Canonical extensions of ordered algebraic structures

and relational completeness of some substructural

logics∗

Michael Dunn Mai Gehrke† Alessandra Palmigiano

Abstract

In this paper we introduce canonical extensions of partially or-dered sets and monotone maps and a corresponding discrete duality.We then use these to give a uniform treatment of completeness of re-lational semantics for various substructural logics with implication asthe residual(s) of fusion.

1 Introduction

Canonical extensions were first introduced by Jonsson and Tarski for Booleanalgebras with operators (BAOs) in their 1950’s papers [14, 15]. Canonicalextensions provide an algebraic formulation of what is otherwise treated viatopological duality or relational methods. The theory of canonical exten-sions has since been simplified and generalized [9, 7, 10], leading to a widelyapplicable and transparent theory which is now ready to be applied evenin the setting of partially ordered algebras. The only restriction is that thebasic operations of the algebras to be considered either preserve or reversethe order in each coordinate. We will call such algebras monotone posetexpansions (MPEs).

Rather than developing a complete and general theory of canonical ex-tensions for MPEs at this stage, we have opted here to develop only what isnecessary to solve a particular problem. In recent years a number of paperson completeness of various substructural logics with respect to relationalsemantics have been published [2, 1, 3, 16, 17]. Relational semantics have

0The second author’s research was partially supported by grant NSF01-4-21760 of theUSA National Science Foundation

0The authors wish to thank M. Dunn’s student Chunlai Zhou for his reading an earlierversion and for his suggestions and corrections.

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proved very useful for modal and intuitionistic logics. In addition, these fitwell with the algebraic theory as they are closely related to topological dual-ity for these varieties. The purpose of searching for relational semantics forsubstructural logics as well is of course that it is hoped that these semanticswill prove a powerful tool in a similar manner in this setting as well. How-ever, topological duality for non-distributive lattice based ordered algebrasis vastly more complicated than its Boolean and distributive counterpartsand this is where canonical extensions come in. Canonical extensions havebeen explored to some extent for non-distributive lattices [7], and the sur-prising fact is that the theory is very similar and essentially as smooth inthis setting as in the Boolean setting.

The purpose of this paper then is to explore the potential of canonicalextensions as a tool for the development of a smooth and transparent the-ory of relational methods for non-distributive lattice based ordered algebraicstructures. To this end we will consider the setting from Dunn’s paper [5].There he attempted to provide a uniform approach to relational semanticsfor various implicative substructural logics by means of representation the-orems for the corresponding algebras suggested by his gaggle theory, see [4](further references and more can be found in [6]). In this work he consideredseveral logics including the Lambeck calculus, linear logic, BCK logic, andrelevance logic. While he achieved a uniform approach in the sense that therelational semantics obtained all arose through concrete representation ofthe algebras, he had to change his method of representation in ad hoc waysto fit the various logics.

We show here that using canonical extensions and the associated dis-crete duality we obtain complete relational semantics for each of the logicsconsidered in [5].

We have organized the paper as follows: In section 2 we define the canon-ical extension of a poset, and derive some of its properties. In section 3 wetreat extension of maps and define canonical extensions for MPEs. In sec-tion 4 we develop a discrete duality for a class of complete lattices thatincludes all lattices that arise as canonical extensions of posets. In section5 we discuss the dual of additional binary operations of the type appropri-ate to our applications. Finally, in section 6 we use the results from theprevious sections to obtain relational semantics for the substructural logicsmentioned above.

It should be noted that apart from solving the problem at hand, the workherein gives the fundamental ideas for how to define topological dualities,possibly for MPEs in general, and definitely for LEs. In particular, the dualcharacterization of arbitrary morphisms in our discrete duality provides the

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relational part of the dual characterizations of arbitrary homomorphisms fortopological duality of bounded lattices, a long standing open problem.

2 The canonical extension of a poset

Let P be a poset. Recall that a filter of P is a non-empty subset F of Psatisfying:

1. F is an up-set, that is, if x ∈ F , y ∈ P , and x ≤ y, then y ∈ F ;

2. F is down-directed, that is, if x and y are in F , then there exists z ∈ Fwith z ≤ x and z ≤ y.

An ideal of P is defined dually. That is, I is an ideal of P provided I is anon-empty up-directed down-set of P . We denote by F and I the familiesof all filters and ideals of P , respectively.

We make the following definitions here:

Definition 2.1. Let P be a poset.

1. An extension of P is an order embedding e : P → Q, i.e., for everyx, y ∈ P , x ≤ y if and only if e(x) ≤ e(y). For ease of notation wewill suppress e and call Q an extension of P and assume that P is asubposet of Q.

2. Given an extension Q of P , an element of Q is called closed providedit is the infimum in Q of some filter F of P . We denote the set of allclosed elements of Q by K(Q). Dually, an element of Q is called openprovided it is the supremum in Q of some ideal I of P . We denote theset of all open elements of Q by O(Q).

3. An extension Q of P is said to be dense provided each element of Q isboth the supremum of all the closed elements below it and the infimumof all the open elements above it.

4. An extension Q of P is said to be compact provided that whenever D isa non-empty down-directed subset of P , U is a non-empty up-directedsubset of P , and

∧Q D ≤ ∨

Q U , then there are x ∈ D and y ∈ U withx ≤ y.

We are now almost ready to give the abstract definition of the canonicalextension of a poset. But first recall that a completion of a poset P is anextension Q of P which also happens to be a complete lattice.

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Definition 2.2. Let P be a poset. A canonical extension of P is a denseand compact completion of P .

Remark 2.3. This definition exactly agrees with the definition of canoni-cal extension for bounded lattices [7], bounded distributive lattices [9], andBoolean algebras [14]. As we shall see next, this definition provides an ab-stract characterization of the very natural completion of P obtained via thepolarity given by the set of filters of P , the set of ideals of P , and the re-lation of non-disjointness between these. One may still question our choiceof filters and ideals, if for no other reason then at least because there areseveral other possible choices that agree with this one in the case of boundedlattices. A note currently in preparation by the two last authors developsextensions of this type in greater generality and shows that the choice whichleads to what is called canonical extension here is unique in allowing certain(algebraically) desirable properties such as preserving the existing finitarylattice structure, commuting with (Cartesian) product, and destroying allpurely infinitary lattice structure. We may also mention that this choicecan be viewed as natural from the point of view of abstract algebraic logic.This is the subject of an article in preparation by the third author. Finally,there certainly are situations in which one does not necessarily want to de-stroy all the infinitary structure of the original poset (e.g. dynamic modallogic) and then one might want to restrict the set of filters and ideals. How-ever, this may lead to loss of preserved algebraic identities. Nevertheless,investigations in this direction would certainly be of interest.

The following proposition tends to cut our work in half when provingvarious results and is of intrinsic importance for applications in logic. Forany poset P we denote by P ∂ the dual poset obtained by reversing the orderon P .

Proposition 2.4. Let P be a poset. If Q is a canonical extension of P ,then Q∂ is a canonical extension of P ∂.

Proof. This follows from the fact that each of the defining properties: dense-ness, compactness, and being a completion are self-dual.

Theorem 2.5. Let P be a poset. If P has a canonical extension then it isunique up to an isomorphism that fixes P .

Proof. First we show that the set of closed elements of any canonical exten-sion Q of P is reverse order isomorphic to the poset (F ,⊆) of filters of P .First of all notice that the definition of closed elements yields a surjective

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order reversing map from F to K(Q) given by F 7→ xF =∧

Q F . Now letF,G ∈ F with xF ≥ xG. Let p ∈ F , then xF ≤ p and thus xG ≤ p. Wenow have D = G is a non-empty down-directed subset of P , U = {p} isa non-empty up-directed subset of P , and

∧Q G = xG ≤ p =

∧Q{p}. By

compactness we get p ∈ G and we have shown that F ⊆ G.By duality we then get that the set of open elements of any canonical

extension Q of P is order isomorphic to the poset (I,⊆) of ideals of P viathe order isomorphism I 7→ yI =

∨Q I.

Next we show that the order structure on K(Q) ∪ O(Q) is uniquelydetermined by showing that for F ∈ F and I ∈ I we have:

1. xF ≤ yI if and only if F ∩ I 6= ∅2. yI ≤ xF if and only if for all x ∈ I and for all y ∈ F , x ≤ y.

To see that this is so notice that the ‘if’ part of 1 is trivial and the ‘onlyif’ part follows by compactness, whereas 2 simply is a reformulation of thedefinitions of supremum and infimum. Thus we have shown that for any twocanonical extensions of P the union of the closed and the open elements ofthe two extensions are isomorphic via an isomorphism that fixes P .

Finally notice that denseness implies that any canonical extension is theDedekind-MacNeille completion of the union of its closed and open elements.Now the uniqueness of the Dedekind-MacNeille completion of a poset com-bined with the uniqueness of the structure of the closed elements union theopen elements yields the desired result.

Theorem 2.6. Let P be a poset, then P has a canonical extension.

Proof. The above uniqueness proof essentially spells out a construction ofthe canonical extension of P : First one defines a quasi-order on F ∪ Iaccording to the two conditions given in the above proof. The correspondingpartially ordered quotient will yield the open elements union the closedelements. Then one takes the Dedekind-MacNeille completion of this poset.

Alternatively, the polarity given by non-empty intersection between thefilters and the ideals of P yields a Galois connection:

Φ : P(F) À P(I) : Ψ

where Φ(X) = {I : F ∈ X implies F ∩ I 6= ∅} and Ψ(X) = {F : I ∈X implies F ∩ I 6= ∅}. The Galois closed subsets G = {X ⊆ F : Ψ(Φ(X)) =X} form a complete lattice. It is not too hard to show that the map e :P → G given by e(p) = {F : p ∈ F} is a canonical extension of P . We leavethe details to the reader.

