LAVINIA CORINA CIUNGU
The notion of a state is an analogue of a probability measure and was first introduced by Kˆopka and Chovanec for MV-algebras and by Rie˘can for BL- algebras. The states have also been studied for different types of non-commutative fuzzy structures such as pseudo-MV algebras, pseudo-BL algebras, bounded R`- monoids, residuated lattices and pseudo-BCK semilattices. In this paper we in- vestigate the states on pseudo-BCK algebras and show that Georgescu’s original problem in [10] for pseudo-BL algebras has a negative solution for good pseudo- BCK algebras. We prove that every Bosbach state on a pseudo-BCK algebra is a Rie˘can state and that every Rie˘can state on a good pseudo-BCK algebra with pseudo-double negation is a Bosbach state. In contrast to the case of pseudo-BL algebras, we show that there exist linearly ordered pseudo-BCK algebras having no Bosbach states.
AMS 2000 Subject Classification: 06B05, 03G25, 28E15.
Key words: Pseudo-BCK algebra, Bosbach state, Rie˘can state, state-morphism, deductive system.
1. INTRODUCTION
The notion of a state is an analogue of a probability measure and has a very important role in the theory of quantum structures (see [8]). The state on MV algebras was first introduced by Kˆopka and Chovanec [16] while the state on BL algebras was introduced by Rie˘can [19]. In the case of non-commutative fuzzy structures, the states were introduced by Dvure˘censkij [5] for pseudo- MV algebras, by Georgescu [10] for pseudo-BL algebras and by Dvure˘censkij and Rach¨unek [9] for bounded non-commutative R`-monoids. In the case of a pseudo-MV algebraM, Dvure˘censkij [4] proved that there is an`-group (G, u) with strong unit u such that M is isomorphic to Γ(G, u) = {g∈G|0≤g≤ u}. This allowed him to define a partial addition +, that is, x+y is defined if x≤y− =u−y and the state is a mapping s:M → [0,1] which preserves the partial addition + and s(1) = 1. We recall that the elementsa and bare orthogonal ifa+bis defined inM. The other non-commutative structures do not have such a group representation and it was more difficult to define the notion of states for these structures. We recall that a state on MV-algebras
MATH. REPORTS10(60),1 (2008), 17–36
always exists in contrast to pseudo-MV algebras (see [5]); on the other hand, in [6] was solved the existence of states for linearly ordered pseudo-BL algebras (see also [7]). In the case of pseudo-BL algebras Georgescu [10] defined the Bosbach and Rie˘can states and, for a good pseudo-BL algebra, he proved that any Bosbach state is also a Rie˘can state. He asked to find an example of Rie˘can state on a good pseudo-BL algebra which is not a Bosbach state.
The notions of Bosbach and Rie˘can states on good pseudo-BL algebras were generalized by Dvure˘censkij and Rach¨unek [9] for bounded R`-monoids.
They also proved that for good bounded R`-monoids the two notions of states coincide. In [1] there were investigated the Bosbach and Rie˘can states on residuated lattices and was proved that there exist Rie˘can states on good residuated lattices which are not Bosbach states. In [18] the above results were also proved in the case of pseudo-BCK semilattices. Inspired by the above mentioned results, in this paper we extend the notion of states to pseudo-BCK algebras. One of the main results consists of proving that any Bosbach state on a good pseudo-BCK algebra is a Rie˘can state, but the converse turns out not to be true. We also prove that every Rie˘can state on a good pseudo- BCK algebra with pseudo-double negation is a Bosbach state. In contrast to the case of pseudo-BL algebras, we show that there exist linearly ordered pseudo-BCK algebras having no Bosbach states. Additionally, we prove some new properties of pseudo-BCK algebras and give examples of pseudo-BCK algebras and good pseudo-BCK algebras.
2. PSEUDO-BCK ALGEBRAS AND THEIR BASIC PROPERTIES
The pseudo-BCK algebras were introduced by Georgescu and Iorgulescu [11] as generalization of BCK algebras in order to give a structure correspond- ing to pseudo-MV algebras, as the bounded commutative BCK algebras cor- responde to MV algebras. Properties of pseudo-BCK algebras and their con- nection with others fuzzy structures were established by Iorgulescu [12], [13], [14], [15],
Definition 2.1 ([12]). A pseudo-BCK algebra (more precisely, reversed left-pseudo-BCK algebra) is a structure A = (A,≤,→, ,1), where ≤ is a binary relation onA,→and are binary operations onAand 1 is an element of A satisfying, for allx, y, z∈A, the axioms below:
(A1) x→y≤(y→z) (x→z),x y≤(y z)→(x z);
(A2) x≤(x→y) y,x≤(x y)→y;
(A3) x≤x;
(A4) x≤1;
(A5) if x≤y andy≤x, thenx=y;
(A6) x≤y iffx→y= 1 iffx y= 1.
Remarks 2.2 ([12]). (1) A pseudo-BCK algebra A = (A,≤,→, ,1) is commutative if→= .
(2) Any commutative pseudo-BCK algebra is a BCK algebra.
Example 2.3. Let A = {o1, a1, b1, c1, o2, a2, b2, c2,1} with o1 < a1, b1 <
c1<1 and a1, b1 incomparable,o2 < a2, b2 < c2 <1 anda2, b2 incomparable.
Assume that any element of the set {o1, a1, b1, c1} is incomparable with any element of the set {o2, a2, b2, c2}. Consider the operations →, given by the tables
→ o1 a1 b1 c1 o2 a2 b2 c2 1 o1 1 1 1 1 o2 a2 b2 c2 1 a1 o1 1 b1 1 o2 a2 b2 c2 1 b1 a1 a1 1 1 o2 a2 b2 c2 1 c1 o1 a1 b1 1 o2 a2 b2 c2 1 o2 o1 a1 b1 c1 1 1 1 1 1 a2 o1 a1 b1 c1 o2 1 b2 1 1 b2 o1 a1 b1 c1 c2 c2 1 1 1 c2 o1 a1 b1 c1 o2 c2 b2 1 1 1 o1 a1 b1 c1 o2 a2 b2 c2 1
o1 a1 b1 c1 o2 a2 b2 c2 1 o1 1 1 1 1 o2 a2 b2 c2 1 a1 b1 1 b1 1 o2 a2 b2 c2 1 b1 o1 a1 1 1 o2 a2 b2 c2 1 c1 o1 a1 b1 1 o2 a2 b2 c2 1 o2 o1 a1 b1 c1 1 1 1 1 1 a2 o1 a1 b1 c1 b2 1 b2 1 1 b2 o1 a1 b1 c1 b2 c2 1 1 1 c2 o1 a1 b1 c1 b2 c2 b2 1 1 1 o1 a1 b1 c1 o2 a2 b2 c2 1
.
