Basic algebras and L-algebras

Abstract

In this paper, we study the relation between L-algebras and basic algebras. In particular, we construct a lattice-ordered effect algebra which improves an example of Chajda et al. (Algebra Univ 60(1), 63–90, 2009).

1 Introduction

Basic algebras, which generalize both MV-algebras and orthomodular lattices, were introduced in Chajda et al. (2009) and Chajda et al. (2007) as a common base for axiomatization of many-valued propositional logics as well as of the logic of quantum mechanics. The relationship between basic algebras, MV-algebras, orthomodular lattices and lattice-ordered effect algebras was considered in Botur (2010), Botur and Halaš (2008), Chajda (2012; 2015), Chajda et al. (2009). One can mention that every MV-algebra is a basic algebra whose induced lattice is distributive (Chajda 2015, P. 18, Lemma 5.2). The sufficient and necessary condition for an orthomodular lattice to be a basic algebra has been obtained in Chajda (2015, P. 17, Theorem 4.3). Relation between lattice-ordered effect algebras and basic algebras was treated in Botur and Halaš (2008), Chajda (2012) by considering their common lattice structure (a lattice with section antitone involutions).

L-algebras, which are related to algebraic logic and quantum structures, were introduced by Rump (2008). Many examples shown that L-algebras are very useful. Yang and Rump (2012), characterized pseudo-MV-algebras and Bosbach’s non-commutative bricks as L-algebras. Wu and Yang (2020) proved that orthomodular lattices form a special class of L-algebras in different ways. It was shown that every lattice-ordered effect algebra has an underlying L-algebra structure in Wu et al. (2019).

In the present paper, we study the relationship between basic algebras and L-algebras. We prove that a basic algebra which satisfies

$$\begin{aligned} (z\oplus \lnot x)\oplus \lnot (y\oplus \lnot x)=(z\oplus \lnot y)\oplus \lnot (x\oplus \lnot y) \end{aligned}$$

can be converted into an L-algebra (Theorem 1). Conversely, if an L-algebra with 0 and relation given by (10) such that it is an involutive bounded lattice can be organized into a basic algebra, it must be a lattice-ordered effect algebra (Theorem 2). Finally, we construct a lattice-ordered effect algebra which improves (Chajda et al. 2009, P. 80, Example 5.3).

2 Preliminaries

Note that basic algebras were introduced in Chajda (2007; 2009), but the axiomatic system was extended by one more axiom which is dependent on the following axioms as shown in Chajda and Kolšík (2009).

Definition 1

A basic algebra is an algebra ${\mathcal {B}}=(B;~\oplus ,~\lnot , 0)$ of type (2, 1, 0) satisfying the following identities:

$$\begin{aligned} (BA1)~&x \oplus 0 = x,\\ (BA2)~&\lnot \lnot x=x,\\ (BA3)~&\lnot (\lnot x\oplus y)\oplus y=\lnot (\lnot y\oplus x )\oplus x,\\ (BA4)~&\lnot (\lnot (\lnot (x\oplus y)\oplus y)\oplus z)\oplus (x\oplus z)=\lnot 0. \end{aligned}$$

For the sake of brevity, we denote by $1=:\lnot 0$.

Let ${\mathcal {B}}=(B,~\oplus ,~\lnot ,~0)$ be a basic algebra. The relation $\le$ defined by

$$\begin{aligned} x\le y~~\text{ if } \text{ and } \text{ only } \text{ if }~\lnot x\oplus y =1 \end{aligned}$$

(1)

is a partial ordering on B such that 0 and 1 are the least and the greatest element of B, respectively.

In what follows, we need the following properties of basic algebras (cf. Chajda 2015; Chajda et al. 2009):

$$\begin{aligned}&x\oplus 1=1=1\oplus x. \end{aligned}$$

(2)

$$\begin{aligned}&~0\oplus x~=~x~. \end{aligned}$$

(3)

$$\begin{aligned}&~\lnot x\oplus x~=~1. \end{aligned}$$

(4)

$$\begin{aligned}&x\le y~\Rightarrow ~\lnot x\ge \lnot y. \end{aligned}$$

(5)

$$\begin{aligned}&x\le y \Rightarrow x\oplus z\le y\oplus z. \end{aligned}$$

(6)

$$\begin{aligned}&y\le x\oplus y. \end{aligned}$$

(7)

Lemma 1

(Chajda 2015, P. 69, Prop. 3.6) For every basic algebra ${\mathcal {B}}=(B,~\oplus ,~\lnot , 0)$, the poset $(B,~\le )$ is a bounded lattice in which the supremum $x \vee y$ and the infimum $x \wedge y$ are given by $x\vee y= \lnot (\lnot x \oplus y) \oplus y$ and $x \wedge y = \lnot (\lnot x \vee \lnot y)$, respectively.

