Expanding FLew with a Boolean connective
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Expanding FL ew with a Boolean connective Rodolfo C. Ertola-Biraben · Francesc Esteva · Llu´ıs GodoAbstract
We expand FL ew with a unary connectivewhose algebraic counterpart is the operation that givesthe greatest complemented element below a given argu-ment. We prove that the expanded logic is conservativeand has the Finite Model Property. We also prove thatthe corresponding expansion of the class of residuatedlattices is an equational class. ACKNOWLEDGEMENT
This paper was published in Soft Computing (Springer-Verlag) on 23 July 2016. The final publication is avail-able at http://dx.doi.org/10.1007/s00500-016-2275-y
In this paper we study the expansion of the substruc-tural logic FL ew , i.e. Full Lambek calculus with ex-change and weakening, with a unary connective B whoseintended algebraic semantics is as follows: given a boundedintegral commutative residuated lattice (or residuatedlattice for short) A , Ba is the maximum, if it exists, ofthe Boolean elements of the universe A below a , whichwe call the greatest Boolean below a , that is, Ba = max { b ∈ A : b ≤ a and b is Boolean } . In fact, this operator is similar to the so-called Baaz-Monteiro ∆ operator, very often used in the context of Rodolfo C. Ertola-BirabenState University of Campinas, Rua S´ergio Buarque deHolanda 251, 13083-859 Campinas, SP, BrazilE-mail: [email protected] Esteva, Llu´ıs GodoArtificial Intelligence Research Institute - CSIC, Campus dela UAB s/n, 08193 Bellaterra, Catalonia, SpainE-mail: { esteva, godo } @iiia.csic.es mathematical fuzzy logic systems that are semilinearexpansions of MTL. Baaz [2] studied it in connectionwith G¨odel logic while H´ajek [10] investigated ∆ in BLlogics in general, see also [8, Chapter 2] for a more gen-eral perspective. Indeed, in such a context of semilinearlogics, i.e. logics that are complete with respect to aclass of linearly ordered algebras, the semantics of ∆ is exactly the above one for B : in a linearly-orderedMTL-algebra, ∆a = 1 if a = 1, and ∆a = 0 otherwise,since the only Boolean elements in a chain are 1 and 0;moreover, from a logical point of view, ∆ϕ representsthe weakest Boolean proposition implying ϕ .The operator B can be also related to the join-complement operation D , also known as dual intuition-istic negation, already considered by Skolem [20] inthe context of lattices with relative meet-complement,and later independently studied by e.g. Moisil [11] andRauszer [16] as well, the latter in the context of ex-pansions of Heyting algebras. It turns out that the op-eration ¬ D and its iterations, where ¬ is the residualnegation, has also very similar properties to B , and insome classes of residuated lattices they even coincide.In this paper we study the operator B in the con-text of FL ew and axiomatize it. We show that the usualaxiomatics of the ∆ operator is actually too strong tocapture the above intended semantics. In fact, the ax-iom ∆ ( ϕ ∨ ψ ) → ( ∆ϕ ∨ ∆ψ )is not sound for B over FL ew any longer. Thus, B isa weaker operator than ∆ . However, as we will see, B keeps most of the properties of ∆ . In particular, theexpansion of FL ew with B is conservative, its corre-sponding class of algebras is an equational class, andhas the same kind of deduction theorem as ∆ . Also, B may also be interesting as ¬ B has a paraconsistent be-haviour. On the negative side, the expansion of a semi- Rodolfo C. Ertola-Biraben et al. linear extension of FL ew with B needs not to remainsemilinear.The paper is structured as follows. In Section 2 weoverview well-known facts about residuated lattices andits Boolean elements, as well as basic facts about thelogic FL ew . Sections 3 and 4 contain an algebraic studyof the operator B . In particular, in Section 3 we studybasic properties and show, among other things, thatthe class RL B of residuated lattices expanded with B is an equational class and state the modalities, while inSection 4 we compare B with the mentioned ∆ and withan operation using the join-complement D . Finally, inSection 5 we focus on logical aspects, introducing thelogic FL Bew , i.e. the expansion of FL ew with the operator B , and show that is a conservative expansion and hasthe Finite Model Property, and hence it is decidable.We conclude with some remarks and open problems.We give appropriate references. However, the paperis self-contained. A = ( A ; ∧ , ∨ , · , → , ,
1) of type (2 , , , , ,
0) suchthat:- ( A ; ∧ , ∨ , ,
1) is a bounded lattice with 0 ≤ a ≤ a ∈ A ,- ( A ; · ,
1) is a commutative monoid (i.e. · is commu-tative, associative, with unit 1), and- → is the residuum of · , i.e., a · b ≤ c iff a ≤ b → c , for all a, b, c ∈ A ,where ≤ is the order given by the lattice structure. Anegation operator is defined as ¬ x = x → RL . It is well known that RL is an equational classand that it constitutes the algebraic semantics of thesubstructural logic FL ew (see Section 2.2). Example 1
In what follows we will have occasion to re-fer several times to the residuated lattice structure de-fined on the five-element lattice of Figure 1 by taking · = ∧ and → its residuum. With these operations, itactually becomes a five element G¨odel algebra, that is,a residuated lattice with · being idempotent and satis-fying the pre-linearity law ( a → b ) ∨ ( b → a ) = 1. rs t Fig. 1
A five element G¨odel algebra
We omit the proof of the following well-known facts,see e.g. [9].
Lemma 1
Let A ∈ RL . For any a, b, c, d ∈ A , the fol-lowing properties hold: (i) if a ∨ b = 1 , a ≤ c , and b ≤ c , then c = 1 , (ii) if a ∨ b = 1 , a · c ≤ d , and b · c ≤ d , then c ≤ d , (iii) if a ≤ b , then ¬ b ≤ ¬ a , (iv) a ∧ ¬ b ≤ ¬ ( a ∧ b ) , (v) a · ¬ b ≤ ¬ ( a → b ) , (vi) if a ∨ b = 1 , then ¬ a ≤ b , (vii) a ≤ ¬¬ a . Special elements in a residuated lattice are thosethat behave as elements in a Boolean algebra.
Definition 1
Let A ∈ RL . An element a of its uni-verse A is called Boolean or complemented iff there isan element b ∈ A such that a ∧ b = 0 and a ∨ b = 1 . In the rest of this section we state several propertiesof Boolean elements that will be useful in what follows.Even if most of them are folklore, we include proofs forall of them for the sake of being self-contained.An equivalent and simpler condition for an elementto be Boolean is the following.
Lemma 2
An element a in the universe of a residuatedlattice is Boolean iff a ∨ ¬ a = 1 .Proof ⇒ ) Suppose there is an element b such that a ∧ b = 0 and a ∨ b = 1. First, using that a ∧ b = 0 and a · b ≤ a ∧ b , we have that a · b = 0. So, b ≤ a → b ≤ ¬ a . Secondly, we have that ¬ a = ¬ a · ¬ a · ( a ∨ b ) = ( ¬ a · a ) ∨ ( ¬ a · b ) = ¬ a · b . So, ¬ a = ¬ a · b .As ¬ a · b ≤ b , it follows that ¬ a ≤ b . So, b = ¬ a . As wehave that a ∨ b = 1, it follows that a ∨ ¬ a = 1. ⇐ ) By hypothesis, we have (i) a ∨ ¬ a = 1. It is enoughto see that a ∧ ¬ a = 0. As a · ¬ a = 0, it is enough toprove that a ∧ ¬ a ≤ a · ¬ a . We have that a ∧ ¬ a ≤ ¬ a .So, by monotonicity of · , we have (ii) a · ( a ∧¬ a ) ≤ a ·¬ a .We also have that a ∧¬ a ≤ a . So, again by monotonicityof · , we have ¬ a · ( a ∧ ¬ a ) ≤ ¬ a · a = a · ¬ a . So, it follows(iii) ¬ a · ( a ∧ ¬ a ) ≤ a · ¬ a . Now, using Lemma 1(ii) with(i), (ii), and (iii), it follows that a ∧ ¬ a ≤ a · ¬ a . (cid:3) xpanding FL ew with a Boolean connective 3 Proposition 1
