00footnotetext: Support of the research by the Austrian Science Fund (FWF),
project I 4579-N, and the Czech Science Foundation (GAČR), project 20-09869L,
entitled “The many facets of orthomodularity”, as well as by ÖAD, project CZ
02/2019, entitled “Function algebras and ordered structures related to logic
and data fusion”, and, concerning the first author, by IGA, project PřF 2020
014, is gratefully acknowledged.

# Residuation in twist products and pseudo-Kleene posets

Ivan Chajda and Helmut Länger

###### Abstract

M. Busaniche, R. Cignoli, C. Tsinakis and A. M. Wille showed that every
residuated lattice induces a residuation on its full twist product. For their
construction they used also lattice operations. We generalize this problem to
left-residuated groupoids which need not be lattice-ordered. Hence, for the
full twist product we cannot use the same construction. We present another
appropriate construction which, however, does not preserve commutativity and
associativity of multiplication. Hence we introduce so-called operator
residuated posets to obtain another construction which preserves the mentioned
properties, but the results of operators on the full twist product need not be
elements, but may be subsets. We apply this construction also to restricted
twist products and present necessary and sufficient conditions under which we
obtain a pseudo-Kleene operator residuated poset.

AMS Subject Classification: 06A11, 06D30, 03G25, 03B47

Keywords: Left-residuated poset, operator residuated poset, twist product,
pseudo-Kleene poset, Kleene poset

##  1 Introduction

M. Busaniche and R. Cignoli ([2]) as well as by C. Tsinakis and A. M. Wille
([7]) showed that if $(L,\leq,\cdot,\rightarrow,1)$ is a residuated lattice
then $\cdot$ and $\rightarrow$ can be used to define binary operations $\odot$
and $\Rightarrow$ on the full twist product of $(L,\leq)$ such that the
resulting structure becomes a residuated lattice again. For the construction
of such operations $\odot$ and $\Rightarrow$ they used the lattice operations
$\vee$ and $\wedge$. When going from lattices to residuated posets, the
natural question arises whether also in this case the corresponding twist
product can be equipped with certain operations $\odot$ and $\Rightarrow$
(without using lattice operations) such that the resulting structure is
residuated again. We solve this problem in the positive. We define suitable
operations $\odot$ and $\Rightarrow$ on the full twist product such that the
arising structure becomes a left-residuated groupoid again. Unfortunately,
this construction does preserve neither commutativity nor associativity of the
original structure $(Q,\leq,\cdot,\rightarrow,1)$. Hence, we try another
approach where instead of operations we use certain operators $\odot$ and
$\Rightarrow$ in such a way that the full twist product becomes an operator
residuated poset and commutativity as well as associativity of the original
operation $\cdot$ are preserved. As the authors already showed in [4], any
poset $\mathbf{Q}=(Q,\leq)$ can be embedded into a pseudo-Kleene poset
$(P_{a}(\mathbf{Q}),\leq,{}^{\prime})$ where $(P_{a}(\mathbf{Q}),\leq)$ is a
certain subposet of the full twist product of $\mathbf{Q}$. This motivated us
to investigate whether our construction of the operators $\odot$ and
$\Rightarrow$ can be extended also to this case, i.e. whether we can determine
for a bounded commutative residuated monoid $(Q,\leq,\cdot,\rightarrow,0,1)$ a
corresponding pseudo-Kleene poset which is operator residuated and into which
$\mathbf{Q}$ can be embedded. We characterize those left-residuated posets for
which our construction is possible.

##  2 Preliminaries

The concept of a Kleene lattice (alias Kleene algebra) was introduced by J. A.
Kalman ([6]), see also [5]. Recall that a Kleene lattice is a distributive
lattice $\mathbf{L}=(L,\vee,\wedge,{}^{\prime})$ with an involution ′
satisfying the so-called normality condition, i.e. the identity

$x\wedge x^{\prime}\leq y\vee y^{\prime}.$

This concept was generalized by the first author in [3]: $\mathbf{L}$ is
called a pseudo-Kleene lattice if it satisfies the above identity, but it need
not be distributive.

Let $(P,\leq)$ be a poset, $a,b\in P$ and $A,B\subseteq P$. Then the lower
cone $L(A)$ of $A$ and the upper cone $U(A)$ of $A$ are defined as follows:

$\displaystyle L(A)$ $\displaystyle:=\\{x\in P\mid x\leq A\\},$ $\displaystyle
U(A)$ $\displaystyle:=\\{x\in P\mid x\geq A\\}.$

Here $x\leq A$ means $x\leq y$ for all $y\in A$ and, similarly, $x\geq A$
means $x\geq y$ for all $y\in A$. The expression $A\leq B$ means $x\leq y$ for
all $x\in A$ and $y\in B$. Instead of $L(\\{a,b\\})$ and $L(U(A))$ we simply
write $L(a,b)$ and $LU(A)$, respectively. Analogously, we proceed in similar
cases. Let $\max A$ denote the set of all maximal elements of $(A,\leq)$. A
unary operation ′ on $P$ is called

  * •

antitone if $x,y\in P$ and $x\leq y$ imply $y^{\prime}\leq x^{\prime}$,

  * •

an involution if it satisfies the identity $x^{\prime\prime}\approx x$.