An involutive bounded lattice (IBL) (Chiara and Giuntini 2002, P. 191, Def. 12.1) is a structure $(L,~\le ,~',~0,~1)$, where $(L,~\le ,~0,~1)$ is a lattice with minimum 0 and maximum 1, $'$ is a unary operation on L such that the following conditions are satisfied:

$$\begin{aligned} \text {(Involutive~law)}~~~&x=x.'' \end{aligned}$$

(8)

$$\begin{aligned} \text {(Antitony)}~~~&\text{ if }~x\le y,~\text{ then }~y'\le x'. \end{aligned}$$

(9)

According to (BA2), (5) and Lemma 1, every basic algebra is an IBL.

Lemma 2

(Chajda 2015, P. 70, Lemma 3.8) The identity

$$\begin{aligned} \lnot (\lnot (x\oplus y)\oplus y)\oplus y=x\oplus y \end{aligned}$$

is true in all basic algebras.

Corollary 1

The identity

$$\begin{aligned} (x\wedge \lnot y)\oplus y=x\oplus y \end{aligned}$$

is true in all basic algebras.

Proof

By Lemmas 1 and 2, $x\oplus y=\lnot (\lnot (x\oplus y)\oplus y)\oplus y=\lnot (\lnot x\vee y)\oplus y=(x\wedge \lnot y)\oplus y$ is true in all basic algebras. $\square$

Definition 2

(Rump and Yang 2012, P. 122) An L-algebra is an algebra $(L,\rightarrow )$ of type (2, 0) satisfying

$$\begin{aligned} (L1)&x\rightarrow x=x\rightarrow 1=1,~1\rightarrow x=x \\ (L2)&(x\rightarrow y)\rightarrow (x\rightarrow z)=(y\rightarrow x)\rightarrow (y\rightarrow z) \\ (L3)&x\rightarrow y =y\rightarrow x =1~~\Rightarrow ~~x=y \end{aligned}$$

for all $x,~y,~z\in L$. 

There is a partial ordering by Rump (2008, P. 2332, Prop. 2)

$$\begin{aligned} x\le y~~\Leftrightarrow ~~x\rightarrow y=1 \end{aligned}$$

(10)

such that 1 is the greatest element of L. If L admits a smallest element 0, we speak of an L-algebra with 0.

Lemma 3

(Rump and Yang 2012, P. 123, Lemma 2.1) Let L be an L-algebra. Then, $x \le y$ implies that $z \rightarrow x \le z \rightarrow y$ for all $x,~y,~z\in L$.

In particular, if L is an L-algebra with 0 and satisfies (8) for every $x\in L$, then

$$\begin{aligned}&x\le x'\rightarrow y,~x'\le x\rightarrow y. \end{aligned}$$

(11)

3 L-algebras and basic algebras

In this section, we are interested in knowing the mutual relation between L-algebras and basic algebras. Assume that they have the same lattice structure. Firstly, we give three types of involutive bounded lattices which can be regarded as both L-algebras and basic algebras: MV-algebras, lattice-ordered effect algebras and orthomodular lattices.

Recall that an MV-algebra Chang (1958) is an algebra $A = (A,~\oplus ,~',~0)$ of type (2,  1,  0) where $(A,~\oplus ,~0)$ is a commutative monoid satisfying (8) and the following identities:

$$\begin{aligned}&x\oplus 0'=0',\\&(x'\oplus y)'\oplus y=(y'\oplus x)'\oplus x. \end{aligned}$$

MV-algebras are both basic algebras and L-algebras (Chajda et al. 2009; Wu et al. 2019).

