ao DOI: 10.2478/s12175-010-0055-1 Math. Slovaca 61 (2011), No. 1, 1 14 MODAL OPERATORS ON COMMUTATIVE RESIDUATED LATTICES M. Kondo (Communicated by Jiří Rachůnek ) ABSTRACT. We prove some fundamental properties of monotone modal operators on bounded commutative integral residuated lattices (CRL). Moreover we give a positive answer to the problem left open in [RACHŮNEK, J. ŠALOU- NOVÁ, D.: Modal operators on bounded commutative residuated Rl-monoids, Math. Slovaca 57 (2007), 321 332]. c 2011 Mathematical Institute Slovak Academy of Sciences 1. Introduction In 1981, Macnab proved ([6]) some properties of modal operators on Heyting algebras which were introduced as algebraic counter-part of the intuitionistic propositional logic. Since that paper has been published, properties of modal operators are considered on other algebras, such as, MV-algebras, bounded commutative residuated Rl-monoids (simply called Rl-monoids) and so on. In this paper, to make a uniform treatment about properties of modal operators on such algebras, we define a monotone modal operators on bounded commutative integral residuated lattices (CRL) and investigate their properties. Those algebras above commonly have our CRL as base algebras, since any Heyting algebra can be considered as a bounded commutative integral residuated lattices if we define an operator by x y = x y. Monotone modal operators are exactly the modal operators on MV-algebras and Rl-monoids and such CRL satisfying the divisibility x y = x (x y). Strictly speaking, when we consider properties of monotone modal operators on CRL with divisibility, since every modal operator is monotone, our results can be directly applied to those algebras. Moreover, 2000 M a t h e m a t i c s Subject Classification: Primary 03B45, 06B05, 06D20, 06D35. K e y w o r d s: commutative residuated lattice, modal operators. This work was supported by Tokyo Denki University Science Promotion Fund (Q08J-06).
M. KONDO as we show below, there is a modal operator which is not monotone on a certain CRL. Thus monotone modal operators are different from modal operators when we consider them on CRL without divisibility. This means that it is worth considering monotone modal operators on CRL. In 2006, Harlenderová andrachůnek proved some algebraic properties of modal operators on MV-algebras in [3]. After that in 2007 algebraic properties on Rl-monoids were investigated by Rachůnek and Šalounová ([9]). Both algebras have our CRL as common base algebras. In [9], the following question left open whether ϕ a is a strong modal operator provided that it is a modal operator on normal bounded commutative integral residuated Rl-monoids. In this paper, we prove some fundamental results of monotone modal operators on bounded commutative integral residuated lattices, especially, a characterization theorem of strong modal operators, that is, a monotone modal operator f is strong if and only if it satisfies x f0 =f(x 0) for all x. As a corollary to one of our results, we give a positive solution to the problem above in [9]. We note that our modal operators in the sense of Macnab and of Harlenderová and Rachůnek are different from those of the theory of modal logics, because, as we prove later, for each modal operator f (in our sense) on any CRL with x = x, wehavef(0) = 0 if and only if f is an identity map. As a consequence of this fact, if we consider a modal operator f (in our sense) on any Boolean algebra B, thenwehavef(x) =x f(0) for all x B. This implies that a modal operator in the sense of this paper is different from that of the theory of modal logics. 2. Commutative integral residuated lattices (CRL) We recall a definition of bounded commutative integral residuated lattice ([2, 4, 5]). An algebraic structure M =(M,,,,, 0, 1) is called a bounded commutative integral residuated lattice if (1) (M,,, 0, 1) is a bounded lattice; (2) (M,, 1) is a commutative monoid with a unit element 1; (3) For all x, y, z M, x y z if and only if x y z. Abinaryoperator is defined by x y =(x y ), where x = x 0. In what follows, by a CRL we mean a bounded commutative integral residuated lattice according to the previous papers ([3, 9]) for the sake of simplicity. 2
