Soft Computing

, Volume 21, Issue 10, pp 2521–2529 | Cite as

Quantum B-algebras: their omnipresence in algebraic logic and beyond

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Abstract

Quantum B-algebras are implicational subreducts of quantales. Their ubiquity and unifying rôle in algebraic logic and beyond is discussed in a brief survey, with old and new examples included. A special section is devoted to CKL-algebras (alias HBCK-algebras). Using a natural embedding of any CKL-algebra into a semibrace, Cornish’s identity is derived, which yields a new syntactic proof of Wroński’s conjecture that CKL-algebras form a variety.

Keywords

Quantum B-algebra CKL-algebra Semibrace Residuated poset Partially ordered group 

1 Introduction

Quantum B-algebras were introduced in Rump (2013) and studied further in Rump and Yang (2014). Their ubiquity and unifying nature in the realm of algebraic logic was established by three main classes of algebras. Prototypes for these classes are BCK-algebras, residuated posets, and effect algebras. On the other hand, it was proved that quantum B-algebras formalize the implicational part of quantales. Recently, the completion\(\widehat{X}\) of a quantum B-algebra X was constructed (Rump 2016), a quantale that can be viewed as an injective envelope of X. Some concepts for quantales survive in the framework of quantum B-algebras. Although the latter need not be multiplicatively closed, it makes sense to speak of a unital quantum B-algebra, with a unit element related to that of the ambient quantale.

Concerning the three prototypes and their generalizations, the main point is that they can be viewed as building blocks for the “generic” (unital) quantum B-algebra. In other words, a quantum B-algebra is like a chameleon: By adding relations, it becomes a pseudo-BCK-algebra, or a residuated poset, or a pseudo-effect algebra, or a further specialization. For example, partially ordered groups can be viewed as quantum B-algebras which satisfy the equations
$$\begin{aligned} (x\rightarrow y)\leadsto y=(x\leadsto y)\rightarrow y=x. \end{aligned}$$
Note that the theory of groups is usually not rated among algebraic logic, though every group can be equipped with a trivial partial order. Therefore, quantum B-algebras also include structures with no obvious logical flavor.

The present paper starts with a brief survey on basic features of not necessarily unital quantum B-algebras, with old and new examples. In a third section, we focus upon a class of quantum B-algebras occurring in connection with Wroński’s conjecture on naturally ordered BCK-algebras.

Roughly speaking, a quantum B-algebra is a partially ordered set with two binary operations \(\rightarrow \) and \(\leadsto \) satisfying three conditions (see Sect. 2). Logically, \(\rightarrow \) and \(\leadsto \) represent the left and right implications, while \(\leqslant \) stands for the entailment relation. In commutative logic, the two implications \(\rightarrow \) and \(\leadsto \) coincide. As already shown in Rump (2013), a quantum B-algebra with trivial partial order is equivalent to a group. Then \(u:=x\rightarrow x\) does not depend on x and gives the unit element of the group. In particular, such a quantum B-algebra is always unital. We also consider the complementary case where the quantum B-algebra reduces to a mere poset. In this way, any partially ordered set with a greatest element u is a unital quantum B-algebra with unit element u.

Apart from the above-mentioned three prototypical specializations and building blocks of a unital quantum B-algebra, the connected components (Rump 2013) of every (not necessarily unital) quantum B-algebra X form a group C(X), which implies, in particular, that X has a distinguished component, according to the unit element of C(X). This can be applied to determine the structure of a pseudo-BCI algebra (see Rump 2013, Sect. 4). Here we exhibit another building block of X, namely, the set G(X) of group-like elements \(x\in X\), which satisfy \(x\rightarrow x=x\leadsto x\) and the above equations \((x\rightarrow y) \leadsto y=(x\leadsto y)\rightarrow y=x\) for all \(y\in X\). We show that G(X) is either empty or a partially ordered group (Theorem 1).

In Sect. 4, we study the relationship between quantum B-algebras and L-algebras (Rump 2008a), which are named by the relation
$$\begin{aligned} (x\rightarrow y)\rightarrow (x\rightarrow z)=(y\rightarrow x)\rightarrow (y\rightarrow z). \end{aligned}$$
(L)
They first occurred in papers of Bosbach (1982) and Traczyk (1988). Apart from characteristic relation (L), an L-algebraX has a logical unit, that is, an element 1 which satisfies
$$\begin{aligned} 1\rightarrow x=x,\qquad x\rightarrow x=x\rightarrow 1=1, \end{aligned}$$
such that the implication
$$\begin{aligned} x\rightarrow y=y\rightarrow x=1\;\Longrightarrow \; x=y \end{aligned}$$
holds for all \(x,y\in X\). An L-algebra is a quantum B-algebra if and only if it is a BCK-algebra. Such algebras are known as HBCK-algebras (Blok and Ferreirim 1993) or CKL-algebras (Rump 2009). Traczyk (1988) studied CKL-algebras as a special class of BCK-algebras. Wroński (1985) conjectured that CKL-algebras coincide with the implicational subreducts of hoops. In particular, this means that Cornish’s identity (Cornish 1982)
$$\begin{aligned}&\bigl (((x\rightarrow y)\rightarrow y)\rightarrow x\bigr )\rightarrow x\\&\quad =\bigl (((y\rightarrow x)\rightarrow x)\rightarrow y\bigr ) \rightarrow y \end{aligned}$$
(J)
holds in any CKL-algebra. Ferreirim (1992), Blok and Ferreirim (1993), Ferreirim (2001) confirmed Wroński’s conjecture by a careful analysis of subdirectly irreducible hoops and by adapting these results to BCK-algebras and CKL-algebras. In this way, they obtained an indirect proof of Cornish’s identity. However, they have not been able to transform their findings into a syntactic proof. This problem was settled by Kowalski (1994), and independently, by Wos and Veroff (1994), using automated reasoning with Otter (McCune 1994). In Theorem 2, we provide a simple, syntactic, self-contained proof of (J) for arbitrary CKL-algebras.

