11footnotetext: Support of the research of both authors by ÖAD, project CZ
04/2017, and by IGA, project PřF 2018 012, as well as support of the second
author by the Austrian Science Fund (FWF), project I 1923-N25, is gratefully
acknowledged.

# Implication in weakly and dually weakly orthomodular lattices

Ivan Chajda and Helmut Länger

###### Abstract

Weakly orthomodular and dually weakly orthomodular lattices were introduced by
the authors in a recent paper. Similarly as for orthomodular lattices we try
to introduce an implication in these lattices which can be easily axiomatized
and which yields a nice lattice structure. As shown in the paper, this can be
realized in several different ways. Moreover, we reveal the connection of
weakly and dually weakly orthomodular lattices to residuated structures.
Moreover, we provide a characterization of these lattices by means of certain
generalized measures.

AMS Subject Classification: 06C15, 03G10, 03G12

Keywords: Weakly orthomodular lattice, dually weakly orthomodular lattice,
complementation, implication, Sasaki implication, residuated l-groupoid

Orthomodular lattices were introduced independently by K. Husimi ([13]) and G.
Birkhoff and J. von Neumann ([4]) as an algebraic semantics of the logic of
quantum mechanics. The reader can find a detailed theory of orthomodular
lattices in the monographs [3] and [14]. Implication in orthomodular lattices
was investigated in [7] – [9]. It was pointed out by the first author and S.
Bonzio in [5] that the orthomodular law may be satisfied also in complemented
lattices which are not orthomodular because their complementation need not be
an orthocomplementation. This fact was the starting point for the study [12]
by the authors where the concepts of a weakly orthomodular respectively dually
weakly orthomodular lattice were introduced and elaborated. It was shown that
these lattices have a number of features similar to those of orthomodular
lattices despite the fact that de Morgan laws do not hold there. The classes
of such lattices form varieties and it was shown that these varieties are
congruence permutable and congruence regular and, of course, congruence
distributive. It is a natural question if weakly orthomodular and/or dually
weakly orthomodular lattices can serve as an algebraic semantics for a
reasonable propositional logic. As usually, when a non-classical logic is
investigated, one is focused on the logical connective implication because
this connective is preferably used in logical deduction and derivations. For
orthomodular lattices, there exist several ways how to introduce implication.
The most popular is the so-called Sasaki implication but also other possible
modifications were treated by several authors. Due to the lack of de Morgan
laws for weakly and dually weakly orthomodular lattices, we have here more
possibilities how to define an implication. Hence, our problem is to choose
those which can be characterized within the theory of weakly and/or dually
weakly orthomodular lattices by simple identities and, moreover, have a
structure with a nice interpretation. This is the aim of the paper.

In the following all algebras considered are assumed to have non-empty
universe.

We start with basic concepts. An orthomodular lattice is an algebra
$\mathbf{L}=(L,\vee,\wedge,{}^{\prime},0,1)$ of type $(2,2,1,0,0)$ such that
$(L,\vee,\wedge,0,1)$ is a bounded lattice and ′ is an orthocomplementation,
i.e. it satisfies the identities $x\vee x^{\prime}\approx 1$, $x\wedge
x^{\prime}\approx 0$, $(x\vee y)^{\prime}\approx x^{\prime}\wedge y^{\prime}$
and $(x\wedge y)^{\prime}\approx x^{\prime}\vee y^{\prime}$ and the double
negation law $(x^{\prime})^{\prime}\approx x$, and moreover, it satisfies the
so-called orthomodular law, i.e.

$x\leq y\text{ implies }y=x\vee(y\wedge x^{\prime})$

or, equivalently,

$x\leq y\text{ implies }x=y\wedge(x\vee y^{\prime}).$

These conditions are equivalent to the following identities:

$(x\wedge y)\vee(x\wedge(x\wedge y)^{\prime})\approx x,$ (1)

respectively

$(x\vee y)\wedge(x\vee(x\vee y)^{\prime})\approx x$ (2)

which are used in the following definition.

######  Definition 1.

([12]) A lattice $\mathbf{L}=(L,\vee,\wedge,{}^{\prime})$ with a unary
operation is called weakly orthomodular if it satisfies identity (1) and
dually weakly orthomodular if it satisfies identity (2). The algebra
$\mathbf{L}$ is said to satisfy the double negation law if it satisfies the
identity $(x^{\prime})^{\prime}\approx x$.

It was shown by the authors in [12] that conditions (1) and (2) are
independent even for lattices with complementation. Of course, every
orthomodular lattice is both weakly and dually weakly orthomodular. In a
(dually) weakly orthomodular lattice the unary operation need neither be
antitone nor an involution, see e.g. [12] for examples. From Definition 1 it
is not clear if such a lattice is bounded and if the unary operation is a
complementation and interchanges $0$ and $1$. This question is answered by the
following lemma.

######  Lemma 2.

([12]) Every weakly orthomodular lattice
$\mathbf{L}=(L,\vee,\wedge,{}^{\prime})$ has a greatest element $1$ and
satisfies the identity $x\vee x^{\prime}\approx 1$. If $\mathbf{L}$ has,
additionally, a smallest element $0$ then $0^{\prime}\approx 1$. Every dually
weakly orthomodular lattice $\mathbf{L}=(L,\vee,\wedge,{}^{\prime})$ has a
smallest element $0$ and satisfies the identity $x\wedge x^{\prime}\approx 0$.
If $\mathbf{L}$ has, additionally, a greatest element $1$ then
$1^{\prime}\approx 0$.

As shown in Lemma 2, if a lattice $\mathbf{L}=(L,\vee,\wedge,^{\prime})$ with
a unary operation is both weakly and dually weakly orthomodular then
$\mathbf{L}$ is bounded and ′ a complementation on $\mathbf{L}$. However, ′
need neither be antitone nor an involution and hence de Morgan laws need not
hold in $\mathbf{L}$. Examples of both weakly and dually weakly orthomodular
lattices whose unary operation is neither antitone nor an involution can be
found in [12].