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Remark 2.7. The description/construction of the canonical extension givenin the proof of Theorem 2.5 is closely related to the construction of thecanonical extension of BAOs given in [13], and the second construction cor-responds to the one used for bounded lattices in [7]. Both constructions areclearly constructive unlike the more traditional constructions via topologicalduality. Below, we will see that the non-constructiveness comes in when wewant to stipulate that the canonical extension has enough completely joinand meet irreducible elements.

We will denote the canonical extension of a poset P by P σ. The set of allcompletely join irreducible elements of a poset P we will denote by J∞(P ),the set of all completely meet irreducible elements of P we will denote byM∞(P ). We now give a few fundamental properties of canonical extension.

Theorem 2.8. Let P , and Q be posets, then

1. (P ∂)σ = (P σ)∂.

2. (P ×Q)σ = P σ ×Qσ.

3. J∞(P σ) is join-dense in P σ and M∞(P σ) is meet-dense in P σ.

4. Any finite meets and joins existing in P are preserved in P σ.

Proof. 1 is just Proposition 2.4. To prove 2 first notice the projection ontoeither coordinate of an up- (down-)directed subset of the product is up-(down-)directed. Conversely, the product of two up- (down-)directed setsin the factors is up- (down-)directed in the product. Thus, viewing P σ ×Qσ as an extension of P × Q, the open (closed) elements are exactly thecoordinate-wise open (closed) elements. Thus it is clear that P σ × Qσ is adense completion of P ×Q. To see that it is also compact, we use again thatthe projection onto either coordinate of an up- (down-)directed subset ofthe product is up- (down-)directed. Then the coordinate-wise compactnessyields the compactness of the product.

Statement 3 holds essentially because filters (ideals) are closed underunions of chains. Also notice that the axiom of choice is needed to prove3. To show that J∞(P σ) is join-dense in P σ, we just need to show thateach closed element is the join of the completely join irreducible elementsbelow it as we already know that the closed elements are join dense in P σ.To this end let x be a closed element of P σ. Let u be the join of 0 andall the completely join irreducible elements of P σ that are below x. Thencertainly u ≤ x. Suppose however that u < x. Now we use the fact that

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the open elements of P σ are meet-dense in P σ to get y ∈ O(P σ) withu ≤ y but x � y. Consider the set S = {v ∈ K(P σ) : v ≤ x but v � y}.Then S is non-empty since x ∈ S. Also if C is a non-empty chain in S,then we claim that c0 =

∧C is again in S. First of all c0 is a closed

element. This is because c0 =∧{p ∈ P : p ≥ c for some c ∈ C} and the

set {p ∈ P : p ≥ c for some c ∈ C} is a filter of P since C is a non-emptychain. Secondly c0 ≤ c ≤ x for any one c ∈ C, so c0 ≤ x. Now supposec0 ≤ y. But then

∧{p ∈ P : p ≥ c for some c ∈ C} ≤ y and by compactnessthere is p ∈ P and c ∈ C so that p ≥ c and p ≤ y. This implies thatc ≤ y which contradicts the fact that c ∈ S. Thus c0 � y and c0 ∈ S andtherefore by Zorn’s Lemma we may conclude that S has minimal elements.Let j be minimal in S. Since j � y we know that j 6= 0. We show thatj is completely join irreducible. To this end suppose that j =

∨T where

T is a subset of P σ. Again because the closed elements are join-dense inP σ, we may assume that T is a set of closed elements. For each t ∈ T witht < j we know that t /∈ S by minimality of j. The only condition that canbe violated is t � y, and thus we know that for each t ∈ T with t < j wehave t ≤ y. It follows that

∨{t ∈ T : t < j} ≤ y, and as j � y, we concludethat

∨{t ∈ T : t < j} 6= j. As j =∨

T there must be t ∈ T with t ≮ j,that is t = j. So we have shown that there is a completely join irreducibleelement in S. But j completely join irreducible and j ≤ x imply that j ≤ uby definition of u. This in turn implies that j ≤ y since u ≤ y by definitionof y. But this contradicts j ∈ S which implies that j � y. We conclude thatu = x and that x is in fact the join of all the completely join irreducibleelements below x.

Finally, to prove 4, let x, y ∈ P and suppose x ∨ y exists in P and callthis element z. Now let u denote the supremum of x and y in P σ. Thenu ≤ z. We show that z ≤ u by showing that that every open element vabove u also is above z. So let v ∈ O(P σ) with u ≤ v. Then x ≤ v andy ≤ v. The fact that v is open means that I = {p ∈ P : p ≤ y} is an idealand that the join of this ideal is v. Since x, y ∈ I there must be p ∈ I withx ≤ p and y ≤ p. Now, since z is the supremum of x and y in P , we musthave that z ≤ p and thus z ∈ I. We conclude that z ≤ v and thus z ≤ u andx∨P y = x∨P σ y. The corresponding statement for binary meets follows byduality. Notice also that if P has a zero then every open element of P σ isabove it since ideals are required to be non-empty. But then the fact thatthe opens are meet dense in P σ implies that the zero of P is also the zeroof P σ. Dually for 1.

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The key fact for the use of canonical extensions in studying relationalsemantics for various logics corresponding to BAOs and DLEs is that canon-ical extension produces a concrete algebra, that is, an algebra which may beviewed as the complex algebra of some relational structure. The correspond-ing fact for the underlying lattice is that it may be viewed as a completefield of sets in the Boolean case and a complete ring of sets in the DL case.This can of course not be true once we are outside the scope of distributivelattices. Nevertheless there is an analogue here, and the complete latticesobtained when taking the canonical extensions of posets are also concretein some sense that we will describe in Section 4. Right here we give theabstract description of these ‘concrete’ lattices.

Definition 2.9. A complete lattice C is perfect provided J∞(C) is join-dense in C and M∞(C) is meet-dense in C, i.e. for every x ∈ C, x =

∨{j ∈J∞(C) : j ≤ x} and x =

∧{m ∈ M∞(C) : x ≤ m}.It should be clear that among Boolean algebras the ones that are perfect

lattices are exactly the complete and atomic Boolean algebras, and amongdistributive lattices the ones that are perfect are exactly the complete, com-pletely distributive lattices that are join generated by their completely joinprime elements, or equivalently the doubly algebraic complete distributivelattices.

Corollary 2.10. Let P be any poset, then P σ is a perfect lattice.

Proof. This is exactly the content of statement 3 in the above theorem.

3 Canonical extensions of maps

In order to be able to define canonical extensions for monotone poset ex-pansions, we now need to describe how we will extend additional operationsfrom a poset to its canonical extension. First we define monotone operationsand monotone poset expansions.

Definition 3.1. Each ε ∈ {1, ∂}n is called a monotonicity type. An n-aryoperation f : Pn → P on a poset P is said to be monotone provided there isε ∈ {1, ∂}n so that the order-variant f : P ε → P ,where P ε = P ε1×. . .×P εn

and P 1 is P and P ∂ is the order-dual of P , is order preserving. For moredetails of this notation see [11].

A monotone poset expansion (MPE) is a tuple (P, (fi)i∈I) where P is aposet and each fi is an ni-ary monotone operation on P . The corresponding

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sequence (εi)i∈I of monotonicity types for the individual operations is calledthe monotonicity type of the MPE (P, (fi)i∈I).

Given a monotone operation f : Pn → P on a poset P , we may considerit as an order preserving map f : P ε → P whose domain is obtained fromP by taking some combination of order duals and finite product. Sincecanonical extension of posets commutes with both of these, all we need todefine is how to extend order preserving maps.

Definition 3.2. Let P and Q be posets, and f : P → Q an order preservingmap. Define maps fσ, fπ : P σ → Qσ by setting:

fσ(u) =∨{∧{f(p) : x ≤ p ∈ P} : u ≥ x ∈ K(P σ)}

fπ(u) =∧{∨{f(p) : y ≥ p ∈ P} : u ≤ y ∈ O(P σ)}.

Remark 3.3. It should be clear to anyone familiar with the theory ofcanonical extensions that these definitions agree with the ones given in thebounded lattice, distributive bounded lattice, and Boolean cases. In thesetting of bounded distributive lattices it was shown in [10] that these ex-tensions satisfy universal properties with respect to certain topologies oncanonical extensions. It was also shown in [10] and [7] that canonical exten-sions (both the sigma or the pi versions) are functorial for monotone DLEsand LEs. Of course the corresponding questions should also be consideredand answered in this more general setting eventually. However, they are notcentral to the problem considered in this paper and we leave them for futurework.

As in the lattice case, the following facts hold for these extensions:

Lemma 3.4. For every order preserving map f : P → Q, both fσ andfπ are order preserving extensions of f . In addition fσ ≤ fπ with equalityholding on both K(P σ) and O(P σ). For u ∈ P σ, x ∈ K(P σ), and y ∈ O(P σ)we have

fσ(u) =∨{fσ(x) : u ≥ x ∈ K(P σ)}

fσ(x) =∧{f(p) : x ≤ p ∈ P}

fπ(u) =∧{fπ(y) : u ≤ y ∈ O(P σ)}

fπ(y) =∨{f(p) : y ≥ p ∈ P}

and fσ and fπ send closed elements to closed elements and open elementsto open elements.

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Proof. Let x, x′ ∈ K(P σ) with x ≤ x′, then as f is order preserving wehave

∧{f(p) : x ≤ p ∈ P} ≤ ∧{f(p) : x′ ≤ p ∈ P} and therefore fσ(x) =∧{f(p) : x ≤ p ∈ P}. Notice also that the set {f(p) : x ≤ p ∈ P} is down-directed so that fσ sends closed elements to closed elements. Furthermorethis description of fσ for closed elements together with the fact that fis order preserving easily yields the fact that fσ extends f . Finally, italso easily implies that fσ is order preserving on K(P σ) and that fσ(u) =∨{fσ(x) : u ≥ x ∈ K(P σ)}. This last fact in turn easily implies that fσ isorder preserving on all of P σ. By duality all the corresponding statementshold for fπ. In terms of the claims about the relationship between fσ andfπ, notice that for x ∈ K(P σ)

fπ(x) =∧{fπ(y) : x ≤ y ∈ O(P σ)}

≤∧{fπ(p) : x ≤ p ∈ P}

=∧{f(p) : x ≤ p ∈ P} = fσ(x).