Then A= (A,≤,→, ,1) is a proper pseudo-BCK algebra.
Remark 2.4 ([17]). Let (Ai,≤,→i, i,1i)i∈I be a collection of pseudo- BCK algebras such that
(i) 1i= 1 for all i∈I,
(ii)Ai∩Aj ={1}for all i, j∈I,i6=j.
Denote A=S
i∈IAi and define x→y=
x→iy ifx, y∈Ai for somei∈I y otherwise,
x y=
x i y ifx, y∈Aifor somei∈I y otherwise.
Then (A,≤,→, ,1) is a pseudo-BCK algebra called theunion of the pseudo- BCK algebras (Ai,≤,→i, i,1i)i∈I.
Proposition 2.5 ([14], [15]). In any pseudo-BCK algebra the following properties hold:
(c1) x≤y implies y →z≤x→z andy z≤x z;
(c2) x≤y, y≤z implies x≤z;
(c3) x→(y z) =y (x→z) and x (y→z) =y→(x z);
(c4) z≤y→x iff y≤z x;
(c5) z→x≤(y→z)→(y→x), z x≤(y z) (y x);
(c6) x≤y→x, x≤y x;
(c7) 1→x=x= 1 x;
(c8) x≤y implies z→x≤z→y andz x≤z y;
(c9) [(y→x) x]→x=y →x, [(y x)→x] x=y x.
Definition 2.6 ([12]). If there is an element 0 of a pseudo-BCK algebra A = (A,≤,→, ,1) such that 0 ≤x (i.e. 0→ x= 0 x = 1) for allx∈A, then it is called the zero of A. A pseudo-BCK algebra with zero is called a bounded pseudo-BCK algebra and is denoted byA= (A,≤,→, ,0,1).
Example 2.7. Let A = {0, a, b, c,1} with 0 < a, b < c < 1 and a, b incomparable. Consider the operations →, given by the tables
→ 0 a b c 1
0 1 1 1 1 1
a 0 1 b 1 1 b a a 1 1 1 c 0 a b 1 1
1 0 a b c 1
0 a b c 1
0 1 1 1 1 1
a b 1 b 1 1 b 0 a 1 1 1 c 0 a b 1 1
1 0 a b c 1
.
Then A= (A,≤,→, ,0,1) is a bounded pseudo-BCK algebra.
Definition 2.8 ([12]). A pseudo-BCK algebra with (pP) condition (i.e.
with pseudo-product condition) or a pseudo-BCK(pP) algebra for short, is a pseudo-BCK algebra A = (A,≤,→, ,1) satisfying the (pP) condition: for all x, y∈A there exist xy= min{z|x≤y→z}= min{z|y≤x z}.
Definition 2.9 ([12]). (1) Let A = (A,≤,→, ,1) be a pseudo-BCK algebra. If the poset (A,≤) is a lattice, then we say thatA is apseudo-BCK lattice.
(2) LetA= (A,≤,→, ,1) be a pseudo-BCK(pP) algebra. If the poset (A,≤) is a lattice, then we say that Ais apseudo-BCK(pP) lattice.
A pseudo-BCK(pP) lattice A = (A,≤,→, ,1) will be denoted A = (A,∨,∧,→, ,1).
Remark 2.10 ([14]). Pseudo-BCK(pP) algebras are caregorically isomor- phic with left-porims (partially ordered, residuated, integral left-monoids).
Remark 2.11 ([12]). (Bounded) pseudo-BCK(pP) lattices are categori- cally isomorphic with (bounded) residuated lattices.
Examples 2.12. (1) If A = (A,≤,→, ,0,1) is the bounded pseudo- BCK lattice from Example 2.7, then min{z|b≤a→z}= min{a, b, c,1} and min{z|a≤b z}= min{a, b, c,1} do not hold. Thus, ba does not exist, soA is not a pseudo-BCK(pP) algebra. Moreover, since (A,≤) is a lattice,A is a pseudo-BCK lattice.
(2) IfA= (A,≤,→, ,0,1) is a reduct of a residuated lattice, then it is obvious that A is a bounded pseudo-BCK(pP) algebra.
Remark 2.13 ([12]). Any bounded linearly ordered pseudo-BCK algebra is with (pP) condition.
Proposition2.14 ([15]). In any pseudo-BCK algebra(pP) the following properties hold:
(c10) xy ≤x, y;
(c11) (x→y)x≤x, y,x(x y)≤x, y;
(c12) y≤x→(yx), y≤x (xy);
(c13) x→y≤(xz)→(yz), x y ≤(zx) (zy);
(c14) x(y→z)≤y→(xz), (y z)x≤y (zx);
(c15) (y →z)(x→y)≤x→z, (x y)(y z)≤x z;
(c16) x→(y→z) = (xy)→z, x (y z) = (yx) z;
(c17) (xz)→(yz)≤x→(z→y),(zx) (zy)≤x (z y);
(c18) x→y≤(xz)→(yz)≤x→(z→y),x y≤(zx) (zy)≤x (z y);
(c19) x≤y implies xz≤yz, zx≤zy.
LetA= (A,≤,→, ,0,1) be a bounded pseudo-BCK algebra. We define two negations − and ∼ ([15]): for all x∈A,x−=x→0,x∼=x 0.