In Boolean algebras $\mathbf{L}=(L,\vee,\wedge,^{\prime},0,1)$ the implication
$x\rightarrow y$ is introduced as the term operation $x^{\prime}\vee y$. This
is the so-called “classical implication”. In orthomodular lattices which are
not Boolean algebras this kind of implication does not have the properties
expected for implication and hence other versions of implication are studied,
in particular the so-called Sasaki implication defined by

$x\rightarrow y:=(x^{\prime}\wedge y^{\prime})\vee y.$

(In an orthomodular lattice the Sasaki projection $p_{b}(a)$ of $a$ onto $b$
is defined by $(a\vee b^{\prime})\wedge b$. Hence the Sasaki implication
$x\rightarrow y$ in orthomodular lattices is nothing else than
$(p_{y^{\prime}}(x))^{\prime}$.) This Sasaki implication was investigated by
J. C. Abbott ([2]). Unfortunately, we cannot use it here in all the cases
because of lack of de Morgan laws. First of all, we consider the implication
defined by $x\rightarrow y:=(x\vee y)^{\prime}\vee y$.

We call a binary operation $\rightarrow$ on a lattice
$\mathbf{L}=(L,\vee,\wedge)$ a D-implication on $\mathbf{L}$ if it satisfies
the following identities:

$\displaystyle(x\vee y)\rightarrow x$ $\displaystyle\approx y\rightarrow x,$
(3) $\displaystyle x\vee y\vee((x\vee z)\rightarrow y)$ $\displaystyle\approx
z\rightarrow(x\vee y),$ (4) $\displaystyle(x\vee y)\wedge(x\rightarrow y)$
$\displaystyle\approx y.$ (5)

(As we will see, D-implications have something to do with dually weak
orthomodularity.)

######  Theorem 3.

Let $\mathbf{L}=(L,\vee,\wedge,0)$ be a lattice with $0$. Then the following
hold:

  1. (i)

If $\rightarrow$ is a D-implication on $\mathbf{L}$ and
$x^{\prime}:=x\rightarrow 0$ for all $x\in L$ then
$(L,\vee,\wedge,{}^{\prime})$ is a dually weakly orthomodular lattice.

  2. (ii)

If $(L,\vee,\wedge,{}^{\prime})$ is a dually weakly orthomodular lattice and
$x\rightarrow y:=(x\vee y)^{\prime}\vee y$ for all $x,y\in L$ then
$\rightarrow$ is a D-implication on $\mathbf{L}$.

  3. (iii)

The correspondence described in (i) and (ii) is one-to-one. Thus a binary
operation $\rightarrow$ on $L$ is a D-implication on $\mathbf{L}$ if and only
if there exists a dually weakly orthomodular lattice
$(L,\vee,\wedge,{}^{\prime})$ such that $x\rightarrow y=(x\vee y)^{\prime}\vee
y$ for all $x,y\in L$.

###### Proof.

  1. (i)

Let $\rightarrow$ be a D-implication on $\mathbf{L}$ and
$x^{\prime}:=x\rightarrow 0$ for all $x\in L$. If $a,b\in L$ and $a\leq b$
then by applying (4) and (5) we compute

$\displaystyle b\wedge(a\vee b^{\prime})$
$\displaystyle=b\wedge(a\vee(b\rightarrow 0))=b\wedge(a\vee 0\vee((a\vee
b)\rightarrow 0))=b\wedge(b\rightarrow(a\vee 0))=$ $\displaystyle=(b\vee
a)\wedge(b\rightarrow a)=a.$

Thus $(L,\vee,\wedge,{}^{\prime})$ is a dually weakly orthomodular lattice.

  2. (ii)

Let $(L,\vee,\wedge,{}^{\prime})$ be a dually weakly orthomodular lattice and
$x\rightarrow y:=(x\vee y)^{\prime}\vee y$ for all $x,y\in L$. Then

    1. (3)

$(x\vee y)\rightarrow x\approx(x\vee y\vee x)^{\prime}\vee x\approx(y\vee
x)^{\prime}\vee x\approx y\rightarrow x$,

    2. (4)

$x\vee y\vee((x\vee z)\rightarrow y)\approx x\vee y\vee(x\vee z\vee
y)^{\prime}\vee y\approx(z\vee x\vee y)^{\prime}\vee x\vee y\approx
z\rightarrow(x\vee y)$,

    3. (5)

$(x\vee y)\wedge(x\rightarrow y)\approx(x\vee y)\wedge((x\vee y)^{\prime}\vee
y)\approx(x\vee y)\wedge(y\vee(x\vee y)^{\prime})\approx y$.

and hence $\rightarrow$ is a D-implication on $\mathbf{L}$.

  3. (iii)

If $\rightarrow$ is a D-implication on $\mathbf{L}$ and
$x^{\prime}:=x\rightarrow 0$ and $x\Rightarrow y:=(x\vee y)^{\prime}\vee y$
for all $x,y\in L$ then

$x\Rightarrow y\approx(x\vee y)^{\prime}\vee y\approx((x\vee y)\rightarrow
0)\vee y\approx y\vee 0\vee((y\vee x)\rightarrow 0)\approx x\rightarrow(y\vee
0)\approx x\rightarrow y.$

If, conversely, $(L,\vee,\wedge,{}^{\prime})$ is a dually weakly orthomodular
lattice and $x\rightarrow y:=(x\vee y)^{\prime}\vee y$ and
$\bar{x}:=x\rightarrow 0$ for every $x,y\in L$ then

$\bar{x}\approx x\rightarrow 0\approx(x\vee 0)^{\prime}\vee 0\approx
x^{\prime}\vee 0\approx x^{\prime}.$

∎

######  Remark 4.

In Boolean algebras $(x\vee y)^{\prime}\vee y$ coincides with the classical
implication since

$(x\vee y)^{\prime}\vee y\approx(x^{\prime}\wedge y^{\prime})\vee
y\approx(x^{\prime}\vee y)\wedge(y^{\prime}\vee y)\approx(x^{\prime}\vee
y)\wedge 1\approx x^{\prime}\vee y.$

The implication introduced before Theorem 3 has a simple axiomatization and
hence we are going to use it in our next investigations. However, we can also
introduce another concept which enables us to assign every such implication a
nice structure similarly as it was done for orthomodular lattices by Abbott
([2]), for Boolean algebras ([1]) or for MV-algebras ([6]).