Also, for each y ∈ O(P σ) with x ≤ y, by compactness, there exists p ∈ Pwith x ≤ p ≤ y. Thus

∧{fπ(y) : x ≤ y ∈ O(P σ)} ≥

∧{fπ(p) : x ≤ p ∈ P}

and we conclude that fπ(x) = fσ(x). By duality fσ(y) = fπ(y) for y ∈O(P σ) as well. Finally we see that for u ∈ P σ

fσ(u) =∨{fσ(x) : u ≥ x ∈ K(P σ)}

=∨{fπ(x) : u ≥ x ∈ K(P σ)}

≤∧{fπ(y) : u ≤ y ∈ O(P σ)} = fπ(u).

Since we consider two, generally, different ways of extending maps, thecanonical extension of an MPE is not uniquely determined. As in the latticeand distributive lattice settings, whether we want to extend a particular ad-ditional operation using the σ- or the π-extension depends on the propertiesof the particular operation to be extended. Here, as in [8] we make a generaldefinition of β-canonical extensions.

Definition 3.5. Let (P, (fi)i∈I) be an MPE with additional operations in-dexed by a set I. Let β be a map from I to the set {σ, π}. Then theβ-canonical extension of (P, (fi)i∈I) is the MPE (P σ, (fβ(i)

i )i∈I).

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In order to obtain relational semantics from a class of algebras it isnecessary that the additional operations are such that they can be encodedby relations on the dual. This will be the case when the extensions of themaps are (possibly up to some turning upside down of coordinates) completeoperators or complete dual operators. Here we show that this happens forresiduated maps.

First we recall some facts about residuation of maps. Let P , Q and Rbe partial orders and let f : P ×Q → R, g : P ×R → Q and h : R×Q → P .The map g is called the right residual of f provided for every p ∈ P , q ∈ Qand r ∈ R,

f(p, q) ≤ r if and only if q ≤ g(p, r).

h is called the left residual of f provided for every p ∈ P , q ∈ Q and r ∈ R,

f(p, q) ≤ r if and only if p ≤ h(r, q).

Further, if g and h are the right and left residuals of f , respectively, then

1. f : P × Q → R, g : P ∂ × R → Q and h : R × Q∂ → P are orderpreserving.

2. If P , Q and R are complete lattices, then f : P × Q → R preservesarbitrary joins in each coordinate, and g : P ∂ × R → Q and h :R×Q∂ → P preserve arbitrary meets in each coordinate.

Proposition 3.6. Let P , Q and R be partial orders and let f : P ×Q → Rbe order preserving. Then for g : P ×R → Q and h : R×Q → P ,

1. if g is the right residual of f , then gπ is the right residual of fσ.

2. If h is the left residual of f , then hπ is the left residual of fσ.

Proof. To prove statement 1, we have to show that for every u ∈ P σ, v ∈ Qσ,and w ∈ Rσ,

fσ(u, v) ≤ w if and only if v ≤ gπ(u,w).

We first show that if s ∈ K(P σ), t ∈ K(Qσ), y ∈ O(Rσ), and fσ(s, t) ≤ y,then t ≤ gπ(s, y). We have fσ(s, t) =

∧{f(p, q) : s ≤ p ∈ P, t ≤ q ∈Q} ≤ m. Since s and t are closed, so is fσ(s, t). In fact

∧{f(p, q) : s ≤p ∈ P, t ≤ q ∈ Q} is a down-directed family in P × Q whose infimum isfσ(s, t). Also, y is open and thus by compactness, f(p, q) ≤ m for somep ∈ P and q ∈ Q with s ≤ p and t ≤ q. Let f(p, q) = r ∈ R. Asf(p, q) ≤ r and g is the right residual of f , q ≤ g(p, r). Now since p ≤ s

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and r = f(p, q) ≤ m, g(p, r) ∈ {g(p′, r′) : s ≤ p′ ∈ P, m ≥ r′ ∈ R} andthus t ≤ q ≤ g(p, r) ≤ ∨{g(p′, r′) : s ≤ p′ ∈ P, m ≥ r′ ∈ R} = gπ(s, y) asdesired. For the general case, assume that fσ(u, v) ≤ w. We need to showthat v ≤ gπ(u,w) =

∧{gπ(s, y) : u ≥ s ∈ K(P σ), w ≤ y ∈ O(Rσ)}, i.e. thatv ≤ gπ(s, y) for every s ∈ K(P σ) and y ∈ O(Rσ), with s ≤ u and w ≤ y.By denseness, v =

∨{t ∈ K(Qσ) : t ≤ v} and it is enough to show thatt ≤ gπ(s, y) for every s ∈ K(P σ)), t ∈ K(Qσ), and y ∈ O(Rσ) with s ≤ u,t ≤ v, and w ≤ y. By assumption,

∨{fσ(s, t) : u ≥ s ∈ K(P σ), v ≥ t ∈K(Qσ)} = fσ(u, v) ≤ w, so fσ(s, t) ≤ y for every s ∈ K(P σ), t ∈ K(Qσ),and y ∈ O(Rσ) such that s ≤ u, t ≤ v and w ≤ y. We have already seenthat this implies t ≤ gπ(s, y).

For the other direction, assume v ≤ gπ(u, w). Then t ≤ gπ(s, y) when-ever s ∈ K(P σ), t ∈ K(Qσ), y ∈ O(Rσ), s ≤ u, t ≤ v, and w ≤ y. Sincegπ(s, y) =

∨{g(p, r) : s ≤ p ∈ P and y ≥ r ∈ R}, by compactness,there are p ∈ P, r ∈ R, with s ≤ p, y ≥ r and t ≤ g(p, r). Using thefact that g is the right residual of f , we get f(p, g(p, r)) ≤ r, and thusfσ(s, t) ≤ f(p, g(p, r)) ≤ r ≤ y. It follows that fσ(u, v) ≤ w as desired.Statement 2 is obtained by interchanging the role of the coordinates of f .

Corollary 3.7. Let P , Q, and R be partial orders and f : P × Q → R amap with right and left residuals g : P × R → Q, and h : R × Q → P ,respectively. Then

1. fσ is a complete operator;

2. gπ : (P σ)∂ × Rσ → Qσ and hπ : Rσ × (Qσ)∂ → P σ are complete dualoperators.

4 Basic Discrete duality

Here we develop a duality between the categories of what we have namedperfect lattices and what we will call perfect posets.

Definition 4.1. A lattice C is called perfect provided J∞(C) is join-densein C and M∞(C) is meet-dense in C. A homomorphism of perfect latticesis a complete lattice homomorphism.

Given a perfect lattice C, we define the dual of C to be C = J∞(C) ∪M∞(C).

Example 4.2. Given a complete and atomic Boolean algebra B, we usuallydefine the discrete dual to be B+, the set of atoms of B, which of course

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is the same as the set of all completely join irreducible elements of B. Theset of completely join irreducible elements is sufficient as long as we staywithin the realm of DL+s. But here we must keep both the completely joinand the completely meet irreducible elements to be able to reconstruct thelattice. Notice that B may be obtained from B+ by taking the disjoint unionB+ ]B+ and defining an order by x ≤ y if and only if either x and y are inthe same copy of B+ and x = y, or x is in the first copy of B+, y is in thesecond copy of B+, and x 6= y.

Definition 4.3. A partial order Z is perfect provided J∞(Z) is join-densein Z, M∞(Z) is meet-dense in Z, and Z = J∞(Z) ∪M∞(Z).

Remark 4.4. The morphisms of this category are not the order preservingmaps or anything like that. The remaining work of this section discoverswhat they are. While it is not the most obvious and simple thing, theremarkable fact here is that they are something that is first order definable(also in the non-surjective case).

We first finish working out the duality on objects. Given a perfect posetZ, we define the dual of Z to be the Dedekind-MacNeille completion Z ofZ.

Example 4.5. Given a complete and atomic Boolean algebra, notice thatthe Dedekind-MacNeille completion of B is isomorphic to the complex alge-bra (namely the power set) of B+.

We will need the following possibly not so well-known but very importantproperty of the Dedekind-MacNeille completion.

Proposition 4.6. For every poset P , J∞(P ) = J∞(P ) and M∞(P ) =M∞(P ).

Proof. Let x ∈ J∞(P ). Then, as x =∨{p : x ≥ p ∈ P} by join-denseness

of P , x ∈ P . If x =∨

P A for some A ⊆ P ⊆ P , then, as the Dedekind-MacNeille completion preserves all existing joins, x =

∨P A, hence, by com-

plete join irreducibility of x in P , x ∈ A. So J∞(P ) ⊆ J∞(P ). Now let x ∈J∞(P ) and assume x =

∨A for some A ⊆ P . Now using the join-denseness

of P in P , we get x =∨

X, where X = {p ∈ P : p ≤ a for some a ∈ A},and since x is completely join irreducible in P , it follows that x = x′ forsome x′ ∈ X. But then, by the definition of X, there is a ∈ A with x′ ≤ a.Finally, since a ≤ x it follows that x = a ∈ A, and x is also completely joinirreducible in P . The statement for completely meet irreducible elementsfollows by order duality.

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Proposition 4.7. For every perfect lattice C,

1. (C) ∼= C;

2. J∞(C) = J∞(C) and M∞(C) = M∞(C);

3. C is a perfect poset.

Proof. The statement 1 follows from the abstract characterization of theDedekind-MacNeille completion of a poset as the complete lattice in whichthe poset is both join- and meet-dense: By the definition of a perfect latticeclearly C = J∞(C) ∪M∞(C) is both join- and meet-dense in C. By 1, thestatement 2 is exactly the content of Proposition 4.6. The statement 3 nowclearly follows by 2 and the definition of C.