Proposition 2.15 ([15]). In a bounded pseudo-BCK algebra the follow- ing properties hold:
(c20) 1−= 0 = 1∼, 0−= 1 = 0∼; (c21) x≤(x−)∼,x≤(x∼)−;
(c22) x→y≤y− x−, x y≤y∼→x∼; (c23) x≤y implies y−≤x− andy∼≤x∼; (c24) x→y∼ =y x− and x y−=y →x∼; (c25) ((x−)∼)−=x−,((x∼)−)∼ =x∼.
Proposition2.16. In a bounded pseudo-BCK algebra the following pro- perties hold:
(c26) x →y−∼=y− x− =x−∼→ y−∼ and x y∼− =y∼ →x∼= x∼− y∼−;
(c27) x → y∼ =y∼− x− = x−∼ → y∼ and x y− =y−∼ → x∼ = x∼− y−;
(c28) (x→y∼−)∼−=x→y∼− and (x y−∼)−∼=x y−∼.
Proof. (c26): By (c24) we have y x− = x → y∼. Replacing y by y− we get y− x− =x→y−∼. Replacing x by x−∼ in the last equality we get y− x−∼− =x−∼ → y−∼. Hence, (c25),we have y− x− =x−∼→ y−∼. Thus, x→ y−∼ =y− x−=x−∼ →y−∼ and, similarly, x y∼− =y∼ → x∼=x∼− y∼−.
(c27) The assertions follow replacing in (c26) y by y∼ and y by y−, and using (c25).
(c28) By (c3) and (c27) we have
1 = (x→y∼−) (x→y∼−) =x→((x→y∼−) y∼−) =
=x→((x→y∼−)∼− y∼−) = (x→y∼−)∼− (x→y∼−).
Hence (x→y∼−)∼− ≤ x → y∼−. On the other hand, by (c21) we have x → y∼− ≤ (x→y∼−)∼−, so (x→y∼−)∼− = x → y∼−. Similarly, (x y−∼)−∼=x y−∼.
Proposition2.17. In a bounded pseudo-BCK(pP) algebra the following properties hold:
(c29) (xn−1 → xn)(xn−2 → xn−1). . .(x1 → x2) ≤ x1 → xn and (x1 x2)(x2 x3). . .(xn−1 xn)≤x1 xn;
(c30) x0 = 0x= 0;
(c31) x1 = 1x=x;
(c32) x−x= 0 and xx∼= 0;
(c33) x≤y− iff xy= 0 and x≤y∼ iff yx= 0;
(c34) x→y− = (xy)− and x y∼= (yx)∼; (c35) x≤y− iff y≤x∼.
Proof. (c29): It follows from (c15) by induction.
(c30): By (c10), fory= 0 and x= 0.
(c31): By (c12), fory= 1 we get 1≤x→(1x), sox→(1x) = 1. It follows by (A6) thatx≤1x. On the other hand, by (c10), we have 1x≤x.
Thus, 1x = x. Similarly, 1 ≤ x (x1), so x (x1) = 1, that is, x≤x1. Therefore, x1 = 1.
(c32): By (c11), fory= 0.
(c33): Using x ≤ y−. Applying (c19) we get xy ≤ y−y = 0, so xy= 0. Conversely, if xy= 0, by (c12) we have x≤y→ (xy) =y→ 0 =y−. Similarly,x≤y∼ iff yx= 0.
(c34): By (c19), taking z= 0.
(c35): It follows from (c33).
Definition 2.18 ([12]). A bounded pseudo-BCK algebra A = (A,≤,→, ,0,1) is with (pDN) (pseudo-Double Negation) condition if it satisfies for all x∈A the condition (pDN) (x−)∼= (x∼)− =x.
Proposition 2.19 ([12]). Let A be a pseudo-BCK algebra with (pDN) condition. Then for all x, y∈A the following properties hold:
(c36) x≤y iff y−≤x− iff y∼≤x∼;
(c37) x→y=y− x−, x y=y∼→x∼; (c38) x∼ →y =y− x;
(c39) (x→y−)∼= (y x∼)−.
Theorem 2.20 ([12]). A bounded pseudo-BCK algebra A = (A,≤,→, ,0,1)with (pDN) condition is with (pP) condition.
Definition 2.21. A bounded pseudo-BCK algebra A is called good if (x−)∼= (x∼)− for all x∈A.
Remark 2.22. It is easy to show that any bounded pseudo-BCK algebra can be embedded into a good one. Indeed, consider the bounded pseudo- BCK algebra A= (A,≤,→, ,0,1) and an element 01 ∈/ A. Consider a new pseudo-BCK algebra A1 = (A1,≤,→1, 1,01,1), where A1 = A∪ {01} and the operations →1 and 1 are defined as
x→1y=
x→y ifx, y∈A, 1 ifx= 01, y∈A1, 01 ifx∈A, y = 01,
x 1y=
x y ifx, y∈A, 1 ifx= 01, y∈A1, 01 ifx∈A, y = 01.
One can easily check that Aas an subalgebra ofA1 andA1 is a good pseudo- BCK algebra.
Example 2.23. Consider the pseudo-BCK lattice A from Example 2.7.
Since (a−)∼= 1 and (a∼)−=a,Ais not good. Ais embedded into the good pseudo-BCK lattice A1 = (A1,≤,→, ,0,1), where A1 = {0, a, b, c, d,1} (in the construction given in Remark 2.22 we replaced c by d, b by c, a by b, 0 by a and 01 by 0, so 0 < a < b, c < d < 1 and b, c are incomparable). The operations → and are defined as it the tables
→ 0 a b c d 1 0 1 1 1 1 1 1 a 0 1 1 1 1 1 b 0 a 1 c 1 1 c 0 b b 1 1 1 d 0 a b c 1 1 1 0 a b c d 1
0 a b c d 1 0 1 1 1 1 1 1 a 0 1 1 1 1 1 b 0 c 1 c 1 1 c 0 a b 1 1 1 d 0 a b c 1 1 1 0 a b c d 1
.
One can easily check that A1 is a good pseudo-BCK algebra. Moreover, we can see that min{z| c≤b → z} = min{b, c, d,1} and min{z |b ≤c z}= min{b, c, d,1} do not hold. Thus,cbdoes not exist, so A1 is without (pP) condition. Since (A1,≤) is a lattice,A1 is a good pseudo-BCK lattice without (pP) condition.