######  Definition 5.

Let $\mathbf{L}=(L,\vee,\wedge)$ be a lattice. A family of compatible dually
weakly orthomodular sublattices of $\mathbf{L}$ is a family
$(([x),\vee,\wedge,{}^{x});x\in L)$ such that for every $x\in L$,
$([x),\vee,\wedge,{}^{x})$ is a dually weakly orthomodular sublattice of
$\mathbf{L}$ and the compatibility condition $z^{x}\vee y=z^{y}$ for all
$x,y,z\in L$ with $x\leq y\leq z$ holds. Here and in the following $[x)$
denotes the set $\\{y\in L\mid x\leq y\\}$.

We are going to show that to every D-implication there can be assigned a
family of compatible dually weakly orthomodular sublattices and that every
D-implication can be constructed just by such a family.

######  Theorem 6.

  1. (i)

Let $\mathbf{L}=(L,\vee,\wedge)$ be a lattice and $\rightarrow$ a
D-implication on $\mathbf{L}$ and put $x^{y}:=x\rightarrow y$ for all $x,y\in
L$. Then $(([x),\vee,\wedge,{}^{x});x\in L)$ is a family of compatible dually
weakly orthomodular sublattices of $\mathbf{L}$.

  2. (ii)

Let $\mathbf{L}=(L,\vee,\wedge)$ be a lattice and
$(([x),\vee,\wedge,{}^{x});x\in L)$ a family of compatible dually weakly
orthomodular sublattices of $\mathbf{L}$ and put $x\rightarrow y:=(x\vee
y)^{y}$ for all $x,y\in L$. Then $\rightarrow$ is a D-implication on
$\mathbf{L}$.

  3. (iii)

The correspondence between D-implications on $\mathbf{L}$ and families of
compatible dually weakly orthomodular sublattices of $\mathbf{L}$ described in
(i) and (ii) is one-to-one.

###### Proof.

Let $a,b,c\in L$.

  1. (i)

At first we show how a D-implication induces a dually weakly orthomodular
lattice $([a),\vee,\wedge,{}^{a})$. Now, if $a\leq b$ then by applying
identities (3) – (5) we obtain

$a=(b\vee a)\wedge(b\rightarrow a)\leq b\rightarrow a=b^{a}.$

Moreover, if $a\leq b\leq c$ then

$\displaystyle c\wedge(b\vee c^{a})$ $\displaystyle=c\wedge(b\vee(c\rightarrow
a))=c\wedge(b\vee a\vee((b\vee c)\rightarrow a))=c\wedge(c\rightarrow(b\vee
a))=$ $\displaystyle=(c\vee b)\wedge(c\rightarrow b)=b$

and hence $([a),\vee,\wedge,{}^{a})$ is a dually weakly orthomodular
sublattice of $\mathbf{L}$. In order to check the compatibility condition we
compute

$c^{a}\vee b=(c\rightarrow a)\vee b=b\vee a\vee((b\vee c)\rightarrow
a)=c\rightarrow(b\vee a)=c\rightarrow b=c^{b}$

which proves the compatibility condition. Thus $(([x),\vee,\wedge,{}^{x});x\in
L)$ is a family of compatible dually weakly orthomodular sublattices of
$\mathbf{L}$.

  2. (ii)

For the binary operation $\rightarrow$ on $L$ defined by $x\rightarrow
y:=(x\vee y)^{y}$ for all $x,y\in L$ we compute

    1. (3)

$(x\vee y)\rightarrow x\approx(x\vee y\vee x)^{x}\approx(y\vee x)^{x}\approx
y\rightarrow x$,

    2. (4)

$x\vee y\vee((x\vee z)\rightarrow y)\approx x\vee y\vee(x\vee z\vee
y)^{y}\approx(x\vee y\vee z)^{y}\vee(x\vee y)\approx(x\vee y\vee z)^{x\vee
y}\approx(z\vee x\vee y)^{x\vee y}=z\rightarrow(x\vee y)$,

    3. (5)

$(x\vee y)\wedge(x\rightarrow y)\approx(x\vee y)\wedge(x\vee y)^{y}\approx y$

showing that $\rightarrow$ is a D-implication on $\mathbf{L}$.

  3. (iii)

If $\rightarrow$ is a D-implication on $\mathbf{L}$,
$([x),\vee,\wedge,{}^{x});x\in L)$ denotes the corresponding family of
compatible dually weakly orthomodular sublattices of $\mathbf{L}$ and
$\Rightarrow$ denotes the D-implication corresponding to this family then
$\Rightarrow$ coincides with $\rightarrow$ since

$x\Rightarrow y\approx(x\vee y)^{y}\approx(x\vee y)\rightarrow y\approx(y\vee
x)\rightarrow y\approx x\rightarrow y.$

Conversely, if $(([x),\vee,\wedge,{}^{x});x\in L)$ is a family of compatible
dually weakly orthomodular sublattices of $\mathbf{L}$, $\rightarrow$ denotes
the corresponding D-implication and $(([x),\vee,$ $\wedge,{}_{x});x\in L)$
denotes the family of compatible dually weakly orthomodular sublattices of
$\mathbf{L}$ corresponding to this D-implication then the two families
coincide since

$b_{a}=b\rightarrow a=(b\vee a)^{a}=b^{a}.$

∎

Now, we turn to weakly orthomodular lattices. At first we investigate the
Sasaki implication.

######  Theorem 7.

Let $\mathbf{L}=(L,\vee,\wedge,\rightarrow,{}^{\prime},0,1)$ be an algebra of
type $(2,2,2,1,0,0)$ such that $(L,\vee,\wedge,0,1)$ is a bounded lattice and
the identities

$\displaystyle x\rightarrow y$ $\displaystyle\approx(x^{\prime}\wedge
y^{\prime})\vee y,$ $\displaystyle(x^{\prime})^{\prime}$ $\displaystyle\approx
x$

are satisfied. Then $(L,\vee,\wedge,{}^{\prime})$ is a weakly orthomodular
lattice if and only if $\mathbf{L}$ satisfies the identities

$\displaystyle 0^{\prime}$ $\displaystyle\approx 1,$ (6)
$\displaystyle(x\rightarrow 0)\rightarrow(x\wedge y)$ $\displaystyle\approx
x.$ (7)

###### Proof.

First assume $(L,\vee,\wedge,{}^{\prime})$ to be a weakly orthomodular
lattice.

  1. (6)

This identity holds according to Lemma 2.