Let A ∈ RL and let a be a Booleanelement of its universe A . Then, for all b, c, d ∈ A thefollowing properties hold: (i) a ∧ b = a · b , (ii) a · a = a , (iii) a ∧ ¬ a = 0 , (iv) a ∧ ( b ∨ c ) = ( a ∧ b ) ∨ ( a ∧ c ) , (v) ¬¬ a = a , (vi) a → b = ¬ a ∨ b , (vii) 0 = ¬ ( a ∨ ¬ a ) , (viii) if a ≤ b ∨ c , a ∧ b ≤ d , a ∧ c ≤ d , then a ≤ d , (ix) if b ∨ c = 1 , a ∧ b ≤ d , a ∧ c ≤ d , then a ≤ d , (x) if a ∧ b ≤ c , then a ∧ ¬ c ≤ ¬ b , (xi) if a ∨ ¬ b = 1 , then b ≤ a .Proof (i) Suppose that (i) a ∨ ¬ a = 1. It is enough tosee that a ∧ b ≤ a · b . We have that a ∧ b ≤ b . So, bymonotonicity of · , we have (ii) a · ( a ∧ b ) ≤ a · b . Wealso have that a ∧ b ≤ a . So, again by monotonicity of · , we have ¬ a · ( a ∧ b ) ≤ ¬ a · a = 0. So, it follows (iii) ¬ a · ( a ∧ b ) ≤ a · b . Now, using Lemma 1(ii) with (i), (ii),and (iii), it follows that a ∧ b ≤ a · b .(ii) Using Part (i), we have a ∧ a = a · a . Also, a ∧ a = a . So, a · a = a .(iii) Using Part(i), we have a ∧ ¬ a = a · ¬ a . Also, a · ¬ a = 0. So, a ∧ ¬ a = 0.(iv) In a residuated lattice it holds that a · ( b ∨ c ) =( a · b ) ∨ ( a · c ). Let a be Boolean. Then, using Part (i)three times, it follows that a ∧ ( b ∨ c ) = ( a ∧ b ) ∨ ( a ∧ c ).(v) Suppose that (i) a ∨ ¬ a = 1. It is enough to seethat ¬¬ a ≤ a . We have that (ii) a · ¬¬ a ≤ a . Also, (iii) ¬ a · ¬¬ a ≤ a , as ¬ a · ¬¬ a = 0. So, using Lemma 1(ii)with (i), (ii), and (iii), ¬¬ a ≤ a .(vi) It is enough to prove (i) ( ¬ a ∨ b ) · a ≤ b and(ii) if x · a ≤ b , then x ≤ ¬ a ∨ b . To see (i), note that ¬ a · a ≤ b and b · a ≤ b , whence ( ¬ a · a ) ∨ ( b · a ) ≤ b . So,using distributivity of · relative to ∨ , we get (i). To see(ii), suppose x · a ≤ b . Then, x ≤ a → b . In order to get x ≤ ¬ a ∨ b , it is enough to derive (iii) a → b ≤ ¬ a ∨ b and use transitivity of ≤ . To get (iii), let us use Lemma1(ii). As a is Boolean, we have a ∨¬ a = 1. Now, a · ( a → b ) ≤ b ≤ ¬ a ∨ b . Also, ¬ a · ( a → b ) ≤ ¬ a ≤ ¬ a ∨ b . So,using Lemma 1(ii), we get (iii).(vii) As we have ¬ c ≤ a ∨ ¬ a , for any c ∈ A , then,using Lemma 1(iii) and Part (iv), we get ¬ ( a ∨ ¬ a ) ≤¬¬ c ≤ c .(viii) As a ≤ b ∨ c , we have a ≤ a ∧ ( b ∨ c ). Now,using Part (iv), it follows that a ≤ ( a ∧ b ) ∨ ( a ∧ c ). Now,as a ∧ b ≤ d and a ∧ c ≤ d , we have, by a basic propertyof ∨ , that ( a ∧ b ) ∨ ( a ∧ c ) ≤ d . So, by transitivity of ≤ , a ≤ d . (ix) As b ∨ c = 1, reason as in Part (viii).(x) Suppose a ∧ b ≤ c . Then a · b ≤ c . By monotonic-ity of · , it follows that ( a · b ) · ¬ c ≤ c · ¬ c = 0. Then, as · is both associative and commutative, ( a · ¬ c ) · b ≤ a · ¬ c ≤ ¬ b . Finally, using that a is Boolean, we get a ∧ ¬ c ≤ ¬ b .(xi) Let a ∨ ¬ b = 1. Then, b · ( a ∨ ¬ b ) = b · b . By distributivity of · relative to ∨ , it follows that( b · a ) ∨ ( b · ¬ b ) = b . As b · ¬ b = 0, we have that b · a = b .So, as a is Boolean, b ∧ a = b , i.e. b ≤ a . (cid:3) Lemma 3
Let A ∈ RL and let a and b be Booleanelements of A . Then, (i) a ∧ b = a · b = ¬ ( ¬ a ∨ ¬ b ) , (ii) ( a · b ) ∨ ( ¬ a · b ) ∨ ( a · ¬ b ) ∨ ( ¬ a · ¬ b ) = 1 .Proof (i) Firstly, we have that ¬ a ≤ ¬ a ∨ ¬ b . So, usingLemma 1(iii), it follows that ¬ ( ¬ a ∨ ¬ b ) ≤ ¬¬ a . Now,using Proposition 1(v) and ≤ -transitivity, we have that ¬ ( ¬ a ∨ ¬ b ) ≤ a . Analogously, we get ¬ ( ¬ a ∨ ¬ b ) ≤ b .Secondly, suppose c ≤ a and c ≤ b , for c ∈ A . Then,using Lemma 1(iii) again, it follows that ¬ a ≤ ¬ c and ¬ b ≤ ¬ c . So, ¬ a ∨ ¬ b ≤ ¬ c . Then, using Lemma 1(iii)once again, ¬¬ c ≤ ¬ ( ¬ a ∨ ¬ b ). Now, using Proposition1(v) and ≤ -transitivity, we get c ≤ ¬ ( ¬ a ∨ ¬ b ).(ii) Suppose that a ∨ ¬ a = b ∨ ¬ b = 1. Then, 1 =( a ∨ ¬ a ) · ( b ∨ ¬ b ) = ( a · b ) ∨ ( ¬ a · b ) ∨ ( a · ¬ b ) ∨ ( ¬ a · ¬ b ). (cid:3) Proposition 2
Let A ∈ RL and let a and b be Booleanelements of A . Then, (i) ¬ a , (ii) a ∨ b , (iii) a ∧ b = a · b ,(iv) a → b , (v) , and (vi) are Boolean.Proof (i) Suppose a ∨ ¬ a = 1. Then, as a ≤ ¬¬ a , weget ¬¬ a ∨ ¬ a = 1.(ii) Use Lemma 3 (ii) and see that a · b ≤ a ≤ a ∨ b ≤ ( a ∨ b ) ∨ ¬ ( a ∨ b ), ¬ a · b ≤ b ≤ a ∨ b ≤ ( a ∨ b ) ∨ ¬ ( a ∨ b ), a · ¬ b ≤ a ≤ a ∨ b ≤ ( a ∨ b ) ∨ ¬ ( a ∨ b ), and ¬ a · ¬ b ≤ ¬ a ∧ ¬ b ≤ ¬ ( a ∨ b ) ≤ ( a ∨ b ) ∨ ¬ ( a ∨ b ).So, a ∨ b is Boolean.(iii) Use Parts (i) and (ii), and Lemma 3(i).(iv) Use Parts (i) and (ii), and Proposition 1(vi).(v) Use Parts (i) and (ii), and Proposition 1(vii).(vi) Use the definition of Boolean element and the factthat ¬ (cid:3) From Proposition 2 it easily follows that, in anyresiduated lattice A , the set of its Boolean elements B ( A ) = { a ∈ A : a is Boolean } is the domain of a sub-algebra of A , which is in fact a Boolean algebra. Indeed, B ( A ) = ( B ( A ); ∧ , ∨ , · , → , ,
1) is the greatest Booleanalgebra contained in A . B(A) is called the Booleanskeleton or the center of A . Rodolfo C. Ertola-Biraben et al. ew The logics we are interested in are extensions or expan-sions of the logic FL ew described below. Definition 2
The language of FL ew has four binaryconnectives, ∧ , ∨ , · , and → , and two constants, and .The axioms of FL ew are: (1) ( ϕ → ψ ) → (( ψ → γ ) → ( ϕ → γ )) , (2) ( γ → ϕ ) → (( γ → ψ ) → ( γ → ( ϕ ∧ ψ ))) , (3) ( ϕ ∧ ψ ) → ϕ and ( ϕ ∧ ψ ) → ψ , (4) ϕ → ( ϕ ∨ ψ ) and ψ → ( ϕ ∨ ψ ) , (5) ( ϕ → γ ) → (( ψ → γ ) → (( ϕ ∨ ψ ) → γ )) , (6) ( ϕ · ψ ) → ( ψ · ϕ ) , (7) ( ϕ · ψ ) → ϕ , (8) ( ϕ → ( ψ → γ )) → (( ϕ · ψ ) → γ ) , (9) (( ϕ · ψ ) → γ ) → ( ϕ → ( ψ → γ )) , (10) 0 → ϕ and ϕ → .The only rule of FL ew is modus ponens : ϕ ϕ → ψψ . We define ¬ ϕ = ϕ → ϕ ↔ ψ = ( ϕ → ψ ) ∧ ( ψ → ϕ ). The following formulas and rules are derivable inFL ew :(11) ϕ → ψ ψ → γϕ → γ ,(12) ϕ → ( ψ → ϕ ),(13) ϕ → ϕ ,(14) ϕ → ( ϕ ·
1) and ( ϕ · → ϕ ,(15) ( ϕ · ( ψ · γ )) ↔ (( ϕ · ψ ) · γ )(16) ( ϕ · ( ϕ → ψ )) → ψ ,(17) (1 → ϕ ) → ϕ , ϕ → (1 → ϕ ),(18) ϕ ψϕ ∧ ψ ,(19) ( ϕ ∨ ¬ ϕ ) → ¬ ( ϕ ∧ ¬ ϕ ).Derivations for (11)-(19) are rather easy. Hence, theyare left to the reader.We will occasionally consider the following exten-sions of FL ew . Definition 3
Consider the following axiomatic exten-sions of FL ew : – Intutitionistic logic IL is FL ew plus the axiom (Contr) ϕ → ( ϕ · ϕ ) . – The logic
MTL is FL ew plus the axiom (Prel) ( ϕ → ψ ) ∨ ( ψ → ϕ ) . – SMTL logic is
MTL plus the axiom (PC) ¬ ( ϕ ∧ ¬ ϕ ) – WNM logic is
MTL plus the axiom (WNM) ¬ ( ϕ & ψ ) ∨ ( ϕ ∧ ψ → ϕ & ψ ) – NM logic is WNM plus the axiom (Inv) ¬¬ ϕ → ϕ – BL logic is MTL plus the axiom (Div) ( ϕ ∧ ψ ) → ( ϕ &( ϕ → ψ )) – Product logic is BL plus (PC) and the axiom (C) ¬ ϕ ∨ (( ϕ → ϕ & ψ ) → ψ ) – G¨odel logic G is BL plus (Contr)– Lukasiewicz logic L is BL plus (Inv) . Remark 1
Note that G¨odel logic G arises also as MTL plus (Contr) or as IL plus (Prel) . In MTL and its extensions we may define φ ∨ ψ :=(( φ → ψ ) → ψ ) ∧ (( ψ → φ ) → φ ). In BL and its exten-sions we may define ϕ ∧ ψ := ϕ &( ϕ → ψ ). Moreover,in IL the formula ( φ · ψ ) ↔ ( φ ∧ ψ ) is derivable, i.e.connectives ∧ and · coincide in IL.All these logics are algebraizable, and hence theyare strongly complete with respect to their correspond-ing classes of algebras. Namely, FL ew is complete withrespect to the variety RL of residuated lattices, MTL iscomplete with respect to the variety of pre-linear resid-uated lattices (MTL-algebras), and IL is complete withrespect to the variety of contractive residuated lattices(Heyting algebras). Moreover, all axiomatic extensionsof MTL are semilinear logics, that is, they are stronglycomplete with respect to the corresponding class of lin-early ordered algebras. For instance, G¨odel logic is com-plete with respect to the class of linearly ordered Heyt-ing algebras, or G¨odel chains. B As explained in the previous section, the set of Booleanelements of a residuated lattice A forms a Booleanalgebra denoted the center or Boolean skeleton of A .Cignoli and Monteiro considered Boolean elements in Lukasiewicz algebras in [6] and [7]. However, as