MODAL OPERATORS ON COMMUTATIVE RESIDUATED LATTICES NextresultsarewellknownconcerningtoCRL([2,3,4,5,9,10]). Wenote the proofs of these results are the same as those of CRL with divisibility. Proposition 1. Let M be a CRL. For all x, y, z M, we have (1) x y x y (2) x y = x z y z,z x z y, y z x z (3) x y z = x (y z) =y (x z) (4) (x y) (y z) x z (5) x x,x = x (6) x y = y x (7) (x y) = x y (8) x y = x y = x y (9) x y = x y (10) x 0=x. Moreover we show the following results which are needed later. Proposition 2. Let M be a CRL. For all x, y, z M, we have (1) x y x z y z (2) x y x z y z (3) x y x z y z (4) x y y x. Proof. Weonlyprovethecaseof(1):x y x z y z. Since(x y) (x z) (x y) x y and (x y) (x z) x z z, wehave (x y) (x z) y z and thus (x y) x z y z. The other cases can be proved similarly. A unary mapping f : M M is called a modal operator on a CRL M in [3, 6] if it satisfies conditions: For all x, y, z M, (1) x fx; (2) f(fx)=fx; (3) f(x y) =fx fy. A modal operator f is called monotone if it satisfies (m) If x y then fx fy. We can show that every modal operator is monotone on any CRL with divisibility x y = x (x y) for all x, y. 3
M. KONDO Indeed, if x y then fx = f(x y) =f(y x) =f(y (y x)) = fy f(y x) fy. Thus the modal operators defined on MV-algebras ([3]) and Rl-monoids ([9]) are monotone modal operators defined here. However, there is a modal operator which is not monotone on a certain CRL. Let X = ( {x/10 : 0 x 10, x Z},,, 0, 1 ) be a bounded lattice where x y =min{x, y} and x y =max{x, y}. If we define operators and on X as x if y =1 x y = y if x =1 0 otherwise 1 if x y and x y = y if x =1 0.9 otherwise, then it is easy to show that the structure (X,,,,, 0, 1) is a bounded commutative integral residuated lattice (CRL). On the CRL we define an operator f : X X by f(x) = 0 if x =0 1 x if 0 <x 0.5 x if x>0.5. By simple calculation, we see that the operator f is the modal operator. But the modal operator f is not monotone. Because, we have 0.2 < 0.4 but f(0.2) = 0.8 0.6 =f(0.4). We have the following results about monotone modal operators. While these are proved as [9, Proposition 3] under the extra condition of divisibility, we can prove them without divisibility. Proposition 3. Let f be any monotone modal operator on a CRL M. Then we have (1) f(x y) fx fy = f(fx fy)=x fy = f(x fy) (2) fx (x f0) f0 (3) fx x f0 (4) fx f(x ) x f0 (5) f(x y) =f(x fy)=f(fx fy) (6) fx fy = f(fx fy). P r o o f. We only show some of cases (1), (2) and (6) for the sake of simplicity. (1) It follows from x (x y) y that f(x (x y)) fy and thus fx f(x y) fy. This means that f(x y) fx fy. Moreover, we have fx fy x fy by x fx. This yields that fx fy x fy f(x fy) fx f(fy)=fx fy. Thus, fx fy f(fx fy) f(fx) f(fy)=fx fy. We get that f(x y) fx fy = f(fx fy) = x fy = f(x fy). (2) We have fx f0 =x f0 by (1) and fx (fx f0) f0, thus fx (fx f0) f0 =(x f0) f0. 4
MODAL OPERATORS ON COMMUTATIVE RESIDUATED LATTICES (6) Since fx fy fx,fy,itisobviousthatf(fx fy) f(fx),f(fy)and thus f(fx fy) fx fy.conversely,sincef is the monotone modal operator, it is clear that fx fy f(fx fy). Thus we have fx fy = f(fx fy). A monotone modal operator f is called strong ([9]) if f(x y) =f(x fy) for all x, y. Proposition 4. Let f be a monotone modal operator on a CRL M. If x f0 = f(x 0) then fx f0 =x f0 for every x M. P r o o f. It follows from the proposition above that fx x f0 and thus fx f0 x f0 f0. Since f0 f0 =f(f0 0) = f(0 f0) = f(f(0 0)) = f(0 0) = f0 by assumption, we have fx f0 x f0. Conversely, it is obvious that x f0 fx f0. Thus we get fx f0 =x f0. We can get a characterization of strong modal operators. Proposition 5. Let f be a monotone modal operator. Then it is strong if and only if it satisfies the condition (s) x f0 =f(x 0) for all x. Proof. Weassumethatf is a strong modal operator. It is obvious from Proposition 3.