The concept of a pseudo-Kleene lattice was generalized by the authors in [4]
for posets as follows:

A pseudo-Kleene poset is a poset $\mathbf{P}=(P,\leq,{}^{\prime})$ with an
antitone involution satisfying the condition

$L(x,x^{\prime})\leq U(y,y^{\prime})$

for all $x,y\in P$. A Kleene poset is a distributive pseudo-Kleene poset.
Recall that a poset $(P,\leq)$ is called distributive if it satisfies one of
the following equivalent LU-identities:

$\displaystyle L(U(x,y),z)$ $\displaystyle\approx LU(L(x,z),L(y,z)),$
$\displaystyle U(L(x,y),z)$ $\displaystyle\approx UL(U(x,z),U(y,z)).$

In [4] it was shown that an arbitrary poset can be embedded into a pseudo-
Kleene poset by means of the so-called twist construction:

The full twist product of a poset $\mathbf{Q}=(Q,\leq)$ is the poset
$(Q^{2},\leq)$ where

$(x,y)\leq(z,v)\text{ if and only if }x\leq z\text{ and }v\leq y$

for all $(x,y),(z,v)\in Q^{2}$. We have

$\displaystyle L((x,y),(z,v))$ $\displaystyle=L(x,z)\times U(y,v),$
$\displaystyle U((x,y),(z,v))$ $\displaystyle=U(x,z)\times L(y,v)$

for all $(x,y),(z,v)\in Q^{2}$.

##  3 Left-residuated groupoids

We will investigate when a residuated poset can be transferred to a residuated
full twist product. For this purpose we will use the twist construction. For
residuated lattices such a transfer was already published in [2] by using a
construction developed in [7].

From now on, let $(Q,\leq,\cdot,\rightarrow,1)$ denote a poset with constant
$1$ endowed with two binary operations $\cdot$ and $\rightarrow$. For our next
investigations, consider the following conditions.

  1. (1)

$x\leq y$ implies $z\cdot x\leq z\cdot y$ (right-isotony),

  2. (2)

$x\leq y$ implies $x\cdot z\leq y\cdot z$ (left-isotony),

  3. (3)

$x\cdot y\leq z$ if and only if $x\leq y\rightarrow z$ (left-adjointness),

  4. (4)

$x\leq y$ implies $z\rightarrow x\leq z\rightarrow y$,

  5. (5)

$x\leq y$ implies $y\rightarrow z\leq x\rightarrow z$,

  6. (6)

$x\cdot 1\approx x$,

  7. (7)

$x\cdot y\leq x,y$

for all $x,y,z\in Q$.

The above mentioned conditions are related as shown in the following Lemmas.

######  Lemma 3.1.

For $(Q,\leq,\cdot,\rightarrow,1)$ the following hold:

  1. (i)

(1) and (3) imply (5).

  2. (ii)

If $\cdot$ is commutative then (1) and (6) imply (7).

###### Proof.

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

  1. (i)

If $a\leq b$ then every one of the following statements implies the next one:

$\displaystyle b\rightarrow c$ $\displaystyle\leq b\rightarrow c,$
$\displaystyle(b\rightarrow c)\cdot b$ $\displaystyle\leq c,$
$\displaystyle(b\rightarrow c)\cdot a$ $\displaystyle\leq c,$ $\displaystyle
b\rightarrow c$ $\displaystyle\leq a\rightarrow c.$

  2. (ii)

We have $a\cdot b\leq a\cdot 1=a$ and $a\cdot b=b\cdot a\leq b$.

∎

######  Lemma 3.2.

Assume $(Q,\leq,\cdot,\rightarrow,1)$ with associative $\cdot$ to satisfy (2)
and (3). Then it satisfies

  1. (8)

$(x\cdot y)\rightarrow z\approx x\rightarrow(y\rightarrow z)$.

###### Proof.

Let $a,b,c\in Q$. Then every of the following statements implies the next one:

$\displaystyle(a\cdot b)\rightarrow c$ $\displaystyle\leq(a\cdot b)\rightarrow
c,$ $\displaystyle((a\cdot b)\rightarrow c)\cdot(a\cdot b)$ $\displaystyle\leq
c,$ $\displaystyle(((a\cdot b)\rightarrow c)\cdot a)\cdot b$
$\displaystyle\leq c,$ $\displaystyle((a\cdot b)\rightarrow c)\cdot a$
$\displaystyle\leq b\rightarrow c,$ $\displaystyle(a\cdot b)\rightarrow c$
$\displaystyle\leq a\rightarrow(b\rightarrow c).$

Moreover, every one of the following statements implies the next one:

$\displaystyle a\rightarrow(b\rightarrow c)$ $\displaystyle\leq
a\rightarrow(b\rightarrow c),$ $\displaystyle(a\rightarrow(b\rightarrow
c))\cdot a$ $\displaystyle\leq b\rightarrow c,$
$\displaystyle((a\rightarrow(b\rightarrow c))\cdot a)\cdot b$
$\displaystyle\leq(b\rightarrow c)\cdot b,$
$\displaystyle(a\rightarrow(b\rightarrow c))\cdot(a\cdot b)$
$\displaystyle\leq(b\rightarrow c)\cdot b.$

Together with $(b\rightarrow c)\cdot b\leq c$ which follows from $b\rightarrow
c\leq b\rightarrow c$ we obtain $(a\rightarrow(b\rightarrow c))\cdot(a\cdot
b)\leq c$ which implies $a\rightarrow(b\rightarrow c)\leq(a\cdot b)\rightarrow
c$. ∎

Now we define one of our main concepts.