far aswe know, the operator defining the greatest Booleanelement below, i.e. the operator B studied in this pa-per, has not yet been studied in the general contextof residuated lattices. One relevant exception is thepaper [18], where Reyes and Zolfaghari define modaloperators (cid:3) and ♦ in the context of Bi-Heyting alge-bras that are shown to correspond respectively to thegreatest and the smallest complemented element belowand above, respectively. Thus, the (cid:3) operation coin-cides with B . In the cited paper, using dual negation(or join-complement) D , always in the context of Bi-Heyting algebras, the authors also study a family of xpanding FL ew with a Boolean connective 5 modal operators (cid:3) n and ♦ n , in a similar way to theone we shall employ in Section 4.2.We will be considering residuated lattices A en-riched with a unary operation B such that, for all a ∈ A , Ba is the greatest Boolean element below a , as definedin the Introduction. It is clear that B can be character-ized by the following three conditions, for a , b in A : (BE1) Ba ≤ a , (BE2) Ba ∨ ¬ Ba = 1, (BI) if b ≤ a and b ∨ ¬ b = 1, then b ≤ Ba .The class of residuated lattices with B will be denotedby RL B . Namely, an RL B -algebra is an algebra A =( A ; ∧ , ∨ , · , → , B, ,
1) such that ( A ; ∧ , ∨ , · , → , ,
1) is aresiduated lattice and B satisfies the above three con-ditions.First of all, note that B is new, that is, B is not ex-pressible by a {∧ , ∨ , · , → , } -term. Indeed, for instance,in the G¨odel algebra G × G (the direct product of thetwo-element Boolean algebra with universe { , } andthe three-element G¨odel algebra with universe { , , } )we have, for any {∧ , ∨ , · , → } -term t , that ta ∈ { , a, } ,where a = (1 , ) is the join reducible coatom, while Ba = (1 ,
0) is the join-irreducible atom, which does notbelong to { , a, } .In the next proposition we see that all operationsremain independent. Proposition 3
The set of operators {∧ , ∨ , · , → , B, } is independent.Proof To see that ∧ is independent of the rest take thedistributive lattice in Figure 1 and define the monoidaloperation · as s · t = 0, s · s = s , and t · t = t , for coatoms s and t . This operation has a corresponding residuum → . Since the only Boolean elements are 0 and 1, theoperator B is defined as B Ba = 0, for all a = 1. Then, note that the set S = { , s, t, } is closedfor ∨ , · , → , 0 and B , but s ∧ t / ∈ S .To see that ∨ is independent of the rest take the algebrathat results from inverting the lattice order in the alge-bra of Example 1 and note that the set S with bottom,both atoms a and a , and top is closed for ∧ , · , → , 0,and B , but a ∨ a / ∈ S .To see that · is independent of the rest take the four-element chain 0 < a < b < a · b = a · a = b · b = a ,for the atom a and the coatom b , and note that the set S = { , b, } is closed for ∧ , ∨ , → , 0, and B , but b · b / ∈ S .To see that → is independent of the rest take the alge-bra of Example 1 and note that the set S = { , r, s, } is closed for ∧ , ∨ , ¬ , 0, and B , but s → r / ∈ S .To see that 0 is independent of the rest take the twoelement Boolean algebra and note that the set S = { } is closed for ∧ , ∨ , · , → , and B , but 0 / ∈ S . The independence of B has already been considered. (cid:3) Lemma 4
Let A ∈ RL B and let a ∈ A . Then, (i) Ba = a iff a is Boolean, (ii) Ba = 1 iff a = 1 , (iii) BBa = Ba .Proof (i) Suppose Ba = a . Using (BE2) it followsthat a ∨ ¬ a = 1. For the other conditional, suppose a ∨ ¬ a = 1. Then, as a ≤ a , using (BI) it follows that a ≤ Ba . The other inequality follows by (BE2) .(ii) Suppose Ba = 1. Using (BE1) , it follows that1 ≤ a , i.e. a = 1. For the other conditional, suppose a = 1. Then, a ≤
1. Using (BI) and the fact that 1 isBoolean (see Proposition 2(vi)), it follows that 1 ≤ Ba .(iii) Considering (BE1) , it is enough to see that Ba ≤ BBa , which follows using (BI) and (BE2) . (cid:3) We also have the following properties.
Lemma 5
Let A ∈ RL B and let a, b ∈ A . Then, (i) B -Monotonicity: if a ≤ b , then Ba ≤ Bb , (ii) B ( a ∧ b ) = Ba ∧ Bb , (iii) B ( a ∧ b ) ≤ a · b , (iv) B ( a · b ) = B ( a ∧ b ) , (v) B ( a · b ) = Ba · Bb , (vi) Ba ∨ Bb ≤ B ( a ∨ b ) , (vii) B ( a → b ) ≤ Ba → Bb , (viii) B , (ix) B ¬ a ≤ ¬ Ba .Proof (i) Suppose a ≤ b . Using (BI) , it is enough tohave Ba ≤ b and Ba ∨¬ Ba = 1. Now, the former followsby (BE1) and the hypothesis, and the latter is (BE2) .(ii) B ( a ∧ b ) ≤ Ba ∧ Bb follows from a ∧ b ≤ a, b us-ing B -monotonicity. The other inequality follows using (BI) , (BE1) , and (iii) in Proposition 2.(iii) By (i) in Proposition 1 and part (ii) we have B ( a ∧ b ) = Ba ∧ Bb = Ba · Bb . The goal follows using Ba ≤ a , Bb ≤ b , and monotonicity of · .(iv) From a · b ≤ a ∧ b by (i), we get B ( a · b ) ≤ B ( a ∧ b ).For the other inequality, using (BI) , it is enough to have B ( a ∧ b ) ≤ a · b and B ( a ∧ b ) Boolean. Now, the formeris (iii) and the latter follows from (BE2) .(v) As Ba is Boolean, by (i) of Proposition 1, wehave Ba ∧ Bb = Ba · Bb . Moreover, by (ii), B ( a ∧ b ) = Ba ∧ Bb . We get our goal using (iv).(vi) It follows using (i) ( B -Monotonicity).(vii) As a → b ≤ a → b , we have ( a → b ) · a ≤ b .Then, by B -monotonicity, B (( a → b ) · a ) ≤ Bb . So,using (v), B ( a → b ) · Ba ≤ Bb . So, B ( a → b ) ≤ Ba → Bb .(viii) It follows because 0 is Boolean.(ix) It follows from (vii), (viii), and ¬ a = a → (cid:3) Rodolfo C. Ertola-Biraben et al.
Regarding the inequalities in the previous lemma,that is, ( iii ) , ( vi ) , ( vii ), and ( ix ), their reciprocals donot hold. Indeed, inequality a · b ≤ B ( a ∧ b ) fails inthe three-element G¨odel algebra G taking the top andthe middle element. Inequality B ( a ∨ b ) ≤ Ba ∨ Bb fails in the algebra of Example 1 taking a and b tobe the coatoms s and t . Also, inequality ¬ Ba ≤ B ¬ a fails in G , taking a to be the middle element. So, alsoinequality Ba → Bb ≤ B ( a → b ) fails.Though B may not exist for every element in a resid-uated lattice, B exists in every finite residuated lattice. Proposition 4
Let A ∈ RL be finite. Then, B existsin A .Proof In a finite residuated lattice A , for any a ∈ A , wehave Ba = W { b ∈ A : b ≤ a and b ∨¬ b = 1 } . It is enoughto see that if b ≤ a , b ∨¬ b = 1, b ≤ a , and b ∨¬ b =1, then (i) b ∨ b ≤ a and (ii) ( b ∨ b ) ∨ ¬ ( b ∨ b ) = 1.Now, (i) follows immediately and (ii) follows using (ii)in Proposition 2. (cid:3) On the other hand, there are infinite residuated lat-tices where B does not exist. Indeed, we have the fol-lowing example due to Franco Montagna (see [1]). Proposition 5
There is an (infinite) G¨odel algebra A and a ∈ A such that Ba does not exist, i.e. where B does not exist.Proof Let [0 , , G be the three-element G¨odel alge-bra. Let us consider A = { a ∈ ([0 , , G ) N such that { i ∈ N : a i = 0 } isfinite } , A = { a ∈ ([0 , , G ) N such that { i ∈ N : a i = 0 } isfinite } , and A = A ∪ A .The set A is the domain of a subalgebra of ([0 , , G ) N .Indeed, if a, b ∈ A , then a ∧ b ∈ A and a → b ∈ A ,if a, b ∈ A , then a ∧ b ∈ A and a → b ∈ A , if a ∈ A and b ∈ A , then a ∧ b ∈ A and a → b ∈ A , and if a ∈ A and b ∈ A , then a ∧ b ∈ A and a → b ∈ A .Also, 0 ∈ A . So, A is the domain of a subalgebra A of([0 , , G ) N .Now, take a to be such that a i = 1 if i is even and a i = if i is odd. Next, consider the set { b ∈ A : b ≤ a and b is Boolean } . It consists of all elements b such that b i = 0 for all odd i and for all but finitely many even i , and b i = 1 otherwise. It can be seen that this set hasno maximum in A . (cid:3) Actually, Montagna’s example of Proposition 5 canbe generalized as follows.