  2. (7)

We have

$\displaystyle(x\rightarrow 0)\rightarrow(x\wedge y)$
$\displaystyle\approx(((x^{\prime}\wedge 0^{\prime})\vee
0)^{\prime}\wedge(x\wedge y)^{\prime})\vee(x\wedge y)\approx$
$\displaystyle\approx(((x^{\prime}\wedge 1)\vee 0)^{\prime}\wedge(x\wedge
y)^{\prime})\vee(x\wedge y)\approx$ $\displaystyle\approx((x^{\prime}\vee
0)^{\prime}\wedge(x\wedge y)^{\prime})\vee(x\wedge
y)\approx((x^{\prime})^{\prime}\wedge(x\wedge y)^{\prime})\vee(x\wedge
y)\approx$ $\displaystyle\approx(x\wedge(x\wedge y)^{\prime})\vee(x\wedge
y)\approx x.$

Conversely, assume $\mathbf{L}$ to satisfy identities (6) and (7). Let $a,b\in
L$ with $a\leq b$. Then

$\displaystyle a\vee(b\wedge a^{\prime})$
$\displaystyle=((b^{\prime})^{\prime}\wedge a^{\prime})\vee a=((b^{\prime}\vee
0)^{\prime}\wedge a^{\prime})\vee a=(((b^{\prime}\wedge 1)\vee
0)^{\prime}\wedge a^{\prime})\vee a=$ $\displaystyle=(((b^{\prime}\wedge
0^{\prime})\vee 0)^{\prime}\wedge a^{\prime})\vee a=(b\rightarrow
0)\rightarrow a=(b\rightarrow 0)\rightarrow(b\wedge a)=b.$

Thus $(L,\vee,\wedge,{}^{\prime})$ is a weakly orthomodular lattice. ∎

Let us note that in a Boolean algebra also the Sasaki implication
$x\rightarrow y=(x^{\prime}\wedge y^{\prime})\vee y$ coincides with the
classical implication $x^{\prime}\vee y$.

For weakly orthomodular lattices we can define implication in a different way.
The reason is that we would like to obtain a connective with nice properties
and, moreover, it can be considered as a dual version of a D-implication in
case the lattice is orthomodular.

We call a binary operation $\rightarrow$ on a lattice
$\mathbf{L}=(L,\vee,\wedge,0)$ with $0$ a W-implication on $\mathbf{L}$ if it
satisfies the following identities:

$\displaystyle(x\rightarrow 0)\rightarrow 0$ $\displaystyle\approx x,$ (8)
$\displaystyle((x\wedge y)\rightarrow 0)\rightarrow x$ $\displaystyle\approx
x,$ (9) $\displaystyle(x\wedge y)\vee(x\rightarrow 0)$ $\displaystyle\approx
x\rightarrow y.$ (10)

(As we will see, W-implications have something to do with weak
orthomodularity.)

######  Theorem 8.

Let $\mathbf{L}=(L,\vee,\wedge,0)$ be a lattice with $0$. Then the following
hold:

  1. (i)

If $\rightarrow$ is a W-implication on $\mathbf{L}$ and
$x^{\prime}:=x\rightarrow 0$ for all $x\in L$ then
$(L,\vee,\wedge,{}^{\prime})$ is a weakly orthomodular lattice satisfying the
double negation law.

  2. (ii)

If $(L,\vee,\wedge,{}^{\prime})$ is a weakly orthomodular lattice satisfying
the double negation law and $x\rightarrow y:=x^{\prime}\vee(x\wedge y)$ for
all $x,y\in L$ then $\rightarrow$ is a W-implication on $\mathbf{L}$.

  3. (iii)

The correspondence described in (i) and (ii) is one-to-one. Thus a binary
operation $\rightarrow$ on $L$ is a W-implication on $\mathbf{L}$ if and only
if there exists a weakly orthomodular lattice $(L,\vee,\wedge,{}^{\prime})$
satisfying the double negation law such that $x\rightarrow
y=x^{\prime}\vee(x\wedge y)$ for all $x,y\in L$.

###### Proof.

  1. (i)

Let $\rightarrow$ be a W-implication on $\mathbf{L}$ and
$x^{\prime}:=x\rightarrow 0$ for all $x\in L$. If $a,b\in L$ and $a\leq b$
then by applying (8) – (10) we compute

$a\vee(b\wedge a^{\prime})=((a\rightarrow 0)\wedge b)\vee((a\rightarrow
0)\rightarrow 0)=(a\rightarrow 0)\rightarrow b=((b\wedge a)\rightarrow
0)\rightarrow b=b.$

Thus $(L,\vee,\wedge,{}^{\prime})$ is a weakly orthomodular lattice. According
to (8) it satisfies the double negation law.

  2. (ii)

Let $(L,\vee,\wedge,{}^{\prime})$ be a weakly orthomodular lattice satisfying
the double negation law and $x\rightarrow y:=x^{\prime}\vee(x\wedge y)$ for
all $x,y\in L$. Then $x\rightarrow 0\approx x^{\prime}\vee(x\wedge 0)\approx
x^{\prime}\vee 0\approx x^{\prime}$. Thus

    1. (8)

$(x\rightarrow 0)\rightarrow 0\approx(x^{\prime})^{\prime}\approx x$,

    2. (9)

$((x\wedge y)\rightarrow 0)\rightarrow x\approx(x\wedge y)^{\prime}\rightarrow
x\approx((x\wedge y)^{\prime})^{\prime}\vee((x\wedge y)^{\prime}\wedge
x)\approx(x\wedge y)\vee(x\wedge(x\wedge y)^{\prime})\approx x$,

    3. (10)

$(x\wedge y)\vee(x\rightarrow 0)\approx(x\wedge y)\vee x^{\prime}\approx
x^{\prime}\vee(x\wedge y)\approx x\rightarrow y$

and hence $\rightarrow$ is a W-implication on $\mathbf{L}$.