Proposition 2.24. Let A = (A,≤,→, ,0,1) be a good pseudo-BCK algebra. Define a binary operation ⊕on A by x⊕y:= y∼ →x∼−. Then for all x, y∈A the following properties hold:
(1)x⊕y=x− y∼−; (2)x, y≤x⊕y;
(3)x⊕0 = 0⊕x=x∼−; (4)x⊕1 = 1⊕x= 1;
(5) (x⊕y)−∼=x⊕y=x−∼⊕y−∼; (6)⊕ is associative.
Proof. (1): It follows from the second identity in (c28), replacingxbyx−. (2): Since x ≤ x∼− ≤ y∼ → x∼−, we have x ≤ x ⊕y. Similarly, y ≤y∼−≤x− y∼−, soy≤x⊕y.
(3): x⊕0 = 0∼ → x∼− = 1 → x∼− =x∼−. Similarly, 0⊕x = x∼ → 0∼−=x∼→0 =x∼−.
(4): 1⊕x=x∼→1∼−=x∼→1 = 1. Similarly, x⊕1 = 1.
(5): (x⊕y)−∼= (y∼ →x∼−)−∼ =y∼→ x∼−=x⊕y (we used (c28)).
We also have x−∼⊕y−∼= (y−∼)∼→(x−∼)−∼=y∼→x−∼=x⊕y.
(6): By (c28) and (c3) we have
(x⊕y)⊕z= (x− y∼−)⊕z=z∼→(x− y∼−)∼−
=z∼→(x− y∼−) =x− (z∼→y∼−) =x− (y⊕z)
=x− (y⊕z)∼−=x⊕(y⊕z).
Remark 2.25. The definition of⊕is inspired from the case of pseudo-BL algebras (see [10]), where x⊕y := (y∼x∼)−. Since in a good residuated lattice we have
(y∼x∼)−= (y−x−)∼, (xy)−=x→y−and (xy)∼=y x∼, the definition of ⊕is valid for the residuated structures, too.
LetA be a pseudo-BCK algebra. For allx, y∈A, define (see [11], [15]) x∨y= (x→y) y, x∪y= (x y)→y.
Proposition2.26 ([11], [15]). In any bounded pseudo-BCK algebra, for all x∈A,
0∨x=x= 0∪x and x∨0 = (x−)∼, x∪0 = (x∼)−.
Proposition2.27. In any bounded pseudo-BCK algebraAthe following properties hold for all x, y∈A:
(1) 1∨x=x∨1 = 1 = 1∪x=x∪1;
(2)x≤y implies x∨y=y andx∪y=y;
(3)x∨x=x∪x=x.
Proof. (1): We have 1∨x= (1→ x) x) = 1 and x∨1 = (x→1) 1) = 1, so 1∨x=x∨1 = 1. Similarly, 1∪x=x∪1 = 1.
(2): x∨y= (x→y) y= 1 y=y. Similarly, x∪y=y.
(3): By definition of ∨and ∪.
Proposition 2.28. In any pseudo-BCK algebra, for all x, y∈A, x∨y→y=x→y and x∪y y=x y.
Proof. It is a consequence of property (c9).
Lemma 2.29. In any pseudo-BCK algebra A:
(1)x∨y (y∨x) is a upper bound of {x, y};
(2)x∪y (y∪x) is a upper bound of {x, y}.
Proof. (1): By (A2) we havex≤(x→y) y. Since, by (c6),y≤(x→ y) y, we conclude thatx, y≤x∨y. Similarly we getx, y≤y∨x.
(2): Similarly to (1).
Definition 2.30 ([11], [15]). Let Abe a pseudo-BCK algebra.
(i) Ifx∨y =y∨xfor all x, y∈A, thenAis called ∨-commutative.
(i0) Ifx∪y=y∪x for allx, y∈A, thenA is called∪-commutative.
Lemma 2.31 ([11], [15]). If A is a pseudo-BCK algebra, then
(i)Ais∨-commutative iff it is a semilattice with respect to∨(under≤);
(i0)Ais∪-commutative iff it is a semilattice with respect to∪(under≤).
Definition 2.32 ([11], [15]). A pseudo-BCK algebra is calledsup-commu- tative if it is both ∨-commutative and∪-commutative.
Theorem 2.33 ([11], [15]). A pseudo-BCK algebra is sup-commutative iff it is a semilattice with respect to both ∨ and∪.
Theorem 2.34 ([12]). Let A = (A,≤,→, ,1) be a sup-commutative pseudo-BCK(pP) algebra. Then
(1) (A,≤) is a lattice, where
x∨y= (y→x) x= (y x)→x, x∧y= ([x→(xy)]∨[y→(xy)]) (xy)
= ([x (xy)]∨[y (xy)])→(xy) for any x, y∈A.
(2) (x→y)∨(y→x) = 1 = (x y)∨(y x) for any x, y∈A.
Corollary 2.35 ([12]). A bounded sup-commutative pseudo-BCK alge- bra A is with (pDN) condition(and hence it is with(pP) condition).
In a bounded sup-commutative pseudo-BCK algebra A for all x, y ∈ A define (see [11], [15]) x∧y= (x−∨y−)∼, x∩y= (x−∪y−)∼.
Theorem2.36 ([12]). A bounded sup-commutative pseudo-BCK algebra is a lattice with respect to both ∨, ∧ and ∪, ∩ (under ≤) and for all x, y we have
x∨y=x∪y, x∧y=x∩y.
Theorem 2.37 ([12]). (1) Let A= (A,≤,→, ,0,1) be a bounded sup- commutative pseudo-BCK algebra. Define Φ0(A) = (A,,⊕,−,∼,0,1) by
x−=x→0, x∼=x 0, xy= (x→y−)∼(= (y x∼)−), y⊕x= (x−y−)∼(= (x∼y∼)−).
Then Φ0(A) is a pseudo-MV algebra.