Proposition 6
Let V be a variety of MTL -algebrassuch that there is a linearly ordered algebra A ∈ V witha proper filter F (i.e. { } ( F ( A ), that is, such that A is not simple. Then, V contains an infinite algebrawhere B does not exist.Proof Let D ∈ V be a chain and F be a filter of A sat-isfying the hypothesis of the proposition. Let us define F ¬ = { x ∈ D | ∃ y ∈ F, x ≤ ¬ y } and let C = F ∪ F ¬ . Itis easy to check that C is the domain of a subalgebraof D . Finally define the following sets: A = { a ∈ C N such that { i ∈ N : a i ∈ F } is finite } , A = { a ∈ C N such that { i ∈ N : a i ∈ F ¬ } is finite } , A = A ∪ A .One can check that again A is the domain of a sub-algebra of C N , taking into account that if x ∈ F and y ∈ F ¬ , then x ∧ y, x ∗ y, x → y ∈ F ¬ , and if x, y ∈ F ¬ ,then x → y ∈ F .Thus, A is a subalgebra and taking an element a such that a i = 1 if i is even and a i = b , for a given b ∈ F \ { } , then the same argument as in Montagna’sexample proves that Ba does not exist. (cid:3) For readers familiar with the main systems of math-ematical fuzzy logic and their algebraic semantics (see[8]), we provide the following corollary with further ex-amples of subvarieties of residuated lattices containingalgebras where B does not exist. Corollary 1
In the following varieties of
MTL -algebras,there is an infinite algebra where B does not exist: – the variety generated by any continuous t-norm, – the varieties generated by either the NM t-norm ora WNM t-norm. Proof
In all these varieties there is an algebra A sat-isfying the conditions of Proposition 6.If the t-norm is either a G¨odel, Product, or a WNMt-norm (including NM), then take as A the standardchain and as F the positive elements respect to ¬ ,i.e., the elements such that ¬ x ≤ x . If the t-norm is Lukasiewicz, then take A as the Chang algebra and F as the set of its positive elements. Finally, if the contin-uous t-norm is a proper ordinal sum, then take A as thestandard chain and F = [ a, a ∈ (0 ,
1) is theend point of a component. It is clear that in all cases F is a proper filter and thus Proposition 6 applies. (cid:3) Actually, this could be generalized in the following sense.In [14] Noguera proves that the variety generated by simple n -contractive MTL-chains is the variety of S n -MTL algebras,i.e. MTL-algebras satisfying the law x ∨¬ x n − = 1. Therefore,any variety of n -contractive MTL-algebras that are not S n -MTL has a chain with a proper filter. In particular, this is thecase for the varieties of WNM and NM-algebras, since theyare 3-contractives and are not S -MTL.xpanding FL ew with a Boolean connective 7 Some papers (e.g. [4]) consider the notion of com-patible operation . Operation B is not compatible, thatis, the congruences of RL and RL B are not the same.To see this, take the three-element Heyting or G¨odelalgebra G with universe { , , } and the equivalencerelation given by θ = { (0 , , ( , ) , ( , , (1 , ) , (1 , } .It holds that θ is a RL -congruence, but not an RL B -congruence, as B = 0 and B modality to mean a finite combination of theunary operators ¬ and B , the next statement shows howmany different modalities there are in RL B . Proposition 7 In RL B there are nine different modal-ities. They may be ordered as follows: on the one hand,the positive modalities B ≤ Id ≤ ¬¬ ≤ ¬ B ¬ (with also B ≤ B ¬¬ ≤ ¬¬ ) and, on the other hand, the negativemodalities B ¬ ≤ ¬ ≤ ¬ B ¬¬ ≤ ¬ B . See Figure 2.Proof The inequalities are immediate. The reverse in-equalities can be seen not to be the case by consideringeither the only atom in the three-element G¨odel algebra G or any of the two non-comparable elements of theHeyting algebra obtained by adding a top element tothe Boolean algebra of 4 elements. There are no othermodalities, because if we apply operations ¬ and B tothe given nine modalities, we do not get anything new,as ¬¬ B = B , BB = B , and B ¬ B = ¬ B . (cid:3) ¬ B ¬¬¬ Id B B ¬¬ ¬ B ¬ B ¬¬¬ B ¬ Fig. 2
The positive and negative modalities of ¬ and B RL B is in factan equational class. To this end, we start focusing ourattention on the following equations, using x y as anabbreviation for x ∨ y ≈ y : (BI1) Bx B ( x ∨ y ), (BI2) B ≈ Lemma 6
Equations (BI1) and (BI2) hold in RL B .Proof The given equations follow immediately fromlemmas 4(ii) and 5(vi), respectively. (cid:3)
We are also interested in the equation (BI3) B ( x ∨ ¬ x ) Bx ∨ ¬ x ,but it is not easy to see that it holds in RL B . Towardsthis goal, we state and prove the following result. Lemma 7 In RL B the following hold: (i) ( B ( x ∨ ¬ x ) ∧ x ) ∨ ¬ ( B ( x ∨ ¬ x ) ∧ x ) ≈ , (ii) B ( x ∨ ¬ x ) ∧ x Bx , (iii) B ( x ∨ ¬ x ) ∧ ¬ Bx ¬ x .Proof (i) Using Lemma 1(i), and using T for the lefthand side of the given equation, it is enough to get(iv) B ( x ∨ ¬ x ) ∨ ¬ B ( x ∨ ¬ x ) ≈ B ( x ∨ ¬ x ) T , and(vi) ¬ B ( x ∨ ¬ x ) T .Part (iv) is immediate because of (BE2) .To see (v), using Proposition 1(viii), note that we havethat B ( x ∨ ¬ x ) x ∨ ¬ x , (immediate using (BE1) ), B ( x ∨ ¬ x ) ∧ x T (also immediate), and B ( x ∨ ¬ x ) ∧¬ x T , which follows from B ( x ∨ ¬ x ) ∧ ¬ x ¬ ( B ( x ∨¬ x ) ∧ x ), which holds because of Lemma 1(iv).To see (vi), note that B ( x ∨ ¬ x ) ∧ x ≤ B ( x ∨ ¬ x ).So, using Lemma 1(iii), it follows that ¬ B ( x ∨ ¬ x ) ¬ ( B ( x ∨ ¬ x ) ∧ x ). And so, ¬ B ( x ∨ ¬ x ) T .(ii) Use (BI) , Part (i), and B ( x ∨ ¬ x ) ∧ x x .(iii) Use Part (ii) and Proposition 1(x). (cid:3) Proposition 8
The equation (BI3) holds in RL B .Proof Using Proposition 1(ix), it is enough to checkthe following three conditions:(i) Bx ∨ ¬ Bx ≈ B ( x ∨ ¬ x ) ∧ Bx Bx ∨ ¬ x , and(iii) B ( x ∨ ¬ x ) ∧ ¬ Bx Bx ∨ ¬ x .Now, (i) is immediate due to (BE2) and (ii) is alsoimmediate as B ( x ∨ ¬ x ) ∧ Bx Bx . Regarding (iii), itfollows from Lemma 7(iii). (cid:3) Remark 2
Arguing as in the proof of Proposition 8and noting that B ( x ∨ ¬ x ) ∧ ¬ Bx B ¬ x follows us-ing (BI) from Lemma 7(iii) and the fact that the term B ( x ∨ ¬ x ) ∧ ¬ Bx is Boolean, it may be seen that alsothe inequality B ( x ∨ ¬ x ) Bx ∨ B ¬ x holds in RL B . Lemma 8 B is monotone just using equations.Proof Suppose x ∨ y ≈ y . Then, (i) B ( x ∨ y ) ≈ By .Now, using (BI1) , we have (ii) Bx ∨ B ( x ∨ y ) ≈ B ( x ∨ y ).From (i) and (ii) we get Bx ∨ By ≈ By . (cid:3) Rodolfo C. Ertola-Biraben et al.
The following theorem answers positively the ques-tion posed above.
Theorem 1 RL B is an equational class. An equationalbasis relative to RL is the following set of equations: (BE1) Bx x , (BE2) Bx ∨ ¬ Bx ≈ , (BI1) Bx B ( x ∨ y ) , (BI2) B ≈ , (BI3) B ( x ∨ ¬ x ) Bx ∨ ¬ x .Proof It is enough to prove (BI) using the given equa-tions. Suppose (i) b ≤ a and (ii) b ∨ ¬ b = 1. From (i),using Lemma 8, it follows (iii) Bb ≤ Ba . From (ii),using (BI2) , it follows B ( b ∨ ¬ b ) = B (BI3) , implies that Bb ∨ ¬ b = 1, which, using(iii) gives Ba ∨ ¬ b = 1, which, using Proposition 1(xi),implies b ≤ Ba . (cid:3) Remark 3
Note that, as expected, the just given proofof (BI) only uses equations (BI1) , (BI2) , and (BI3) . It is also natural to inquire whether the given equa-tions are independent.