(4) that f(x 0) = f(x ) x f0 andx f0 f(x f0) = f(x 0). Thus x f0 =f(x 0). Conversely, suppose that x f 0 = f(x 0). In general, wehavex y = x y 0 for all x, y M. This yields that f(x fy)=f((x fy) 0) = x fy f0 = x y f0 (by Prop. 4) = f(x y 0) = f(x y). In [9], a characterization theorem is proved, which have the divisibility x y = x (x y) as an extra axioms of our CRL. We can also get the same result on CRL as in [9]. This means that the divisibility is not essential for characterizing the monotone modal operators. Theorem 1. Let M be a CRL and f : M M a mapping. Then f is a monotone modal operator on M if and only if it satisfies the following two conditions: (a) fx fy = x fy (b) f(x y) fx fy. 5
M. KONDO P r o o f. It is easy to prove that if f is a monotone modal operator then it satisfies the conditions. We only show the converse, that is, if a mapping f satisfies the conditions then it is the monotone modal operator. Since 1 = fx fx = x fx by (a), we have x fx. Next if x y, sincey fy which is just proved above, then x fy and thus 1 = x fy = fx fy. This means that fx fy and that f is an order preserving mapping. The fact that 1 = fx fx = f(fx) fx by (a) implies f(fx)=fx. Finally, since x y f(x y) and thus y x f(x y) =fx f(x y), we have fx y f(x y) =fy f(x y) andfx fy f(x y). It follows from (b) that f(x y) =fx fy. Corollary 1. Let f : M M be a mapping on a CRL M. Thenf is a strong modal operator if and only if it satisfies (a) (b) fx fy = x fy f(x y) fx fy (c) x f0 =f(x 0). Let M be a bounded commutative integral residuated lattice and f : M M be a monotone modal operator. The next result is proved in [9] in the case of Rl-monoid, that is, bounded commutative integral residuated lattices with divisibility. We also show the similar result without divisibility. Theorem 2. Let M be a CRL and f : M M a monotone modal operator. Then the image (f(m),,,,,f0, 1) of M by f is also a CRL, where fx fy is defined by fx fy = f(fx fy). P r o o f. For all fx,fy f(m), it is easy to show that sup {fx,fy} = fx fy = f(m) f(fx fy). Moreover, since fx fy = f(fx fy) f(m), f(m) isclosed under the operations,, and. Itisobviousthat(f(M),,,f0,f1) is a bounded lattice and (f(m),, 1) is a commutative monoid. Moreover, we also have fx fy fz if and only if fx fy fz = f(fy fz). Thus we can conclude that (f(m),,,,,f0, 1) is the bounded commutative residuated lattice. In the statement above, f0 = 0 does not hold in general. For example, let M = {0,a,1} with an order 0 a 1. Concerning to this order, the set M is a Heyting algebra, thus, of course it is a CRL. If we define a mapping f : M M by f0 =fa = a and f1 = 1, then it is easy to show that f is a monotone modal operator on M and f0 0. So it is interesting to consider the case of f0 =0. Corollary 2. Let M be a CRL and f a monotone modal operator. Then we have f0 =0if and only if f(x )=x for all x M. 6
MODAL OPERATORS ON COMMUTATIVE RESIDUATED LATTICES P r o o f. Suppose that f0 = 0. It follows from Proposition 3.(2) that fx (x f0) f0 =(x 0) 0=x and hence that fx x for all x M. This means that f(x ) (x ) = x. On the other hand, it is clear that x f(x ). Thus we have f(x )=x for all x M. Conversely, if we assume that f(x )=x then f0 =f(1 )=1 =0. From the above, in case of CRL with meeting the condition x = x, we have the following result. Corollary 3. Let M be a CRL with x = x and f a mapping on M with f0 =0. Then f is a monotone modal operator if and only if it is an identity map. We get another characterization of f0 =0. Corollary 4. Let M be a CRL with x = x and f a monotone modal operator on M. Thenf0 =0if and only if x y = fx (fy). P r o o f. From the corollary above, we have x y =(x y) = f((x y) )= f(x y ) fx f(y )=fx (fy). On