2 The category of quantum B-algebras

A quantum B-algebra (Rump 2013; Rump and Yang 2014) is defined to be a partially ordered set X with two binary operations \(\rightarrow \) and \(\leadsto \) satisfying
$$\begin{aligned}&x\rightarrow (y\leadsto z)=y\leadsto (x\rightarrow z) \end{aligned}$$
(1)
$$\begin{aligned}&x\leqslant y\rightarrow z\;\Longleftrightarrow \; y\leqslant x\leadsto z \end{aligned}$$
(2)
$$\begin{aligned}&y\leqslant z\;\Longrightarrow \; x\rightarrow y\leqslant x\rightarrow z \end{aligned}$$
(3)
for all \(x,y,z\in X\). The first two axioms express the relationship between \(\rightarrow \) and \(\leadsto \) on two levels, namely, by an Eq. (1), and as an equivalence (2). Both \(\rightarrow \) and \(\leadsto \) stand for logical implication, a left one and a right one, which have to be distinguished in a non-commutative framework. The partial order relation \(\leqslant \) stands for entailment. So Eq. (1) states an equivalence between logical propositions, while (2) connects two entailment analogues on a higher level. Note that instead of the two implications \(\rightarrow \) and \(\leadsto \), there is only one type of entailment, which brings to mind that non-commutativity of logic disappears on a higher level. So it is not surprising that Eq. (1) cannot be dropped, as the following example shows.

Example 1

Let G be a quasigroup. Thus, for \(x,y\in G\), there are unique elements \(x\rightarrow y\) and \(x\leadsto y\) in G with \((x\rightarrow y)x=y= x(x\leadsto y)\). With the trivial partial order, we have \(x=y\rightarrow z\Longleftrightarrow xy=z \Longleftrightarrow y=x\leadsto z\). This proves (2), while (3) is trivial. To check what Eq. (1) is about, we set \(u:=y\leadsto z\) and \(v:=x\rightarrow z\). Then \(yu=z=vx\), and Eq. (1) is equivalent to \(x\rightarrow u=y\leadsto v\). With \(t:=x\rightarrow u=y\leadsto v\), this gives \(y(tx)=yu=vx=(yt)x\). Thus Eq. (1) implies that G is associative, whence it cannot be dropped, in general.

To see that (2) is not redundant, too, consider a set X with trivial partial order and \(x\rightarrow y=x\leadsto y=y\) for all \(x,y\in X\). Then (1) and (3) hold, but (2) fails if \(|X|>1\).

The implication (3) has a symmetric counterpart
$$\begin{aligned} y\leqslant z\;\Longrightarrow \; x\leadsto y\leqslant x\leadsto z \end{aligned}$$
(4)
which holds for every quantum B-algebra (see (19) in Rump and Yang 2014). Furthermore, the inequalities
$$\begin{aligned}&y\rightarrow z \leqslant (x\rightarrow y)\rightarrow (x\rightarrow z) \end{aligned}$$
(5)
$$\begin{aligned}&y\leadsto z \leqslant (x\leadsto y)\leadsto (x\leadsto z), \end{aligned}$$
(6)
which represent the lower-level versions of (3) and (4), are valid for any quantum B-algebra. Associated with the letter “B” in BCK-algebras, they gave birth to the name “quantum B-algebra,” while the prefix “quantum” refers to a close relationship to quantales. One can show (Rump and Yang 2014) that (2)–(6) are equivalent to the axioms of a quantum B-algebra.
Statements like (3)–(6) are familiar in algebraic logic. There are similar versions (see (20) in Rump and Yang 2014)
$$\begin{aligned} x\leqslant y\;&\Longrightarrow \; y\rightarrow z\leqslant x\rightarrow z \end{aligned}$$
(7)
$$\begin{aligned} x\leqslant y\;&\Longrightarrow \; y\leadsto z\leqslant x\leadsto z, \end{aligned}$$
(8)
and
$$\begin{aligned} x\rightarrow y&\leqslant (y\rightarrow z)\leadsto (x\rightarrow z)\\ x\leadsto y&\leqslant (y\leadsto z)\rightarrow (x\leadsto z), \end{aligned}$$
which also hold in a quantum B-algebra. Recall that a partially ordered semigroup Q is said to be a quantale  (Mulvey 1986; Rosenthal 1990) if Q is a complete lattice satisfying
$$\begin{aligned} a\cdot \left( \bigvee A\right) =\bigvee aA,\qquad \left( \bigvee A\right) \cdot a=\bigvee Aa \end{aligned}$$
for \(a\in Q\) and \(A\subset Q\). By Rump (2013), Theorem 2.3, a quantum B-algebra is the same as a system \((X;\rightarrow ,\leadsto ,\leqslant )\) which can be embedded into a quantale Q such that the operations \(\rightarrow \) and \(\leadsto \) and the relation \(\leqslant \) are induced by the residuals
$$\begin{aligned} a\rightarrow b:= & {} \bigvee \{ x\in Q\,|\, xa\leqslant b\},\\ a\leadsto b:= & {} \bigvee \{ x\in Q\,|\, ax\leqslant b\}, \end{aligned}$$
and the partial order in Q.
Note that a quantale has a multiplication \(\cdot \) which is related to the residuals by the adjointness relation
$$\begin{aligned} x\leqslant y\rightarrow z\;\Longleftrightarrow \; xy\leqslant z\;\Longleftrightarrow \; y\leqslant x\leadsto z. \end{aligned}$$
(9)
Of course, every quantale is a quantum B-algebra. More generally, every residuated poset, that is, a partially ordered semigroup with binary operations \(\rightarrow \) and \(\leadsto \) satisfying (9), is a quantum B-algebra.
A multiplicative unit element of a residuated poset X can be characterized as an element \(u\in X\) which satisfies
$$\begin{aligned} u\rightarrow x=u\leadsto x=x \end{aligned}$$
for all \(x\in X\). For an arbitrary quantum B-algebra X, such an element u is unique. If it exists, X is called unital. On the other hand, a residuated poset X is commutative if and only if it satisfies
$$\begin{aligned} x\rightarrow y=x\leadsto y \end{aligned}$$
for all \(x,y\in X\). Quantum B-algebras with this property are called commutative.
Quantum B-algebras form a category \(\mathbf{qBAlg}\). Morphisms \(f:X\rightarrow Y\) are defined to be monotone maps satisfying
$$\begin{aligned} f(x\rightarrow y)\leqslant f(x)\rightarrow f(y) \end{aligned}$$
(10)
for all \(x,y\in X\). Note that (10) is equivalent to
$$\begin{aligned} f(x\leadsto y)\leqslant f(x)\leadsto f(y). \end{aligned}$$
(11)
For example, if (11) holds, then (2) implies \(x\leqslant (x\rightarrow y) \leadsto y\), which yields
$$\begin{aligned} f(x)\leqslant f\bigl ((x\rightarrow y)\leadsto y\bigr )\leqslant f(x\rightarrow y) \leadsto f(y). \end{aligned}$$
Whence (10) follows by (2). By Rump (2016), Theorem 1, injective envelopes exist in the category \(\mathbf{qBAlg}\), and injective objects are just the quantales.