Proposition 4.8. The following are equivalent for every poset P :

1. P is a perfect poset;

2. P = J∞(P ) ∪M∞(P ) and P is a perfect lattice.

Proof. If P is a perfect poset, then P = J∞(P )∪M∞(P ) by definition. Also,P is join-dense in P , and J∞(P ) is join-dense in P , and Dedekind-MacNeillecompletion preserves existing joins, so J∞(P ) is join-dense in P . Finally,by Proposition 4.6, it follows that J∞(P ) = J∞(P ). The fact that M∞(P )is meet-dense in P follows by order duality. For the converse, assume thatP = J∞(P ) ∪M∞(P ) and that P is a perfect lattice. By Proposition 4.7,J∞(P ) = J∞(P ), and since P is assumed to be a perfect lattice this set isjoin-dense in P . But then it is certainly also join-dense in the smaller setP . By order duality we conclude that M∞(P ) is meet-dense in P . We havethus shown that 2 implies 1.

Corollary 4.9. For every perfect poset Z, Z is a perfect lattice and (Z) = Z.

Now that we have defined the dual of objects we turn to maps. We willwant to apply this discrete duality not only to perfect lattices but actually toperfect lattices with additional monotone operations that are either completeoperators or complete dual operators. Thus, eventually at least, we need tofind duals not only for homomorphisms but also for complete operatorsand complete dual operators. We will do this, for the binary case, in thenext section. However, here we already treat the unary case as a completehomomorphism may be seen as a complete operator and a complete dualoperator with the additional equation that says the two are equal.

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Definition 4.10. Let f : C → D be a completely join preserving mapbetween perfect lattices. Define Rf ⊆ J∞(C)×M∞(D) as follows:

xRf n if and only if f(x) ≤ n.

Remark 4.11. Notice that this definition is a departure from the way Rf

is defined in the distributive and Boolean settings. There, Rf is a binaryrelation from J∞(D) to J∞(C) given by y ≤ f(x). This definition is alsoavailable to us here and would also work. In fact, it is an easy ‘toggle’between these two and it may very well be that one in general wants to keepboth available. However, for the work we will do in this paper, the abovedefinition makes the exposition slightly smoother, and it is all we need toname explicitely.

In the distributive and Boolean settings, we have a tight relationshipbetween J∞(C) and M∞(C) as they are isomorphic. This means that allthe structure can be moved to just one of these sets, as it traditionally is.In the general setting that we treat here, J∞(C) and M∞(C) are no longerisomorphic in general, and both sets must be kept in play. Accordingly,we will need some machinery to toggle back and forth between J∞(C) andM∞(C).

Definition 4.12. Given a poset P in which J∞(P ) is join-dense and M∞(P )is meet-dense, we define

()u : P(J∞(P )) −→ P(M∞(P ))

A 7−→ {m : a ∈ A implies a ≤ m}.

()l : P(M∞(P )) −→ P(J∞(P ))

B 7−→ {x : b ∈ B implies x ≤ b}.Remark 4.13. It is well known that these maps form a Galois connection.It is easy to see that the Dedekind-MacNeille completion of P , which is equalto the Dedekind-MacNeille completion of J∞(P ) ∪ M∞(P ), is isomorphicto the lattice of the Galois closed sets of this Galois connection. This isof course the restriction of the upper/lower Galois connection on P , whichgives the standard construction of the Dedekind-MacNeille completion. Theabove construction is sufficient exactly because J∞(P ) is join-dense in Pand M∞(P ) is meet-dense in P . Thus for a perfect poset Z this Galoisconnection gives Z, and for a perfect lattice C this Galois connection gives

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C = C. We only prove the following two facts about this Galois connection,which will be used repeatedly in what follows.

Lemma 4.14. If C is a perfect lattice, then for every X ⊆ J∞(C) andevery Y ⊆ M∞(C),

1.∧

Xu =∨

X.

2.∨

Y l =∧

Y .

Proof. We just prove 1. Let∨

X = x0. As C is a perfect lattice, x0 =∧{m ∈ M∞(C) : m ≥ x0}, so∨

X =∧{m ∈ M∞(C) : m ≥ x0}

=∧{m ∈ M∞(C) : x ∈ X implies m ≥ x}

=∧

Xu.

Proposition 4.15. For every completely join preserving map f : C → Dbetween perfect lattices, every x ∈ J∞(C) and every n ∈ M∞(D),

R1. (Rf [x])lu = Rf [x].

R2. (R−1f [n])ul = R−1

f [n].

Proof. In order to prove 1, as ( )lu is a closure operator, we just need toshow that (Rf [x])lu ⊆ Rf [x]. Let n ∈ (Rf [x])lu. We have to show thatf(x) ≤ n. As n ∈ (Rf [x])lu, then n ≥ ∨

R[x]l =∧

Rf [x], so it is enough toshow that f(x) =

∧Rf [x]. As D is perfect, then f(x) =

∧{n ∈ M∞(D) :f(x) ≤ n} =

∧Rf [x]. The proof of 2 is similar.

Definition 4.16. Let Y and Z be perfect posets, and let R ⊆ J∞(Y ) ×M∞(Z). We say that R is an (unary) operator relation provided for everyx ∈ J∞(Y ) and every n ∈ M∞(Z), the following conditions are satisfied:

R1. (R[x])lu = R[x].

R2. (R−1[n])ul = R−1[n].

Given an operator relation R ⊆ J∞(Y ) ×M∞(Z), define fR : Y → Zby setting:

1. For every x ∈ J∞(Y ), fR(x) =∧

R[x];

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2. For every u ∈ Y , fR(u) =∨{fR(x) : u ≥ x ∈ J∞(Y )}.

Lemma 4.17. Let Y and Z be perfect posets and R ⊆ J∞(Y )×M∞(Z) anoperator relation. For every u ∈ Y we have

fR(u) =∧

(⋂{R[x] : u ≥ x ∈ J∞(Y )})

=∧{m ∈ M∞(Z) : u ≥ x ∈ J∞(Y ) implies xRm}.

Proof. For every u ∈ Y ,

fR(u) =∨{fR(x) : u ≥ x ∈ J∞(Y )}

=∨{∧R[x] : u ≥ x ∈ J∞(Y )}

=∨{∧(R[x])lu : u ≥ x ∈ J∞(Y )}

=∨{∨(R[x])l : u ≥ x ∈ J∞(Y )}

=∨

(⋃{(R[x])l : u ≥ x ∈ J∞(Y )})

=∨

(⋂{R[x] : u ≥ x ∈ J∞(Y )})l

=∧

(⋂{R[x] : u ≥ x ∈ J∞(Y )}).

Proposition 4.18. Given Y and Z perfect posets and an operator relationR ⊆ J∞(Y )×M∞(Z), we have:

1. ≤ ◦R ◦ ≤ = R.

2. fR(x) ≤ n if and only if xRn.

3. fR is order preserving.

4. fR preserves arbitrary joins.

Proof. To prove 1 assume that x ≤ x′Rn, and let us show that xRn. By R2,it is enough to show that x ∈ (R−1[n])ul. As x′Rn, x′ ∈ R−1[n], so x′ ≤ mfor every m ∈ (R−1[n])u. Hence x ≤ x′ ≤ m for every m ∈ (R−1[x])u,and so x ∈ (R−1[n])ul. This shows that ≤ ◦ R ⊆ R. Now assume thatxRn′ ≤ n, and let us show that xRn. By R1, it is enough to show thatn ∈ (R[x])lu. As xRn′, n′ ∈ R[x], so y ≤ n′ for every y ∈ (R[x])l. Hence

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y ≤ n′ ≤ n for every y ∈ (R[x])l, and so n ∈ (R[x])lu. This shows that≤ ◦ R ◦ ≤ ⊆ R. The other containment is trivial as the order is reflexive.For 2 first notice that if xRn, then n ∈ R[x] and hence fR(x) =

∧R[x] ≤ n.

For the converse, assume that fR(x) ≤ n. By R1, it is enough to showthat n ∈ (R[x])lu. For every y ∈ (R[x])l, y ≤ ∧

R[x] = fR(x) ≤ n, hencen ∈ (R[x])lu. To prove 3 it is enough to show that fR is order preservingon J∞(Y ). Let x, x′ ∈ J∞(Y ) with x ≤ x′. If we show that R[x′] ⊆ R[x]then fR(x) =

∧R[x] ≤ ∧

R[x′] = fR(x′) and we are done. But if x′Rnthen x ≤ x′Rn and so by 1 xRn. To prove 4 let u ∈ Y and supposeu =

∨A. Since fR is order preserving we have fR(u) ≥ ∨

fR(A). Thuswe need to show that fR(u) ≤ ∨

fR(A). To this end first notice that asY is join-generated by J∞(Y ) we can suppose that A ⊆ J∞(Y ). But thenfR(a) =

∧R[a] and then, as in the proof of Lemma 4.17, we get

∨fR[A] =∧

(⋂{R[a] : a ∈ A}). Finally since each a ∈ A satisfies u ≥ a ∈ J∞(Y ),

we have∧

(⋂{R[a] : a ∈ A}) ≥ ∧

(⋂{R[x] : u ≥ x ∈ J∞(Y )}) and thus∨

fR[A] ≥ fR(u) as desired.

Proposition 4.19. For every completely join preserving map f : C → Dbetween perfect lattices, fRf

= f .

Proof. For every x ∈ J∞(C) = J∞(C), fRf(x) =

∧{n ∈ M∞(D) : xRfn} =∧{n ∈ M∞(D) : f(x) ≤ n} = f(x) and then the result follows as bothfunctions are completely join preserving.

Proposition 4.20. For any two perfect posets Y and Z and operator rela-tion R ⊆ J∞(Y )×M∞(Z), RfR

= R.

Proof. For x ∈ J∞(Y ) and n ∈ M∞(Z), we have xRfRn if and only if

fR(x) ≤ n if and only if xRn.