(2)Conversely, letA= (A,,⊕,−,∼,0,1)be a pseudo-MV algebra. De- fine Ψ0(A) = (A,≤,→, ,0,1) by
x≤y iff x⊕y−= 1 iff y∼⊕x= 1, x→y=y⊕x−= (xy∼)−, x y=x∼⊕y= (y−x)∼.
Then Ψ0(A) is a bounded sup-commutative pseudo-BCK algebra.
(3)The maps Φ0 and Ψ0 are mutually inverse.
Corollary 2.38 ([12]). Bounded sup-commutative pseudo-BCK alge- bras are categorically isomorphic with pseudo-MV algebras.
Definition 2.39. Let A be a pseudo-BCK algebra. A subset D⊆A is a deductive system of Aif it satisfies the conditions
(DS1) 1∈D;
(DS2) for allx, y∈A, ifx, x→y∈D, theny∈D.
Condition (DS2) is equivalent to the condition (DS02) for allx, y ∈A, if x, x y∈D, theny∈D.
A deductive system D of a pseudo-BCK algebra A is called proper if D6=A.
Definition 2.40. A deductive sistem D of a pseudo-BCK algebra A is called normal if it satisfies the condition
(DS3) for allx, y∈A,x→y∈D iffx y∈D.
Definition 2.41. A deductive system is calledmaximal if it is proper and not strictly contained in any other deductive system.
Examples 2.42. Consider the pseudo-BCK latticeA from Example 2.7.
(1) The deductive systems ofA areD1 ={a, c,1},D2 ={b, c,1},D3 = {c,1} and D4 ={1}.
(2)D1 and D2 are maximal deductive systems.
(3)D3 is a normal deductive system.
(4)D1 andD2 are not normal deductive systems (c→a=a∈D1, while c a=b /∈D1 anda 0 =b∈D2, while a→0 = 0∈/D2).
Example 2.43. In the pseudo-BCK lattice A from Example 2.23, D = {a, b, c, d,1} is a maximal normal deductive system.
3. STATES ON PSEUDO-BCK ALGEBRAS
Similarly by to [10], we introduce the notion of a Bosbach state on a bounded pseudo-BCK algebra A.
Definition 3.1. ABosbach stateon a bounded pseudo-BCK algebraAis a function s:A→[0,1] such that, for anyx, y∈A,
(B1) s(x) +s(x→y) =s(y) +s(y→x);
(B2) s(x) +s(x y) =s(y) +s(y x);
(B3) s(0) = 0 ands(1) = 1.
Example 3.2. Consider the bounded pseudo-BCK latticeA from Exam- ple 2.23. The function s:A→ [0,1] defined by s(0) = 0, s(a) = 1, s(b) = 1, s(c) = 1,s(d) = 1, s(1) = 1, is the unique Bosbach state onA.
However, there are bounded pseudo-BCK algebras, that have no Bosbach state on them.
Example 3.3. Consider the bounded pseudo-BCK latticeA from Exam- ple 2.7. Let us prove thatAhas no Bosbach states on it. Indeed, assume that A admits a Bosbach state ssuch that s(0) = 0,s(a) =α,s(b) =β,s(c) =γ, s(1) = 1. Froms(x) +s(x→y) =s(y) +s(y→x), takingx=a,y= 0,x=b, y = 0 and then x=c, y= 0 we get α= 1, β = 0, γ = 1. On the other hand, takingx=b,y= 0 ins(x) +s(x y) =s(y) +s(y x), we getβ+ 0 = 0 + 1, so 0 = 1, which is a contradiction. Hence A does not admit a Bosbach state.
Proposition3.4. LetAbe a bounded pseudo-BCK algebra andsa Bos- bach state on it. Then for all x, y∈A the following properties hold:
(1)y ≤ximpliess(y)≤s(x)ands(x→y) =s(x y) = 1−s(x)+s(y);
(2)s(x→y) = 1−s(x∨y) +s(y) ands(x y) = 1−s(x∪y) +s(y);
(3)s(x∨y) =s(y∨x) and s(x∪y) =s(y∪x);
(4)s(x−) =s(x∼) = 1−s(x);
(5)s(x−∼) =s(x∼−) =s(x−−) =s(x∼∼) =s(x);
(6) s(x− y−) = s(y−∼ → x−∼) = s(y → x−∼) and s(x∼ → y∼) = s(y∼− x∼−) =s(y x∼−);
(7) s(x → y∼) = s(y∼− x−) = s(x−∼ → y∼) and s(x y−) = s(y−∼→x∼) =s(x∼− y−).
Proof. (1): By (B1) and (A6) we have s(x) +s(x→y) =s(y) + 1, so s(x→y) = 1−s(x) +s(y).
Similarly, by (B2) and (A6) we have s(x y) = 1−s(x) +s(y). Thus, s(x→y) =s(x y) = 1−s(x) +s(y). Sinces(y)−s(x) =s(x→y)−1≤0, we have s(y)≤s(x).
(2): Sincey ≤x∨y, using (1), we gets(x∨y→y) = 1−s(x∨y) +s(y).
Hence, by Proposition 2.28, we haves(x→y) = 1−s(x∨y) +s(y). Similarly, from y ≤ x∪y and (1), we get s(x ∪y y) = 1−s(x∪y) +s(y). By Proposition 2.28, s(x y) = 1−s(x∪y) +s(y).
(3): From the identities s(x→y) = 1−s(x∨y) +s(y) and s(y→x) = 1−s(y∨x)+s(x) we gets(x∨y)−s(y∨x) =s(y →x)−s(x→y)−s(x)+s(y) = 0 (in the later equation we used (B1)). Thus,s(x∨y) =s(y∨x) and, similarly, s(x∪y) =s(y∪x).
(4): Takey= 0 in (B1) and (B2).
(5): It follows from (4).
(6): It follows by (c26).
(7): It follows by (c27).