the other hand, since x fx and y fy, fx (fy) x y. This implies that x y = f(x y ) fx (fy) x y and thus x y = f(x y )=fx (fy). Conversely, suppose that x y = fx (fy) and take y = x.thenwe have 1 = x x = fx (f(x )). This means that fx (f(x )). Thus if we put x =0thenf0 (f(0 )) =(f1) =1 =0. A non-empty subset F of a bounded commutative integral residuated lattice M =(M,,,,, 0, 1) is called a filter if it satisfies the conditions: (f1) If x, y F then x y F ; (f2) If x F and x y then y F. Arelationθ F on M is defined from a filter F as follows. For all x, y M, (x, y) θ F x y, y x F. 3. Normal CRL A bounded commutative integral residuated lattice M is called normal if (x y) = x y. We note that, for any CRL M, itisnormalifand only if (x y) = x y and (x y) = x y. Indeed, if M is a normal CRL then (x y) =(x y ) = x y = x y and x y =(x y ) =(x y) =(x y). Conversely, if two equations hold then (x y) =(x y ) = x y. This means that M is normal. 7
M. KONDO In case of normal CRL, we show that the double negation operator is a strong modal operator. Proposition 6. Let M be a CRL. M is normal if and only if is a strong modal operator. P r o o f. Suppose that M is a normal CRL. It is easy to verify that (1) x x ; (2) If x y then x y ; (3) x = x ; (4) (x y) = x. This means that the double negation operator is the monotone modal operator. Moreover, since x 0 = x 0=x =(x ), it follows that is the strong modal operator. Conversely, if the double negation is a strong modal operator, then it is clear that (x y) = x y.namely,m is normal. Corollary 5. For any CRL M, If M is normal then M is a bounded commutative integral residuated lattice. Let I(M) be the set of all idempotent elements of M with respect to, that is, I(M) ={a : a a = a}. It is familiar that if a I(M) thena x = a x for all x M in case of CRL with divisibility ([9]). In case of normal CRL, we can characterize the set I(M) ofidempotent. Proposition 7. Let M be a normal CRL. Then a I(M) if and only if a a = a. Proof. If a I(M) thenwehavea a =(a a ) =(a ) = a. Conversely, if a = a a then a = a =(a a) = a a. Thus a I(M). Concerning to a relation between the operator ϕ a and the set I(M), we have the next result. While it is proved in [9] that ϕ a is a strong modal operator on a normal Rl-monoid if and only if a,a I(M), ϕ a is not a monotone modal operator on any normal CRL in general, because CRL does not have divisibility. The following means that if ϕ a is a strong modal operator on a normal CRL M provided that it is a monotone modal operator and a I(M). This result can be used to solve the open problem in [9] as Theorem 4, which is one of main results in the paper. Corollary 6. Let M be a normal CRL. Then a (x y) =(a x) (a y)= a (x (a y)) if and only if a I(M). That is, ϕ a (x y) =ϕ a (x ϕ a (y)) if and only if a I(M). 8
MODAL OPERATORS ON COMMUTATIVE RESIDUATED LATTICES P r o o f. Suppose that a (x y) =(a x) (a y) holds for all x, y M. If we take x = y =0,thenwehavea (0 0) = (a 0) (a 0) and thus a = a a by a 0 =a and 0 0 = 0. It follows from a a = a a that a = a a and hence a I(M) bytheabove. Conversely, if a I(M), since a a = a,thenwehavea (x y) = a (x y) =(a a) (x y) =(a x) (a y). 4. Divisibility A condition x y = x (x y) for all x, y is called divisibility. It says that an operator can be represented by. The condition yields interesting results to the structure theory of bounded commutative integral residuated lattices. The following result is proved as [9, Lemma 1] (cf. [1]). Proposition 8. Let M be a CRL with divisibility, that is, Rl-monoid ([9]). Then we have (1) a I(M) if and only if a x = a x (2) If a I(M) then a x =(a x) (3) (x y) = x y (4) (x y) = x y. In [8], one proved that, for any DRl 1(i) -semigroup M (i.e., bounded commutative integral residuated lattice with x (y z) =(x y) (x z) and divisibility), if M satisfies the condition x = x then it is an MV-algebra. It follows from a similar proof that