Residuation in lattice effect algebras

TL;DR

This paper establishes a correspondence between lattice effect algebras and quasiresiduated lattices, providing a new structural perspective and reconstruction method for these algebraic systems, including pseudoeffect algebras.

Contribution

It introduces quasiresiduated lattices and demonstrates their equivalence and reconstructive relationship with lattice effect algebras and pseudoeffect algebras.

Findings

Every lattice effect algebra can be organized into a commutative quasiresiduated lattice.

Every such lattice can be converted back into a lattice effect algebra.

Good lattice pseudoeffect algebras can be organized into quasiresiduated lattices with divisibility.

Abstract

We introduce the concept of a quasiresiduated lattice and prove that every lattice effect algebra can be organized into a commutative quasiresiduated lattice with divisibility. Also conversely, every such a lattice can be converted into a lattice effect algebra and every lattice effct algebra can be reconstructed form its assigned quasiresiduated lattice. We apply this method also for lattice pseudoeffect algebras introduced recently by Dvurecenskij and Vetterlein. We show that every good lattice pseudoeffect algebra can be organized into a (possibly non-commutative) quasiresiduated lattice with divisibility and conversely, every such a lattice can be converted into a lattice pseudoeffect algebra. Moreover, also a good lattice pseudoeffect algebra can be reconstructed from the assigned quasiresiduated lattice.

Equations

$$x\le y\text{ if there exists some }z\in E\text{ with }x+z=y$$

$$x\le y\text{ implies }y\odot (y\to x)=x$$

$$x\odot y\le z\text{ if and only if }x\le y\to z$$

$$x\odot y$$

$$x\to y$$

$$a\vee b^{\prime }=b^{\prime }+((a\vee b^{\prime }{)}^{\prime }+b^{\prime }{)}^{\prime }\le b^{\prime }+(b\wedge c)=b\to c.$$

$$(a\vee b^{\prime })\odot b=(b^{\prime }+(a\vee b^{\prime }{)}^{\prime }{)}^{\prime }\le (b^{\prime }+((b\wedge c)+b^{\prime }{)}^{\prime }{)}^{\prime }=(b\wedge c{)}^{\prime \prime }=b\wedge c$$

$$b\odot (b\to a)=(b^{\prime }+(a+b^{\prime }{)}^{\prime }{)}^{\prime }=a^{\prime \prime }=a$$

$$x\otimes y:=(x\vee y^{\prime })\odot y$$

$$x\otimes y\le y\wedge z\text{ if and only if }x\vee y^{\prime }\le y\to z.$$

$$y\to z=(y\wedge z)+y^{\prime }=y\to (y\wedge z),$$

$$(x\vee y^{\prime })\odot y\le y\wedge z\text{ if and only if }x\vee y^{\prime }\le y\to (y\wedge z)$$

$$X\odot y\le Z\text{ if and only if }X\le y\to Z.$$

$$x+y$$

$$x^{\prime }$$

$$0\vee a^{\prime }=a^{\prime }\le a^{\prime }=a\to 0$$

$$a\odot a^{\prime }=a\odot (0\vee a^{\prime })\le a\wedge 0=0,$$

$$a\odot (b\vee a^{\prime })=a\odot b=0\le a\wedge 0$$

$$b=b\vee a^{\prime }\le a\to 0=a^{\prime }$$

$$a+b$$

$$a^{\prime }\odot b^{\prime }$$

$$a$$

$$b$$

$$a\le b\text{ holds in }\mathbb{E}(\mathbf{R}),$$

$$a+b^{\prime }\text{ is defined},$$

$$a^{\prime }\odot b\text{ is defined},$$

$$a\le b\text{ holds in }\mathbf{R}.$$

$$\mathbf{E}$$

$$\mathbb{C}(\mathbf{E})$$

$$\mathbb{E}(\mathbb{C}(\mathbf{E}))$$

$$a^{\ast }=a\to 0=(a\wedge 0)+a^{\prime }=0+a^{\prime }=a^{\prime }.$$

$$a\oplus b\text{ is defined},$$

$$a\le b^{\prime }\text{ in }\mathbb{C}(\mathbf{E}),$$

$$a\le b^{\prime }\text{ in }\mathbf{E}$$

$$a\oplus b=(a^{\ast }\odot b^{\ast }{)}^{\ast }=(a^{\prime }\odot b^{\prime }{)}^{\prime }=(a^{\prime \prime }+b^{\prime \prime }{)}^{\prime \prime }=a+b.$$