An effect algebra (Foulis and Bennett 1994, P. 1333, Def. 2.1) is a system $(E,~+,~0, 1)$ consisting of a set E with two special elements $0,~1\in E$, called the zero and the unit, and with a partially defined binary operation $+$ satisfying the following conditions for all $p,~q,~r\in E$.

(E1) (Commutative law)   If $p+ q$ is defined, then $q + p$ is defined and $p + q = q + p.$

(E2) (Associative law)   If $p + q$ is defined and $(p + q ) + r$ is defined, then $q + r$ and $p + (q + r)$ are defined and $p + (q + r) = (p + q) + r$.

(E3) (Orthosupplement law)   For every $p \in E$, there exists a unique $q \in E$ such that $p + q$ is defined and $p + q = 1.$ The unique element q is written as $p'$ and called the orthosupplement of p.

(E4) (Zero-one law)    If $p + 1$ is defined, then $p=0.$

Let $(E,~+,~0,~1)$ be an effect algebra. Define a binary relation on E by

$$\begin{aligned}&a\le b~~\text{ if } \text{ for } \text{ some }~c\in E,~c+ a=b \end{aligned}$$

(12)

which is a partial ordering on E such that 0 and 1 are the smallest element and the greatest element of E, respectively. If the poset $(E,~\le )$ is a lattice, then E is called a lattice-ordered effect algebra. 

Lemmas 4 and 5 show that there is a mutual correspondence between lattice-ordered effect algebras, basic algebras and L-algebras.

Lemma 4

(Chajda 2012, P. 8, Thm. 12) Let ${\mathcal {E}}=(E,~+,~0,~1)$ be a lattice-ordered effect algebra. Define

$$\begin{aligned}&x\oplus y:=(x\wedge y')+y~~and~~\lnot x:=x'. \end{aligned}$$

(13)

Then, ${\mathcal {B}}(E)=(E,~\oplus ,~\lnot ,~0)$ is a basic algebra (whose lattice order coincides with the original one).

Define $x\rightarrow y:=(x\wedge y)+ x'$.

Lemma 5

(Wu et al. 2019, P106, Thm. 3.3) Every lattice-ordered effect algebra $(E,~+,~0,~1)$ gives rise to an L-algebra $(E,~\rightarrow )$ with negation such that $x'=x\rightarrow 0$ is exactly the orthosupplement of x in $(E,~+,~0,~1)$.

Let $(L,~+,~0,~1)$ be a lattice-ordered effect algebra. Define

$$\begin{aligned} x\oplus y:=(x\wedge y')+ y, \end{aligned}$$

and then, $(L,~\oplus ,~\lnot ,~0)$ is a basic algebra by Lemma 4. By Lemma 5, $(L,~\rightarrow ,~0,~1)$ is an L-algebra, where

$$\begin{aligned} x\rightarrow y:=(x\wedge y)+x'. \end{aligned}$$

Then, $x\oplus y=y'\rightarrow x$.

An orthomodular lattice (OML) Kalmbach (1983) is an algebra ${\mathcal {L}}=(L,~\vee ,~\wedge ,~',~0,~1)$ of type (2, 2, 1, 0, 0) satisfying (8), (9) and the following axioms: (i) $(L,~\vee ,~\wedge ,~0,~1)$ is a bounded lattice. (ii) $x\le y$ implies $y=x\vee (y\wedge x').$

In Chajda (2015), the author uses

$$\begin{aligned}&x\oplus y:=(x\wedge y')\vee y~~and~~\lnot x:= x' \end{aligned}$$

(14)

to convert an orthomodular lattice $(L,~\vee ,~\wedge ,~',~0,~1)$ into a basic algebra $(L,~\oplus ,~\lnot ,~0).$

Define

$$\begin{aligned} ~&x\rightarrow y:=x'\vee (x\wedge y), \end{aligned}$$

(15)

then every orthomodular lattice L gives rise to an L-algebra $(L,\rightarrow )$ in [16]. Then, $x\oplus y=y'\rightarrow x$.

Now, we will give a basic algebra which is also an L-algebra.

Example 1

Let ${\mathcal {B}}=(\{0,~a,~\lnot a,~1\},~\oplus ,~\lnot ,~0)$ be a basic algebra, where $\oplus$ is given in Table 1.