The above takes care of dualizing unary completely join preserving maps.By order duality we then also are able to dualize completely meet preservingmaps: If g : C → D is a completely meet preserving map between perfectlattices, then g∂ : C∂ → D∂ is a completely join preserving map betweenperfect lattices. The associated relation Sg ⊆ M∞(C) × J∞(D) is definedby setting for every m ∈ M∞(C) and every y ∈ J∞(D)

mSg y if and only if g(m) ≥ y,

and the dual relations, which we will refer to as (unary) dual operator rela-tions, are characterized by the following properties:

S1. (Sg[m])ul = Sg[m].

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S2. (S−1g [y])lu = S−1

g [y].

Notice that S1 is R2 for R−1, where R is a relation corresponding to acompletely join preserving map, and similarly for S2 and R1.

Now that we have found duals for completely join preserving as well ascompletely meet preserving maps we are ready to dualize homomorphisms.As mentioned above, we will do this by finding the dual characterization ofan equational theory where we have one complete operator f , one completedual operator g, and one equation f = g – or actually rather two inequalities:f ≤ g and g ≤ f . Since we have already described the relations dual to fand g, what we still need to do is to find first order correspondents of the twoinequalities f ≤ g and g ≤ f . Notice that, using modal notation, these are ofthe form 3 ≤ ¤ and ¤ ≤ 3, respectively. The first is the most basic type ofSahlqvist equation, namely the strictly positive kind, and the second is themore involved kind. Nevertheless, finding the first order duals in this settingis not much more involved than in Sahlqvist theory for classical modal logicbased on Boolean algebras. Thus this certainly indicates the basic buildingblocks of a Sahlqvist-type correspondence theory for the partially orderedsetting. Working this out further is the subject of current work by the twolast authors and H. Priestley.

Proposition 4.21. Let f, g : C → D be maps between perfect lattices with fcompletely join preserving and g completely meet preserving. Furthermore,let R be the operator relation corresponding to f , and let S be the dualoperator relation corresponding to g. Then the following are equivalent:

1. For all u ∈ C f(u) ≤ g(u);

2. For all x ∈ J∞(C) and m ∈ M∞(C), if x ≤ m then f(x) ≤ g(m);

3. For all x ∈ J∞(C) and m ∈ M∞(C), if x ≤ m then (R[x])l ⊆ S[m];

4. For all x ∈ J∞(C) and m ∈ M∞(C), if x ≤ m then (S[m])u ⊆ R[x].

Proof. That 1 implies 2 is clear. On the other hand if 2 holds and u ∈ C,then

f(u) =∨{f(x) : u ≥ x ∈ J∞(C)}

≤∧{g(m) : u ≤ m ∈ M∞(C)} = g(u)

and thus 1 holds. To show that 2 is equivalent to 3, notice that f(x) =∧R[x] =

∨(R[x])l and g(m) =

∨S[m] =

∨(S[m])ul. Finally, the equiva-

lence of 3 and 4 follows simply from the fact that S[m] and R[x] are closedunder ul and lu, respectively.

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The following lemma is the essential content of the Sahlqvist mechanismfor equations like ¤ ≤ 3. It is just the fact that completely join preservingand completely meet preserving maps are residuated combined with theobservation that the relation for the residual is the converse of the relationfor the original maps.

Lemma 4.22. Let f : C → D be a completely join preserving map betweenperfect lattices, and let n ∈ M∞(D). Then

For all u ∈ C f(u) ≤ n if and only if u ≤ un

where un =∨

R−1[n]. Dually, let g : C → D be a completely meet preservingmap between perfect lattices, and let y ∈ J∞(D). Then

For all u ∈ C g(u) ≥ y if and only if u ≥ vy

where vy =∧

S−1[y].

Proof. This is because f is completely join preserving and g is completelymeet preserving.

Proposition 4.23. Let f, g : C → D be maps between perfect lattices withf completely join preserving and g completely meet preserving. Then thefollowing are equivalent:

1. For all u ∈ C g(u) ≤ f(u);

2. For all n ∈ M∞(D) g(un) ≤ f(un);

3. For all y ∈ J∞(D) and n ∈ M∞(D), if vy ≤ un then y ≤ f(un);

4. For all y ∈ J∞(D) and n ∈ M∞(D),

if (S−1[y])l ⊆ R−1[n] then⋂{R[x] : x ∈ R−1[n]} ⊆ y ↑;

5. For all y ∈ J∞(D) and n ∈ M∞(D), if vy ≤ un then g(vy) ≤ n;

6. For all y ∈ J∞(D) and n ∈ M∞(D),

if (S−1[y])l ⊆ R−1[n] then⋂{S[m] : m ∈ S−1[y]} ⊆ n ↓.

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Proof. The fact that 1 implies 2 is clear. To see that 2 implies 1, let u ∈ Cand n ∈ M∞(D) with n ≥ f(u). Then u ≤ un and thus we have g(u) ≤g(un) ≤ f(un) ≤ n. Since this holds for every n ∈ M∞(D) g(u) ≤ f(u)and 1 holds. The fact that 2 is equivalent to 3 follows from the fact thatg(un) ≤ f(un) holds if and only if for all y ∈ J∞(D) y ≤ g(un) impliesy ≤ f(un), and from the fact that y ≤ g(un) holds if and only if vy ≤ un.The statement 4 is equivalent to 3 as it just uses the definitions of the variousentities in 3 in terms of R and S. Finally 5 and 6 follow by order dualityfrom 3 and 4, respectively.

Definition 4.24. Given perfect posets Y and Z, we define a morphism fromZ to Y to be a pair (R, S) of binary relations R ⊆ J∞(Y )×M∞(Z) and S ⊆M∞(Y )×J∞(Z) satisfying R1 and R2, and S1 and S2, respectively, as wellas the equivalent conditions of both Proposition 4.21 and Proposition 4.23.

Remark 4.25. While it is not clear at this point how to best understandthese morphisms, the main point is that all morphisms, not just surjectiveones, have first-order definable duals.

One alternative option which may be better than 4 and 6 in Proposition4.23 is:4. ((R−1[n])u)S ⊆ ((R−1[n])R)l;6. ((S−1[y])S)u ⊆ ((S−1[y])l)R

where AR = {n : for all x if x ∈ A then xRn}. Note that in this notationAu = A≤.

For surjective homomorphisms we get a somewhat simpler description ofthe dual, thus explaining why these were easier to discover.

Lemma 4.26. Let h : C → D be a complete homomorphism between perfectlattices that is surjective, then

1. For all n ∈ M∞(D) we have un ∈ M∞(C);

2. For all y ∈ J∞(D) we have vy ∈ J∞(C).

Proof. To prove 1, recall that un is the largest element of C that is mappedbelow n by h. Also, by surjectivity, we must have h(un) = n. Now we getn = h(un) =

∧{h(m) : un ≥ m ∈ M∞(C)} since h is also completely meetpreserving. But then, by complete meet irreducibility of n, we get n = h(m)for some m with un ≥ m ∈ M∞(C). But then, by maximality of un withrespect to h( ) ≤ n we have that un = m ∈ M∞(C). The proof of 2 is(order) dual.

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Thus for h : C → D a surjective complete homomorphism betweenperfect lattices, define

rh : M∞(D) → M∞(C)

n 7→ un =∨{x : h(x) ≤ n}

and

sh : J∞(D) → J∞(C)

y 7→ vy =∧{m : h(m) ≥ y}.

Proposition 4.27. Let h : C → D be a surjective complete homomorphismbetween perfect lattices, then

1. For all n, n′ ∈ M∞(D) n ≤ n′ if and only if rh(n) ≤ rh(n′);

2. For all y, y′ ∈ J∞(D) y ≤ y′ if and only if sh(y) ≤ sh(y′);

3. For all y ∈ J∞(D) and n ∈ M∞(D) y ≤ n if and only if sh(y) ≤ rh(n);

4. For all x ∈ J∞(C) (r−1h (x ↑))l = s−1

h (x ↓) and (s−1h (x ↓))u = r−1

h (x ↑);5. For all m ∈ M∞(C) (r−1

h (m ↑))l = s−1h (m ↓) and (s−1

h (m ↓))u =r−1h (m ↑).

Proof. For 1, if n ≤ n′ then clearly rh(n) ≤ rh(n′). For the converse, sinceh is onto, we have n = h(rh(n)) ≤ h(rh(n′)) = n′. The statement 2 is dualto 1. For 3, if y ≤ n, then h(sh(y)) = y ≤ n so sh(y) ≤ rh(n). Conversely,if sh(y) ≤ rh(n), then y = h(sh(y)) ≤ h(rh(n)) = n. Statements 4 and 5follow from the fact that for any u ∈ C r−1

h (u ↑) = {n : h(u) ≤ n} ands−1h (u ↓) = {y : h(u) ≥ y}.

Proposition 4.28. Let C and D be perfect lattices. For a pair of maps(r : M∞(D) → M∞(C), s : J∞(D) → J∞(C)) satisfying the five propertiesin Proposition 4.27, the map

h = h(r,s) : C → D

where h(u) =∧{n : u ≤ rh(n)} =

∨{y : u ≥ sh(y)} is a surjective completehomomorphism with rh = r and sh = s.

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Proof. Define R ⊆ J∞(C) ×M∞(D) by xRn if and only if x ≤ rh(n) andS ⊆ M∞(C)× J∞(D) by mRy if and only if m ≥ sh(y). Then we have foru ∈ C

fR(u) =∧{n : For all x if x ≤ u then xRn}

=∧{n : For all x if x ≤ u then x ≤ r(n)}

=∧{n : u ≤ r(n)}

and duallygS(u) =

∨{y : u ≥ s(y)}.

Thus we will have that h is a well-defined complete homomorphism if wecan show that (R, S) is a morphism of perfect posets. The fact that h issurjective and that rh = r and sh = s is easy to check.

Let n ∈ M∞(D), then R−1[n] = {r(n)}l and thus (R−1[n])ul = R−1[n].Dually we have (S−1[y])lu = S−1[y]. For x ∈ J∞(C),

(R[x])lu = ({n : x ≤ r(n)})lu

= (r−1(x ↑))lu

= r−1(x ↑)= R[x]

by property 4 in Proposition 4.27, and dually (S[m])ul = S[m] by property5 in Proposition 4.27. Thus fR is completely join preserving and gS iscompletely meet preserving. We now check that R and S give the samefunction. Let x ∈ J∞(C) and m ∈ M∞(C) with x ≤ m then

(R[x])l = s−1(x ↓)= {y : s(y) ≤ x}⊆ {y : s(y) ≤ x} = S[m].