$$x\le y\text{ if there exists some }z\in E\text{ with }x+z=y$$

Full text

AMS Subject Classification: 03G25,03G12,06D35

Keywords: lattice effect algebra, lattice pseudoeffect algebra, quasiresiduated lattice, quasiadjointness, divisibility

In order to axiomatize quantum logic effects in a Hilbert space, Foulis and Bennett ([6]) introduced the so-called effect algebras. These are partial algebras with one partial binary operation which can be converted into bounded posets in general and into lattices in particular cases. It turns out that effect algebras form a successful axiomatization of the logic of quantum mechanics, but we suppose that there exists a connection with a kind of substructural logics whose algebraic semantics is based on residuated lattices. An attempt in this direction was already done in [2] where the so-called conditional residuation was introduced. A disadvantage of this approach is that the axioms of residuated structures are reflected only in the case when the terms used in adjointness are defined. This is an essential restriction which prevents the development of this theory. The aim of the present paper is to introduce the more general concept of quasiresiduation and to show that lattice effect algebras and lattice pseudoeffect algebras satisfy this concept. Pseudoeffect algebras were introduced recently by Dvurečenski and Vetterlein ([5]).

We start with the following definition.

1. (E1)

$x+y$* is defined if and only if so is $y+x$ and in this case $x+y=y+x$,* 2. (E2)

$(x+y)+z$* is defined if and only if so is $x+(y+z)$ and in this case $(x+y)+z=x+(y+z)$,* 3. (E3)

$x^{\prime}$* is the unique $u\in E$ with $x+u=1$,* 4. (E4)

if $1+x$ is defined then $x=0$.

On $E$ a binary relation $\leq$ can be defined by

[TABLE]

($x,y\in E$). Then $(E,\leq,0,1)$ is a bounded poset and $\leq$ is called the induced order of $\mathbf{E}$. If $(E,\leq)$ is a lattice then $\mathbf{E}$ is called a lattice effect algebra.

In the sequel we will use the properties of effect algebras listed in the following lemma.

1. (i)

$a^{\prime\prime}=a$, 2. (ii)

$a\leq b$* implies $b^{\prime}\leq a^{\prime}$,* 3. (iii)

$a+b$* is defined if and only if $a\leq b^{\prime}$,* 4. (iv)

if $a\leq b$ and $b+c$ is defined then $a+c$ is defined and $a+c\leq b+c$, 5. (v)

if $a\leq b$ then $a+(a+b^{\prime})^{\prime}=b$, 6. (vi)

$a+0=0+a=a$, 7. (vii)

$0^{\prime}=1$* and $1^{\prime}=0$.*

A partial monoid is a partial algebra $\mathbf{A}=(A,\odot,1)$ of type $(2,0)$ where $1\in A$ and $\odot$ is a partial operation satisfying the following conditions for all $x,y,z\in A$:

1. (i)

$(x\odot y)\odot z$ is defined if and only if so is $x\odot(y\odot z)$ and in this case $(x\odot y)\odot z=x\odot(y\odot z)$, 2. (ii)

$x\odot 1=1\odot x=x$.

The partial monoid $\mathbf{A}$ is called commutative if it satisfies the following condition for all $x,y\in A$:

1. (iii)

$x\odot y$ is defined if and only if so is $y\odot x$ and in this case $x\odot y=y\odot x$,

The authors already introduced a certain modification of residuation for sectionally pseudocomplemented lattices, see [3]. For lattice effect algebras, we introduce another version of residuation called quasiresiduation.