Proposition 9
The set { (BE1) , (BE2) , (BI1) , (BI2) , (BI3) } is independent.Proof To see that (BE1) is independent of the rest,take the three-element G¨odel algebra G and define B Ba = 1, if a is not 0.To see that (BE2) is independent of the rest, take thefour-element G¨odel chain G and define B B Ba = 0, for the only atom a of the chain, and Bc = c ,for the remaining element c .To see that (BI1) is independent of the rest, take theG¨odel algebra G × G . If a is any of the four Booleanelements, then put Ba = a , else put Ba = 0.To see that (BI2) is independent of the rest, take again G , but now define Ba = 0, for every a .Finally, to see that (BI3) is independent of the rest,take the four-element Boolean G × G algebra anddefine, for any a , if a = 1, then Ba = 1, else Ba = 0. (cid:3) RL B -algebrasIn this section we show the subdirectly irreducible mem-bers of RL B are those whose Boolean elements are onlythe top and bottom elements. Definition 4
Let A ∈ RL B . A set F contained in A is said to be a RL B -filter iff for all a, b ∈ A it satisfies (1) 1 ∈ F , (2) if a ∈ F and a ≤ b , then b ∈ F , (3) if a, b ∈ F , then a · b ∈ F , (4) if a ∈ F , then Ba ∈ F . Proposition 10
Let A ∈ RL B . The lattice of RL B -congruences is isomorphic to the set of RL B -filters. In-deed, let f : Con ( A ) −→ F il ( A ) be defined by: if ≡ isa RL B -congruence, then f ( ≡ ) is the RL B -filter F ≡ = { a ∈ A : a ≡ } . Then, the function f is an isomor-phism such that if F is a RL B -filter, then f − ( F ) is the RL B -congruence ≡ F defined by a ≡ F b iff a → b, b → a ∈ F .Proof It is obvious that F ≡ is a RL B -filter. In order toprove that ≡ F is a congruence we need to prove that if a ≡ F b , then Ba ≡ F Bb , since the other conditions areknown to be true for any residuated lattice. So, suppose a → b and b → a ∈ F . Then, by the fourth condition inthe definition of filter, B ( a → b ) ∈ F . Now, using (vi)in Lemma 5 and the second condition in the definitionof filter, it follows that Ba → Bb ∈ F . Analogously, weobtain that Bb → Ba ∈ F . Finally, it is also obviousthat f − ◦ f = Id . (cid:3) Now we can characterize a family of RL B -filters. Proposition 11
Let A ∈ RL B . If a ∈ B ( A ) , then F a = [ a,
1] = { x ∈ A : a ≤ x ≤ } is a RL B -filter.Proof It is obvious that F a satisfies the first two condi-tions of a RL B -filter. The third is an easy consequenceof the fact that if a ∈ B ( A ), then a ∗ x = a ∧ x and thusif x, y ∈ F a , then a = a ∧ y ≤ x ∗ y and thus x ∗ y ∈ F a .Finally, if x ∈ F , then a = Ba ≤ Bx . (cid:3) From now on, F a denotes the principal filter definedby a ∈ B ( A ).In order to characterize the subdirectly irreducible RL B -algebras, we will use the result of [22, Theorem97]: an algebra A is subdirectly reducible iff there existsa family of non-trivial congruences σ i such that theirintersection is the identity. In our case, this means that A is subdirectly irreducible iff there is a unique coatomin the lattice of RL B -congruences of A . Proposition 12
Let A ∈ RL B . Then, A is subdirectlyirreducible iff B ( A ) = { , } .Proof Observe first that if F is a RL B -filter of A , then F contains a Boolean element a (by the third conditionof RL B -filter) and, thus, F contains F a . So, to obtainthe intersection of the non-trivial RL B -filters of A itis enough to compute the intersection of the filters F a .However, this intersection is not the identity iff there ex-ists a unique Boolean element a such that a is a coatomof B ( A ). So, being B ( A ) a Boolean algebra, this impliesthat B ( A ) = { , } . (cid:3) xpanding FL ew with a Boolean connective 9 B with other operations Operation B is strongly related to other operations con-sidered in the literature, e.g. the Monteiro-Baaz ∆ andan operation defined with the join-complement D . Inthis section we study these relationships.4.1 Comparing B with ∆ The operation ∆ was already considered by Monteiro inhis paper on symmetric Heyting algebras in 1980 (see[13]). Monteiro considered the same definitions of pos-sibility and necessity operations given by Moisil in [11](see p. 67 in [13]). However, instead of using Moisil’snotation, Monteiro used ∇ and ∆ , respectively. Whenso doing, he did not explicitly mention Moisil. However,many works by Moisil appear in the list of referencesof [13], including [11] and [12]. Monteiro also consid-ered the ∆ operator in the setting of linear symmetricHeyting algebras and studied the properties of ∆ inthe totally linear case (see [13, Ch. 5, Sect. 3]). In 1996,independently, Baaz in [2] considered an expansion ofG¨odel logic with a connective he also called ∆ satis-fying certain axioms and the rule ϕ/∆ϕ . Although hedid not cite Monteiro, the proposed axioms are equiv-alent to the properties that Monteiro proved for his ∆ operator in the framework of totally linear symmetricHeyting algebras. Baaz also provided a deduction theo-rem using ∆ : Γ, ϕ ⊢ ψ iff Γ ⊢ ∆ϕ → ψ . In 1998, H´ajekconsidered Baaz’s ∆ in the context of BL-algebras andBL logic (see pp. 57-61 [10]). He gave for ∆ exactlythe same axioms as Baaz presented in [2] for G¨odellogic. He observed that all ∆ axioms make it behavelike a necessity operator, with the exception of the ax-iom ∆ ( ϕ ∨ ψ ) → ( ∆ϕ ∨ ∆ψ ), that is characteristic ofpossibility operations (see Remark 2.4.7 of [10]). The ∆ operation has also been studied in the more general con-text of Mathematical fuzzy logic, see several chaptersin the handbook [8]. More recently, in [1] the authorsstudy, among other things, the expansion of FL ew withthe ∆ operator and show that it is conservative.In MTL, ∆ can always be defined over chains, namelyas ∆ ∆x = 0 for all x = 0, and thus, ∆ and B over MTL-chains coincide. But there are (non-linearly)MTL-algebras where ∆ does not exist. Nevertheless,this is not a problem because MTL is semilinear, andthe semantics of ∆ over chains is clear. However, thereis not a clear semantical interpretation of the axioms of ∆ in the general context of residuated lattices.In the context of a residuated lattice, the operator ∆ is introduced e.g. in [1] by the same equations as inMTL or BL (cf.[10, p. 58]): ( ∆ E1) ∆x x , ( ∆ E2) ∆x ∨ ¬ ∆x ≈ ( ∆ I1) ∆ ( x ∨ y ) ∆x ∨ ∆y , ( ∆ I2) ∆ ≈ ( ∆ I3) ∆x ∆∆x , ( ∆ I4) ∆ ( x → y ) ∆x → ∆y ,where, again, x y abbreviates x ∨ y ≈ y . Note that ( ∆ I3) may be derived from the rest: it is enough tocheck that an operator satisfying the rest of the equa-tions, satisfies all the equations in Theorem 1, and hencethe quasi-equation (BI) as well; then use (iii) of Lemma4. Also, regarding their defining equations, the only dif-ference between ∆ and B is that ∆ satisfies ∆ ( x ∨ y ) ∆x ∨ ∆y , whereas B only satisfies the particular case y = ¬ x , that is, B only satisfies B ( x ∨¬ x ) Bx ∨ B ¬ x ,as stated in Remark 2.We will denote by RL ∆ the class of residuated lat-tices expanded with ∆ .It will be useful to bear in mind the following fact. Lemma 9
Let A ∈ RL ∆ and a ∈ A . Then, ∆a = a iff a is Boolean.Proof Supposing ∆a = a , using ( ∆ E2) it follows that a is Boolean. On the other hand, suppose a is Boolean.Considering ( ∆ E1) , it is enough to prove that a ≤ ∆a . By Lemma 1(ii), it is enough in turn to prove ∆a ∨ ∆ ¬ a = 1, a · ∆a ≤ ∆a , and a · ∆ ¬ a ≤ ∆a . Thefirst condition holds using ( ∆ I1) and ( ∆ I2) , since a isBoolean. The second condition is immediate. For thethird, observe that a · ∆ ¬ a ≤ a · ¬ a = 0 ≤ ∆a .Actually, ∆ is somewhat stronger than B in the fol-lowing sense. Proposition 13
Let A ∈ RL . If ∆ exists in A , thenso does B , with B = ∆ .Proof Considering Theorem 1, all we have to see is that ∆ satisfies the equational basis given for B . This is im-mediate excepting (BI1) . Let us see that the equation ∆x ∆ ( x ∨ y ) also holds. As we have x → ( x ∨ y ) ≈ ( ∆ I2) and ( ∆ I4) we get 1 ∆x → ∆ ( x ∨ y ),which gives ∆x ∆ ( x ∨ y ). (cid:3) On the other hand, we have the following result.