######  Definition 3.3.

$(Q,\leq,\cdot,\rightarrow,1)$ is called a left-residuated groupoid if it
satisfies (3) and (6). It is called

  * •

bounded if $(Q,\leq)$ is bounded ($0$ is the bottom and $1$ the top element),

  * •

commutative if $\cdot$ is commutative,

  * •

associative if $\cdot$ is associative.

A commutative residuated monoid is a commutative and associative left-
residuated
groupoid.

An example of a bounded residuated monoid which is not a lattice is as
follows:

######  Example 3.4.

The poset visualized in Figure 1:

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

together with the operations given by

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

is a bounded commutative residuated monoid which is not a lattice.

The next lemma shows some elementary properties of left-residuated groupoids.

######  Lemma 3.5.

The following hold:

  1. (i)

Every left-residuated groupoid $(Q,\leq,\cdot,\rightarrow,1)$ satisfies (2),
(4) and

    1. (9)

$1\rightarrow x\approx x$.

  2. (ii)

If $(Q,\leq,\cdot,\rightarrow,1)$ satisfies (5) and (9) then it satisfies

    1. (10)

$x\leq y\rightarrow x$.

###### Proof.

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

  1. (i)

    1. (2)

If $a\leq b$ then every one of the following statements implies the next one:

$\displaystyle b\cdot c$ $\displaystyle\leq b\cdot c,$ $\displaystyle b$
$\displaystyle\leq c\rightarrow(b\cdot c),$ $\displaystyle a$
$\displaystyle\leq c\rightarrow(b\cdot c),$ $\displaystyle a\cdot c$
$\displaystyle\leq b\cdot c.$

    2. (4)

If $a\leq b$ then every one of the following statements implies the next one:

$\displaystyle c\rightarrow a$ $\displaystyle\leq c\rightarrow a,$
$\displaystyle(c\rightarrow a)\cdot c$ $\displaystyle\leq a,$
$\displaystyle(c\rightarrow a)\cdot c$ $\displaystyle\leq b,$ $\displaystyle
c\rightarrow a$ $\displaystyle\leq c\rightarrow b.$

    3. (9)

We have $1\rightarrow a\leq 1\rightarrow a$ implies $1\rightarrow
a=(1\rightarrow a)\cdot 1\leq a$, and $a\cdot 1\leq a$ implies $a\leq
1\rightarrow a$.

  2. (ii)

We have $a=1\rightarrow a\leq b\rightarrow a$.

∎

Now we show that every left-residuated groupoid naturally induces a left-
residuated groupoid on its full twist product.

######  Theorem 3.6.

Let $(Q,\leq,\cdot,\rightarrow,1)$ be a poset with binary operations $\cdot$
and $\rightarrow$ and a constant $1$, let $a,b\in Q$ and $f,g$ be surjective
mappings from $Q^{2}$ to $Q$ satisfying $f(a,b)=g(a,b)=1$ and consider the
full twist product $(Q^{2},\leq,\odot,\Rightarrow,(a,b))$ of $(Q,\leq)$ with
binary operations $\odot$ and $\Rightarrow$ defined by

$\displaystyle(x,y)\odot(z,v)$ $\displaystyle:=(x\cdot
f(z,v),g(z,v)\rightarrow y),$ $\displaystyle(x,y)\Rightarrow(z,v)$
$\displaystyle:=(f(x,y)\rightarrow z,v\cdot g(x,y))$

for all $(x,y),(z,v)\in Q^{2}$ and the constant $(a,b)$. Then
$(Q,\leq,\cdot,\rightarrow,1)$ is a left-residuated groupoid if and only if
$(Q^{2},\leq,\odot,\Rightarrow,$ $(a,b))$ has this property.

###### Proof.

We investigate when $(Q^{2},\leq,\odot,\Rightarrow,(a,b))$ satisfies (3) and
(6). The following are equivalent:

$\displaystyle(Q^{2},\leq,\odot,\Rightarrow,(a,b))\text{ satisfies (3)},$
$\displaystyle(x,y)\odot(z,v)\leq(t,w)\text{ is equivalent to
}(x,y)\leq(z,v)\Rightarrow(t,w),$ $\displaystyle(x\cdot
f(z,v),g(z,v)\rightarrow y)\leq(t,w)\text{ is equivalent to
}(x,y)\leq(f(z,v)\rightarrow t,w\cdot g(z,v)),$ $\displaystyle(x\cdot
f(z,v)\leq t\text{ and }w\leq g(z,v)\rightarrow y)\text{ is equivalent to}$
$\displaystyle\quad\quad(x\leq f(z,v)\rightarrow t\text{ and }w\cdot
g(z,v)\leq y),$ $\displaystyle(Q,\leq,\cdot,\rightarrow,1)\text{ satisfies
(3)}.$

Moreover, the following are equivalent:

$\displaystyle(Q^{2},\leq,\odot,\Rightarrow,(a,b))\text{ satisfies (6)},$
$\displaystyle(x,y)\odot(a,b)\approx(x,y),$ $\displaystyle(x\cdot
f(a,b),g(a,b)\rightarrow y)\approx(x,y),$ $\displaystyle(x\cdot 1,1\rightarrow
y)\approx(x,y),$ $\displaystyle x\cdot 1\approx x\text{ and }1\rightarrow
y\approx y,$ $\displaystyle(Q,\leq,\cdot,\rightarrow,1)\text{ satisfies (6)
and (7)}.$

Now Lemma 3.5 completes the proof. ∎

######  Corollary 3.7.

Let $(Q,\leq,\cdot,\rightarrow,1)$ be a poset with binary operations $\cdot$
and $\rightarrow$ and a constant $1$ and consider the full twist product
$(Q^{2},\leq,\odot,\Rightarrow,(1,1))$ of $(Q,\leq)$ with binary operations
$\odot$ and $\Rightarrow$ defined by

$\displaystyle(x,y)\odot(z,v)$ $\displaystyle:=(x\cdot z,v\rightarrow y),$
$\displaystyle(x,y)\Rightarrow(z,v)$ $\displaystyle:=(x\rightarrow z,v\cdot
y)$

for all $(x,y),(z,v)\in Q^{2}$ and the constant $(1,1)$. Then
$(Q,\leq,\cdot,\rightarrow,1)$ is a left-residuated groupoid if and only if
$(Q^{2},\leq,\odot,\Rightarrow,$ $(1,1))$ has this property.

###### Proof.

This is the special case of Theorem 3.6 where $a=b=1$, $f$ is the first and
$g$ the second projection. ∎

######  Corollary 3.8.

Let $(Q,\leq,\cdot,\rightarrow,1)$ be a poset with binary operations $\cdot$
and $\rightarrow$ and a constant $1$ and consider the full twist product
$(Q^{2},\leq,\odot,\Rightarrow,(1,1))$ of $(Q,\leq)$ with binary operations
$\odot$ and $\Rightarrow$ defined by

$\displaystyle(x,y)\odot(z,v)$ $\displaystyle:=(x\cdot v,z\rightarrow y),$
$\displaystyle(x,y)\Rightarrow(z,v)$ $\displaystyle:=(y\rightarrow z,v\cdot
x)$

for all $(x,y),(z,v)\in Q^{2}$ and the constant $(1,1)$. Then
$(Q,\leq,\cdot,\rightarrow,1)$ is a left-residuated groupoid if and only if
$(Q^{2},\leq,\odot,\Rightarrow,$ $(1,1))$ has this property.

###### Proof.

This is the special case of Theorem 3.6 where $a=b=1$, $f$ is the second and
$g$ the first projection. ∎

##  4 Operator residuated posets

One can easily see that the left-residuated groupoid
$(Q^{2},\leq,\odot,\Rightarrow,(a,b))$ from Theorem 3.6 need neither be
commutative nor associative even if $(Q,\leq,\cdot,\rightarrow,1)$ has this
property. Hence, we define the next concept.

######  Definition 4.1.

An operator residuated poset is an ordered six-tuple
$(Q,\leq,\odot,\Rightarrow,0,1)$ such that

  1. (i)

$(Q,\leq,0,1)$ is a bounded poset,

  2. (ii)

$\odot$ and $\Rightarrow$ are mappings from $Q^{2}$ to $2^{Q}$ (so-called
operators),

  3. (iii)

$x\odot y\approx y\odot x$,

  4. (iv)

$\bigcup\limits_{u\in x\odot y}(u\odot z)=\bigcup\limits_{u\in y\odot
z}(x\odot u)$ (operator associativity),

  5. (v)

$x\odot y\leq z$ if and only if $x\leq y\Rightarrow z$

for all $x,y,z\in Q$.

The following result shows that when using an operator residuated structure on
the full twist product, commutativity and associativity of the original
bounded left-residuated groupoid are preserved.

######  Theorem 4.2.

Let $(Q,\leq,\cdot,\rightarrow,0,1)$ be a bounded commutative residuated
monoid and $a_{0}\in Q$. Then $(Q^{2},\leq,\odot,\Rightarrow,(0,1),(1,0))$
where $(Q^{2},\leq)$ is the full twist product of $(Q,\leq)$ and the operators
$\odot$ and $\Rightarrow$ on $Q^{2}$ are defined by

$\displaystyle(x,y)\odot(z,v)$ $\displaystyle:=\\{(x\cdot z,x\rightarrow
v),(x\cdot z,z\rightarrow y)\\},$ $\displaystyle(x,y)\Rightarrow(z,v)$
$\displaystyle:=\\{(x\rightarrow z,x\cdot v),(v\rightarrow y,x\cdot v)\\}$

for all $(x,y),(z,v)\in Q^{2}$ is an operator residuated poset and the mapping
$x\mapsto(x,a_{0})$ an embedding of $(Q,\leq)$ into $(Q^{2},\leq)$.

###### Proof.

Let $a,b,c,d,e,f\in Q$. According to Lemmas 3.1, 3.2 and 3.5,
$(Q,\leq,\cdot,\rightarrow,0,1)$ satisfies (1) – (10).