  3. (iii)

If $\rightarrow$ is a W-implication on $\mathbf{L}$ and
$x^{\prime}:=x\rightarrow 0$ and $x\Rightarrow y:=x^{\prime}\vee(x\wedge y)$
for all $x,y\in L$ then

$x\Rightarrow y\approx x^{\prime}\vee(x\wedge y)\approx(x\wedge
y)\vee(x\rightarrow 0)\approx x\rightarrow y.$

If, conversely, $(L,\vee,\wedge,{}^{\prime})$ is a weakly orthomodular lattice
satisfying the double negation law and $x\rightarrow y:=x^{\prime}\vee(x\wedge
y)$ and $\bar{x}:=x\rightarrow 0$ for every $x,y\in L$ then

$\bar{x}\approx x\rightarrow 0\approx x^{\prime}\vee(x\wedge 0)\approx
x^{\prime}\vee 0\approx x^{\prime}.$

∎

######  Remark 9.

In an orthomodular lattice $x^{\prime}\vee(x\wedge
y)=(p_{x}(y^{\prime}))^{\prime}$. Moreover, in Boolean algebras
$x^{\prime}\vee(x\wedge y)$ coincides with the classical implication since

$x^{\prime}\vee(x\wedge y)\approx(x^{\prime}\vee x)\wedge(x^{\prime}\vee
y)\approx 1\wedge(x^{\prime}\vee y)\approx x^{\prime}\vee y.$

Besides the previous results, we can list some elementary but important
properties of the implications mentioned above.

######  Proposition 10.

Let $\mathbf{L}=(L,\vee,\wedge,{}^{\prime})$ be a dually weakly orthomodular
lattice and $x\rightarrow y=(x\vee y)^{y}$. Then

  1. (i)

$x\rightarrow 0\approx x^{\prime}$,

  2. (ii)

if $\mathbf{L}$ has a $1$ then $x\rightarrow 1\approx 1$,

  3. (iii)

if $\mathbf{L}$ has a $1$ then $1\rightarrow x\approx x$,

  4. (iv)

if $\mathbf{L}$ has a $1$ then $x\rightarrow y=1$ implies $x\leq y$.

###### Proof.

  1. (i)

$x\rightarrow 0\approx(x\vee 0)^{\prime}\vee 0\approx x^{\prime}$,

  2. (ii)

if $\mathbf{L}$ has a $1$ then $x\rightarrow 1\approx(x\vee 1)^{\prime}\vee
1\approx 1$,

  3. (iii)

if $\mathbf{L}$ has a $1$ then $1\rightarrow x\approx(1\vee x)^{\prime}\vee
x\approx 1^{\prime}\vee x\approx 0\vee x\approx x$,

  4. (iv)

if $\mathbf{L}$ has a $1$ then $x\rightarrow y=1$ implies $x\leq x\vee
y=(x\vee y)\wedge 1=(x\vee y)\wedge(x\rightarrow y)=(x\vee y)\wedge((x\vee
y)^{\prime}\vee y)=y$.

∎

Similarly, we can list several interesting but elementary properties for the
W-implication in weakly orthomodular lattices. According to Theorem 8, if
$\mathbf{L}=(L,\vee,\wedge,{}^{\prime})$ is a weakly orthomodular lattice then
a W-implication can be expressed as

$x\rightarrow y:=x^{\prime}\vee(x\wedge y)$

for all $x,y\in L$.

We can prove

######  Proposition 11.

Let $\mathbf{L}=(L,\vee,\wedge,{}^{\prime})$ be a weakly orthomodular lattice
and define $x\rightarrow y:=x^{\prime}\vee(x\wedge y)$ for all $x,y\in L$.
Then

  1. (i)

$x\rightarrow x\approx 1$,

  2. (ii)

$x\rightarrow 1\approx 1$,

  3. (iii)

$x\vee(x\rightarrow y)\approx 1$,

  4. (iv)

$x\rightarrow(x\wedge y)\approx x\rightarrow y$,

  5. (v)

$x\leq y$ implies $x\rightarrow y=1$,

  6. (vi)

if $\mathbf{L}$ has a $0$ then $x\rightarrow 0\approx x^{\prime}$.

###### Proof.

  1. (i)

$x\rightarrow x\approx x^{\prime}\vee(x\wedge x)\approx x^{\prime}\vee
x\approx 1$,

  2. (ii)

$x\rightarrow 1\approx x^{\prime}\vee(x\wedge 1)\approx x^{\prime}\vee
x\approx 1$,

  3. (iii)

$x\vee(x\rightarrow y)\approx x\vee x^{\prime}\vee(x\wedge y)\approx 1$,

  4. (iv)

$x\rightarrow(x\wedge y)\approx x^{\prime}\vee(x\wedge x\wedge y)\approx
x^{\prime}\vee(x\wedge y)\approx x\rightarrow y$,

  5. (v)

$x\leq y$ implies $x\rightarrow y=x^{\prime}\vee(x\wedge y)=x^{\prime}\vee
x=1$,

  6. (vi)

if $\mathbf{L}$ has a $0$ then $x\rightarrow 0\approx x^{\prime}\vee(x\wedge
0)\approx x^{\prime}\vee 0\approx x^{\prime}$.

∎

Now, we want to investigate the question how to characterize weakly
orthomodular or dually weakly orthomodular lattices via the systems of their
principal filters or principal ideals, respectively, which are determined by
special mappings of the lattice into the interval $[0,1]$ of the real numbers.
Hence, such mappings can be considered as generalized measures.

######  Definition 12.

For a lattice $\mathbf{L}=(L,\vee,\wedge,{}^{\prime})$ with a unary operation
we define

$\displaystyle S_{1}(\mathbf{L}):=\\{s\colon L\rightarrow[0,1]\mid
s(x\vee(y\wedge x^{\prime}))=s(y)\text{ for all }x,y\in L\text{ with }x\leq
y\\},$ $\displaystyle S_{2}(\mathbf{L}):=\\{s\colon L\rightarrow[0,1]\mid
s(y\wedge(x\vee y^{\prime}))=s(x)\text{ for all }x,y\in L\text{ with }x\leq
y\\}.$

Let us mention that the sets $S_{1}(\mathbf{L})$ and $S_{2}(\mathbf{L})$ are
non-empty since they contain every constant function from $L$ to $[0,1]$, in
particular the constant functions with value $0$ respectively $1$.

######  Theorem 13.

Let $\mathbf{L}=(L,\vee,\wedge,{}^{\prime})$ be a lattice with a unary
operation. Then the following hold:

  1. (i)

$\mathbf{L}$ is weakly orthomodular if and only if for every $x\in L$ which is
not the smallest element of $\mathbf{L}$ there exists some $s\in
S_{1}(\mathbf{L})$ satisfying $s^{-1}(\\{1\\})=[x)$.