Finally we show that for all n ∈ M∞(D) we have gS(un) ≤ fR(un). Noticethat

fR(un) = fR(r(n))

=∧{n′ : r(n) ≤ r(n′)}

=∧{n′ : n ≤ n′} = n,

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so we just need to show that gs(un) ≤ n. We have

gS(un) = gS(r(n))

=∨{y : s(y) ≤ r(n)}

=∨{y : y ≤ n} = n.

This completes the proof.

Thus one may take the duals of surjective perfect lattice morphisms tobe pairs of maps (r : M∞(Z) → M∞(Y ), s : J∞(Z) → J∞(Y )), where Yand Z are perfect posets, satisfying the five properties in Proposition 4.27.

5 Discrete duals of binary operators and dual op-erators

We now return to the task of developing duals of complete operators andcomplete dual operators.

Lemma 5.1. Let C1, C2 and D be perfect lattices, f : C1 × C2 → Dbe completely join preserving in each coordinate, g : C∂

1 × C2 → D andh : C1 × C∂

2 → D be completely meet preserving in each coordinate. Thenfor every ui ∈ Ci, i = 1, 2,

1. f(u1, u2) =∨{f(x1, x2) : ui ≥ xi ∈ J∞(Ci)}.

2. g(u1, u2) =∧{g(x1,m2) : u1 ≥ x1 ∈ J∞(C1) and u2 ≤ m2 ∈

M∞(C2)}.3. h(u1, u2) =

∧{h(m1, x2) : u2 ≥ x2 ∈ J∞(C2) and u1 ≤ m1 ∈M∞(C1)}.

Proof. To prove 1, let ui ∈ Ci, i = 1, 2. As Ci is perfect, ui =∨

Ai, whereAi = {xi ∈ J∞(Ci) : xi ≤ ui}. As f preserves arbitrary joins in eachcoordinate, we get that

f(u1, u2) = f(∨

A1, u2)

=∨{f(x1, u2) : x1 ∈ A1}

=∨{∨{f(x1, x2) : x2 ∈ A2} : x1 ∈ A1}

=∨{f(x1, x2) : xi ∈ Ai, i = 1, 2}.

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To prove 2, let ui ∈ Ci, i = 1, 2. As Ci is perfect, u1 =∨

A1, whereA1 = {x1 ∈ J∞(C1) : x1 ≤ u1}, and u2 =

∧B2, where B2 = {m2 ∈

M∞(C2) : m2 ≥ u2}. As g preserves arbitrary meetss in each coordinate,we get that

g(u1, u2) = g(∨

A1, u2)

=∧{g(x1, u2) : x1 ∈ A1}

=∧{∧{f(x1,m2) : m2 ∈ B2} : x1 ∈ A1}

=∧{f(x1,m2) : x1 ∈ A1 and m2 ∈ B2}.

The proof of 3 is similar to the proof of statement 2.

Remark 5.2. Notice that the result and the arguments given in the abovelemma and its proof depend on the fact that f(u1, u2) = 0 as soon as atleast one of u1 and u2 is 0 (and corresponding statements may be made forg and h). This will indeed be the case for all the basic operations consideredin this paper, but it is of course a little restrictive. A duality theory caneasily be developed without this restriction, however then 0 must be addedto the set of completely join irreducibles and 1 must be added to the set ofcompletely meet irreducibles. For work of this kind in the DL setting see[12].

Let C1, C2 and D be perfect lattices, and let f : C1 × C2 → D becompletely join preserving in each coordinate. Let us define Rf ⊆ J∞(C1)×J∞(C2)×M∞(D) as follows:

Rf (x1, x2, n) if and only if f(x1, x2) ≤ n.

Proposition 5.3. For every f : C1×C2 → D completely join preserving ineach coordinate, evey xi ∈ J∞(Ci), i = 1, 2, and every n ∈ M∞(D),

R1. (Rf (x1, x2, )lu = Rf (x1, x2, ).

R21. Rf ( , x2, n)ul = Rf ( , x2, n).

R22. Rf (x1, , n)ul = Rf (x1, , n).

Proof. To prove 1 let n ∈ (Rf (x1, x2, ))lu. We have to show that f(x1, x2) ≤n. As n ∈ (Rf (x1, x2, ))lu, then n ≥ ∨

(Rf (x1, x2, ))l =∧

Rf (x1, x2, )

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by lemma 4.14, so it is enough to show that f(x1, x2) =∧

Rf (x1, x2, ).Using the fact that D is a perfect lattice, we get the desired equality:

f(x1, x2) =∧{n ∈ M∞(D) : f(x1, x2) ≤ n}

=∧{n ∈ M∞(D) : Rf (x1, x2, n)}

=∧

Rf (x1, x2, ).

To prove 2 let x2 ∈ (Rf (x1, , n))ul, then x2 ≤∧

(Rf (x1, , n))u. By lemma4.14,

∧(Rf (x1, , n))u =

∨Rf (x1, , n). As f is completely join preserving,

f is order preserving, and so

f(x1, x2) ≤ f(x1,∨

Rf (x1, , n))

=∨{f(x1, x

′) : x′ ∈ Rf (x1, , n)}=

∨{f(x1, x

′) : f(x1, x′) ≤ n} ≤ n.

The proof of 3 is like the proof of statement 2.

Definition 5.4. Let Y1, Y2, and Z be perfect posets, and let R ⊆ J∞(Y1)×J∞(Y2)×M∞(Z). We call R a (binary) operator relation provided, for everyxi ∈ J∞(Yi), i = 1, 2 and for every n ∈ M∞(Z)

R1. (R(x1, x2, )lu = R(x1, x2, ).

R21. R( , x2, n)ul = R( , x2, n).

R22. R(x1, , n)ul = R(x1, , n).

Given such a binary operator relation, define fR : Y 1 × Y 2 → Z by setting:

1. For xi ∈ J∞(Y i), i = 1, 2, fR(x1, x2) =∧

R(x1, x2, ).

2. For ui ∈ Y i, i = 1, 2,

f(u1, u2) =∨{f(x1, x2) : ui ≥ xi ∈ J∞(Y i), i = 1, 2}

.

Proposition 5.5. Let Y1, Y2, and Z be perfect posets and R ⊆ J∞(Y1) ×J∞(Y2)×M∞(Z) be an operator relation, then

1. R( , x2, n) and R(x1, , n) are downsets, and R(x1, x2, ) is an upset.

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2. fR(x1, x2) ≤ n if and only if R(x1, x2, n).

3. fR is order preserving.

4. fR is completely join preserving in each coordinate.

Proof. To prove 1 we show that R( , x2, n) is a downset: Assume x1 ≤ x′

and R(x′, x2, n). We show that x1 ∈ (R( , x2, n))ul. As R(x′, x2, n), x′ ∈R( , x2, n), so x′ ≤ m for every m ∈ (R( , x2, n))u. So x1 ≤ x′ ≤ mfor every m ∈ (R( , x2, n))u and consequently x1 ∈ (R( , x2, n))ul. Wealso show that R(x1, x2, ) is an upset: Assume that n′ ≤ n and thatn′ ∈ R(x1, x2, ). We show that n ∈ (R(x1, x2, ))lu. As n′ ∈ R(x1, x2, ),y ≤ n′ for every y ∈ (R(x1, x2, ))l. Hence y ≤ n′ ≤ n for every y ∈(R(x1, x2, ))l and n ∈ (R(x1, x2, ))lu. The proof of the last statement issimilar. To prove 2, first notice that if R(x1, x2, n), then n ∈ R(x1, x2, )and thus fR(x1, x2) ≤ n. Conversely, if fR(x1, x2) ≤ n, then y ≤ n foreach y ∈ (R(x1, x2, ))l. That is, n ∈ (R(x1, x2, ))lu = R(x1, x2, ) asdesired. To prove 3, it is enough to show that fR is order preserving onJ∞(Y1) × J∞(Y2). To this end, let xi, x

′i ∈ J∞(Yi) with xi ≤ x′i. Let

n ∈ R(x′1, x′2, ), then x1 ≤ x′1 ∈ R( , x′2, n), and so by 1 R(x1, x

′2, n), and

so x2 ≤ x′2 ∈ R(x1, , n) and we get R(x1, x2, n) by 1. It now follows that

fR(x1, x2) =∧

R(x1, x2, ) ≤∧

R(x′1, x′2, ) = fR(x′1, x

′2).

To prove 4 we show that fR is completely join preserving in the first co-ordinate: Let ui ∈ Y i and assume that u1 =

∨A. That fR(u1, u2) ≥∨{fR(a, u2) : a ∈ A} follows from 3. For the other direction we may assume

that A ⊆ J∞(Y1) since J∞(Y1) is join-dense in Y 1. Also, since Z is meet-generated by M∞(Z), it is enough to show that if n ∈ M∞(Z) and n ≥∨{fR(a, u2) : a ∈ A} then n ≥ fR(u1, u2) =

∨{fR(x1, x2) : xi ≤ ui}, i.e.fR(x1, x2) ≤ n whenever xi ≤ ui, for i = 1, 2. But n ≥ ∨{fR(a, u2) : a ∈ A}implies that n ≥ ∨{fR(a, x2) : a ∈ A} and this means that R(a, x2, n) foreach a ∈ A. Now let m ∈ (R( , x2, n))u, then m ≥ a for each a ∈ A, andthus x1 ≤ u1 =

∨A ≤ m. So x1 ≤ m for each m ∈ (R( , x2, n))u. That

is, x1 ∈ (R( , x2, n))ul = R( , x2, n) and R(x1, x2, n) or fR(x1, x2) ≤ n asdesired.