3 Further examples

In classical algebra, most structures are based on semigroups. When partial orders arise, residuals are notably absent. In every first course on group theory, however, subgroups are characterized as non-empty subsets H for which \(a,b\in H\) implies that H also contains \(ab^{-1}\). Well, this is a residual!

Example 2

Let G be a group. With
$$\begin{aligned} a\rightarrow b:=ba^{-1},\qquad a\leadsto b:=a^{-1}b, \end{aligned}$$
(12)
G is a quantum B-algebra with trivial partial order.

In Rump (2013), Theorem 4.2, we proved the converse:

Proposition 1

A quantum B-algebra X is a group if and only if its partial order is trivial.

This was obtained as a consequence of a more general theorem. For convenience, we give a direct proof.

Proof

Let X be a quantum B-algebra with trivial partial order. Assume that the equation \(z\rightarrow x=z\rightarrow y\) holds for given \(x,y,z\in X\). Then (5) implies that
$$\begin{aligned} x\rightarrow y= & {} (z\rightarrow x)\rightarrow (z\rightarrow y)\\= & {} (z\rightarrow y)\rightarrow (z\rightarrow y)=y\rightarrow y. \end{aligned}$$
Hence (2) yields \(y=(x\rightarrow y)\leadsto y=x\). So the map \(x\mapsto (z\rightarrow x)\) is injective. On the other hand, (1) and (2) yield \(x=\bigl (x\rightarrow (z\rightarrow z))\leadsto (z\rightarrow z)=z\rightarrow \bigl ((x\rightarrow (z\rightarrow z)) \leadsto z\bigr )\), which shows that \(x\mapsto (z\rightarrow x)\) is bijective for all \(z\in X\). \(\square \)
Thus, for given \(x,y\in X\), there is a unique element \(xy\in X\) with \(y\rightarrow xy= x\). For \(x,y,z\in X\), inequality (5) gives \(x\rightarrow (y\rightarrow z)=(y\rightarrow xy) \rightarrow (y\rightarrow z)=xy\rightarrow z\). Hence
$$\begin{aligned} x\rightarrow (y\rightarrow z)=xy\rightarrow z. \end{aligned}$$
(13)
Therefore, \(y\rightarrow (z\rightarrow x(yz))=yz\rightarrow x(yz)=x\). Hence \(z\rightarrow x(yz)=xy\), and thus \((xy)z=x(yz)\). So the product is associative. Furthermore, (5) gives \(y=x\rightarrow yx=(x\rightarrow x)\rightarrow (x\rightarrow yx)=(x\rightarrow x)\rightarrow y\), hence \(x\rightarrow x=y\leadsto y\) for all \(x,y\in X\). Let us write u for this unique element \(x\rightarrow x= y\leadsto y\). Then \(x\rightarrow ux=u=x\rightarrow x\) yields \(ux=x\), and \(x=(x\leadsto x)\rightarrow x= u\rightarrow x\) implies that \(xu=x\) for all \(x\in X\). Finally, \(x\rightarrow u=x\rightarrow (x\rightarrow u)x\) gives \((x\rightarrow u)x=u\), which shows that every element of X is invertible. \(\square \)

Example 3

The other extreme of a quantum B-algebra occurs when there is just a partial order. Thus, let \(\Omega \) be a partially ordered set with a greatest element 1. Define
$$\begin{aligned} x\rightarrow y:={\left\{ \begin{array}{ll} 1 &{} \text { for } x\leqslant y\\ y &{} \text { for } x\not \leqslant y\end{array}\right. } \end{aligned}$$
for \(x,y\in \Omega \). This makes \(\Omega \) into a commutative unital quantum B-algebra with unit element \(u=1\). Note that for any pair \(x,y\in \Omega \), there is a unique element \(x\leadsto y\in \Omega \) which satisfies (2), namely, \(x\leadsto y=x\rightarrow y\).