Remark 3.5 ([18]). In the case of a pseudo-BCK semilattice, properties (2), (6) and (7) become
(20) s(x→y) =s(x y) = 1−s(x∨y) +s(y);
(60) s(x → y) = s(y− x−) = s(x−∼ → y−∼) = s(x → y−∼) = s(x−∼→y) and s(x y) =s(y∼ →x∼) =s(x∼− y∼−) =s(x y∼−) = s(x∼− y);
(70) s(x∼→y) =s(y− x∼−) =s(x∼ →y−∼) =s(y− x).
Proposition 3.6. Let A be a bounded pseudo-BCK algebra and a func- tion s : A → [0,1] such that s(0) = 0. Then the following properties are equivalent:
(a)s is a Bosbach state on A;
(b)y≤x impliess(x→y) =s(x y) = 1−s(x) +s(y)for allx, y∈A;
(c) s(x → y) = 1−s(x∨y) +s(y) and s(x y) = 1−s(x∪y) +s(y) for all x, y∈A.
Proof. (a) implies (b): This is proved in Proposition 3.4(1).
(b) implies (c): Assertion (c) is in fact assertion (2) of Proposition 3.4 which is proved using Proposition 3.4(1), so (b) implies (c).
(c) implies (a): From (c) and Proposition 3.4(3) we have s(x) +s(x → y) =s(x) + 1−s(x∨y) +s(y) = 1−s(y∨x) +s(x) +s(y) =s(y) +s(y→x).
Similarly, s(x) +s(x y) =s(x) + 1−s(x∪y) +s(y) = 1−s(y∪x) +s(x) + s(y) = s(y) +s(y x). Moreover, by (c) and Proposition 2.27(3) we have s(1) =s(x→x) = 1−s(x) +s(x) = 1. Thus,sis a Bosbach state onA.
Consider the bounded sup-commutative BCK(P) algebra (i.e. the MV algebra) A L = ([0,1],≤,→ L,0,1), where → L is the Lukasiewicz implication x→ L y= min{1−x+y,1}.
Definition 3.7. Let Abe a bounded pseudo-BCK algebra. A state-mor- phism on Ais a functionm:A→[0,1] such that
(SM1) m(0) = 0;
(SM2) m(x→y) =m(x y) =m(x)→ Lm(y).
Proposition 3.8. A state-morphism on a bounded pseudo-BCK algebra A is a Bosbach state on A.
Proof. It is obvious thatm(1) =m(x→ x) = m(x) → L m(x) = 1. We also have
m(x)+m(x→y) =m(x)+m(x)→ Lm(y) =m(x)+min{1−m(x)+m(y),1}=
= min{1 +m(y),1 +m(x)}=m(y) + min{1−m(y) +m(x),1}=
=m(y) +m(y)→ Lm(x) =m(y) +m(y→x).
Similarly, m(x) +m(x y) =m(y) +m(y x). Thus, sis a Bosbach state
on A.
Proposition 3.9. Let A be a bounded pseudo-BCK algebra. A Bosbach state m onA is a state-morphism if and only if
m(x∨y) =m(x∪y) = max{m(x), m(y)}
for all x, y∈A.
Proof. If m is a state-morphism on A, then by Proposition 3.8 m is a Bosbach state. According to Proposition 3.4(2) we have
m(x∨y) = 1 +m(y)−m(x→y) = 1 +m(y)−(m(x)→ L m(y)) =
= 1 +m(y)−min{1−m(x) +m(y),1}=
= 1 +m(y) + max{−1 +m(x)−m(y),−1}= max{m(x), m(y)}.
Similarly,
m(x∪y) = 1 +m(y)−m(x y) = 1 +m(y)−(m(x)→ L m(y)) =
= 1 +m(y)−min{1−m(x) +m(y),1}=
= 1 +m(y) + max{−1 +m(x)−m(y),−1}= max{m(x), m(y)}.
For the converse, assume thatm is a Bosbach state on Asuch that m(x∨y) =m(x∪y) = max{m(x), m(y)} for all x, y∈A.
Then, by Proposition 3.4(2), we have
m(x→y) = 1−m(x∨y) +m(y) = 1 +m(y)−max{(m(x), m(y)}=
= 1+m(y)+min{−m(x),−m(y)}= min{1−m(x)+m(y),1}=m(x)→ Lm(y).
Similarly,
m(x y) = 1−m(x∪y) +m(y) = 1 +m(y)−max{(m(x), m(y)}=
= 1+m(y)+min{−m(x),−m(y)}= min{1−m(x)+m(y),1) =m(x)→ L m(y).
Thus,m is a state-morphism on A.
Example 3.10. Consider A={0, a, b, c,1} with 0< a < b, c <1 and b, c incomparable. Assume the operations →, given by the tables
→ 0 a b c 1
0 1 1 1 1 1
a 0 1 1 1 1 b 0 a 1 c 1 c 0 b b 1 1
1 0 a b c 1
0 a b c 1
0 1 1 1 1 1
a 0 1 1 1 1 b 0 c 1 c 1 c 0 a b 1 1
1 0 a b c 1
.
Then A = (A,≤,→, ,0,1) is a bounded pseudo-BCK lattice. Since (A,≤) is a lattice, A is a pseudo-BCK lattice. Moreover, we can see that cb = min{z|c≤b→ z}= min{b, c,1} does not exist. Hence, A is a pseudo-BCK lattice without (pP) condition and the function m:A→[0,1] defined by
m(0) = 0, m(a) = 1, m(b) = 1, m(c) = 1, m(d) = 1, m(1) = 1 is the unique state-morphism on A. Moreover, m(x ∨ y) = m(x ∪y) = max{m(x), m(y)}for all x, y∈A, hence m also is a Bosbach state onA.
Similarly to [10], for the case of good pseudo-BL algebras, we define the notion below.
Definition 3.11. Let A be a good bounded pseudo-BCK algebra. Two elements x, y∈A are called orthogonal, and we write x⊥y, iff x−∼≤y∼. If x, y∈Aare orthogonal, we define a partial operation + onAbyx+y :=x⊕y.
Lemma3.12. LetAbe a good bounded pseudo-BCK algebra andx, y∈A.