  2. (ii)

$\mathbf{L}$ is dually weakly orthomodular if and only if for every $x\in L$
which is not the greatest element of $\mathbf{L}$ there exists some $s\in
S_{2}(\mathbf{L})$ satisfying $s^{-1}(\\{0\\})=(x]$.

###### Proof.

Let $a,b\in L$ and assume $a\leq b$.

  1. (i)

If $\mathbf{L}$ is weakly orthomodular then for every $x\in L$ every $s\colon
L\rightarrow[0,1]$ satisfying $s(y)=1$ if $y\geq x$ and $s(y)<1$ otherwise is
an element of $S_{1}(\mathbf{L})$ satisfying $s^{-1}(\\{1\\})=[x)$.
Conversely, assume that for every $x\in L$ which is not the smallest element
of $\mathbf{L}$ there exists some $s\in S_{1}(\mathbf{L})$ satisfying
$s^{-1}(\\{1\\})=[x)$. If $b$ is the smallest element of $\mathbf{L}$ then
$a\vee(b\wedge a^{\prime})=b$ since $a\vee(b\wedge a^{\prime})\leq b$. Now
assume $b$ not to be the smallest element of $\mathbf{L}$. Then there exists
some $s\in S_{1}(\mathbf{L})$ with $s^{-1}(\\{1\\})=[b)$. Now $s(a\vee(b\wedge
a^{\prime}))=s(b)=1$ and hence $a\vee(b\wedge a^{\prime})\geq b$. Together
with $a\vee(b\wedge a^{\prime})\leq b$ we obtain $a\vee(b\wedge
a^{\prime})=b$.

  2. (ii)

If $\mathbf{L}$ is dually weakly orthomodular then for every $x\in L$ every
$s\colon L\rightarrow[0,1]$ satisfying $s(y)=0$ if $y\leq x$ and $s(y)>0$
otherwise is an element of $S_{2}(\mathbf{L})$ satisfying
$s^{-1}(\\{0\\})=(x]$. Conversely, assume that for every $x\in L$ which is not
the greatest element of $\mathbf{L}$ there exists some $s\in
S_{2}(\mathbf{L})$ satisfying $s^{-1}(\\{0\\})=(x]$. If $a$ is the greatest
element of $\mathbf{L}$ then $b\wedge(a\vee b^{\prime})=a$ since
$b\wedge(a\vee b^{\prime})\geq a$. Now assume $a$ not to be the greatest
element of $\mathbf{L}$. Then there exists some $s\in S_{2}(\mathbf{L})$ with
$s^{-1}(\\{0\\})=(a]$. Now $s(b\wedge(a\vee b^{\prime}))=s(a)=0$ and hence
$b\wedge(a\vee b^{\prime})\leq a$. Together with $b\wedge(a\vee
b^{\prime})\geq a$ we obtain $b\wedge(a\vee b^{\prime})=a$.

∎

One can ask why the function $s\colon L\rightarrow[0,1]$ was introduced in
this way. The right answer is that it was mentioned as a generalized measure
on a lattice with complementation. The problem is that we cannot assume de
Morgan laws in our case. Thus relatively strong assumptions are necessary as
stated in Definition 12. However, we are able to get a more detailed way how
to introduce such functions.

For a lattice $(L,\vee,\wedge,{}^{\prime})$ with a unary operation ′ and a
mapping $s\colon L\rightarrow[0,1]$ we define the following conditions:

  1. (i)

$s(1)=1$,

  2. (ii)

if $x,y\in L$ and $x\leq y$ then $s(x\vee y^{\prime})=s(x)+s(y^{\prime})$,

  3. (iii)

if $x,y\in L$ and $x\leq y$ then $s((x\vee y^{\prime})\vee(x^{\prime}\wedge
y))=1$,

  4. (iv)

if $x,y\in L$ and $x\leq y$ then $s(x\vee(y\wedge(x\vee
y^{\prime}))^{\prime})=1$.

It should be mentioned that conditions (iii) and (iv) are not so strong as it
looks at the first glance. Namely, if $(L,\vee,\wedge,{}^{\prime})$ satisfies
de Morgan laws then

$\displaystyle(x\vee y^{\prime})\vee(x^{\prime}\wedge y)$
$\displaystyle\approx(x^{\prime}\wedge y)^{\prime}\vee(x^{\prime}\wedge
y)\approx 1,$ $\displaystyle x\vee(y\wedge(x\vee y^{\prime}))^{\prime}$
$\displaystyle\approx x\vee(y^{\prime}\vee(x\vee
y^{\prime})^{\prime})\approx(x\vee y^{\prime})\vee(x\vee
y^{\prime})^{\prime}\approx 1.$

Thus (iii) and (iv) follow directly by the simple condition (i). In other
words, this generalized measure $s$ does not distinguish certain terms that
coincide whenever de Morgan laws hold.

Now we can state

######  Proposition 14.

Let $\mathbf{L}=(L,\vee,\wedge,{}^{\prime})$ be a lattice with a
complementation and $s\colon L\rightarrow[0,1]$ satisfy (i) and (ii). Then
$s\in S_{1}(\mathbf{L})$ if $s$ satisfies (iii) and $s\in S_{2}(\mathbf{L})$
if $s$ satisfies (iv).

###### Proof.

Let $a,b\in L$. Since $a\leq a$ we have

$s(a)+s(a^{\prime})=s(a\vee a^{\prime})=s(1)=1$

according to (i) and (ii) whence $s(a^{\prime})=1-s(a)$. Now assume $a\leq b$.
Then $a\vee(b\wedge a^{\prime})\leq b$ and

$s(a\vee(a^{\prime}\wedge b))+s(b^{\prime})=s((a\vee(a^{\prime}\wedge b))\vee
b^{\prime})=s((a\vee b^{\prime})\vee(a^{\prime}\wedge b))=1$

according to (ii) and (iii) and hence

$s(a\vee(b\wedge a^{\prime}))=1-s(b^{\prime})=1-(1-s(b))=s(b),$

i.e. $s\in S_{1}(\mathbf{L})$. Similarly, we have $a\leq b\wedge(a\vee
b^{\prime})$ and

$s(a)+s((b\wedge(a\vee b^{\prime}))^{\prime})=s(a\vee(b\wedge(a\vee
b^{\prime}))^{\prime})=1$

according to (ii) and (iv) whence

$s(b\wedge(a\vee b^{\prime})=1-(1-s(b\wedge(a\vee
b^{\prime})))=1-s((b\wedge(a\vee b^{\prime}))^{\prime})=s(a)$

proving $s\in S_{2}(\mathbf{L})$. ∎

######  Example 15.

Consider the lattice M3 depicted in Fig. 1

$0$$a$$b$$c$$1$Fig. 1

where the unary operation ′ is defined as follows:

$\begin{array}[]{c|ccccc}x&0&a&b&c&1\\\ \hline\cr
x^{\prime}&1&b&c&a&0\end{array}$

Let $s\colon{\rm M}_{3}\rightarrow[0,1]$ be defined by

$\begin{array}[]{c|ccccc}x&0&a&b&c&1\\\ \hline\cr
s(x)&0&1/2&1/2&1/2&1\end{array}$

One can easily check that (i) – (iv) are satisfied. Thus, $s\in
S_{1}(\mathbf{L})\cap S_{2}(\mathbf{L})$.