Let us discuss some generalizations of Examples 2 and 3.

Example 4

Generalizing Example 2, every partially ordered group is a quantum B-algebra with residuals (12). Conversely, Proposition 3.6 of Rump (2013) states that a quantum B-algebra \(X\not =\varnothing \) is a partially ordered group if and only if it satisfies
$$\begin{aligned} (x\rightarrow y)\leadsto y=(x\leadsto y)\rightarrow y=x \end{aligned}$$
(14)
for all \(x,y\in X\).

Based on this example, we give the following

Definition 1

Let X be a quantum B-algebra. We call an element \(x\in X\)group-like if it satisfies \(x\rightarrow x=x\leadsto x\) and Eq. (14) for all \(y\in X\).

Theorem 1

Let X be a quantum B-algebra. The set G(X) of group-like elements is either empty or a partially ordered subgroup of X.

Proof

By (2), we have \(x\rightarrow y\leqslant \bigl ((x\rightarrow y)\leadsto y\bigr )\rightarrow y\) and
$$\begin{aligned} x\leqslant (x\rightarrow y)\leadsto y\leqslant \bigl (((x\rightarrow y)\leadsto y)\rightarrow y\bigr )\leadsto y, \end{aligned}$$
hence \(\bigl ((x\rightarrow y)\leadsto y\bigr )\rightarrow y\leqslant x\rightarrow y\). Thus, by symmetry, we obtain
$$\begin{aligned} x\rightarrow y= & {} \bigl ((x\rightarrow y)\leadsto y\bigr )\rightarrow y\nonumber \\ x\leadsto y= & {} \bigl ((x\leadsto y)\rightarrow y\bigr )\leadsto y. \end{aligned}$$
(15)
For the moment, let us call \(x\in X\)right admissible if
$$\begin{aligned} (x\leadsto y)\rightarrow y=x \end{aligned}$$
holds for all \(y\in X\). Assume that \(y\in X\) is right admissible. For \(x,z,t \in X\), this gives
$$\begin{aligned} \begin{aligned} \bigl ((x\rightarrow y)\leadsto z\bigr )\rightarrow z&\leqslant \bigl (t\rightarrow ((x\rightarrow y)\leadsto z)\bigr ) \rightarrow (t\rightarrow z) \\&= \bigl ((x\rightarrow y)\leadsto (t\rightarrow z)\bigr )\rightarrow (t\rightarrow z). \end{aligned} \end{aligned}$$
With \(t:=y\leadsto z\), the first equation of (15) then yields \(x\rightarrow y\leqslant \bigl ((x\rightarrow y)\leadsto z\bigr )\rightarrow z\leqslant \bigl ((x\rightarrow y)\leadsto y\bigr ) \rightarrow y=x\rightarrow y\). Thus \(x\rightarrow y\) is again right admissible.
Similarly, we call \(x\in X\)left admissible if \((x\rightarrow y)\leadsto y=x\) holds for all \(y\in X\). Assume that \(x,y\in X\) are group-like. Then \(x\rightarrow y\) is right admissible. By the dual of the above argument, \(u:=x\rightarrow x=x\leadsto x\) is left admissible. Hence
$$\begin{aligned} x\rightarrow y= & {} x\rightarrow \bigl ((y\rightarrow x)\leadsto x\bigr )=(y\rightarrow x)\leadsto (x\rightarrow x)\\= & {} (y\rightarrow x)\leadsto u \end{aligned}$$
is left admissible. This proves Eq. (14) for \(x\rightarrow y\) instead of x.
Next we show that \((x\rightarrow y)\rightarrow (x\rightarrow y)=(x\rightarrow y)\leadsto (x\rightarrow y)\). With
$$\begin{aligned} z:=x\rightarrow \bigl ((y\leadsto u)\rightarrow x\bigr ), \end{aligned}$$
we have
$$\begin{aligned} \begin{aligned} x\leadsto z&= x\rightarrow \bigl (x\leadsto ((y\leadsto u)\rightarrow x)\bigr )\\&=x\rightarrow \bigl ((y\leadsto u)\rightarrow (x\leadsto x)\bigr )\\&= x\rightarrow \bigl ((y\leadsto u)\rightarrow u\bigr )=x\rightarrow y. \end{aligned} \end{aligned}$$
Hence
$$\begin{aligned} \begin{aligned} (x\rightarrow y)\rightarrow (x\rightarrow y)&= (x\leadsto z)\rightarrow (x\leadsto z)\\&=x\leadsto \bigl ( (x\leadsto z)\rightarrow z\bigr )\\&=x\leadsto x=x\rightarrow x\\&= x\rightarrow \bigl ((x\rightarrow y)\leadsto y\bigr )\\&=(x\rightarrow y)\leadsto (x\rightarrow y). \end{aligned} \end{aligned}$$
Thus \(x\rightarrow y\) is group-like. By symmetry, G(X) is a quantum B-subalgebra of X. Now Rump (2013), Proposition 3.6, implies that G(X) is a partially ordered group if \(G(X)\not =\varnothing \). \(\square \)
Next we turn our attention to Example 3. Let X be a set with a binary operation \(\rightarrow \). An element \(1\in X\) is said to be a logical unit (Rump 2008a) if
$$\begin{aligned} 1\rightarrow x=x,\qquad x\rightarrow x=x\rightarrow 1=1 \end{aligned}$$
(16)
holds for all \(x\in X\). Note that a logical unit must be unique. If 1 is a simultaneous logical unit for two binary operations \(\rightarrow \) and \(\leadsto \), then X is said to be a pseudo-BCK-algebra (Georgescu and Iorgulescu 2001; van Alten 2006; Dvurečenskij and Kühr 2009) if for \(x,y,z\in X\),
$$\begin{aligned}&(x\rightarrow y)\leadsto \bigl ((y\rightarrow z)\leadsto (x\rightarrow z)\bigr )=1\\&(x\leadsto y)\rightarrow \bigl ((y\leadsto z)\rightarrow (x\leadsto z)\bigr )=1\\&x\rightarrow y=y\leadsto x=1\;\Longrightarrow \; x=y. \end{aligned}$$