Then
(1)x⊥y iff y−∼≤x−; (2)x⊥x− andx+x−= 1;
(3)x∼⊥x andx∼+x= 1;
(4)x⊥0 andx+ 0 =x∼−;
(5) 0⊥x and0 +x=x∼−;
(6) if x ≤ y, then x ⊥y−, y∼ ⊥ x and x+y− = y → x−∼, y∼+x = y x∼−.
Proof. (1): x−∼≤y∼ iffy−∼≤x−∼−=x−.
(2): Sincex−∼≤x−∼= (x−)∼, we have x ⊥x− and x+x−=x−∼→ x−∼= 1.
(3): Similarly, from x−∼ ≤x−∼ = (x∼)− we get x∼ ⊥x and x∼+x= x∼→x∼−∼ =x∼→x∼= 1.
(4): Since x−∼≤1 = 0∼, we havex⊥0 andx+ 0 = 0∼ →x∼−= 1→ x∼−=x∼−.
(5): Sincex−∼≤1 = 0−, we have 0⊥xand 0 +x=x∼→0∼−=x∼→ 0 =x∼−.
(6): Since x ≤y, we have y− ≤ x−, that is, (y−)−∼ ≤ x−, so x ⊥y−. Moreover, x+y− = y−∼ → x−∼ = y → x−∼ (by (c26)). Similarly, we have y∼≤x∼, so (y∼)−∼≤x∼, that is,y∼⊥x andy∼+x=x∼→y∼−∼=x∼→ y∼=y x∼− (by (c26)).
Definition3.13. LetAbe a good bounded pseudo-BCK algebra. ARie˘can state on A is a functions:A→[0,1] such that, for all x, y∈A,
(R1) if x⊥y, thens(x+y) =s(x) +s(y);
(R2) s(1) = 1.
Example3.14. Consider again the good bounded pseudo-BCK algebraA from Example 2.23. We claim that the Bosbach state s:A→[0,1] defined by
s(0) = 0, s(a) = 1, s(b) = 1, s(c) = 1, s(d) = 1, s(1) = 1 also is a Rie˘can state onA. Indeed, the pairs (x, y) of orthogonal elements of A are given in the table
x y x−∼ y∼ x+y
0 0 0 1 0
0 a 0 0 1
0 b 0 0 1
0 c 0 0 1
0 d 0 0 1
0 1 0 0 1
a 0 1 1 1
b 0 1 1 1
c 0 1 1 1
d 0 1 1 1
1 0 1 1 1
One can easily check that sis a Rie˘can state.
Proposition3.15. Ifsis a Rie˘can state on a good bounded pseudo-BCK algebra A, then for all x, y∈A the following properties hold:
(1)s(x−) =s(x∼) = 1−s(x);
(2)s(0) = 0;
(3)s(x−∼) =s(x∼−) =s(x−−) =s(x∼∼) =s(x);
(4) if x ≤ y, then s(x) ≤ s(y) and s(y → x−∼) = s(y x∼−) = 1 +s(x)−s(y);
(5) s((x∨y) → x−∼) = s((x∨y) x−∼) = 1−s(x∨y) +s(x) and s((x∪y)→x−∼) =s((x∪y) x−∼) = 1−s(x∪y) +s(x);
(6) s((x∨y) → y−∼) = s((x∨y) y−∼) = 1−s(x∨y) +s(y) and s((x∪y)→y−∼) =s((x∪y) y−∼) = 1−s(x∪y) +s(y).
Proof. (1): Sincex⊥x−andx+x−= 1, we haves(x)+s(x−) =s(1) = 1, so s(x−) = 1−s(x). Similarly, s(x∼) = 1−s(x).
(2): It follows from the fact that 0⊥0 and 0 + 0 = 0.
(3): Using the fact that x⊥0 and x+ 0 =x∼−, we gets(x) =s(x∼−) and, similarly, the other equalities.
(4): Since x ≤ y, we have x ⊥ y− and x+y− = y → x−∼. Hence s(x) +s(y−) = s(y → x−∼), so s(x)−s(y) = s(y → x−∼)−1 ≤ 0, that is, s(x)≤s(y). We get s(y→x−∼) = 1 +s(x)−s(y). Similarly, from x≤y we havey∼⊥x andy∼+x=y x∼−and we gets(y x∼−) = 1 +s(x)−s(y).
(5): We use Lemma 3.12. It follows fromx ≤x∨y, that x ⊥(x∨y)− and x+ (x∨y)−= (x∨y)→x−∼. Hences(x+ (x∨y)−) =s((x∨y)→x−∼).
Thus, s((x∨y) →x−∼) = 1−s(x∨y) +s(x). Similarly, from x ≤x∨y we get (x∨y)∼⊥x and (x∨y)∼+x= (x∨y) x−∼. Hences((x∨y)∼+x) = s((x∨y) x−∼) and we get 1−s(x∨y) +s(x) = s(x∨y x−∼). Thus, s((x∨y) → x−∼) = s((x∨y) x−∼) = 1−s(x∨y) +s(x) and, similarly, s((x∪y)→x−∼) =s((x∪y) x−∼) = 1−s(x∪y) +s(x).
(6): This follows similarly to (5).
Remark 3.16 ([18]). In the case of a pseudo-BCK semilattice, properties (5) and (6) become
(50) s(x→y−∼) =s(x y∼−) = 1−s(x∨y) +s(y).
Theorem 3.17. Let A be a good bounded pseudo-BCK algebra. Any Bosbach state on Ais a Rie˘can state.
Proof. Let s be a Bosbach state on A. Assume x ⊥ y, i.e. x−∼ ≤y∼. By Proposition 3.4(4) and (1) we have 1 +s(x−∼) =s(y∼) +s(y∼ → x−∼).
It follows that 1 +s(x) = 1−s(y) +s(y∼ → x−∼), hence s(y∼ → x−∼) = s(x) +s(y). Therefore, s(x+y) =s(x) +s(y). Since by hypothesis s(1) = 1, s is a Rie˘can state onA.
In the next example we show that there exists a Rie˘can state which is not a Bosbach state.