1. (C1)

$(C,\odot,1)$* is a partial commutative monoid where $x\odot y$ is defined if and only if $x^{\prime}\leq y$,* 2. (C2)

$x^{\prime\prime}=x$, and $x\leq y$ implies $y^{\prime}\leq x^{\prime}$, 3. (C3)

$(x\vee y^{\prime})\odot y\leq y\wedge z$* if and only if $x\vee y^{\prime}\leq y\rightarrow z$.*

Here $x^{\prime}$ is an abbreviation for $x\rightarrow 0$. The commutative quasiresiduated lattice $\mathbf{C}$ is called divisible if

[TABLE]

for all $x,y\in C$.

Note that the terms in (C3) are everywhere defined.

In case $y^{\prime}\leq x$ and $z\leq y$ condition (C3) reduces to

[TABLE]

which is usual adjointness. Therefore condition (C3) will be called commutative quasiadjointness. Hence, contrary to the similar concept in [2], in commutative quasiadjointness we have only everywhere defined terms in $\mathbf{C}$ although $\mathbf{C}$ is a partial algebra.

Our aim is to show that every lattice effect algebra can be organized into a commutative quasiresiduated lattice.

[TABLE]

($x,y\in E$). Then $\mathbb{C}(\mathbf{E}):=(E,\vee,\wedge,\odot,\rightarrow,0,1)$ is a divisible commutative quasiresiduated lattice.

[TABLE]

If, conversely, $a\vee b^{\prime}\leq b\rightarrow c$ then $a\vee b^{\prime}\leq(b\wedge c)+b^{\prime}$ and $b^{\prime}\leq(b\wedge c)^{\prime}$ and hence

[TABLE]

proving (C3). If $a\leq b$ then $b^{\prime}\leq a^{\prime}$ and hence

[TABLE]

proving divisibility. ∎

[TABLE]

for all $x,y\in C$ then $\otimes$ is everywhere defined and satisfies the identities $x\otimes 1\approx 1\otimes x\approx x$, and commutative quasiadjointness can then be expressed in the form

[TABLE]

This means that our definition of commutative quasiresiduation differs from that of usual residuation only in the point that $y$ occurs on the right-hand side of $x\otimes y\leq y\wedge z$ and $y^{\prime}$ on the left-hand side of $x\vee y^{\prime}\leq y\rightarrow z$. On the other hand, using this version, divisibility cannot be easily defined. Moreover, since in lattice effect algebras we have

[TABLE]

commutative quasiadjointness can be rewritten in the form

[TABLE]

which corresponds to usual adjointness if we abbreviate $x\vee y^{\prime}$ by $X$ and $y\wedge z$ by $Z$, i.e.

[TABLE]

We can prove also the converse.

[TABLE]

($x,y\in C$). Then $\mathbb{E}(\mathbf{C}):=(C,+,{}^{\prime},0,1)$ is a lattice effect algebra whose order coincides with that in $\mathbf{C}$.

[TABLE]

we have

[TABLE]

i.e. $a\odot a^{\prime}=0$. If, conversely, $a\odot b=0$ then $a^{\prime}\leq b$ and hence

[TABLE]

whence

[TABLE]

showing $b=a^{\prime}$. Hence $a\odot b=0$ if and only if $b=a^{\prime}$. Now the following are equivalent:

[TABLE]

This shows (E3). Moreover, the following are equivalent:

[TABLE]

Since $(C,\vee,\wedge)$ is a lattice and the partial order relations in $\mathbf{C}$ and $\mathbb{E}(\mathbf{C})$ coincide, $\mathbb{E}(\mathbf{C})$ is a lattice effect algebra. ∎

Moreover, every lattice effect algebra can be reconstructed from the assigned quasiresiduated lattice as shown in the following result.

[TABLE]

and $a,b\in E$. Then

[TABLE]

Moreover, the following are equivalent:

[TABLE]

and in this case

[TABLE]

∎

Now we turn our attention to a more general case. The following concept was introduced by Dvurečenskij and Vetterlein ([5]).