  1. (i)

It is evident that $(Q^{2},\leq,(0,1),(1,0))$ is a bounded poset.

  2. (ii)

$\odot$ and $\Rightarrow$ are mappings from $(Q^{2})^{2}$ to $2^{(Q^{2})}$.

We must prove (iii) – (v) of Definition 4.1.

  1. (iii)

We have

$\displaystyle(a,b)\odot(c,d)$ $\displaystyle=\\{(a\cdot c,a\rightarrow
d),(a\cdot c,c\rightarrow b)\\}=\\{(c\cdot a,c\rightarrow b),(c\cdot
a,a\rightarrow d)\\}=$ $\displaystyle=(c,d)\odot(a,b).$

  2. (iv)

We have

$\displaystyle\bigcup_{(x,y)\in(a,b)\odot(c,d)}((x,y)\odot(e,f))$
$\displaystyle=\bigcup_{(x,y)\in\\{(a\cdot c,a\rightarrow d),(a\cdot
c,c\rightarrow b)\\}}((x,y)\odot(e,f))=$ $\displaystyle=((a\cdot
c,a\rightarrow d)\odot(e,f))\cup((a\cdot c,c\rightarrow b)\odot(e,f))=$
$\displaystyle=\\{((a\cdot c)\cdot e,(a\cdot c)\rightarrow f),((a\cdot c)\cdot
e,e\rightarrow(a\rightarrow d)),$ $\displaystyle\hskip 17.07164pt((a\cdot
c)\cdot e,(a\cdot c)\rightarrow f),((a\cdot c)\cdot
e,e\rightarrow(c\rightarrow b))\\}=$ $\displaystyle=\\{(a\cdot(c\cdot
e),a\rightarrow(c\rightarrow f)),(a\cdot(c\cdot e),(c\cdot e)\rightarrow b),$
$\displaystyle\hskip 17.07164pt(a\cdot(c\cdot e),a\rightarrow(e\rightarrow
d)),(a\cdot(c\cdot e),(c\cdot e)\rightarrow b)\\}=$
$\displaystyle=((a,b)\odot(c\cdot e,c\rightarrow f))\cup((a,b)\odot(c\cdot
e,e\rightarrow d))=$ $\displaystyle=\bigcup_{(x,y)\in\\{(c\cdot e,c\rightarrow
f),(c\cdot e,e\rightarrow d)\\}}((a,b)\odot(x,y))=$
$\displaystyle=\bigcup_{(x,y)\in(c,d)\odot(e,f)}((a,b)\odot(x,y)).$

  3. (v)

The following are equivalent:

$\displaystyle(a,b)\odot(c,d)\leq(e,f),$ $\displaystyle\\{(a\cdot
c,a\rightarrow d),(a\cdot c,c\rightarrow b)\\}\leq(e,f),$ $\displaystyle
a\cdot c\leq e,f\leq a\rightarrow d\text{ and }f\leq c\rightarrow b,$
$\displaystyle a\leq c\rightarrow e,a\leq f\rightarrow d\text{ and }c\cdot
f\leq b,$ $\displaystyle(a,b)\leq\\{(c\rightarrow e,c\cdot f),(f\rightarrow
d,c\cdot f)\\},$ $\displaystyle(a,b)\leq(c,d)\Rightarrow(e,f).$

Finally, $(a,a_{0})\leq(b,a_{0})$ is equivalent to $a\leq b$. ∎

######  Example 4.3.

If $(Q,\leq,\cdot,\rightarrow,0,1):=(\\{0,1\\},\leq,\cdot,(x,y)\mapsto
1-x+xy,0,1)$ (where $+$, $-$ and $\cdot$ denote addition, subtraction and
multiplication of the reals, respectively) then the tables for $\odot$ and
$\Rightarrow$ look as follows:

$\begin{array}[]{c|c|c|c|c}\odot&(0,0)&(0,1)&(1,0)&(1,1)\\\
\hline\cr(0,0)&\\{(0,1)\\}&\\{(0,1)\\}&\\{(0,0),(0,1)\\}&\\{(0,0),(0,1)\\}\\\
(0,1)&\\{(0,1)\\}&\\{(0,1)\\}&\\{(0,1)\\}&\\{(0,1)\\}\\\
(1,0)&\\{(0,0),(0,1)\\}&\\{(0,1)\\}&\\{(1,0)\\}&\\{(1,0),(1,1)\\}\\\
(1,1)&\\{(0,0),(0,1)\\}&\\{(0,1)\\}&\\{(1,0),(1,1)\\}&\\{(1,1)\\}\end{array}$
$\begin{array}[]{c|c|c|c|c}\Rightarrow&(0,0)&(0,1)&(1,0)&(1,1)\\\
\hline\cr(0,0)&\\{(1,0)\\}&\\{(0,0),(1,0)\\}&\\{(1,0)\\}&\\{(0,0),(1,0)\\}\\\
(0,1)&\\{(1,0)\\}&\\{(1,0)\\}&\\{(1,0)\\}&\\{(1,0)\\}\\\
(1,0)&\\{(0,0),(1,0)\\}&\\{(0,1)\\}&\\{(1,0)\\}&\\{(0,1),(1,1)\\}\\\
(1,1)&\\{(0,0),(1,0)\\}&\\{(0,1),(1,1)\\}&\\{(1,0)\\}&\\{(1,1)\\}\end{array}$

##  5 Pseudo-Kleene posets

It was shown by the authors in [4] that every poset $\mathbf{Q}=(Q,\leq)$ can
be embedded into a pseudo-Kleene one. For this we use a certain modification
of the full twist product construction as follows.