we have the same result for CRL with divisibility, that is, for any CRL M with divisibility, if it satisfies the condition x = x then it is an MV-algebra. From the above, we can verify the following result without difficulty. Theorem 3. Let M be a normal CRL with divisibility, that is, normal Rl-monoid ([9]). If we define an operator g on M by gx = x for all x M, thenit is a strong modal operator on M and it is also a homomorphism from M onto g(m) =(g(m),,,,, 0, 1) which is an MV-algebra, that is, M/ker g = g(m), whereker g = { (x, y) : gx = gy }. Proof. If we take gx = x, then it is easy to show that g is a strong modal operator by Proposition 6. It follows from Theorem 2 that g(m) = (g(m),,,,, 0, 1) is a CRL with divisibility, where gx gy = g(gx gy). Moreover, since (gx) = x = x = gx in g(m), we have a = a for every a g(m). Thus we can conclude that g(m) isthemv-algebra by the argument above. Since gx gy = g(gx gy) =(x y ) =(x y ) = 9
M. KONDO (x y ) =(x y) = g(x y), the mapping g is a homomorphism by Proposition 8 from M onto g(m). The kernel ker g = {(x, y) : gx = gy} of the homomorphism g is a congruence on M by Proposition 2. This means that M/ker g = g(m). 5. Other operators In this section we study properties of three operators ϕ a (x) = a x, ψ a (x) =a x and χ a (x) =(x a) a for a M defined on a CRL M. Some properties of these operators are studied on stronger algebras than ours ([3, 6, 9]). At first we consider a case of ϕ a.in[9],itisprovedthatϕ a isastrongmodal operator if and only if a,a I(M) in any normal CRL bounded commutative residuated l-monoid, that is, CRL with divisibility ([9, Theorem 11]). The following question lefts open whether ϕ a is a strong modal operator provided that it is a modal operator on a normal bounded commutative residuated l-monoid. Since our monotone modal operator is the same as the modal operator on any CRL with divisibility, the question above can be represented by the following general statement: Is ϕ a strong modal operator provided that it is a monotone modal operator on any normal CRL with divisibility? We can give a positive answer to the question. Theorem 4. Let M be a normal CRL with divisibility. Then ϕ a is a strong modal operator provided that it is a monotone modal operator on M. Proof. Let M be a normal CRL with divisibility and ϕ a a monotone modal operator on it. Since x ϕ a 0=x a 0 =x a and ϕ a (x )=a x = a x, we have x ϕ a 0=ϕ a (x ). It follows from Proposition 5 above that ϕ a is the strong modal operator. Next, we consider the case of ψ a. The only if part of following results and the next one are proved in [9], where divisibility is supposed. We can show the stronger results without divisibility. Proposition 9. Let M be a CRL. For any element a M, we have a I(M) if and only if ψ a (x) ψ a (y) ψ a (x y). P r o o f. Suppose that a I(M). Since a (a x) (a y) =a a (a x) (a y) =(a (a x)) (a (a y)) x y, 10
MODAL OPERATORS ON COMMUTATIVE RESIDUATED LATTICES we have (a x) (a x) a x y, thatis,ψ a (x) ψ a (y) ψ a (x y). Conversely, we assume that ψ a (x) ψ a (y) ψ a (x y) for all x, y M. If we take x = y = a then we have ψ a (a) ψ a (a) ψ a (a a). and hence (a a) (a a) a a a. This means that a = a a and a I(M). Proposition 10. For any CRL M, a I(M) if and only if x ψ a (y) = ψ a (x) ψ a (y) for all x, y M. P r o o f. Suppose that a I(M). Since (a x y) a (a x) =(a x y) a a (a x) (a x y) a x y, we have a x y a (a x) y and thus x ψ a (y) =x (a y) = a x y a (a x) y =(a x) (a y) =ψ a (x) ψ a (y). On the other hand, since x a x, it is easy to show that a x a (a x) and hence that a (a x) y a x y. This means that ψ a (x) ψ a y x ψ a (y). Thus we have x ψ a (y) =ψ a (x) ψ a (y). Conversely, if x ψ a (y) =ψ a (x) ψ a (y) holds for all x, y M, thenby putting x = a and y = a a we have a (a a a) =(a a) (a a a). Since a (a a a) =a a a a =1,wehave1=1 (a a a) = a a a. This implies that a a a and hence that a = a a, thatis, a I(M). Corollary 7. Let M be a CRL and a I(M). Thenψ a is a monotone modal operator if and only if ψ a (x) ψ a (y) ψ a (x y). Proposition 11. Let M be a CRL. If a I(M), thenψ a (x y) =ψ a (x ψ a y). Thus, if ψ a is a monotone modal operator and a I(M) then it is a strong modal operator. Proof. Since y a y = ψ a y,wegetx y x ψ a y and thus ψ a (x y) ψ a (x ψ a y). On the other hand, we note from ψ a (u )=a u =(a u) and a u (a u) u u =0thata u (a u).sinceψ a (x ψ a y) = ψ a ((x (ψ a y) ) )=(a x (ψ a y) ),wehavea x (ψ a y) = a x (a y) a x a y = a x y by a I(M). 11