1. (P1)

If $x+y$ is defined then there exist $u,w\in P$ with $u+x=y+w=x+y$, 2. (P2)

$(x+y)+z$* is defined if and only if $x+(y+z)$ is defined, and in this case $(x+y)+z=x+(y+z)$,* 3. (P3)

$\bar{x}$* is the unique $u\in P$ with $u+x=1$, and $\tilde{x}$ is the unique $w\in P$ with $x+w=1$,* 4. (P4)

if $1+x$ or $x+1$ is defined then $x=0$.

The pseudoeffect algebra $\mathbf{P}$ is called good if $\widetilde{\bar{x}+\bar{y}}=\overline{\tilde{x}+\tilde{y}}$ for all $x,y\in P$ with $\tilde{x}\leq y$.

On $P$ a binary relation $\leq$ can be defined by

[TABLE]

($x,y\in P$). Then $(P,\leq,0,1)$ is a bounded poset and $\leq$ is called the induced order of $\mathbf{P}$. If $(P,\leq)$ is a lattice then $\mathbf{P}$ is called a lattice pseudoeffect algebra.

For our investigation we need the following results taken from [5].

1. (i)

$\bar{\tilde{a}}=\tilde{\bar{a}}=a$, 2. (ii)

the following are equivalent: $a\leq b$, $\bar{b}\leq\bar{a}$, $\tilde{b}\leq\tilde{a}$, 3. (iii)

$a+b$* is defined if and only if $a\leq\bar{b}$,* 4. (iv)

if $a\leq b$ and $b+c$ is defined then $a+c$ is defined and $a+c\leq b+c$, 5. (v)

if $a\leq b$ and $c+b$ is defined then $c+a$ is defined and $c+a\leq c+b$, 6. (vi)

if $a\leq b$ then $a+\widetilde{\bar{b}+a}=\overline{a+\tilde{b}}+a=b$, 7. (vii)

$a+0=0+a=a$, 8. (viii)

$\bar{0}=\tilde{0}=1$* and $\bar{1}=\tilde{1}=0$,* 9. (ix)

the following are equivalent: $a\leq b$, there exists some $d\in P$ with $a+d=b$, there exists some $e\in P$ with $e+a=b$.

Since pseudoeffect algebras are more general than effect algebras, we must define quasiresiduated lattice for the case when the partial operation $\odot$ is not commutative and the mapping $x\mapsto\bar{x}$ is not an involution.

1. (Q1)

$(Q,\odot,1)$* is a partial monoid where $x\odot y$ is defined if and only if $\tilde{x}\leq y$,* 2. (Q2)

$\tilde{\bar{x}}=\bar{\tilde{x}}=x$, and $x\leq y$ implies $\bar{y}\leq\bar{x}$ and $\tilde{y}\leq\tilde{x}$, 3. (Q3)

$(x\vee\bar{y})\odot y\leq y\wedge z$* if and only if $x\vee\bar{y}\leq y\rightarrow z$,* 4. (Q4)

$y\odot(x\vee\tilde{y})\leq y\wedge z$* if and only if $x\vee\tilde{y}\leq y\leadsto z$,* 5. (Q5)

$\widetilde{\bar{x}\odot\bar{y}}=\overline{\tilde{x}\odot\tilde{y}}$.

Here $\bar{x}$ and $\tilde{x}$ are abbreviations for $x\rightarrow 0$ and $x\leadsto 0$, respectively. The quasiresiduated lattice $\mathbf{Q}$ is called divisible if

[TABLE]

for all $x,y\in Q$.

Note that the terms in (Q3) and (Q4) are everywhere defined.

In case $\bar{y}\leq x$ and $z\leq y$ condition (Q3) reduces to

[TABLE]

which is usual adjointness. Analogously, in case $\tilde{y}\leq x$ and $z\leq y$ condition (Q4) reduces to

[TABLE]

which is usual adjointness if $\odot$ is commutative. Therefore conditions (Q3) and (Q4) will be called quasiadjointness. Hence, contrary to the similar concept in [2], in quasiadjointness we have only everywhere defined terms in $\mathbf{Q}$ although $\mathbf{Q}$ is a partial algebra.