Example 3.18. Consider A = {0, a, b, c,1} with 0 < a < b < c <1 and the operations →, given by the tables
→ 0 a b c 1
0 1 1 1 1 1
a b 1 1 1 1 b b c 1 1 1 c 0 a b 1 1
1 0 a b c 1
0 a b c 1
0 1 1 1 1 1
a b 1 1 1 1 b b b 1 1 1 c 0 b b 1 1
1 0 a b c 1
.
Then (A,≤,→, ) is a good pseudo-BCK lattice. (Moreover, since (A,≤,→, ) is a chain, it also is a good pseudo-BCK(pP) lattice, that is, a good residuated lattice). The function s:A→[0,1] defined by
s(0) = 0, s(a) = 1/2, s(b) = 1/2, s(c) = 1, s(1) = 1
is a Rie˘can state. Indeed, the orthogonal elements ofA are the pairs (x, y) in the table
x y x−∼ y∼ x+y
0 0 0 1 0
0 a 0 b b
0 b 0 b b
0 c 0 0 1
0 1 0 0 1
a 0 b 1 b
a a b b 1
a b b b 1
b 0 b 1 b
b a b b 1
b b b b 1
c 0 1 1 1
1 0 1 1 1
One can easily check thatsis a Rie˘can state, but the functionsdefined above is not a Bosbach state. Indeed, checking condition (B2) we obtain
s(a) +s(a b) =s(a) +s(1) = 1/2 + 1 = 3/2, s(b) +s(b a) =s(b) +s(b) = 1/2 + 1/2 = 1,
so condition (B2) does not hold. We conclude that s is not a Bosbach state.
Moreover, the pseudo-BCK algebra A has no Bosbach state on it. Indeed, assume thatAadmits a Bosbach statessuch thats(0) = 0,s(a) =α,s(b) =β,
s(c) =γ, s(1) = 1. From s(x) +s(x → y) = s(y) +s(y →x), taking x =a, y = 0, x =b,y = 0 and then x =c, y= 0, we get α = 1/2,β = 1/2, γ = 1.
But, we already proved that sis not a Bosbach state.
Remarks 3.19. (1). In the case of good pseudo-BL algebras, Georgescu [10] asked to find an example of Rie˘can state which is not a Bosbach state.
The above example gives a positive answer to this problem in the case of good pseudo-BCK algebras.
(2) In ([9], Theorem 3.8) it is proved that in the case of good bounded non-commutative R`-monoids, the Bosbach and Rie˘can states coincide. The proof is based on the fact that (x→y)−∼ =x−∼ → y−∼ and (x y)−∼= x−∼ y−∼ in any bounded non-commutative R`-monoid (see [9], Lem- ma 2.1). But these properties are not satisfied in any good pseudo-BCK algebra. Indeed, consider again the good pseudo-BCK(pP) lattice from Ex- ample 3.18. We have (b→a)−∼ = 0, but b−∼ → a−∼ = 1. Therefore, the above mentioned result for good bounded non-commutative R`-monoids is not valid in the case of good pseudo-BCK algebras, as we can see in the above example.
(3) Dvure˘censkij [6] proved that every linearly ordered pseudo-BL algebra admits a Bosbach state. Since the pseudo-BCK algebraA from the above ex- ample is linearly ordered, we conclude that in contrast to pseudo-BL algebras, there exist linearly ordered pseudo-BCK algebras having no Bosbach state.
(4) In the case of pseudo-BL algebras Georgescu [10] proved that the existence of a state-morphism is equivalent to the existence of a maximal filter which is normal. In the case of pseudo-BCK algebras, this is not true. Indeed, in the pseudo-BCK algebra A from Example 3.18, D = {c,1} is a maximal normal deductive system, but we proved that there is no Bosbach state on A.
Thus, there is no state-morphism on A.
Theorem3.20. Every Rie˘can state on a good pseudo-BCK(pDN)algebra is a Bosbach state.
Proof. Letsbe a Rie˘can state on a good pseudo-BCK(pDN) algebraA.
By Proposition 3.15(2) we have s(0) = 0. Since y−∼ = y for all y ∈ A, by Proposition 3.15(6) we gets(x∨y→y) = 1−s(x∨y)+s(y) ands(x∪y→y) = 1−s(x∪y) +s(y). By Proposition 2.28, we haves(x→y) = 1−s(x∨y) +s(y) and s(x y) = 1−s(x∪y) +s(y). Finally, by Proposition 3.6,sis a Bosbach state on A.
Theorem 3.21. If A is a good pseudo-BCK algebra satisfying the iden- tities
(x→y)−∼=x∨y→y−∼ and (x y)−∼=x∪y y−∼, then the Bosbach and Rie˘can states coincide.
Proof. Let s be a Rie˘can state on a good pseudo-BCK algebra A. By Proposition 3.15(2) we have s(0) = 0. By Proposition 3.15(3), (5), (6) we get
s(x→y) =s((x→y)−∼) =s(x∨y→y−∼) = 1−s(x∨y) +s(y) and
s(x y) =s((x y)−∼) =s(x∨y y−∼) = 1−s(x∨y) +s(y).
Thus, by Proposition 3.6, sis a Bosbach state on A.
Example 3.22. Consider again the good bounded pseudo-BCK latticeA from Example 2.23. Since x−∼ = 1 and x → y−∼ = 1 for all x, y ∈ A, A satisfies the identities from Theorem 3.21. Hence the Bosbach and Rie˘can states coincide, which is in accordance with Example 3.14.
Remark 3.23. In the case of a good pseudo-BCK semilattice it was proved in [18] that x∨y → y−∼ = x → y−∼ and x∨y y−∼ = x y−∼ for all x, y∈A. Hence the identities from Theorem 3.21 become
(x→y)−∼=x→y−∼, (x y)−∼=x y−∼. Thus, Theorem 3.21 becomes Theorem 6.11 from [18].
Acknowledgement. The author would like to thank Professor Afrodita Iorgulescu for her useful remarks and suggestions that helped to improve the presentation.
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Received 15 September 2007 Polytechnical University of Bucharest Department of Mathematics
Splaiul Independent¸ei 313 Bucharest, Romania lavinia ciungu@math.pub.ro