A connection between orthomodular lattices and residuated structures was
investigated by the authors in [10] and [11]. Despite the fact that the
complementation in lattices being both weakly and dually weakly orthomodular
need not be antitone, we are still able to find a residuated structure which
can be assigned to every such lattice. This is important from the point of
view of algebraic semantics because residuated structures are applied in
algebraic semantics of several fuzzy logics. Thus, although we started with a
generalization of the semantics of the logic of quantum mechanics, we can
provide a bridge between these rather different approaches. For this reason,
the W-implication introduced before Theorem 8 turns out to be more
appropriate.

######  Lemma 16.

Let $\mathbf{L}=(L,\vee,\wedge,{}^{\prime})$ be a dually weakly orthomodular
lattice and $a,b,c\in L$ and define

$\displaystyle x\odot y$ $\displaystyle:=(x\vee y^{\prime})\wedge y,$
$\displaystyle x\rightarrow y$ $\displaystyle:=x^{\prime}\vee(x\wedge y)$

for all $x,y\in L$. If $a\leq b\rightarrow c$ then $a\odot b\leq c$.

###### Proof.

Using dual weak orthomodularity, if $a\leq b\rightarrow c$ we compute

$\displaystyle a\odot b$ $\displaystyle=(a\vee b^{\prime})\wedge
b\leq((b\rightarrow c)\vee b^{\prime})\wedge b=(b^{\prime}\vee(b\wedge c)\vee
b^{\prime})\wedge b=b\wedge((b\wedge c)\vee b^{\prime})=$
$\displaystyle=b\wedge c\leq c.$

∎

Consider the lattice ${\rm M}_{3}$ depicted in Fig. 1 where the unary
operation ′ is defined as follows:

$\begin{array}[]{c|ccccc}x&0&a&b&c&1\\\ \hline\cr
x^{\prime}&1&b&c&b&0\end{array}$

It is easy to check that $(M_{3},\vee,\wedge,{}^{\prime})$ is both weakly and
dually weakly orthomodular, but not orthomodular since
$(a^{\prime})^{\prime}=b^{\prime}=c\neq a$. Hence it does not satisfy the
double negation law. On the other hand,
$((x^{\prime})^{\prime})^{\prime}\approx x^{\prime}$. This is e.g. the
condition for negation in intuitionistic logic. For such lattices we can state
the following theorem.

######  Theorem 17.

Let $\mathbf{L}=(L,\vee,\wedge,{}^{\prime})$ be a weakly and dually weakly
orthomodular lattice satisfying $((x^{\prime})^{\prime})^{\prime}\approx
x^{\prime}$ and $a,b,c\in L$ and define

$\displaystyle x\odot y$ $\displaystyle:=(x\vee y^{\prime})\wedge y,$
$\displaystyle x\rightarrow y$ $\displaystyle:=x^{\prime}\vee(x\wedge y)$

for all $x,y\in L$. Then $\mathbf{L}$ satisfies the identity $(x\rightarrow
y)\odot x\approx x\wedge y$. Moreover,

$a\odot(b^{\prime})^{\prime}\leq c\text{ if and only if
}a\leq(b^{\prime})^{\prime}\rightarrow c.$

###### Proof.

We have

$(x\rightarrow y)\odot x\approx(x^{\prime}\vee(x\wedge y)\vee
x^{\prime})\wedge x\approx x\wedge((x\wedge y)\vee x^{\prime})\approx x\wedge
y.$

If $a\leq(b^{\prime})^{\prime}\rightarrow c$ then
$a\odot(b^{\prime})^{\prime}\leq c$ according to Lemma 16. Conversely, assume
$a\odot(b^{\prime})^{\prime}\leq c$. Using weak orthomodularity and the
identity $((x^{\prime})^{\prime})^{\prime}\approx x^{\prime}$ we compute

$\displaystyle a$ $\displaystyle\leq a\vee b^{\prime}=b^{\prime}\vee((a\vee
b^{\prime})\wedge(b^{\prime})^{\prime})=b^{\prime}\vee((b^{\prime})^{\prime}\wedge(a\vee((b^{\prime})^{\prime})^{\prime})\wedge(b^{\prime})^{\prime})=$
$\displaystyle=b^{\prime}\vee((b^{\prime})^{\prime}\wedge(a\odot(b^{\prime})^{\prime}))\leq
b^{\prime}\vee((b^{\prime})^{\prime}\wedge
c)=((b^{\prime})^{\prime})^{\prime}\vee((b^{\prime})^{\prime}\wedge
c)=(b^{\prime})^{\prime}\rightarrow c.$

∎

For the reader’s convenience, we repeat the following concept which is
frequently used in the theory of residuated structures.

######  Definition 18.

A left residuated l-groupoid is an algebra
${\mathbf{L}}=(L,\vee,\wedge,\odot,\rightarrow,0,1)$ of type $(2,2,2,2,0,0)$
such that $(L,\vee,\wedge,0,1)$ is a bounded lattice, $x\odot 1\approx 1\odot
x\approx x$ and for all $x,y,z\in L$, $x\odot y\leq z$ is equivalent to $x\leq
y\rightarrow z$.

For lattices being both weakly and dually weakly orthomodular and satisfying
the double negation law we can prove a result which is stronger than that of
Theorem 17, see the following theorem.