Proposition 5.6. Let C1, C2, D be perfect lattices, and f : C1 × C2 → D amap that is completely join preserving in each coordinate, then fRf

= f .

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Proof. For xi ∈ J∞(Ci) for i = 1, 2,

fRf(x1, x2) =

∧{n ∈ M∞(Z) : Rf (x1, x2, n)}

=∧{n ∈ M∞(Z) : f(x1, x2) ≤ n}f(x1, x2).

So fRf= f on J∞(C1)× J∞(C2), and for (u1, u2) ∈ C1 × C2 we get

fRf(u1, u2) =

∨{fRf(x1, x2) : ui ≥ xi ∈ J∞(Ci)}

=∨{f(x1, x2) : ui ≥ xi ∈ J∞(Ci)}

=∨{∨{f(x1, x2) : u1 ≥ x1 ∈ J∞(C1)} : u2 ≥ x2 ∈ J∞(Y2)}

=∨{f(u1, x2) : u2 ≥ x2 ∈ J∞(C2)}

= f(u1, u2).

Proposition 5.7. Let Y1, Y2, and Z be perfect posets, and let R ⊆ J∞(Y1)×J∞(Y2)×M∞(Z) be a operator relation, then RfR

= R.

Proof. We have RfR(x1, x2, n) if and only if fR(x1, x2) ≤ n if and only if

R(x1, x2, n).

The above treatment also takes care of binary maps g : C∂1 ×C2 → D and

h : C1 × C∂2 → D that are completely meet preserving in each coordinate:

If g : C∂1 × C2 → D is completely meet preserving in each coordinate, then

g∂ : C1 × C∂2 → D∂ is a completely join preserving map. Given x1 ∈

J∞(C1), m2 ∈ M∞(C2), and x ∈ J∞(D), the relation associated with gSg ⊆ J∞(C1)×M∞(C2)× J∞(D) is defined by

Sg(x1,m2, x) if and only if g(x1,m2) ≥ x,

and these dual relations are characterized by the following properties:

S1. (Sg(x1,m2, ))ul = Sg(x1,m2, ).

S2∂1 . (Sg( ,m2, x))ul = Sg( ,m2, x).

S22. (Sg(x1, , x))lu = Sg(x1, , x).

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Notice that if R ⊆ J∞(Y1)×J∞(Y2)×M∞(Z) is an operator relation, thenthe relation S ⊆ J∞(Y1) × M∞(Z) × J∞(Y2) obtained by simply movingaround the order of the coordinates: R(x1, x2, n) if and only if S(x1, n, x2)is a relation of the type just described, that is, it satisfies S1, S2∂

1 , and S2∂2 .

As the next proposition shows this is no coincidence:

Proposition 5.8. Let C be a perfect lattice, f : C × C → C a completeoperator, and g : C∂ × C → C a complete dual operator. The following twostatement are equivalent:

1. g is the right residual of f ;

2. For all x, y ∈ J∞(C) and for all m ∈ M∞(C) we have

Rf (x, y,m) if and only if Sg(x,m, y)

Proof. Suppose g is the right residual of f , and let x, y ∈ J∞(C) and m ∈M∞(C), then Rf (x, y, m) if and only if f(x, y) ≤ m if and only if y ≤g(x,m) if and only if Sg(x,m, y). Conversely assume 2, and let u, v, w ∈ C.If f(u, v) ≤ w, then for each x, y ∈ J∞(C) with x ≤ u, y ≤ v we havef(x, y) ≤ w. Consequently, for each m ∈ M∞(C) with m ≥ w, we havef(x, y) ≤ m, that is, Rf (x, y, m). But then, by assumption, Sg(x,m, y)holds, that is, g(x,m) ≥ y. It now follows that

v =∨{y : v ≥ y ∈ J∞(C)}

≤∨{g(x,m) : u ≥ x ∈ J∞(C), w ≤ m ∈ M∞(C)} = g(u, w).

Similarly v ≤ g(u,w) implies f(u, v) ≤ w.

Analogously one can define the relations Th dual to complete dual oper-ators of the type of h, and one can prove corresponding statements aboutthese.

6 Applications to substructural logic

We will now obtain relational semantics for the fragment given by impli-cation and fusion for each of the following logics: Lambeck calculus, linearlogic, relevance logic, BCK logic, and intuitionistic logic. Our set-up is asin [5], and we refer to that paper for details.

As explained there, each of these logics correspond to classes of orderedalgebras. In all cases the appropriate algebras are MPEs of the form (P, ; ,→

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,←) where ; ,→,← are binary operations. The operation ; is known asfusion, and the two arrows are known as implication, and it is assumedthat the fusion operation is residuated with → and ← as its right and leftresidual, respectively. We call these algebras residuated algebras:

Definition 6.1. A residuated algebra is an MPE (P, f, g, h) where P is apartially ordered set, f, g, and h are binary operations on P , and g and hare the right and left residual of f , respectively.

Each of the logics we will consider corresponds to a subclass of thesealgebras given by some combination of the following inequations:

(α;β); γ ≈ α; (β; γ) (associative)α; β ≈ β;α (commutative)

α - α; α (square-increasing)α; β - β (right-lower-bounded)

The Lambeck calculus corresponds to the class of associative residuatedalgebras, the fragment of linear logic to associative and commutativeresiduated algebras, the fragment of relevance logic to square-increasingassociative and commutative residuated algebras, and the fragment ofBCK logic corresponds to right-lower-bounded associative and com-mutative residuated algebras. Finally, the fragment of intuitionistic logiccorresponds to the class of associative and commutative residuated al-gebras that satisfy both square-increasing and right-lower-bounded.

Here we will first show that each of these equations are canonical forresiduated algebras, that is, if any one of them is satisfied in a residuatedalgebra, then it is also satisfied in the canonical extension of that residuatedalgebra. This of course implies that the class of algebras correspondingto each of these logics is generated by its perfect members, where perfectresiduated algebras are defined by:

Definition 6.2. A perfect residuated algebra is an MPE (C, f, g, h) where Cis a perfect lattice, f, g, and h are binary operations on C, and g and h arethe right and left residual of f , respectively. In particular f is a completeoperator on C, and g : C∂ ×C → C and h : C ×C∂ → C are complete dualoperators.

Thus we are in a situation where the lattices obtain as canonical exten-sions fall under the discrete duality as described in the previous two sections,we may define:

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Definition 6.3. Given a perfect residuated algebra, C = (C, f, g, h), itsdiscrete dual is the structure C = (C,R), where R ⊆ J∞(C) × J∞(C) ×M∞(C) is given by

R(x, y, m) if and only if f(x, y) ≤ m

if and only if y ≤ g(x,m)if and only if x ≤ h(m, y)

Proposition 6.4. Let C = (C, f, g, h) be a perfect residuated algebra, andC = (C,R) its dual structure. Then C is a perfect poset, and R satisfies R1and R2.

Proof. This follows readily from the results in Sections 4 and 5.

Definition 6.5. Let X = (X, R) be a structure where X is a perfect poset,and R ⊆ J∞(X)× J∞(X)×M∞(X) satisfies R1 and R2. We will call sucha structure a Kripke structure for now. Define relations S and T by

R(x, y, m) if and only if S(x,m, y)if and only if T (m, y, x).

Then the dual of X is the residuated algebra X = (X, fR, gS , hT ).

Proposition 6.6. For any perfect residuated algebra C, (C) = C, and forany Kripke structure X = (X, R), we have (X) = X.

Proof. Again this follows from the results in sections 4 and 5, and we leavethe details to the reader.

Secondly, after showing that each of the equations associative, com-mutative, square-increasing, and right-lower-bounded are canonicalfor residuated algebras, we show that each of them has a first order corre-spondent, that is, it holds in a perfect residuated algebra if and only if itsfirst order correspondent holds on the dual structure of that perfect residu-ated algebra.

The consequence of these two facts then is that the class of dual struc-tures satisfying the appropriate first order correspondents provides a com-plete relational semantics for the given logic.

We start by proving the canonicity of the equations:

Proposition 6.7. Let f : P × P → P be such that fσ preserves arbitraryjoins in each coordinate. If f satisfies associative, then so does fσ.

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Proof. Let t : P × P × P → P be defined by

t(p1, p2, p3) = f(p1, f(p2, p3)) = f(f(p1, p2), p3).

It is enough to show that for all u1, u2, u3 ∈ P σ,

fσ(u1, fσ(u2, u3)) = tσ(u1, u2, u3) = fσ(fσ(u1, u2), u3).

We first show it for x1, x2, x3 ∈ K(P σ):

tσ(x1, x2, x3) =∧{t(p1, p2, p3) : xi ≤ pi ∈ P, i = 1, 2, 3}

=∧{f(p1, f(p2, p3)) : xi ≤ pi ∈ P, i = 1, 2, 3}

≥ ∧{f(p1, q) : x1 ≤ p1 ∈ P, fσ(x2, x3) ≤ q ∈ P}.The reverse inequality tσ(x1, x2, x3) ≤

∧{f(p1, q) : x1 ≤ p1 ∈ P, fσ(x2, x3) ≤q ∈ P} follows from the fact that {f(p2, p3) : x1 ≤ p1, i = 1, 2} is down-directed in the set of elements from P that are above fσ(x2, x3). As for thegeneral case, since fσ preserves arbitrary joins in each coordinate, we get:

tσ(u1, u2, u3) =∨{tσ(x1, x2, x3) : ui ≥ xi ∈ K(P σ), i = 1, 2, 3}

=∨{fσ(x1, f

σ(x2, x3)) : ui ≥ xi ∈ K(P σ), i = 1, 2, 3}=

∨{fσ(∨{x1 : u1 ≥ x1 ∈ K(P σ)}, fσ(x2, x3)) : ui ≥ xi ∈ K(P σ), i = 2, 3}

= fσ(u1,∨{fσ(x2, x3)) : ui ≥ xi ∈ K(P σ), i = 2, 3})

= fσ(u1,∨{fσ(u2, x3)) : u3 ≥ x3 ∈ K(P σ)})

= fσ(u1, fσ(u2, u3))

Similarly, one can show that tσ(u1, u2, u3) = fσ(fσ(u1, u2), u3).