Proposition 14
There exist finite residuated latticeswhere B exists, but ∆ does not.Proof Using Proposition 4, it follows that B exists inthe G¨odel algebra of Example 1, as the algebra is finite.Now, take its coatoms s and t . To see that ∆ does notexist, note that ( ∆ E1) and ( ∆ E2) imply that ∆s = ∆t = 0. So, ∆s ∨ ∆t = 0. However, ∆ ( s ∨ t ) = 1, dueto ( ∆ I2) . Then, ( ∆ I1) is not satisfied. (cid:3)
Example 1 makes clear the basic difference between ∆ and B , when we define them over MTL-algebras. Itis well known that MTL ∆ , the expansion of MTL with ∆ , is semilinear, i.e. each algebra of the variety is asubdirect product of lineraly ordered MTL ∆ -algebras.Moreover, we have seen that ∆ and B coincide overchains. Thus, Example 1 proves that MTL B , the ex-pansion of MTL with B , is not semilinear. In fact, thiswas already clear from Proposition 12, since there ex-ist subdirectly irreducible MTL B -algebras (like the onedefined in Example 1) that are not linearly ordered.4.2 Comparing B with an operation using thejoin-complementThe join-complement operation D has a long history.In 1919, Skolem considered lattices expanded with bothmeet and join relative complements (see § a , he used the notations a and 1 − a , respectively.In 1942, Moisil defined possibility as ¬¬ and neces-sity as DD in a logical setting where he had both intu-itionistic negation ¬ and its dual D (see § D form anequational class (see [19]). In fact, the meet is not needed,as the class with join and join-complement D is alreadyan equational class. In 1974, Rauszer, mainly consider-ing algebraic aspects, studied a logic with conjunction,disjunction, conditional, and its dual (see [16]). She alsoincluded both intuitionistic negation ¬ and its dual D ,though these can be easily defined. Her axiomatizationincluded the expected axioms plus the rules modus po-nens and ϕ/ ¬ Dϕ . She also provided a deduction the-orem using ( ¬ D ) n . She neither mentioned Skolem norMoisil.In the context of a join semi-lattice A , it is possibleto postulate the existence of the join-complement Da =min { b ∈ A : for all c ∈ A , c ≤ a ∨ b } , for a ∈ A . This isequivalent to the following two conditions: (DI) b ≤ a ∨ Da , for all a, b ∈ A , (DE) for any a, b ∈ A , if for all c ∈ A , c ≤ a ∨ b ,then Da ≤ b .In a join semi-lattice the existence of D implies theexistence of both top 1 = a ∨ Da , for any a , and bot-tom 0 = D ( a ∨ Da ), for any a . Moreover, D can beequationally characterized by the following three equa-tions, where, again, we use x y as an abbreviation for x ∨ y ≈ y : (DI) y x ∨ Dx , (DE1) D ( x ∨ Dx ) y , (DE2) Dy x ∨ D ( x ∨ y ).In what follows, RL D will denote the class of resid-uated lattices expanded with an operation D satisfyingthese equations. Obviously, by definition, RL D is anequational class. Notice that in a residuated lattice A ,having in the signature the symbols 0 and 1 for the bot-tom and top elements, respectively, the above definitionof D can be simplified to Da = min { b ∈ A : a ∨ b = 1 } ,and the condition (DE) simplifies to be (DE ′ ) for any a, b ∈ A , if a ∨ b = 1, then Da ≤ b .Moreover the equations (DI) and (DE1) can also besimplified to: (DI ′ ) x ∨ Dx ≈ (DE1 ′ ) D ≈ Remark 4
Note that, while x ¬¬ x holds in RL ,from (DE ′ ) and (DI ′ ) it follows that DDx x holdsin RL D . Note also that in a Heyting algebra D is thedual of ¬ , since in that case ¬ coincides with the meetcomplement. As in the case of B , D may not exist in some resid-uated lattices, but it always exists in the finite ones. Proposition 15
Let A be a finite residuated lattice.Then D exists in A .Proof It is enough to prove that V { b ∈ A : a ∨ b = 1 } exists in A . For that, it is enough to see that if a ∨ b = 1and a ∨ b = 1, then a ∨ ( b ∧ b ) = 1. Now, from theantecedent it follows that ( a ∨ b ) · ( a ∨ b ) = 1 and usingtwice the distributive law of · with respect to ∨ , we havethat ( a · a ) ∨ ( a · b ) ∨ ( b · a ) ∨ ( b · b ) = 1. Any subterm t of the left-hand term is such that t ≤ a ∨ ( b ∧ b ). (cid:3) Following [16] and [18], we consider now the com-pound operation ¬ D and its relation to B . First, let usstate the following fact. Lemma 10
Let A ∈ RL D and a, b ∈ A . Then, (i) ¬ Da ≤ a , (ii) if a ≤ b , then Db ≤ Da and ¬ Da ≤ ¬ Db .Proof (i) follows from a ∨ Da = 1 using Lemma 1(vi).(ii) Assume a ≤ b . Then, 1 = a ∨ Da ≤ b ∨ Da . Hence,by (DE) we have Db ≤ Da . Now, apply Lemma 1(iii)to get ¬ Da ≤ ¬ Db . (cid:3) In [5, Section 5] the authors prove a result aboutiterations of the operation ¬ D in the context of meet-complemented distributive lattices with D . Once triv-ially adapted to RL D , it is the following fact. xpanding FL ew with a Boolean connective 11 Proposition 16 (i)
For any natural n > , let RL D n be the subvariety of RL D defined by adding to those of RL D the following equation: ( ¬ D ) n +1 x ≈ ( ¬ D ) n x. Then, the sequence of varieties RL D ⊂ RL D ⊂ . . . ⊂ RL D n ⊂ . . . is strictly increasing. (ii) There are algebras of RL D where none of theequations given in (i) hold. Next, consider the following example of a Heytingalgebra where B exists but D does not. Example 2 B exists in the Heyting algebra 1+( N × N ) ∂ of Figure 3, where ( N × N ) ∂ is obtained ‘turning upsidedown’ the partial order N × N . In that Heyting algebra, Ba = 1 if a = 1 else Ba = 0. However, D does not existfor the elements (0 , n ) and ( n, n . (0,2) (0,1) (1,0) (2,0) Fig. 3
The residuated lattice 1 + ( N × N ) ∂ , where B existsbut D does not There are also RL -algebras where D exists, but B does not (see the end of Section 2 of [5] for an exampleof a Heyting algebra). Note that in Franco Montagna’sexample, neither B nor D exist.In the following case, existence of D implies exis-tence of B . Proposition 17
Let A ∈ RL and ¬ a ∨ ¬¬ a = 1 , forall a ∈ A . Then, if D exists in A , then B also exists in A , with B = ¬ D .Proof Let a ∈ A . Then, ¬ Da exists in A . We have tosee (i) ¬ Da ≤ a , (ii) ¬ Da ∨¬¬ Da = 1, and (iii) if b ≤ a and b ∨ ¬ b = 1, then b ≤ ¬ Da . Now, (i) holds as seen inLemma 10(i) and (ii) follows from the hypothesis that ¬ a ∨ ¬¬ a = 1, for any a ∈ A . To see (iii), suppose (iv) b ≤ a and (v) b ∨ ¬ b = 1. Due to Lemma 10(ii), we havethat (iv) implies Da ≤ Db and, using (DE) , (v) implies Db ≤ ¬ b . So, by ≤ -transitivity it follows that Da ≤ ¬ b .Then, in a residuated lattice we have b ≤ ¬ Da . (cid:3) Remark 5
Given the conditions of Proposition 17, tak-ing any of the coatoms in the algebra of Example 1, itis easy to see that Dx ≈ ¬ Bx does not hold. Also, thereciprocal of Proposition 17 is not the case, as the alge-bra in Example 2 satisfies the equation ¬ x ∨ ¬¬ x ≈ and B exists in that algebra, but D does not exist. Taking into account De Morgan laws valid in anyMTL-algebra , we can easily obtain the following con-sequence of the previous proposition. Corollary 2
Let A be a SMTL-algebra, i.e. an MTL-algebra such that for all a ∈ A , a ∧ ¬ a = 0 . Then, forany a ∈ A , if Da exists, then so does Ba , and Ba = ¬ Da . Lemma 11
Let A ∈ RL D and a ∈ A . Then, the fol-lowing are equivalent: (i) a is Boolean, (ii) ¬ Da = a , (iii) Da = ¬ a .Proof (i) ⇒ (ii) Suppose a ∨¬ a = 1. Then, using (DE) , Da ≤ ¬ a . Then, a ≤ ¬ Da . Now, by Lemma 10(i) wehave ¬ Da ≤ a . So, ¬ Da = a .(ii) ⇒ (iii) Suppose ¬ Da = a . Then, ¬¬ Da = ¬ a .As Da ≤ ¬¬ Da , we have that Da ≤ ¬ a . Now, by (DI) , a ∨ Da = 1. So, also ¬ a ≤ Da . Then, Da = ¬ a .(iii) ⇒ (i) Suppose Da = ¬ a . As using (DI) wehave a ∨ Da = 1, it follows that a ∨ ¬ a = 1. (cid:3) As a direct consequence we have the following fact.