######  Theorem 19.

Let $\mathbf{L}=(L,\vee,\wedge,{}^{\prime})$ be a weakly and dually weakly
orthomodular lattice satisfying the double negation law and $a,b\in L$ and
define

$\displaystyle x\odot y$ $\displaystyle:=(x\vee y^{\prime})\wedge y,$
$\displaystyle x\rightarrow y$ $\displaystyle:=x^{\prime}\vee(x\wedge y)$

for all $x,y\in L$. Then $\mathbf{L}$ is bounded and
$(L,\vee,\wedge,\odot,\rightarrow,0,1)$ a left residuated l-groupoid
satisfying the identity

$(x\rightarrow y)\odot x\approx x\wedge y.$

Moreover,

$\displaystyle a\odot b=a$ $\displaystyle\text{ if and only if }a\leq b,$
$\displaystyle a\rightarrow b=b$ $\displaystyle\text{ if and only if
}a^{\prime}\leq b.$

###### Proof.

According to Lemma 2, $\mathbf{L}$ is bounded and $1^{\prime}\approx 0$. Let
$c\in L$. If $a\odot b\leq c$ then, due to weak orthomodularity of
$\mathbf{L}$ and to the double negation law,

$\displaystyle a$ $\displaystyle\leq a\vee b^{\prime}=b^{\prime}\vee((a\vee
b^{\prime})\wedge(b^{\prime})^{\prime})=b^{\prime}\vee((a\vee
b^{\prime})\wedge b)=b^{\prime}\vee(b\wedge(a\vee b^{\prime})\wedge b)=$
$\displaystyle=b^{\prime}\vee(b\wedge(a\odot b))\leq b^{\prime}\vee(b\wedge
c)=b\rightarrow c.$

Conversely, if $a\leq b\rightarrow c$ then $a\odot b\leq c$ according to Lemma
16. Moreover,

$\displaystyle x\odot 1$ $\displaystyle\approx(x\vee 1^{\prime})\wedge
1\approx(x\vee 0)\wedge 1\approx x\wedge 1\approx x,$ $\displaystyle 1\odot x$
$\displaystyle\approx(1\vee x^{\prime})\wedge x\approx 1\wedge x\approx x,$
$\displaystyle(x\rightarrow y)\odot x$
$\displaystyle\approx(x^{\prime}\vee(x\wedge y)\vee x^{\prime})\wedge x\approx
x\wedge((x\wedge y)\vee x^{\prime})\approx x\wedge y.$

Now the following are equivalent: $a\odot b=a$; $(a\vee b^{\prime})\wedge
b=a$; $b\wedge(a\vee b^{\prime})=a$; $a\leq b$. Finally, the following are
equivalent: $a\rightarrow b=b$; $a^{\prime}\vee(a\wedge b)=b$;
$a^{\prime}\vee(b\wedge(a^{\prime})^{\prime})=b$; $a^{\prime}\leq b$. ∎

Of course, every orthomodular lattice is both weakly and dually weakly
orthomodular and satisfies the double negation law. In the following we
present a lattice being both weakly and dually weakly orthomodular and
satisfying the double negation law, but not being orthomodular.

######  Example 20.

The lattice whose Hasse diagram is depicted in Fig. 2

$0$$a$$b$$c$$d$$e$$f$$g$$h$$1$Fig. 2

whose unary operation ′ is defined by

$\begin{array}[]{c|cccccccccc}x&0&a&b&c&d&e&f&g&h&1\\\ \hline\cr
x^{\prime}&1&g&h&f&e&d&c&a&b&0\end{array}$

is both weakly and dually weakly orthomodular and satisfies the double
negation law, but it is not orthomodular since ′ is not monotone since $a\leq
f$, but $f^{\prime}=c\not\leq g=a^{\prime}$.

We can prove also the converse of Theorem 19.

######  Theorem 21.

Let $\mathbf{L}=(L,\vee,\wedge,\odot,\rightarrow,0,1)$ be a left residuated
l-groupoid satisfying the identities

$\displaystyle(x\rightarrow 0)\rightarrow 0$ $\displaystyle\approx x,$
$\displaystyle x\odot y$ $\displaystyle\approx(x\vee(y\rightarrow 0))\wedge
y,$ $\displaystyle x\rightarrow y$ $\displaystyle\approx(x\rightarrow
0)\vee(x\wedge y)$

and define $x^{\prime}:=x\rightarrow 0$ for all $x\in L$. Then
$(L,\vee,\wedge,{}^{\prime})$ is a weakly and dually weakly orthomodular
lattice satisfying the double negation law.

###### Proof.

Let $a,b\in L$ and $a\leq b$. Then $a\leq b^{\prime}\vee(b\wedge
a)=b\rightarrow a$ and hence $b\wedge(a\vee b^{\prime})=a\odot b\leq a$
according to left-adjointness. This shows $b\wedge(a\vee b^{\prime})\leq a$.
Since the converse inequality is obvious, $(L,\vee,\wedge,{}^{\prime})$ is
dually weakly orthomodular. Now $a\vee(b\wedge a^{\prime})\leq b$. To prove
the converse inequality, we must use the double negation law. We have

$b\odot a^{\prime}=(b\vee(a^{\prime})^{\prime})\wedge a=(b\vee a)\wedge
a^{\prime}=b\wedge a^{\prime}\leq b.$

Thus, using left-adjointness, we obtain that $b\odot a^{\prime}\leq b$ is
equivalent to

$b\leq a^{\prime}\rightarrow b=(a^{\prime})^{\prime}\vee(b\wedge
a^{\prime})=a\vee(b\wedge a^{\prime})$

proving $a\vee(b\wedge a^{\prime})\geq b$. Together we have $b=a\vee(b\wedge
a^{\prime})$. Thus $(L,\vee,\wedge,{}^{\prime})$ is also weakly orthomodular
and, obviously, satisfies the double negation law. ∎

## References

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Authors’ addresses:

Ivan Chajda
Palacký University Olomouc
Faculty of Science
Department of Algebra and Geometry
17\. listopadu 12
771 46 Olomouc
Czech Republic
<EMAIL_ADDRESS>
Helmut Länger
TU Wien
Faculty of Mathematics and Geoinformation
Institute of Discrete Mathematics and Geometry
Wiedner Hauptstraße 8-10
1040 Vienna
Austria, and
Palacký University Olomouc
Faculty of Science
Department of Algebra and Geometry
17\. listopadu 12
771 46 Olomouc
Czech Republic
<EMAIL_ADDRESS>