Let $a\in Q$ and define

$\displaystyle P_{a}(\mathbf{Q})$ $\displaystyle:=\\{(x,y)\in Q^{2}\mid
L(x,y)\leq a\leq U(x,y)\\},$ $\displaystyle(x,y)\leq(z,v)$
$\displaystyle:\Leftrightarrow(x\leq z\text{ and }v\leq y),$
$\displaystyle(x,y)^{\prime}$ $\displaystyle:=(y,x)$

for all $(x,y),(z,v)\in Q^{2}$. The following was proved in [4]:

  * •

$(P_{a}(\mathbf{Q}),\leq,{}^{\prime})$ is a pseudo-Kleene poset,

  * •

the mapping $x\mapsto(x,a)$ is an embedding of $\mathbf{Q}$ into
$(P_{a}(\mathbf{Q}),\leq)$,

  * •

$(P_{a}(\mathbf{Q}),\leq,{}^{\prime})$ is a Kleene poset if and only if
$\mathbf{Q}$ is distributive.

Since $P_{a}(\mathbf{Q})$ is a subset of the full twist product of
$\mathbf{Q}$, it is a question if residuation from
$(Q,\leq,\cdot,\rightarrow,1)$ can be transferred to $P_{a}(\mathbf{Q})$ as
shown in Theorem 3.6. Unfortunately, this is not possible in general since
$P_{a}(\mathbf{Q})$ need not be closed under the operators $\odot$ and
$\Rightarrow$ defined in Theorem 3.6. However, we can get necessary and
sufficient conditions under which $P_{a}(\mathbf{Q})$ is closed under these
operators and hence becomes a pseudo-Kleene operator residuated poset.

If $\mathbf{Q}=(Q,\leq)$ is a poset, $a,b\in Q$ and every element of
$P_{a}(\mathbf{Q})$ is comparable with $(a,a)$ then $(a,b)\in
P_{a}(\mathbf{Q})$ and hence $(a,a)\leq(a,b)$ or $(a,b)\leq(a,a)$ whence
$b\leq a$ or $a\leq b$, i.e. $b$ is comparable with $a$.

######  Theorem 5.1.

Let $(Q,\leq,\cdot,\rightarrow,0,1)$ be a bounded commutative residuated
monoid and $a\in Q$ with $a\cdot a=a$, put $\mathbf{Q}:=(Q,\leq)$ and assume
that all elements of $P_{a}(\mathbf{Q})$ are comparable with $(a,a)$. Then

  * •

$(P_{a}(\mathbf{Q}),\leq,\odot,\Rightarrow,(0,1),(1,0))$ where the operators
$\odot$ and $\Rightarrow$ are defined by

$\displaystyle(x,y)\odot(z,v)$ $\displaystyle:=\\{(x\cdot z,x\rightarrow
v),(x\cdot z,z\rightarrow y)\\},$ $\displaystyle(x,y)\Rightarrow(z,v)$
$\displaystyle:=\\{(x\rightarrow z,x\cdot v),(v\rightarrow y,x\cdot v)\\}$

for all $(x,y),(z,v)\in P_{a}(\mathbf{Q})$ is an operator residuated poset if
and only if the following two conditions hold:

    1. (11)

$a\cdot x<a$ implies $a\cdot x=0$,

    2. (12)

$a<x$ implies $x\rightarrow a=a$.

  * •

$(P_{a}(\mathbf{Q}),\leq,{}^{\prime})$ where $(x,y)^{\prime}:=(y,x)$ for all
$(x,y)\in P_{a}(\mathbf{Q})$ is a pseudo-Kleene poset.

  * •

The mapping $x\mapsto(x,a)$ is an embedding of $\mathbf{Q}$ into
$(P_{a}(\mathbf{Q}),\leq)$.

  * •

$(x,y)^{\prime}\in(x,y)\Rightarrow(0,1)$ for all $(x,y)\in P_{a}(\mathbf{Q})$.

###### Proof.

Let $b,c,d,e\in Q$. According to Lemmas 3.1, 3.2 and 3.5,
$(Q,\leq,\cdot,\rightarrow,0,1)$ satisfies (1) – (10). If $a\leq x,y$ then
$a=a\cdot a\leq a\cdot y\leq x\cdot y$ according to (1) and (2), i.e. we have

  1. (13)

$a\leq x,y$ implies $a\leq x\cdot y$.

  * •

Assume $(b,c),(d,e)\leq(a,a)$. Then the following hold:
$(b\cdot d,b\rightarrow e)\leq(a,a)$ because of (7) and (10).
$(b\cdot d,d\rightarrow c)\leq(a,a)$ because of (7) and (10).
Since $b\leq a$ we have $b\cdot e\leq a$ according to (7).
If $b\cdot e=a$ then $(b\rightarrow d,b\cdot e)$ is comparable with $(a,a)$.
If $b\cdot e<a$ then $(b\rightarrow d,b\cdot e)$ is comparable with $(a,a)$ if
and only if $a\leq b\rightarrow d$.
$(e\rightarrow c,b\cdot e)\geq(a,a)$ because of (7) and (10).