Corollary 3
Let A be a residuated lattice where both B and D exist. Then, Da ≤ ¬ Ba , for all a ∈ A . Equiv-alently, Ba ≤ ¬ Da , for all a ∈ A .Proof Let a ∈ A . We have that Ba ≤ a . Hence, usingLemma 10(ii), Da ≤ DBa . Now, since Ba is Boolean,using Lemma 11 it follows that Da ≤ ¬ Ba . (cid:3) Remark 6
The equality B ≈ ¬ D does not hold. In-deed, consider the join-irreducible coatom c in the Heyt-ing algebra in Figure 4, where Bc < ¬ Dc . Lemma 12
Let A be a RL D -algebra and a, b ∈ A . Wehave that if b ≤ a and b ∨ ¬ b = 1 , then b ≤ ¬ Da .Proof Suppose b ≤ a . Then, ¬ Db ≤ ¬ Da . Now, usingthe hypothesis b ∨ ¬ b = 1 and Lemma 11, we have that ¬ Db = b . So, b ≤ ¬ Da . (cid:3) Notice that not all instances of De Morgan laws are validin the variety of MTL-algebras, for instance the equations ¬ ( x ∧ y ) ≈ ¬ x ∨ ¬ y and ¬ ( x ∨ y ) ≈ ¬ x ∧ ¬ y are valid, but x ∧ y ≈ ¬ ( ¬ x ∨ ¬ y ) is not.2 Rodolfo C. Ertola-Biraben et al. c ¬ DcBcDc
Fig. 4
Behaviour of B and D in a coatom of a residuatedlattice Proposition 18
Let A be a RL D -algebra, let a ∈ A ,and let us have a finite number of elements b ∈ A suchthat b ≤ a . Then, Ba = ( ¬ D ) n a , for some n ∈ N .Proof In the case a is Boolean, Ba = ( ¬ D ) a . In thecase a is not Boolean, take ¬ Da . Now, Lemma 12 sayswe will not be missing Boolean elements. Repeating theprocedure we will find the first Boolean below a . (cid:3) Proposition 19 (i) In every algebra of RL D n , for anynatural number n ≥ , B exists, with B = ( ¬ D ) n .(ii) There are RL D -algebras where B does not exist.Proof (i) It suffices to see that ( ¬ Dx ) n satisfies (BE1) , (BE2) , and (BI) . It always satifies (BE1) , as it is eas-ily seen by induction using ¬ Da ≤ a and ≤ -transitivity.We get (BE2) applying Lemma 11 on the hypothesisthat ( ¬ D ) n +1 = ( ¬ D ) n . Finally, to get (BI) , supposeboth (i) b ≤ a and (ii) b ∨¬ b = 1. From (i) it follows (iii) ¬ Db ≤ ¬ Da . From (ii), using Lemma 11, we get (iv) ¬ Db = b . From (iii) and (iv) we get b ≤ ¬ Da . Repeatthe argument n times to get b ≤ ( ¬ D ) n a .(ii) cf. end of Section 2 in [5]. (cid:3) In the next proposition we will use the following DeMorgan properties for ¬ and D . In the proof we use theabreviation { x i } for { x i ∈ A : i ∈ I } . Lemma 13
Let A be a complete RL D -algebra. Then,both (i) ¬ W { x i } = V {¬ x i } and (ii) D V { x i } = W { Dx i } .Proof We prove only (ii), (i) is already known. By (DI) we have that x j ∨ Dx j = 1. Then, x j ∨ W { Dx i } = 1 forall j ∈ I and so, V { x i } ∨ W { Dx i } = 1. So, by (DE) weobtain that D V { x i } ≤ W { Dx i } .For the other inequality, using (DI ′ ) , we have that V { x i } ∨ D V { x i } = 1. Now, from V { x i } ≤ x j and x j ≤ x j ∨ D V { x i } we obtain V { x i } ≤ x j ∨ D V { x i } ,and taking into account that D V { x i } ≤ x j ∨ D V { x i } ,we have that V { x i } ∨ D V { x i } ≤ x j ∨ D ( V { x i } . Now,by (DI ′ ) we obtain that { x i } ∨ D V { x i } = 1, which, by (DE) , implies Dx j ≤ D V { x i } , for all j ∈ I . Thus, wefinally get W { Dx i } ≤ D V { x i } . (cid:3) In the next proposition, N and N + denote the set ofnatural numbers including 0 and excluding 0, respec-tively. Proposition 20
Let A be a complete RL D -algebra.Then, Ba exists, with Ba = V { ( ¬ D ) n a : n ∈ N } , forany a ∈ A .Proof Considering the definition of B , it is enough toprove, for a ∈ A , (i) V { ( ¬ D ) n a : n ∈ N } ≤ x , (ii) V { ( ¬ D ) n a : n ∈ N } ∨ ¬ V { ( ¬ D ) n a : n ∈ N } = 1, and(iii) if b ≤ a and b ∨ ¬ b = 1, then b ≤ V { ( ¬ D ) n a : n ∈ N } . Now, (i) follows, because a ∈ { ( ¬ D ) n a : n ∈ N } ,as a = ( ¬ D ) a . Regarding (ii) and using Lemma 11, itis enough to prove that ¬ D ( V { ( ¬ D ) n a : n ∈ N } ) = V { ( ¬ D ) n a : n ∈ N } . As it is always the case, forany b ∈ A , that ¬ Db ≤ b , it suffices to prove that V { ( ¬ D ) n a : n ∈ N } ≤ ¬ D ( V { ( ¬ D ) n a : n ∈ N } ). Now,using both properties of Lemma 13, we have that theright hand side of the just given inequality is equal to V { ( ¬ D ) n a : n ∈ N + } . It is clear that V { ( ¬ D ) n a : n ∈ N } ≤ V { ( ¬ D ) n a : n ∈ N + } , because V { ( ¬ D ) n a : n ∈ N } ≤ ( ¬ D ) m a , for m ∈ N + . Regarding (iii), suppose(iv) b ≤ a and b ∨ ¬ b = 1, the last of which implies,by Lemma 11, that (v) ¬ Db = b . In order to provethat b ≤ V { ( ¬ D ) n a : n ∈ N } , it is enough to provethat b ≤ ( ¬ D ) n a , for all n ∈ N , which easily followsby induction, as b ≤ a = ( ¬ D ) a , by (iv), and suppos-ing that b ≤ ( ¬ D ) n a , it follows, using Lemma 10, that ¬ Db ≤ ( ¬ D ) n +1 a , and, using (v), b ≤ ( ¬ D ) n +1 a . (cid:3) In Example 2 we saw that the existence of B ina residuated lattice does not force the existence of D .Now, let us see that operation ∆ is stronger than B inthis respect. Proposition 21
Let A ∈ RL . If ∆ exists in A , thenalso D exists in A , with D = ¬ ∆ and ∆ = ¬ D .Proof Suppose A is a residuated lattice where ∆ ex-ists. Then, also ¬ ∆ exists. We have to prove that Da = ¬ ∆a , for any a ∈ A . We have that a ∨ ¬ ∆a = 1, as ∆a ∨ ¬ ∆a = 1 and ∆a ≤ a . Now, suppose a ∨ b = 1.Then, ∆ ( a ∨ b ) = ∆ ∆ ( a ∨ b ) = ∆a ∨ ∆b . So, ∆a ∨ ∆b = 1. Using Lemma1(vi), ¬ ∆a ≤ ∆b . Moreover, ∆b ≤ b . So, ¬ ∆a ≤ b .Let us also see that ∆a = ¬ Da , for all a . Fromthe first part it follows that ¬ Da = ¬¬ ∆a . Now, from ( ∆ E2) , using Lemma 1(vi), it follows that ¬¬ ∆a ≤ ∆a .And using Lemma 1(vii) we have that ∆a ≤ ¬¬ ∆a . (cid:3) Remark 7
The reciprocal of Proposition 21 is not thecase, as D exists in Example 1, but ∆ does not. Corollary 4
Let A ∈ RL . If ∆ exists in A , then also B and D exist, and we have ∆ = B = DD = ¬ D . xpanding FL ew with a Boolean connective 13 Proof
Considering Propositions 13 and 21, it is enoughto prove that ∆ = DD , which follows from ¬ ∆ ¬ ∆ = ∆ .As for any a ∈ A , ∆a is Boolean due to ( ∆ E2) , usingLemma 9 it is enough to check that ¬¬ ∆ = ∆ , whichfollows again from ( ∆ E2) and Lemma 1(vi). (cid:3)
Bew
In this section we introduce an expansion of FL ew witha unary connective B , whose intended algebraic seman-tics is the variety of RL B -algebras studied in Section 3.Indeed, we define FL Bew as the expansion of FL ew with the following axiom schemas: (B1) Bϕ → ϕ , (B2) Bϕ ∨ ¬ Bϕ , (B3) B ( ϕ ∨ ¬ ϕ ) → ( Bϕ ∨ ¬ ϕ ), (B4) B ( ϕ → ψ ) → ( Bϕ → Bψ ).and the following additional rule: (B) From ϕ derive Bϕ .We denote (finitary) derivability in FL Bew by ⊢ .Note that we have the following facts. Lemma 14 (i) ⊢ Bϕ → BBϕ and(ii) Bϕ → ψ ⊢ Bϕ → Bψ .Proof For (i) check the following derivation:1. Bϕ ∨ ¬ Bϕ (B2) B ( Bϕ ∨ ¬ Bϕ ) 1, rule (B) BBϕ ∨ ¬ Bϕ (B3) , 2, mp4. BBϕ → ( Bϕ → BBϕ ) FL ew ¬ Bϕ → ( Bϕ → BBϕ ) FL ew Bϕ → BBϕ
3, 4, 5, FL ew (ii) follows easily using (i). (cid:3) Clearly, FL
Bew is a
Rasiowa implicative logic (cf. [17]).Then, it follows that it is algebraizable in the sense ofBlok and Pigozzi [3]. It is also straightforward to checkthat the variety RL B is its equivalent algebraic seman-tics. Algebraizability immediately implies strong com-pleteness of FL Bew with respect to RL B . Theorem 2
For every set Γ ∪ { ϕ } of formulas, Γ ⊢ ϕ iff for every A ∈ RL B and every A -evaluation e , e ( ϕ ) = 1 , whenever e [ Γ ] ⊆ { } . In FL
Bew the usual form of the deduction theoremdoes not hold. Indeed, ϕ ⊢ Bϕ , but ϕ → Bϕ , Whichfails in the three-element G¨odel algebra [0 , , G , where B B = B e in thisalgebra, if e ( ϕ ) = 1, then e ( Bϕ ) = 1, but for e ( ϕ ) = we have e ( Bϕ ) = 0, and thus e ( ϕ → Bϕ ) = 0.Actually, FL Bew enjoys the same form of deductiontheorem holding for logics with the ∆ operator (cf. [2,Proposition 2.2]). Theorem 3