Table 1 $\oplus$ of Example 1

Define $x\rightarrow y:=y\oplus \lnot x$ and $x'=x\rightarrow 0:=\lnot x$; then, we have Table 2.

Table 2 $\rightarrow$ of Example 1

An easy computation shows that ${\mathcal {B}}$ is also an L-algebra.

Next, we will give a characterization of basic algebras to be L-algebras.

Theorem 1

Let $(B,~\oplus ,~\lnot ,~0)$ be a basic algebra which satisfies the following condition:

Then, $(B,~\rightarrow )$ is an L-algebra.

Proof

Define $x\rightarrow y:= y\oplus \lnot x.$

By (2), $x\rightarrow 1=1\oplus \lnot x=1$.$~1\rightarrow x=x\oplus \lnot 1=x\oplus ~0=x$. By (4), $x\rightarrow x=x\oplus \lnot x=1.$ This verifies (L1).

$(x\rightarrow y)\rightarrow (x\rightarrow z) =(x\rightarrow z)\oplus \lnot (x \rightarrow y)=(z\oplus \lnot x)\oplus \lnot (y\oplus \lnot x).$ Similarly, $(y\rightarrow x)\rightarrow (y\rightarrow z)=(z\oplus \lnot y)\oplus \lnot (x\oplus \lnot y).$ By (LB), we have verified (L2) in the definition of an L-algebra.

Assume that $x\rightarrow y=y\rightarrow x=1,$ then $y\oplus \lnot x=x\oplus \lnot y=1.$ Since $y\oplus \lnot x=1\Leftrightarrow \lnot y\le \lnot x\Leftrightarrow x\le y$ by (5) and (BA2), then $x\le y,~y\le x.$ Hence, $x=y.$ This verifies (L3).

Then, $(B,~\rightarrow )$ is an L-algebra. $\square$

There are many basic algebras which are not L-algebras with respect to the original involutive bounded lattice structure.

Example 2

Let us consider the ortholattice $O_6$ with the following Hasse diagram.

By Corollary 1 and the properties of basic algebras, it is routine to verify that $(O_6,~\oplus ,~\lnot ,~0)$ is a basic algebra, where $\lnot x=x'$ and $\oplus$ is given in Table 3.

Table 3 $\oplus$ of Example 2

Assume $O_6$ can be converted into an L-algebra with the operation $\rightarrow .$ By (L2),  $(b\rightarrow a)\rightarrow b'=(a\rightarrow b)\rightarrow a'=1\rightarrow a'=a'.$ Then, $b\rightarrow a=a'$, since $b'\le b\rightarrow a$, whence $a'\rightarrow b'=a'$. However, $a\le a'\rightarrow b'=a'$, which is a contradiction. Thus, $O_6$ is not an L-algebra.

Conversely, under what conditions can an L-algebra be regarded as a basic algebra? Since every basic algebra is an IBL, we are interested in the L-algebra L with 0 and relation given by (10) such that the L is an IBL. Define $x\oplus y:=y'\rightarrow x,$ and we have the following theorem:

Theorem 2

Let $(L,~\rightarrow )$ be an L-algebra with 0 and relation given by (10) such that L is an involutive bounded lattice, where $x'=x\rightarrow 0$. Define

$$\begin{aligned} x\oplus y:=y'\rightarrow x. \end{aligned}$$

If $(L,~\oplus ,~\lnot ,~0)$ is a basic algebra, then L must be a lattice-ordered effect algebra.

Proof

Since L is an involutive bounded lattice, then $x''=x$ and $x\le y\Rightarrow x'\ge y'$ for every $x,~y\in L.$ Define $x\oplus y:=y'\rightarrow x$, and then, $x\vee y=y'\rightarrow (y'\rightarrow x')'$ by Lemma 1. Assume $x\le y,$ then

$$\begin{aligned} \begin{aligned} y\rightarrow x=&(y\vee x)\rightarrow x\\ =\,&(x'\rightarrow (x'\rightarrow y')')\rightarrow x\\ =\,&(x'\rightarrow (x'\rightarrow y')')\rightarrow (x'\rightarrow 0)\quad by~(L2)\\ =\,&((x'\rightarrow y')'\rightarrow x')\rightarrow ((x'\rightarrow y')'\rightarrow 0)\\&by~(11)~and~(9)\\ =\,&1\rightarrow (x'\rightarrow y') \quad by~(L1)\\ =\,&x'\rightarrow y'. \end{aligned} \end{aligned}$$