Example 5

It can be shown Rump and Yang (2014) that every pseudo-BCK-algebra X satisfies \(x\rightarrow y=1\Longleftrightarrow x\leadsto y=1\). Therefore, we can introduce a partial order
$$\begin{aligned} x\leqslant y\;:\Longleftrightarrow \; x\rightarrow y=1\Longleftrightarrow x\leadsto y=1 \end{aligned}$$
which makes X into a quantum B-algebra. Conversely, by Rump and Yang (2014), Corollary 2 of Proposition 12, a quantum B-algebra X is pseudo-BCK if and only if X has a greatest element which is a unit element. In particular, every partially ordered set with greatest element 1 is a pseudo-BCK-algebra unit element 1.
Note that Examples 4 and 5 are disjoint: A partially ordered group X is pseudo-BCK if and only if \(|X|=1\). There is a common generalization. Namely, an algebra \((X;\rightarrow ,\leadsto )\) with a partial order \(\leqslant \) and a constant u is called a pseudo-BCI algebra (Jun et al. 2006) if it satisfies
$$\begin{aligned}&x\leqslant (x\rightarrow y)\leadsto y,\qquad x\leqslant (x\leadsto y)\rightarrow y\\&x\rightarrow y\leqslant (y\rightarrow z)\leadsto (x\rightarrow z)\\&x\leadsto y\leqslant (y\leadsto z)\rightarrow (x\leadsto z)\\&x\leqslant y\;\Longleftrightarrow \; x\rightarrow y=u\;\Longleftrightarrow \; x\leadsto y=u \end{aligned}$$
for \(x,y,z\in X\).

Example 6

Every pseudo-BCI algebra is a unital quantum B-algebra. Conversely, by Rump (2013), Proposition 4.4, a unital quantum B-algebra is pseudo-BCI if and only if its unit element is maximal.

More generally, Theorem 1 yields

Corollary 1

Let \(X\not =\varnothing \) be a commutative quantum B-algebra. Assume that every element of X is contained in a maximal element. Then the maximal elements form a subgroup of X which coincides with G(X).

Proof

Every maximal element of X is group-like. Hence G(X) is non-empty and contains a maximal element t. Suppose that \(x,y\in G(X)\) satisfy \(x\leqslant y\). Then \(t\leqslant tx^{-1}y\), which yields \(t=tx^{-1}y\). Hence \(x=y\). So the partial order of G(X) is trivial. Therefore, G(X) must coincide with the set of maximal elements in X. \(\square \)

Corollary 2

Let X be a quantum B-algebra with a smallest element 0. Then X has a greatest element 1, and \(G(X)= \{1\}\). In particular, \(1\rightarrow 1=1\leadsto 1=1\).

Proof

For all \(x,y\in X\), we have \(0\leqslant x\rightarrow y\). Hence \(x\leqslant 0\leadsto y\), which shows that \(1:=0\leadsto y\) is a greatest element. Thus \(1\rightarrow 1=1\rightarrow (0 \leadsto y)=0\leadsto (1\rightarrow y)=1\), and similarly, \(1\leadsto 1=1\). So 1 is group-like, and the argument in the preceding proof shows that the partial order of G(X) is trivial. Whence \(G(X)=\{1\}\). \(\square \)

One may ask whether a quantum B-algebra with a greatest element is pseudo-BCI. The following example shows that this need not be the case.

Example 7

Consider the finite quantum B-algebra \(X=\{0,x,y,z, t,1\}\) given by the following tables and Hasse diagram

Example 8

X can be obtained as a quantum B-subalgebra of the quantale given by the power set of the semigroup

Here \(G(X)=\{1\}\), but 1 is not a unit element of X. Thus X is neither pseudo-BCK nor even pseudo-BCI.

Example 9

The prototype of a pseudo-BCI algebra is a semidirect product \(G\ltimes X\) of a group G by a pseudo-BCK-algebra X (see Rump 2013, Sect. 4). For a group homomorphism \(\gamma :G\rightarrow \text{ Aut }(X)\), the semidirect product \(G\ltimes X\) consists of the formal products ax with \(a\in G\) and \(x\in X\), with partial order
$$\begin{aligned} ax\leqslant by\;:\Longleftrightarrow \; a=b,\, x\leqslant y \end{aligned}$$
and residuals
$$\begin{aligned} \begin{aligned} ax\rightarrow by&:= ba^{-1} (x\rightarrow y)^{a^{-1}}\\ ax\leadsto by&:= a^{-1}b(x^{a^{-1}b}\leadsto y), \end{aligned} \end{aligned}$$
where \(x^a:=\gamma (a^{-1})(x)\). More generally, twisted semidirect products \(G\ltimes _\delta X\) of partially ordered groups with pseudo-BCK-algebras form a class of quantum B-algebras. These so-called quantum BL-algebras (Rump and Yang 2014) are unital quantum B-algebras which are characterized by the property that \(x\rightarrow u\) and \(x\leadsto u\) are invertible for all x in the sense of Rump and Yang (2014), Definition 8. For any quantum B-algebra X, these invertible elements form a subgroup of X, the unit group\(X^\times \). In general, \(X^\times \) is a proper subgroup of G(X).