  * •

Assume $(b,c)\leq(a,a)\leq(d,e)$. Then the following hold:
Since $b\leq a$ we have $b\cdot d\leq a$ according to (7).
If $b\cdot d=a$ then $(b\cdot d,b\rightarrow e)$ is comparable with $(a,a)$.
If $b\cdot d<a$ then $(b\cdot d,b\rightarrow e)$ is comparable with $(a,a)$ if
and only if $a\leq b\rightarrow e$.
$(b\cdot d,d\rightarrow c)\leq(a,a)$ because of (7) and (10).
$(b\rightarrow d,b\cdot e)\geq(a,a)$ because of (7) and (10).
$(e\rightarrow c,b\cdot e)\geq(a,a)$ because of (7) and (10).

  * •

Assume $(d,e)\leq(a,a)\leq(b,c)$. Then the following hold:
$(b\cdot d,b\rightarrow e)\leq(a,a)$ because of (7) and (10).
Since $d\leq a$ we have $b\cdot d\leq a$ according to (7).
If $b\cdot d=a$ then $(b\cdot d,d\rightarrow c)$ is comparable with $(a,a)$.
If $b\cdot d<a$ then $(b\cdot d,d\rightarrow c)$ is comparable with $(a,a)$ if
and only if $a\leq d\rightarrow c$.
Since $a\leq b,e$ we have $a\leq b\cdot e$ according to (13).
If $b\cdot e=a$ then $(b\rightarrow d,b\cdot e)$ is comparable with $(a,a)$.
If $a<b\cdot e$ then $(b\rightarrow d,b\cdot e)$ is comparable with $(a,a)$ if
and only if $b\rightarrow d\leq a$.
If $b\cdot e=a$ then $(e\rightarrow c,b\cdot e)$ is comparable with $(a,a)$.
If $a<b\cdot e$ then $(e\rightarrow c,b\cdot e)$ is comparable with $(a,a)$ if
and only if $e\rightarrow c\leq a$.

  * •

Assume $(a,a)\leq(b,c),(d,e)$. Then the following hold:
Since $a\leq b,d$ we have $a\leq b\cdot d$ according to (13).
If $b\cdot d=a$ then $(b\cdot d,b\rightarrow e)$ is comparable with $(a,a)$.
If $a<b\cdot d$ then $(b\cdot d,b\rightarrow e)$ is comparable with $(a,a)$ if
and only if $b\rightarrow e\leq a$.
If $b\cdot d=a$ then $(b\cdot d,d\rightarrow c)$ is comparable with $(a,a)$.
If $a<b\cdot d$ then $(b\cdot d,d\rightarrow c)$ is comparable with $(a,a)$ if
and only if $d\rightarrow c\leq a$.
$(b\rightarrow d,b\cdot e)\geq(a,a)$ because of (7) and (10),
Since $e\leq a$ we have $b\cdot e\leq a$ according to (7).
If $b\cdot e=a$ then $(e\rightarrow c,b\cdot e)$ is comparable with $(a,a)$.
If $b\cdot e<a$ then $(e\rightarrow c,b\cdot e)$ is comparable with $(a,a)$ if
and only if $a\leq e\rightarrow c$.

Hence $(x,y)\odot(z,v)\subseteq P_{a}(\mathbf{Q})$ and
$(x,y)\Rightarrow(z,v)\subseteq P_{a}(\mathbf{Q})$ for all $(x,y),(z,v)\in
P_{a}(\mathbf{Q})$ if and only if the following statements hold:

  1. (a)

$b,d\leq a\leq c,e$ and $b\cdot e<a$ imply $a\leq b\rightarrow d$.

  2. (b)

$b,e\leq a\leq c,d$ and $b\cdot d<a$ imply $a\leq b\rightarrow e$.

  3. (c)

$c,d\leq a\leq b,e$ and $b\cdot d<a$ imply $a\leq d\rightarrow c$.

  4. (d)

$c,d\leq a\leq b,e$ and $a<b\cdot e$ imply $b\rightarrow d\leq a$.

  5. (e)

$c,d\leq a\leq b,e$ and $a<b\cdot e$ imply $e\rightarrow c\leq a$.

  6. (f)

$c,e\leq a\leq b,d$ and $a<b\cdot d$ imply $b\rightarrow e\leq a$.

  7. (g)

$c,e\leq a\leq b,d$ and $a<b\cdot d$ imply $d\rightarrow c\leq a$.

  8. (h)

$c,e\leq a\leq b,d$ and $b\cdot e<a$ imply $a\leq e\rightarrow c$.