Then by Theorem 3.9 in Wu et al. (2019), L is a lattice-ordered effect algebra. $\square$

By Rump (2008, P. 2346, Example 1), every partially ordered set with the greatest element 1 can be regarded as an L-algebra. We have already known that every basic algebra $(B,~\oplus ,~\lnot ,~0)$ is an IBL such that 1 is the greatest element of B, so it can be regarded as an L-algebra. But we are focused on the L-algebra with 0 and relation given by (10) such that it is an involutive bounded lattice, where $x'=x\rightarrow 0.$

In conclusion, we get an interesting relationship diagram as follows:

An involutive bounded lattice which is neither a basic algebra nor an L-algebra (relation given by (10) such that it is an involutive bounded lattice) is given in the following.

Example 3

Let us consider the involutive bounded lattice $G_6$.

Assume that $G_6$ can be converted into an L-algebra with 0 such that $x':=x\rightarrow 0$. By (11), $x'\le x\rightarrow y,~y\le y'\rightarrow x$, then $x\rightarrow y= x'$ or $y'(x\ge y,x\rightarrow y \ne 1)$ and the possible values of $y'\rightarrow x$ are $y,~x,~x',~y'.$

By (L2) and (L1),

$$\begin{aligned} (x\rightarrow y)\rightarrow x'=(y\rightarrow x)\rightarrow (y\rightarrow 0)=1\rightarrow y'=y' \end{aligned}$$

(16)

and

$$\begin{aligned} x'=1\rightarrow x'=(x\rightarrow y')\rightarrow x'=(y'\rightarrow x)\rightarrow y. \end{aligned}$$

(17)

If $x\rightarrow y=x',$ then $1=x'\rightarrow x'=y'$ by (16), a contradiction. Thus, $x\rightarrow y=y'$ which implies that $y'\rightarrow x'=y'$ by (16).

There are only four possible values of $y'\rightarrow x$: $y,~x,~x',~y'.$

• (i) If $y'\rightarrow x=y,$ then $y\rightarrow y=x'$ by (17). However, $y\rightarrow y=1.$ Hence, $y'\rightarrow x\ne y.$

• (ii) Assume $y'\rightarrow x=x$, then $x\rightarrow y=x'$ by (17), which contradicts $x\rightarrow y=y'.$

• (iii) If $y'\rightarrow x= x'$, then $x'\rightarrow y=x'$ by (17). Since $x\le x'\rightarrow y$, then $x\le x'\rightarrow y=x'$. However, x is uncomparable with $x'$, and  then, $y'\rightarrow x\ne x'.$

• (iv) Assume $y'\rightarrow x=y'$, then $y'\rightarrow y=x'$ by (17). Nevertheless, $x=1\rightarrow x=(x'\rightarrow y')\rightarrow x=(y'\rightarrow x')\rightarrow y=y'\rightarrow y=x',$ which is a contradiction.

The above shows that no matter how we define $\rightarrow$ on $G_6,$ it cannot be converted into an L-algebra (the induced partial ordering binary relation by (10) is an involutive bounded lattice).

We will verify that $G_6$ can also not be a basic algebra in the following.

Assume $G_6$ can be converted into a basic algebra with operation $\oplus$ such that $x'=\lnot x$. By Lemma 1,

$$\begin{aligned} x=x\vee y=\lnot (\lnot x\oplus y)\oplus y. \end{aligned}$$

(18)

Since $y\le \lnot x~\oplus y$, $y\le x$ and $y\oplus \lnot y=1,$ then the possible values of $\lnot x~\oplus y$ are $x,~\lnot x,~\lnot y.$

By Lemma 1, we can obtain

$$\begin{aligned} \lnot x=\lnot x\vee y=\lnot (x\oplus y)\oplus y \end{aligned}$$

(19)

and

$$\begin{aligned} \lnot y=\lnot y\vee y=\lnot (y\oplus y)\oplus y. \end{aligned}$$

(20)

Thus, we get the possible values of $x\oplus y$ and $y\oplus y$ which are also $x,~\lnot x,~\lnot y.$

We will divide into three cases to discuss the values of $\lnot x\oplus y.$

• (i) If $\lnot x\oplus y=\lnot x,$ then $x\oplus y=x$ by (18). Since $y\le x,$ then $y \oplus y \le x \oplus y=x$ by (6). Then, $y\oplus y=x.$ By (20), $\lnot x\oplus y=\lnot y\ne \lnot x$, a contradiction.