More examples and classes of quantum B-algebras can be found in Rump (2013), Rump and Yang (2014), Rump (2016). In the following section, we deal with a special class of commutative quantum B-algebras.

4 Quantum B-algebras versus L-algebras

A set X with a binary operation \(\rightarrow \) is said to be a cycloid (Rump 2008a) if the equation
$$\begin{aligned} (x\rightarrow y)\rightarrow (x\rightarrow z)=(y\rightarrow x)\rightarrow (y\rightarrow z) \end{aligned}$$
(L)
holds for \(x,y,z\in X\). If, in addition, X has a logical unit 1 with
$$\begin{aligned} x\rightarrow y=y\rightarrow x=1\;\Longrightarrow \; x=y \end{aligned}$$
(17)
for \(x,y\in X\), then X is called an L-algebra (Rump 2008a). Every L-algebra is partially ordered by
$$\begin{aligned} x\leqslant y\;:\Longleftrightarrow \; x\rightarrow y=1. \end{aligned}$$
A special class of L-algebras are left hoops, that is, monoids H with an additional binary operation \(\rightarrow \) satisfying
$$\begin{aligned} a\rightarrow a=1 \end{aligned}$$
(E)
$$\begin{aligned} ab\rightarrow c=a\rightarrow (b\rightarrow c) \end{aligned}$$
(A)
$$\begin{aligned} (a\rightarrow b)a=(b\rightarrow a)b \end{aligned}$$
(H)
for \(a,b,c\in H\). If H is commutative as a semigroup, then H is said to be a hoop (Büchi and Owens 1975; Blok and Raftery 1997; Blok and Ferreirim 1993).

There is an intermediate structure between L-algebras and left hoops.

Definition 2

A commutative monoid \((A,\wedge )\) with a second binary operation \(\rightarrow \) is said to be a semibrace (Rump 2008b) if
$$\begin{aligned} a\rightarrow (b\wedge c)&= (a\rightarrow b)\wedge (a\rightarrow c)&\qquad a\rightarrow 1&= 1 \end{aligned}$$
(18)
$$\begin{aligned} (a\wedge b)\rightarrow c&= (a\rightarrow b)\rightarrow (a\rightarrow c)&\qquad 1\rightarrow a&= a \end{aligned}$$
(19)
holds for \(a,b,c\in A\).

A semibrace is equivalent to a brace (Rump 2007) if and only if \((A,\wedge )\) is a group. Braces are ring-like structures which are closely related to set-theoretic solutions of the quantum Yang–Baxter equation Etingof et al. (1999), Lu et al. (2000), Rump (2005). Every semibrace X is a cycloid, but not necessarily an L-algebra. If X is an L-algebra, we call X an L-semibrace. Then 1 is a logical unit, and X is a \(\wedge \)-semilattice. By Rump (2008a), Proposition 4, every left hoop is an L-semibrace.

Every L-algebra satisfies the identity
$$\begin{aligned} (y\rightarrow z)\rightarrow \bigl ((x\rightarrow y)\rightarrow (x\rightarrow z)\bigr )=1. \end{aligned}$$
(B)
An L-algebra X which satisfies
$$\begin{aligned} x\rightarrow (y\rightarrow z)=y\rightarrow (x\rightarrow z) \end{aligned}$$
(C)
for \(x,y,z\in X\) is called a CKL-algebra (Rump 2009). Note that any CKL-algebra also satisfies
$$\begin{aligned} x\leqslant y\rightarrow x \end{aligned}$$
(K)
for \(x,y\in X\). Every CKL-algebra is a commutative unital quantum B-algebra with unit element 1.
CKL-algebras are equivalent to naturally ordered BCK-algebras, that is, BCK-algebras satisfying (L). As an implicational version of the basic equation for a hoop, Eq. (L) has also been denoted by (H). Accordingly, CKL-algebras have been called HBCK-algebras (Blok and Ferreirim 1993). A conjecture of Wroński (1985) states that CKL-algebras form a variety and coincide with the implicational subreducts of hoops. These conjectures were proved by Ferreirim (1992), Blok and Ferreirim (1993), Ferreirim (2001). In particular, they proved that the variety of CKL-algebras is given by identities (16), (B), (C), (L) together with Cornish’s identity (Cornish 1982)
$$\begin{aligned}&\bigl (((x\rightarrow y)\rightarrow y)\rightarrow x\bigr )\rightarrow x\\&\quad =\bigl (((y\rightarrow x)\rightarrow x)\rightarrow y\bigr ) \rightarrow y. \end{aligned}$$
(J)
Note that implication (17) follows by this equation. A syntactic proof of (J) was given by Kowalski (1994).

There are two natural ways to embed a CKL-algebra X into a larger structure. Firstly, as an L-algebra, X has a self-similar closure (Rump 2008a), that is, X can be represented as an L-subalgebra of a self-similar left hoop S(X) such that X generates S(X) as a monoid. By Rump (2008a), Theorem 3, the self-similar closure S(X) is unique, up to isomorphism. Furthermore, Rump (2008a), Proposition 10, implies that S(X) satisfies (K). However, (C) does not carry over to S(X), or equivalently, S(X) need not be commutative. Secondly, as a quantum B-algebra, X has a completion (Rump 2016), that is, X embeds densely into a quantale \(\widehat{X}\). In particular, this means that every element of \(\widehat{X}\) is of the form \(\bigwedge A\) for a subset \(A\subset X\). Hence \(\widehat{X}\) is commutative.