Proposition 6.8. Let f : P × P → Q be order preserving. If f satisfiescommutative, then so does fσ.

Proof. Let x1, x2 ∈ K(P σ),

fσ(x1, x2) =∧{f(p1, p2) : xi ≤ pi ∈ P, i = 1, 2}

=∧{f(p2, p1) : xi ≤ pi ∈ P, i = 1, 2}

= fσ(x2, x1).

So for any u1, u2 ∈ P σ,

fσ(u1, u2) =∨{fσ(x1, x2) : ui ≥ xi ∈ K(P σ), i = 1, 2}

=∨{fσ(x2, x1) : ui ≥ xi ∈ K(P σ), i = 1, 2}

= fσ(u2, u1).

Proposition 6.9. Let f : P × P → P be order preserving. If f satisfiessquare-increasing, then so does fσ.

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Proof. Let x ∈ K(P σ) then

x =∧{p : x ≤ p ∈ P}

≤∧{f(p, p) : x ≤ p ∈ P}

≤∧{f(p, q) : x ≤ p, q ∈ P} = fσ(x, x)

The second inequality holds because the set {(p, p) : x ≤ p ∈ P} is filteringin {(p, q) : x ≤ p, q ∈ P}, and this is because the projections of the secondset is a filter, namely the filter of P corresponding to the closed element x.Now for u ∈ P σ

u =∨{x : u ≥ x ∈ K(P σ)}

≤∨{fσ(x, x) : u ≥ x ∈ K(P σ)}

≤∨{f(x, y) : u ≥ x, y ∈ K(P σ)} = fσ(u, u).

Here the second ≤ simply holds because {f(x, y) : u ≥ x, y ∈ K(P σ)}contains {fσ(x, x) : u ≥ x ∈ K(P σ)}.Proposition 6.10. Let f : P × Q → Q be order preserving. If f satisfiesright-lower-bounded, then so does fσ.

Proof. Suppose f(p, q) ≤ q for every p ∈ P , and q ∈ Q, then for everyx ∈ K(P σ), y ∈ K(Qσ)

fσ(x, y) =∧{f(p, q) : x ≤ p ∈ P, y ≤ q ∈ Q}

≤ ∧{q : y ≤ q ∈ Q} = y

And then for u ∈ P σ and v ∈ Qσ,

fσ(u, v) =∨{fσ(x, y) : u ≥ x ∈ K(P σ), v ≥ y ∈ K(Qσ)}

≤ ∨{y : v ≥ y ∈ K(Qσ)} = v

This completes the canonicity proofs. Now we turn to correspondence.

Proposition 6.11. Let C be a perfect lattice and f : C×C → C a completeoperator. Then the following statements are equivalent:

1. For all u1, u2, u3 ∈ C, f(u1, f(u2, u3)) = f(f(u1, u2), u3).

2. For all x1, x2, x3 ∈ J∞(C), f(x1, f(x2, x3)) = f(f(x1, x2), x3).

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Proof. Clearly 1 implies 2. The converse follows from the fact that foru1, u2, u3 ∈ C

f(u1, f(u2, u3)) =∨{f(x1, f(x2, x3)) : ui ≥ xi ∈ J∞(C), i = 1, 2, 3}

and

f(f(u1, u2), u3)) =∨{f(f(x1, x2), x3)) : ui ≥ xi ∈ J∞(C), i = 1, 2, 3}.

This is obtained by using the fact that f preserves joins in each coordinaterepeatedly.

The point is that the second statement in the above proposition onlyrefers to elements of the dual structure. These kind of statements easilytranslate to first order statements on the dual. Here we do the transla-tion in an algorithmic way not worrying about getting the simplest possiblestatements.

Proposition 6.12. Let f : C ×C → C be a complete operator on a perfectlattice. Let xi ∈ J∞(C), i = 1, 2, 3 and m ∈ M∞(C).

1. The following are equivalent:

(a) f(x1, f(x2, x3)) ≤ m.

(b) ∀x′2([∀m′(Rf (x2, x3,m′) → x′2 ≤ m′)] → Rf (x1, x

′2,m)).

2. The following are equivalent:

(a) f(f(x1, x2), x3) ≤ m.

(b) ∀x′1([∀m′(Rf (x1, x2,m′) → x′1 ≤ m′)] → Rf (x′1, x3,m)).

Proof. We just prove the first equivalence, and we resort to formal firstorder statements in doing so. We assume that any element named x withany super- or subscript comes from J∞(C) whereas any element named mwith any superscript comes from M∞(C).

f(x1, f(x2, x3)) ≤ m

⇔∀x′2(x′2 ≤ f(x2, x3) ⇒ Rf (x1, x′2,m))

⇔∀x′2(∀m′[Rf (x2, x3,m′) ⇒ x′2 ≤ m′] ⇒ Rf (x1, x

′2,m))

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Definition 6.13. Let Φa denote the first order statement:

∀x1, x2, x3 ∀m([∀x′2(∀m′[R(x2, x3,m

′) ⇒ x′2 ≤ m′] ⇒ R(x1, x′2,m))]

⇔[∀x′1([∀m′′(R(x1, x2,m′′) → x′1 ≤ m′′)] → R(x′1, x3,m))])

When interpreting this statement in a structure we assume the variablesnamed by x’s range over completely join irreducibles and the variablesnamed by m’s range over completely meet irreducibles.

Corollary 6.14. Let (X, R) be a Kripke structure, then the following state-ments are equivalent:

1. (X, R) satisfies Φa;

2. (X, R) satisfies associative.

Proposition 6.15. Let C be a perfect lattice, and f : C × C → D be acomplete operator. Then the following statements are equivalent:

1. For all u1, u2 ∈ C, f(u1, u2) = f(u2, u1).

2. For all x1, x2 ∈ J∞(C), f(x1, x2) = f(x2, x1).

Proof. Similar to the proof of Proposition 6.11.

Definition 6.16. Let Φc denote the first order statement:

∀x1, x2 ∀m (R(x1, x2,m) ⇔ R(x2, x1,m))

When interpreting this statement in a structure we assume the variablesnamed by x’s range over completely join irreducibles and the variable namedby m ranges over completely meet irreducibles.

Corollary 6.17. Let (X, R) be a Kripke structure, then the following state-ments are equivalent:

1. (X, R) satisfies Φc;

2. (X, R) satisfies commutative.

Proposition 6.18. Let C be a perfect lattice, and f : C × C → C be acomplete operator. Then the following statements are equivalent:

1. For all u ∈ C, u ≤ f(u, u).

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2. For all x ∈ J∞(C), x ≤ f(x, x).

Proof. Similar to Prop 6.11.

Definition 6.19. Let Φsi denote the first order statement:

∀x ∀m (R(x, x,m) ⇒ x ≤ m)

When interpreting this statement in a structure we assume the variablenamed by x ranges over completely join irreducibles and the variable namedby m ranges over completely meet irreducibles.

Corollary 6.20. Let (X, R) be a Kripke structure, then the following state-ments are equivalent:

1. (X, R) satisfies Φsi;

2. (X, R) satisfies square-increasing.

Proposition 6.21. Let C be a perfect lattice, and f : C × C → C be acomplete operator. Then the following statements are equivalent:

1. For all u1, u2 ∈ C, f(u1, u2) ≤ u1.

2. For all x1, x2 ∈ J∞(C), f(x1, x2) ≤ x1.

Proof. Similar to the proof of Proposition 6.11.

Definition 6.22. Let Φrlb denote the first order statement:

∀x1, x2 ∀m (x1 ≤ m ⇒ R(x1, x2,m))

When interpreting this statement in a structure we assume the variablesnamed by x’s range over completely join irreducibles and the variable namedby m ranges over completely meet irreducibles.

Corollary 6.23. Let (X, R) be a Kripke structure, then the following state-ments are equivalent:

1. (X, R) satisfies Φrlb;

2. (X, R) satisfies right-lower-bounded.

We conclude:

Theorem 6.24. The class of Kripke structures satisfying Φa is a completesemantics for the Lambeck calculus.

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Theorem 6.25. The class of Kripke structures satisfying Φa and Φc is acomplete semantics for the fragment of linear logic given by implication andfusion.

Theorem 6.26. The class of Kripke structures satisfying Φa, Φc, and Φsi isa complete semantics for the fragment of relevance logic given by implicationand fusion.

Theorem 6.27. The class of Kripke structures satisfying Φa, Φc, and Φrlb

is a complete semantics for the fragment of linear logic given by implicationand fusion.

Theorem 6.28. The class of Kripke structures satisfying Φa, Φc, Φsi, andΦrlb is a complete semantics for the fragment of intuitionistic logic given byimplication and fusion.

References

[1] Allwein, G. and Dunn, J. M., A Kripke semantics for linear logic, TheJournal of Symbolic Logic 58, 514-545, 1993.

[2] Allwein, G. and MacCaull, W., A Kripke semantics for the logic ofGelfand quantales, Studia Logica 68, no. 2, 173–228, 2001.

[3] Bucciarelli, A. and Ehrhard, T., On phase semantics and denotationalsemantics in multiplicative-additive linear logic. Ann. Pure Appl. Logic102, no. 3, 247–282, 2000.

[4] Dunn, J. M., Gaggle theory, an abstraction of galois connections andresiduation, with applications to negation, implication, and various log-ical operators, in Logics in AI (European Workshop JELIA, Amster-dam), ed. J. Van Eijck, Lecture Notes in AI, Springer Verlag, 31-51,1990.

[5] Dunn, J. M., Partial gaggles applied to logics with restricted struc-tural rules, Substructural logics (Tbingen, 1990), 63–108, Stud. LogicComput., 2, Oxford Univ. Press, New York, 1993.

[6] Dunn, J. M. and Hardegree, G., Algebraic Methods in PhilosophicalLogic, Oxford University Press, 2001.

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