M. KONDO This implies that ψ a (x ψ a y)=(a x (ψ a y) ) (a x y ) = ψ a ((x y ) ) = ψ a (x y). Thus we can conclude that ψ a (x ψ a y)=ψ a (x y) ifa I(M). Next result follows from the above immediately. Corollary 8. Let M be a CRL and a I(M). Then ψ a is a monotone modal operator if and only if ψ a (x y) ψ a x ψ a y. Next we consider the case of χ a (x) =(x a) a. fundamental result concerning to the operator. At first we show a Proposition 12. For any a, x M, we have ((x a) a) a = x a. P r o o f. It follows from x (x a) a that x (x a) a and thus ((x a) a) a x a. On the other hand, since (x a) (((x a) a) a) =((x a) a) ((x a) a) =1,wehavex a ((x a) a) a. This implies ((x a) a) a = x a. Lemma 1. Let M be a CRL and a, x, y M. We have x χ a y = χ a x χ a y. P r o o f. It is enough to prove (a) x χ a y χ a x χ a y, (b) χ a x χ a y x χ a y. For the case of (a), since (x ((y a) a)) ((x a) a) (y a) =((y a) (x a)) ((x a) a) (y a) (x a) ((x a) a) a, we get (x ((y a) a)) ((x a) a) (y a) a, that is, x χ a y χ a x χ a y. 12
MODAL OPERATORS ON COMMUTATIVE RESIDUATED LATTICES For the other case (b), since (((x a) a) (((y a) a)) x (y a) =((y a) (((x a) a) a)) x (y a) =((y a) (x a)) x (y a) (x a) x a, we have, (((x a) a) ((y a) a)) x (y a) a and thus ((x a) a) ((y a) a) x ((y a) a). This means that χ a x χ a y x χ a y. It follows from the two cases above that we can get x χ a y = χ a x χ a y. We note the result above holds for any CRL, so we have the following. Corollary 9. Let M be a CRL and χ a : M M a mapping defined by χ a x = (x a) a for all x M. Then χ a is a monotone modal operator if and only if χ a (x y) χ a x χ a y. Remark 1. The result above is proved in [9] as Corollary 19. But the proof requires the assumption of divisibility and a B(M), that is, a has a lattice complement in M. Our result does not require neither condition. REFERENCES [1] DVUREČENSKIJ, A. RACHŮNEK, J.: Bounded commutative residuated l-monoids with general comparability and states, Soft Comput. 10 (2006), 212 218. [2] GALOTOS, N. JIPSEN, P. KOWALSKI, T. ONO, H.: Residuated Lattices: An Algebraic Glimpse at Substructural Logics. Stud. Logic Found. Math. 151, Elsevier, Amsterdam, 2007. [3] HARLENDEROVÁ, M. RACHŮNEK, J.: Modal operators on MV-algebras, Math.Bohem. 131 (2006), 39 48. [4] HART, J. B. RAFTER, L. TSINAKIS, C.: The structure of commutative residuated lattices, Internat. J. Algebra Comput. 12 (2002), 509 524. [5] JIPSEN, P. TSINAKIS, C.: A Survey of residuated lattices. In: Proceedings of the Conference on Lattice-ordered Groups and f-rings (J. Martinez, ed.). Dev. Math. 7, Kluwer Academic Publ., Dordrecht, 2002, pp. 19 56. [6] MACNAB, D. S.: Modal operators on Heyting algebras, Algebra Universalis 12 (1981), 5 29. [7] RACHŮNEK, J.: DRl-semigroups and MV-algebras, Czechoslovak Math. J. 48 (1998), 365 372. 13
M. KONDO [8] RACHŮNEK, J.: MV-algebras are categorically equivalent to a class of DRl 1(i)-semigroups, Math. Bohem. 123 (1998), 437 441. [9] RACHŮNEK, J. ŠALOUNOVÁ, D.: Modal operators on bounded commutative residuated Rl-monoids, Math. Slovaca, 57 (2007), 321 332. [10] RACHŮNEK, J. SLEZÁK, V.: Negation in bounded commutative DRl-monoids, Czechoslovak Math. J. 56 (2006), 755 763. Received 7. 12. 2008 Accepted 14. 5. 2009 School of Information Environment Tokyo Denki University JAPAN E-mail: kondo@sie.dendai.ac.jp 14