Now (a) is equivalent to the following statements:

$\displaystyle b\cdot a<a\text{ implies }a\leq b\rightarrow 0,$ $\displaystyle
a\cdot b<a\text{ implies }a\cdot b\leq 0,$ $\displaystyle a\cdot b<a\text{
implies }a\cdot b=0,$ $\displaystyle(11).$

In the same way one can see that (b), (c) and (h) are equivalent to (11).
Moreover, (d) is equivalent to the following statements:

$\displaystyle a\leq b,e\text{ and }a<b\cdot e\text{ imply }b\rightarrow a\leq
a,$ $\displaystyle a<b\text{ implies }b\rightarrow a\leq a,$ $\displaystyle
a<b\text{ implies }b\rightarrow a=a,$ $\displaystyle(12).$

In the same way one can see that (e), (f) and (g) are equivalent to (12).
Moreover, we have

$(b,c)^{\prime}=(c,b)\in\\{(b\rightarrow 0,b),(c,b)\\}=\\{(b\rightarrow
0,b\cdot 1),(1\rightarrow c,b\cdot 1)\\}=(b,c)\Rightarrow(0,1).$

The rest follows from Theorem 4.2. ∎

######  Example 5.2.

Consider the bounded commutative residuated semigroup
$(Q,\leq,\cdot,\rightarrow,0,$ $1)$ with $Q=\\{0,a,1\\}$, $0<a<1$ and

$\begin{array}[]{c|ccc}\cdot&0&a&1\\\ \hline\cr 0&0&0&0\\\ a&0&a&a\\\
1&0&a&1\end{array}\quad\quad\quad\begin{array}[]{c|ccc}\rightarrow&0&a&1\\\
\hline\cr 0&1&1&1\\\ a&0&1&1\\\ 1&0&a&1\end{array}$

and put $\mathbf{Q}:=(Q,\leq)$. It is easy to check that $\mathbf{Q}$ is a
distributive lattice and $(Q,\leq,\cdot,\rightarrow,0,1)$ is a bounded
commutative residuated monoid satisfying all the assumptions of Theorem 5.1.
The poset $(P_{a}(\mathbf{Q}),\leq)$ is depicted in Figure 2:

$(0,1)$$(0,a)$$(a,1)$$(a,a)$$(1,a)$$(a,0)$$(1,0)$Fig. 2

Then the operators $\odot$ and $\Rightarrow$ have the following tables:

$\begin{array}[]{c|c|c|c|c|c|c|c}\odot&0a&01&a0&aa&a1&10&1a\\\ \hline\cr
0a&01&01&01&01&01&0a,01&0a,01\\\ 01&01&01&01&01&01&01&01\\\
a0&01&01&a0&a0,a1&a0,a1&a0&a0,a1\\\ aa&01&01&a0,a1&a1&a1&a0,aa&aa,a1\\\
a1&01&01&a0,a1&a1&a1&a0,a1&a1\\\ 10&0a,01&01&a0&a0,aa&a0,a1&10&10,1a\\\
1a&0a,01&01&a0,a1&aa,a1&a1&10,1a&1a\end{array}$
$\begin{array}[]{c|c|c|c|c|c|c|c}\Rightarrow&0a&01&a0&aa&a1&10&1a\\\ \hline\cr
0a&10&a0,10&10&10&a0,10&10&10\\\ 01&10&10&10&10&10&10&10\\\
a0&0a&0a&10&0a,1a&0a,1a&10&0a,1a\\\ aa&0a,1a&0a,aa&10&1a&aa,1a&10&1a\\\
a1&0a,1a&0a,1a&10&1a&1a&10&1a\\\ 10&0a&01&a0,10&0a,aa&01,a1&10&0a,1a\\\
1a&0a,1a&01,a1&a0,10&aa,1a&a1&10&1a\end{array}$

Hence $(P_{a}(\mathbf{Q}),\leq,\odot,\Rightarrow,(0,1),(1,0))$ is an operator
residuated poset $(P_{a}(\mathbf{Q}),\leq,{}^{\prime})$ is a Kleene lattice.

######  Example 5.3.

On the other hand, the bounded residuated monoid
$(Q,\leq,\cdot,\rightarrow,1)$ from Example 3.4 has only two idempotents,
namely $0$ and $1$. Every element of $P_{0}(\mathbf{Q})=(\\{0\\}\times
Q)\cup(Q\times\\{0\\})$ is comparable with $(0,0)$. But if $x\neq 1$ then
$0<x$, but $x\rightarrow 0\neq 0$ contradicting (12). Similarly, every element
of $P_{1}(\mathbf{Q})=(\\{1\\}\times Q)\cup(Q\times\\{1\\})$ is comparable
with $(1,1)$. But if $x\neq 0,1$ then $0<1\cdot x<1$ contradicting (11).

## References

  * [1] 9
  * [2] M. Busaniche and R. Cignoli, The subvariety of commutative residuated lattices represented by twist-products. Algebra Universalis 71 (2014), 5–22.
  * [3] I. Chajda, A note on pseudo-Kleene algebras, Acta Univ. Palack. Olomuc. Fac. Rerum Natur., Math. 55 (2016), 39–45.
  * [4] I. Chajda and H. Länger, Kleene posets and pseudo-Kleene posets. Miskolc Math. Notes (submitted). http://arxiv.org/abs/2006.04417.
  * [5] R. Cignoli, Injective De Morgan and Kleene algebras. Proc. Amer. Math. Soc. 47 (1975), 269–278.
  * [6] J. A. Kalman, Lattices with involution. Trans. Amer. Math. Soc. 87 (1958), 485–491.
  * [7] C. Tsinakis and A. M. Wille, Minimal varieties of involutive residuated lattices. Studia Logica 83 (2006), 407–423.

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>