• (ii) If $\lnot x\oplus y=x$ and $x\oplus y=x$, then $\lnot x\oplus y=\lnot x$ by (19). This contradicts the assumption. If $x\oplus y=\lnot x,$ since $y\oplus y\le x\oplus y=\lnot x$, then $y\oplus y=\lnot x.$ Thus by (20), $x\oplus y=\lnot y\ne \lnot x$. So $x\oplus y=\lnot y$, which implies $y\oplus y =\lnot x.$ But $x=\lnot x\oplus y\ge y\oplus y=\lnot x$, which is impossible.

• (iii) If $\lnot x\oplus y=\lnot y,$ then $y\oplus y=x$ by (18). Suppose that $x\oplus y=x$, then $\lnot x\oplus y=\lnot x\ne \lnot y$ by (19). If $x\oplus y=\lnot y$, then $y\oplus y=\lnot x\ne x.$ So $x\oplus y=\lnot x.$ However, $\lnot x=x\oplus y\ge y\oplus y=x$, which is absurd.

None of the above cases is satisfied, which means $G_6$ can also not be considered as a basic algebra.

4 A lattice-ordered effect algebra with different basic algebra structures

In this section, we construct a lattice-ordered effect algebra with two different basic algebra structures and improve (Chajda et al. 2009, P. 80,Example 5.3) which stated as follows:

Fig. 1

Lattice of Example 5.3 in Chajda et al. (2009)

Let us consider the lattice from Fig. 1 with the antitone involution on the section [b,  1] defined by $b^b=1,~(\lnot b)^b=\lnot b,~(\lnot a)^b=\lnot a,~1^b=b$.

An easy inspection shows that the derived basic algebra ${\mathcal {A}}=(A,~\oplus ,~\lnot ,~0)$ is not a lattice-ordered effect algebra [because it does not fulfill (21)], where $A=\{0,~a,~b,~\lnot a,~\lnot b,~1\}$ and the addition $\oplus$ is given in Table 4:

$$\begin{aligned} x\le \lnot y~~and~~x\oplus y\le \lnot z \Rightarrow x\oplus (z\oplus y)=(x\oplus y)\oplus z.\nonumber \\ \end{aligned}$$

(21)

Table 4 $\oplus$ of Example 5.3 in Chajda et al. (2009)

It is easily seen that when $x=0,~y=b$ and $z=a,$$x\oplus (z\oplus y)=0\oplus (a\oplus b)=a\oplus b=\lnot a\ne \lnot b=b\oplus a=(0\oplus b)\oplus a=(x\oplus y)\oplus z$. Hence, A does not fulfill (21).

However, using the same $(A,~\oplus ,~\lnot ,~0)$ as in Fig. 1 and Table 4, we consider

Example 4

The basic algebra ${\mathcal {A}}=({A,~\oplus ,~\lnot ,~0})$ can be converted into a lattice-ordered effect algebra $(\{0,~a,~b,~a',~b', ~1\},~+,~',~0)$ whose operation is given in Table 5. If $x+y$ is undefined for $x,~y\in \{0,~a,~b,~a',~b',~1\}$, we denote it by “−.”

Table 5 $\oplus$ of Example 4

Remark 1

In Chajda et al. (2009) [P. 75, Prop. 4.5], lattice-ordered effect algebras can be viewed as basic algebras. We can obtain the derived basic algebra of the lattice-ordered effect algebra $(A,~+)$ from Example 4.

Define $x\oplus ^{*} y:=(x\wedge y')\oplus y$ and $\lnot x:=x'$. Then, $\mathcal {A^*}=(A^*,~\oplus ^{*},~\lnot ,~0)$ is a basic algebra with $\oplus ^{*}$ given in Table 6.

Table 6 $\oplus ^{*}$ of Remark 1

Hence, we obtain two different basic algebra structures whose operations are given in Tables 4 and 6 from the same lattice-ordered algebra from Example 4.