Similarly as for effect algebras, we prove that every good lattice pseudoeffect algebra can be organized into a quasiresiduated lattice which, however, need not be commutative.

[TABLE]

($x,y\in P$). Then $\mathbb{Q}(\mathbf{P}):=(P,\vee,\wedge,\odot,\rightarrow,\leadsto,0,1)$ is a divisible quasiresiduated lattice.

[TABLE]

If, conversely, $a\vee\bar{b}\leq b\rightarrow c$ then $a\vee\bar{b}\leq\bar{b}+(b\wedge c)$ and $\bar{b}\leq\overline{b\wedge c}$ and hence

[TABLE]

roving (Q3). If $b\odot(a\vee\tilde{b})\leq b\wedge c$ then $\overline{\tilde{b}+\widetilde{a\vee\tilde{b}}}\leq b\wedge c$ and $\tilde{b}\leq a\vee\tilde{b}$ and hence

[TABLE]

If, conversely, $a\vee\tilde{b}\leq b\leadsto c$ then $a\vee\tilde{b}\leq(b\wedge c)+\tilde{b}$ and $\tilde{b}\leq\widetilde{b\wedge c}$ and hence

[TABLE]

proving (Q4). If $a\leq b$ then $\bar{b}\leq\bar{a}$ and $\tilde{b}\leq\tilde{a}$ and hence

[TABLE]

proving divisibility. ∎

We can prove also the converse.

[TABLE]

($x,y\in Q$). Then $\mathbb{P}(\mathbf{Q}):=(Q,+,\,\bar{}\,,\,\tilde{}\,,0,1)$ is a good lattice pseudoeffect algebra whose order coincides with that in $\mathbf{Q}$.

[TABLE]

we have

[TABLE]

i.e. $a\odot\tilde{a}=0$. If, conversely, $a\odot b=0$ then $\tilde{a}\leq b$ and hence

[TABLE]

whence

[TABLE]

showing $b=\tilde{a}$. Hence $a\odot b=0$ if and only if $b=\tilde{a}$. Since

[TABLE]

we have

[TABLE]

i.e. $\bar{a}\odot a=0$. If, conversely, $b\odot a=0$ then $\tilde{b}\leq a$, i.e. $\bar{a}\leq b$, and hence

[TABLE]

whence

[TABLE]

showing $b=\bar{a}$. Hence $b\odot a=0$ if and only if $b=\bar{a}$. Now the following are equivalent:

[TABLE]

This shows (P3). Since

[TABLE]

we have

[TABLE]

i.e. $a\leadsto a=1$. If $a\leq\bar{b}$ then because of $\bar{b}\vee a\leq\bar{a}\leadsto\bar{a}$ we have

[TABLE]

whence

[TABLE]

showing that $a+\widetilde{a+b}$ is defined. Now in case $a\leq\bar{b}$ the following are equivalent:

[TABLE]

Since

[TABLE]

we have

[TABLE]

i.e. $a\rightarrow a=1$. If $b\leq\tilde{a}$ then because of $\tilde{a}\vee b\leq\tilde{b}\rightarrow\tilde{b}$ we have

[TABLE]

whence

[TABLE]

showing that $\overline{a+b}+b$ is defined. Now in case $b\leq\tilde{a}$, i.e. $a\leq\bar{b}$ the following are equivalent:

[TABLE]

This shows (P1). Now the following are equivalent:

[TABLE]

Since $(Q,\vee,\wedge)$ is a lattice and the partial order relations in $\mathbf{Q}$ and $\mathbb{P}(\mathbf{Q})$ coincide, $\mathbb{P}(\mathbf{Q})$ is a lattice pseudoeffect algebra. ∎

As in the case of effect algebras, also every good lattice pseudoeffect algebra can be reconstructed from its assigned quasiresiduated lattice.

[TABLE]

and $a,b\in E$. Then

[TABLE]

Moreover, the following are equivalent:

[TABLE]

and in this case

[TABLE]

∎