Γ, ϕ ⊢ ψ iff Γ ⊢ Bϕ → ψ .Proof ⇒ ) We prove by induction on every formula χ i (1 ≤ i ≤ n ) of the given derivation of ψ from Γ ∪ { ϕ } that Γ ⊢ Bϕ → χ i . If χ i = ϕ , then the result followsdue to axiom schema (B1) . If χ i belongs to Γ or isan instance of an axiom, then the result follows using modus ponens and the derivability of the schema χ i → ( Bϕ → χ i ). If χ i comes by application of modus ponens on previous formulas in the derivation, then the resultfollows, because from Bϕ → χ k and Bϕ → ( χ k → χ i )we may derive ( Bϕ & Bϕ ) → ( χ k &( χ k → χ i )) and thenalso Bϕ → χ i , using transitivity of → applied to thederivable formulas Bϕ → ( Bϕ & Bϕ ) and ( χ k & ( χ k → χ i )) → χ i . Finally, if χ i = Bχ k comes using rule (B) from formula χ k , then from Bϕ → χ k we may derive Bϕ → Bχ k using Lemma 14(ii). ⇐ ) To the derivation given by the hypothesis add astep with ϕ . In the next step put Bϕ , which follows fromthe previous formula using rule (B) . Finally, derive ψ using modus ponens . (cid:3) Thanks to this B -deduction theorem, the logic FL Bew has the following property: if we expand FL
Bew with anyfurther rule ϕ , . . . , ϕ n /ϕ , then it is possible to disposeof the rule just adding the axiom ( Bϕ ∧· · ·∧ Bϕ n ) → ϕ .This property also holds for the logics FL ∆ew and FL Dew . Proposition 22 FL Bew is a conservative expansion of FL ew .Proof Use Proposition 4 and the Finite Model Prop-erty of FL ew (see [15]). (cid:3) One could analogously define the expansion of MTL(which is in turn the extension of FL ew with the pre-linearity axiom ( ϕ → ψ ) ∨ ( ψ → ϕ )) with B , withthe same additional axioms and rule, yielding the logicMTL B , which is algebraizable and strongly completewith respect to the variety MTL B of MTL B -algebras.However, unlike the case of expansion with ∆ , MTL B is not a semilinear logic, that is, it is not complete withrespect to the class of MTL B -chains. The reason is thatthe ∨ -form of rule (B) , “from ψ ∨ ϕ derive ψ ∨ Bϕ ”,is not derivable in MTL B . Indeed, taking the coatoms s and t in the G¨odel algebra of Example 1, it is clearthat s ∨ t = 1, while s ∨ Bt = s ∨ s .As a final result, we can show that FL Bew inheritsfrom FL ew the Finite Model Property (FMP). Beforeproving this, we introduce some preliminary notation.For a logic L ∈ { FL ew or FL Bew } , let us denoteby F m ( L, V ar ) the set of L -formulas built from a set V ar of propositional variables. Now let us define theenlarged set of propositional variables
V ar ∗ = V ar ∪{ “ Bϕ ” | Bϕ ∈ F m (FL
Bew , V ar ) } , where “ Bϕ ” is in-tended to denote a fresh propositional variable, one for each formula Bϕ ∈ F m (FL
Bew , V ar ). Then, we candefine a one-to-one translation of every formula ϕ ∈ F m (FL
Bew , V ar ) into a formula ϕ ∗ ∈ F m (FL ew , V ar ∗ ),by just inductively defining:- 0 ∗ = 0,- if ϕ = p ∈ V ar , then ϕ ∗ = p ,- if ϕ = Bψ , then ϕ ∗ = “ Bψ ”,- if ϕ = ψ ⊙ χ , then ϕ ∗ = ψ ∗ ⊙ χ ∗ , for ⊙ ∈ {∧ , ∨ , & , →} .If Γ is a set of formulas, we write Γ ∗ = { ϕ ∗ | ϕ ∈ Γ } . Note that for any ψ ∈ F m (FL ew , V ar ∗ ), there is aformula ϕ ∈ F m (FL
Bew , V ar ) such that ϕ ∗ = ψ .Moreover, we need the following result that will al-low us to reduce proofs in FL Bew to proofs in FL ew . Lemma 15
Let T be the set of all instances of axiomsof FL Bew . For each set Γ ∪ { ϕ } ⊆ F m (FL
Bew , V ar ) , itholds that Γ ⊢ F L
Bew ϕ iff Γ ∗ ∪ Cg ∗ ∪ T ∗ ⊢ F L ew ϕ ∗ ,where Cg = { Bϕ ↔ Bψ | Γ ⊢ F L
Bew ϕ ↔ ψ } . The proof is quite straightforward and analogous tothose of similar results that can be found in the lit-erature in slightly different contexts.
Theorem 4 FL Bew enjoys the FMP, that is, if Γ F L
Bew ϕ , then there is a finite A ∈ RL B and an A -evaluation e such that e ( Γ ) = 1 and e ( ϕ ) < .Proof If Γ F L
Bew ϕ , by Lemma 15, it holds that Γ ∗ ∪ Cg ∗ ∪ T ∗ F L ew ϕ ∗ , and by strong completeness andFMP of FL ew , there is a finite algebra C ∈ RL and C -evaluation v such that v ( Γ ∗ ∪ Cg ∗ ∪ T ∗ ) = 1 and v ( ϕ ∗ ) <
1. Then, the result will follow from the follow-ing facts:
Claim 1 : G = { v (“ Bϕ ”) | Bϕ ∈ F m (FL
Bew , V ar ) } is aset of Boolean elements of C . “You were a little grave,” Proof of the claim:
It is enough to check that v (( Bϕ ) ∗ ) ∨¬ v (( Bϕ ) ∗ ) = v (( Bϕ ) ∗ ∨ ¬ ( Bϕ ) ∗ ) = v (( Bϕ ∨ ¬ Bϕ ) ∗ ) =1, where the latter holds because Bϕ ∨¬ Bϕ is the axiom (B2) of FL Bew . ⊣ Claim 2 : Let A be the RL -algebra generated by theset X = { v ( ϕ ) | ϕ ∈ F m (FL ew , V ar ∗ ) } , which is finitesince A is a subalgebra of C . Then, B exists in A and B ( A ) = G . Therefore, A is indeed an RL B -algebra. Proof of the claim:
That A is finite is obvious, and thus,by Proposition 5, B exists. On the other hand, the el-ements of G keep being Boolean in A . Hence, the onlymissing thing to check is that any Boolean element of A already belongs to G . This is also clear since Booleanelements are closed by propositional combinations withconnectives. ⊣ Claim 3 : Let us define the A -evaluation (taking A as RL B -algebra) e : V ar → A defined by e ( p ) = v ( p ).Then, for any ϕ , e ( ϕ ) = v ( ϕ ∗ ), in particular, e ( Bϕ ) = v (“ Bϕ ”). Proof of the claim:
We prove that e ( ϕ ) = v ( ϕ ∗ ) bystructural induction.- if ϕ is a propositional variable, it holds by construc-tion- if ϕ = ψ ⊙ χ for ⊙ ∈ {∧ , ∨ , & , →} , by inductionhypothesis we have e ( ψ ) = v ( ψ ∗ ) and e ( χ ) = v ( χ ∗ ),and hence e ( ϕ ) = e ( ψ ⊙ ψ ) = e ( ψ ) ⊙ e ( χ ) = v ( ψ ∗ ) ⊙ v ( χ ∗ ) = v ( ψ ∗ ⊙ χ ∗ ) = v (( ψ ⊙ χ ) ∗ ) = v ( ϕ ∗ ).- If ϕ = Bψ , then we have to prove that v (“ Bψ ”) = B ( e ( ψ )), the latter being equal to e ( Bψ ) by def-inition. Therefore, we have to prove in turn thatthe three defining conditions (BE1) , (BE2) , and (BI) are satisfied by v (“ Bψ ”) = v (( Bψ ) ∗ ) to be thegreatest Boolean below e ( ψ ), assuming by inductionthat v ( ψ ∗ ) = e ( ψ ). (BE1) Since Bψ → ψ is axiom (BE1) of F L
Bew , we havethat 1 = v (( Bψ → ψ ) ∗ ) = v (( Bψ ) ∗ ) → v ( ψ ∗ ) = v (( Bψ ) ∗ ) → e ( ψ ). Hence, v (( Bψ ) ∗ ) ≤ e ( ψ ). (BE2) is clear from Claim 1. (BI) We have to check that if b ∈ B ( A ) is such that b ≤ e ( ψ ) = v ( ψ ∗ ), then b ≤ v (( Bψ ) ∗ ). If b ∈ B ( A ), by construction of A , then there existsa formula χ such that b = v (( Bχ ) ∗ ). On theother hand, by (ii) of Lemma 12, we know that Bχ → ψ, Bχ ∨ ¬ Bχ ⊢ Bχ → Bψ . Thus, we alsoknow that if v (( Bχ ) ∗ ) ≤ v ( ψ ) ∗ and v (( Bχ ) ∗ ) ∨¬ v (( Bχ ) ∗ ) = 1, then v (( Bχ ) ∗ ≤ v (( Bψ ) ∗ ). Now,the two conditions are satisfied, hence we have b = v (( Bχ ) ∗ ≤ v (( Bψ ) ∗ ).This closes the proof of Claim 3. ⊣ Finally, from these claims it readily follows that e ( Γ ) = v ( Γ ∗ ) = 1 and e ( ϕ ) = v ( ϕ ∗ ) <
1, as required. (cid:3)
In this paper we have considered the expansion of FL ew with the operator B , that in algebraic terms providesthe greatest Boolean below a given element of a residu-ated lattice. Among other things, we have axiomatizedit and shown that the resulting logic is a conservativeexpansion enjoying the Finite Model Property. The ax-ioms for B turn out to be very close to those of theMonteiro-Baaz ∆ operator, in fact only one axiom is aweaker version of the one for ∆ . Even if the propertiesare very similar, that small difference causes, e.g. thatin the context of MTL, the expansion with B is not any xpanding FL ew with a Boolean connective 15 longer a semilinear logic, in contrast to the expansionwith ∆ .As a matter of fact, we have chosen this topic forour humble contribution to honour the memory of ourbeloved and late friend Franco Montagna, because itwas suggested by Franco to the first author during thepreparation of their joint manuscript [1], together withAmidei, where they study the expansion of FL ew andother substructural logics with ∆ . Acknowledgments and Compliance with EthicalStandards
The authors are thankful to an anonymous reviewer forhis/her comments that have helped to improve the fi-nal layout of this paper. The authors have been fundedby the EU H2020-MSCA-RISE-2015 project 689176–SYSMICS. Esteva and Godo have been also fundedby the FEDER/MINECO Spanish project TIN2015-71799-C2-1-P. The authors declare that they have noconflict of interest. This article does not contain anystudies with human participants or animals performedby any of the authors.
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