U/F . This will define a

homomorphism from U toU/F . U is an Ba , so by Lemma 2.6, U/F is a chain, for every prime filter F .To show that is one to one, let |A|, |B| U and (|A|) = (|B|). Then for every prime filter F , we have(|A| |B|) (|B| |A|) F . So by Corollary 1.7, |A| |B| = |B| |A| = 1. Then BPC A B andBPC B A. Therefore BPC A B, by Theorem 2.11. That means |A| = |B|.

The two following Lemmas are Ba versions of Horn [4]:Lemma 2.14 Let B be the direct product of Bas Bi, i I , let B = B, f be an algebraic model and let

Bi = Bi, fi be algebraic models for every i I such that fi = i f , where i is the projection of B on Bi.Then for any sequent A B we have for every i I , i(f(A)) i(f(B)) iff fi(A) fi(B). Hence, A Bis valid in B iff it is valid in each Bi.

Lemma 2.15 If B is a subalgebra of A, then for every algebraic modelB = B, f and every sequent A Bwe have f(A) f(B) in B iff f(A) f(B) in A. Hence, if A B is valid in A, it is valid in B.

Theorem 2.16 For any sequent A B, the followings are equivalent:1. BPC A B2. For every Ba , B, B A B.3. For every chain B, B A B.4. U A B.P r o o f. By Theorem 2.8, 1. implies 2. and 4. implies 1. Since every chain is an Ba , 2. implies 3. By

Lemmas 2.14, 2.15 and Theorem 2.13, 3. implies 4.

As applications of the above Theorem 2.16, we have the following corollaries:Corollary 2.17 BPC (A ( B)) ((A B) A).P r o o f. Let B be any chain and a, b B. Then either (a b) a, which yields (a b) a = 1, or

a < a b, which with a 1 a implies a (1 a) (a b) (1 b). So a (1 b) = 1. In eithercase, (a (1 b)) ((a b) a)) = 1. The result follows from Theorem 2.16.

On the other hand, Example 2.2 shows:Corollary 2.18 BPC (A B) ((A B) A).The next corollary is Proposition 1.7 in [6]:Corollary 2.19 BPC is not complete with respect to the class of all linear Kripke models.P r o o f. We note that both of the axiom schemas (A B) ((A B) A) and (A B) (B A)

are valid in every linear Kripke model. The result follows from Corollary 2.18.



Our last theorem shows that Bas may be characterized by the fact that a certain subset of the set of theirprime filters has a tree-like property.

Theorem 2.20 A Ba is an Ba iff for every prime filter F the set S = {G : F G and G is a prime filter}is linearly ordered by inclusion.

P r o o f. Suppose B is an Ba and F is a prime filter on B. Suppose G, H S are incomparable. Then thereexist x GH and y H G. Since (x y) (y x) = 1, then either (x y) F or (y x) F . Sowe have y G or x H , a contradiction.

Conversely, suppose the condition holds but (x y) (y x) = 1 for some x, y B. By Corollary 1.7there is a prime filter F on B which dose not contain (x y) (y x). So x y / F and y x / F . ByLemma 1.5, there exist G, H S such that x G, y / G and y H , x / H . Contradiction.

Corollary 2.21 Let B be an Ba . Then for every x, y, z B,1. x y z = (x z) (y z),2. z x y = (z x) (z y).P r o o f.1. Clearly (x z) (y z) x y z. Suppose x y z (x z) (y z). Then by Lemma

1.5, there is a prime filter F such that (x y z) F and (x z) (y z) / F . So x z, y z / F .Again by Lemma 1.5 there are prime filters F1, F2 such that F F1, F F2, and x F1, z / F1, y F2, andz / F2. Since B is an Ba, by Theorem 2.20, F1 F2 or F2 F1. Suppose F1 F2. Then x y F1 andz / F1. On the other hand, from F F1 and x y F1 we have z F1. A contradiction.

2. Similar to 1.

Acknowledgements The first author would like to thank the Institute for Studies in Theoretical Physics and Mathematics(IPM) for the financial support.

References[1] M. Ardeshir, Aspects of Basic Logic. PhD thesis, Department of Mathematics, Statistics and Computer Science, Mar-

quette University, 1995.[2] M. Ardeshir and W. Ruitenburg, Basic propositional calculus I. Math. Logic Quarterly 44, 317 343 (1998).[3] M. Dummett, A propositional calculus with denumerable matrix. J. Symbolic Logic 24, 97 106 (1959).[4] A. Horn, Logic with truth values in a linearly ordered Heyting algebra. J. Symbolic Logic 34, 395 408. (1969)[5] H. Rasiowa and R. Sikorski, The Mathematics of Metamathematics (PWN, Warsaw 1963).[6] Y. Suzuki, F. Wolter, and M. Zakharyaschev, Speaking about transitive frames in propositional languages. Journal of

Logic, Language and Information 7, 317 339 (1998).[7] A. S. Troelstra and D. van Dalen, Constructivism in Mathematics, Vol. 1 (North-Holland Publ. Comp., Amsterdam 1988).[8] A. Visser, A propositional logic with explicit fixed points. Studia Logica 40, 155 175 (1981).


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Alizadeh-Ardeshir-On the Linear Lindenbaum Algebra of Basic Propositional Logic