None of the embeddings of X into S(X) or \(\widehat{X}\) immediately yields an embedding into a hoop: S(X) is a left hoop, not necessarily commutative, while \(\widehat{X}\) is a commutative quantale, but not necessarily a hoop. Thus it is tempting to believe that a natural embedding into a hoop might still be found between the Scylla of non-commutative hoops and the Charybdis of commutative non-hoops. This would be a strong improvement since Blok and Ferreirim’s construction is highly non-canonical. Roughly speaking, they reduce the problem to finitely generated subdirectly irreducible CKL-algebras and show that every n-generated subdirectly irreducible CKL-algebra splits into an ordinal sum \(F\vee S\) where F is \((n-1)\)-generated and S is totally ordered. So F comes within the inductive hypothesis, while S is a subreduct of an MV-algebra.

As (J) holds in any hoop, the result of Blok and Ferreirim yields an indirect proof that (J) is valid for any CKL-algebra. Kowalski (1994) obtained a (rather intricate) syntactic proof. Independently, Veroff got an automated proof (see Wos and Veroff 1994, Sect. 7.3) using Otter (McCune 1994). In what follows, we give an alternative, hopefully more intelligible proof.

By Rump (2008b), Theorem 3, every L-algebra naturally embeds into an L-semibrace C(L) which is uniquely determined, up to isomorphism, by the property that every element of C(X) is a finite meet \(x_1\wedge \cdots \wedge x_n\) with \(x_i\in X\). By Rump (2008b), Corollary 2 of Theorem 3, C(X) coincides with the \(\wedge \)-subsemilattice of S(X) generated by X. Note, however, that existing meets in X need not be respected by the embedding \(X\hookrightarrow C(X)\).

Proposition 2

Let X be a CKL-algebra. Then C(X) is a CKL-algebra.

Proof

By Rump (2008a), Proposition 10, S(X) satisfies (K). Hence (K) holds in C(X). So it remains to verify Eq. (C) in C(X). By (18), we can assume that \(z\in X\). So we have to verify
$$\begin{aligned} a\rightarrow (b\rightarrow z)=b\rightarrow (a\rightarrow z) \end{aligned}$$
for \(a=x_1\wedge \cdots \wedge x_n\) with \(x_1,\ldots ,x_n,z\in X\) and \(b\in C(X)\). By induction and symmetry, we can assume that this has been shown for \(a= y_1\wedge \cdots \wedge y_m\) with \(m<n\) and \(y_1,\ldots ,y_m\in X\). With \(c:=x_1\wedge \cdots \wedge x_{n-1}\) and \(x:=x_n\), we obtain \(a\rightarrow (b\rightarrow z)=(c\wedge x) \rightarrow (b\rightarrow z)=(c\rightarrow x)\rightarrow \bigl (c\rightarrow (b\rightarrow z)\bigr )=(c\rightarrow x)\rightarrow \bigl (b\rightarrow (c\rightarrow z)\bigr )\). Since \(c\rightarrow x\in X\), this gives \(a\rightarrow (b\rightarrow z)=b\rightarrow \bigl ((c\rightarrow x)\rightarrow (c\rightarrow z)\bigr )=b\rightarrow \bigl ((c\wedge x)\rightarrow z\bigr )=b\rightarrow (a\rightarrow z)\). \(\square \)
To save brackets, we henceforth abbreviate
$$\begin{aligned} x_1\rightarrow \cdots \rightarrow x_n:=(x_1\rightarrow \cdots \rightarrow x_{n-1})\rightarrow x_n \end{aligned}$$
for \(n>2\). The following result is essentially due to Cignoli and Torrens (see Cignoli and Torrens Torrell 2004, Lemmas 1.6 and 3.3).

Proposition 3

Let X be a CKL-algebra and \(x,y,z\in X\) with \(z\leqslant y\). For \(u\in X\), we set \(u':=u\rightarrow z\). Then \(x\rightarrow y''=x''\rightarrow y''=(x\rightarrow y)''\).

Proof

We show first that
$$\begin{aligned} (y''\rightarrow y)''=1. \end{aligned}$$
(20)
Equation (15) give \(y'=y''\rightarrow z\leqslant y''\rightarrow y\). So (7) implies that \(t:=(y''\rightarrow y)'\leqslant y''\). Hence \(y''\rightarrow t=y''\rightarrow (y''\rightarrow y)'= (y''\rightarrow y)\rightarrow (y''\rightarrow z)=(y\rightarrow y'')\rightarrow (y\rightarrow z)=y\rightarrow z\), and thus \(t\rightarrow z=(t\rightarrow y'')\rightarrow (t\rightarrow z)=(y''\rightarrow t)\rightarrow (y''\rightarrow z)=(y\rightarrow z)\rightarrow y'=1\). This proves Eq. (20). Furthermore, \(x\rightarrow y''=y'\rightarrow x'=y'\rightarrow x'''=x''\rightarrow y''\). Hence \((x\rightarrow y'')''\rightarrow (x\rightarrow y'')=x\rightarrow \bigl ((x\rightarrow y'')''\rightarrow y''\bigr )= x\rightarrow \bigl ((x\rightarrow y'')\rightarrow y''\bigr )=1\), and thus
$$\begin{aligned} (x\rightarrow y'')''=x\rightarrow y''. \end{aligned}$$
Finally, Eq. (20) and inequality (5) give
$$\begin{aligned} \begin{aligned} 1&= (y''\rightarrow y)''\leqslant ((x\rightarrow y'')\rightarrow (x\rightarrow y))''\\&\leqslant ((x\rightarrow y'')\rightarrow (x\rightarrow y)'')''\\&= (x\rightarrow y'')\rightarrow (x\rightarrow y)''. \end{aligned} \end{aligned}$$
Whence \(x\rightarrow y''\leqslant (x\rightarrow y)''\leqslant (x\rightarrow y'')''=x\rightarrow y''\). \(\square \)

Now we are ready to prove

Theorem 2

Every CKL-algebra satisfies (J).

Proof

We show first that (J) holds for \(x\leqslant y\). With \(z:=x\), Eq. (J) then becomes
$$\begin{aligned} y''=y''\rightarrow y\rightarrow y. \end{aligned}$$
The inequality “\(\leqslant \)” is obvious. Conversely, Proposition 3 yields \((y''\rightarrow y\rightarrow y)\rightarrow y''=(y''\rightarrow y\rightarrow y\rightarrow y)''=(y''\rightarrow y)''=y''\rightarrow y''=1\).
Now let \(x,y\in X\) be arbitrary. By Proposition 2, C(X) is a CKL-algebra. We set \(z:=x\wedge y\in C(X)\). With the abbreviation of Proposition 3, this gives
$$\begin{aligned} \begin{aligned} x''&= (x\rightarrow y)'=(x\rightarrow y\rightarrow x)\wedge (x\rightarrow y\rightarrow y)\\&= \bigl ((x\rightarrow y\rightarrow y)\rightarrow (x\rightarrow y\rightarrow x)\bigr )(x\rightarrow y\rightarrow y)\\&= \bigl (y\rightarrow (x\rightarrow y))\rightarrow (y\rightarrow x)\bigr )(x\rightarrow y\rightarrow y) \end{aligned} \end{aligned}$$
in S(X). Thus (K) yields
$$\begin{aligned} x''=(y\rightarrow x)(x\rightarrow y\rightarrow y). \end{aligned}$$
and
$$\begin{aligned} x''\rightarrow x= & {} (y\rightarrow x)\rightarrow (x\rightarrow y\rightarrow y\rightarrow x)\\= & {} (x\rightarrow y\rightarrow y)\rightarrow (y\rightarrow x\rightarrow x). \end{aligned}$$
Furthermore,
$$\begin{aligned} \begin{aligned} (x\rightarrow y)\rightarrow (x\rightarrow y\rightarrow y\rightarrow x)&= (x\rightarrow y\rightarrow y\rightarrow y)\\&\rightarrow (x\rightarrow y\rightarrow y\rightarrow x)\\&= \bigl (y\rightarrow (x\rightarrow y\rightarrow y)\bigr )\\&\rightarrow (y\rightarrow x)\\&= y\rightarrow x. \end{aligned} \end{aligned}$$
(21)
Thus (K) gives \(x\rightarrow y\rightarrow y\rightarrow x\leqslant (x\rightarrow y)\rightarrow (x\rightarrow y\rightarrow y\rightarrow x)=y\rightarrow x\). Hence
$$\begin{aligned} \begin{aligned}&x\rightarrow y\rightarrow y\rightarrow x\rightarrow x =\bigl ((x\rightarrow y\rightarrow y\rightarrow x)\\&\quad \rightarrow (y\rightarrow x)\bigr )\rightarrow (x\rightarrow y\rightarrow y\rightarrow x\rightarrow x)\\&\quad = \bigl ((y\rightarrow x)\rightarrow (x\rightarrow y\rightarrow y\rightarrow x)\bigr )\rightarrow (y\rightarrow x\rightarrow x)\\&\quad = \bigl ((x\rightarrow y\rightarrow y)\rightarrow (y\rightarrow x\rightarrow x)\bigr )\rightarrow (y\rightarrow x\rightarrow x)\\&\quad = (x''\rightarrow x)\rightarrow (y\rightarrow x\rightarrow x)=(y\rightarrow x)\\&\quad \rightarrow (x''\rightarrow x\rightarrow x)\\&\quad = (y\rightarrow x)\rightarrow (x\rightarrow z\rightarrow z\rightarrow x\rightarrow x)\\&\quad = (y\rightarrow x)\rightarrow (z{\rightarrow } x{\rightarrow } x\rightarrow z\rightarrow z)=(y\rightarrow x)\rightarrow x''\\&\quad = (y\rightarrow x\rightarrow x)'' \end{aligned} \end{aligned}$$
by Proposition 3. In S(X) we have
$$\begin{aligned} \begin{aligned} (y\rightarrow x\rightarrow x)'&= (y{\rightarrow } x{\rightarrow } x\rightarrow x)\wedge (y\rightarrow x\rightarrow x\rightarrow y)\\&= (y\rightarrow x)\wedge (y\rightarrow x\rightarrow x\rightarrow y)\\&= \bigl ((y\rightarrow x)\rightarrow (y{\rightarrow } x{\rightarrow } x{\rightarrow } y)\bigr )(y\rightarrow x)\\&= (x\rightarrow y)(y\rightarrow x) \end{aligned} \end{aligned}$$
by Eq. (21). Hence
$$\begin{aligned} (y\rightarrow x\rightarrow x)''= & {} (x\rightarrow y)(y\rightarrow x)\rightarrow z =(x\rightarrow y)\\&\rightarrow \bigl ((y\rightarrow x)\rightarrow z\bigr ). \end{aligned}$$
So (C) implies that the left-hand side of (J) is symmetric in x and y. \(\square \)

Notes

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Conflict of interest

The authors confirm that there is no conflict of interest.

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Copyright information

© Springer-Verlag Berlin Heidelberg 2017

Authors and Affiliations

  1. 1.Institute for Algebra and Number TheoryUniversity of StuttgartStuttgartGermany

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