Tense Logic Based on Finite Orthomodular Posets

Abstract

It is widely accepted that the logic of quantum mechanics is based on orthomodular posets. However, such a logic is not dynamic in the sense that it does not incorporate time dimension. To fill this gap, we introduce certain tense operators on such a logic in an inexact way, but still satisfying requirements asked on tense operators in the classical logic based on Boolean algebras or in various non-classical logics. Our construction of tense operators works perfectly when the orthomodular poset in question is finite. We investigate the behaviour of these tense operators, e.g. we show that some of them form a dynamic pair. Moreover, we prove that if the tense operators preserve one of the inexact connectives conjunction or implication as defined by the authors recently in another paper, then they also preserve the other one. Finally, we show how to construct the binary relation of time preference on a given time set provided the tense operators are given, up to equivalence induced by natural quasi-orders.

1 Introduction

At the beginning of the twentieth century it was recognized that the logic of quantum mechanics differs essentially from classical propositional calculus based algebraically on Boolean algebras. Husimi [12] and Birkhoff together with von Neumann [3] introduced orthomodular lattices in order to serve as an algebraic base for the logic of quantum mechanics, see [1]. These lattices incorporate many aspects of this logic with one exception. Namely, within the logic of quantum mechanics the disjunction of two propositions need not exist in the case when these propositions are neither orthogonal nor comparable. This fact motivated a number of researchers to consider orthomodular posets instead of orthomodular lattices within their corresponding investigations, see e.g. [14] and references therein.

A propositional logic, either classical or non-classical, usually does not incorporate the dimension of time. In order to organize this logic as a so-called tense logic (or time logic in another terminology, see [10, 11, 15, 16]) we usually construct the so-called tense operators P, F, H and G. Their meaning is as follows:

$$ \begin{array}{@{}rcl@{}} P & \ldots&\text{``It has at some time been the case that''}, \\ F & \ldots&\text{``It will at some time be the case that''}, \\ H & \ldots&\text{``It has always been the case that''}, \\ G & \ldots&\text{``It will always be the case that''}. \end{array} $$

As the reader may guess, we need a certain time scale. For this reason a time frame(T,R) was introduced. By T≠∅ is meant a set of time, either finite or infinite, and $R\subseteq T^{2}$, R≠∅, is the relation of time preference, i.e. for s,t ∈ T we say that s R t means s is “before” t, or t is “after” s.

For our purposes in this paper we will consider only so-called serial relations (see [7]), i.e. relations R such that for each s ∈ T there exist some r,t ∈ T with r R s and s R t.

Of course, if R is reflexive then it is serial. Usually, R is considered to be a partial order relation or a quasi-order, see e.g. [4, 9, 10].

Another important task in tense logics is to construct for given tense operators a relation R on a given time set T such that the new tense operators constructed by means of this relation R coincide with the given tense operators.

The problem arises when our logic is based on a poset (A,≤) where joins and meets need not exist. In [6] such a situation has been solved by embedding of A^T into a complete lattice. Then we can use the aforementioned definitions of tense operators P, F, H and G, but the disadvantage is that the results of these operators need not belong to A^T again. This motivated us to try another approach, see Section 3.

Moreover, if ⊙ denotes conjunction and → implication within the given propositional logic, we usually ask our tense operators to satisfy the following inequalities and equalities:

• (1) P(x) ⊙ H(y) ≤ P(x ⊙ y),

• (2) H(x) ⊙ P(y) ≤ P(x ⊙ y),

• (3) H(x) ⊙ H(y) = H(x ⊙ y),

• (4) F(x) ⊙ G(y) ≤ F(x ⊙ y),

• (5) G(x) ⊙ F(y) ≤ F(x ⊙ y),

• (6) G(x) ⊙ G(y) = G(x ⊙ y),

• (7) $P(x\rightarrow y)\leq H(x)\rightarrow P(y)$,

• (8) $H(x\rightarrow y)\leq P(x)\rightarrow P(y)$,

• (9) $H(x\rightarrow y)\leq H(x)\rightarrow H(y)$,

• (10) $F(x\rightarrow y)\leq G(x)\rightarrow F(y)$,

• (11) $G(x\rightarrow y)\leq F(x)\rightarrow F(y)$,

• (12) $G(x\rightarrow y)\leq G(x)\rightarrow G(y)$.

For our considerations we need the connectives ⊙ and → introduced in an orthomodular poset. In general this task is ambiguous, see [6, 8, 10, 13], but for finite orthomodular posets (in fact for orthomodular posets of finite height) it was solved by the authors in [5] where these connectives are introduced in such a way that one gets an adjoint pair. This is the reason why we will restrict ourselves to finite orthomodular posets only.

The connectives introduced in [5] are everywhere defined, but their results need not be elements of the given orthomodular poset, but may be subsets of this poset containing (incomparable) elements of the same (maximal) truth value. This may cause a problem for the composition of tense operators. Namely, we want to show that our couple (P,G) of tense operators forms a dynamic pair on the given orthomodular poset $\mathbf A=(A,\leq ,{}^{\prime },0,1)$, i.e. the following axioms (P1) – (P3) hold (see [7]):

• (P1) G(1) = 1 and P(0) = 0,

• (P2) p ≤ q implies G(p) ≤ G(q) and P(p) ≤ P(q),

• (P3) q ≤ (G ∘ P)(q) and (P ∘ G)(q) ≤ q.

If $G,P\colon A^{T}\rightarrow (\mathcal P_{+}A)^{T}$ for some time frame (T,R) then we must solve the problem how to define the compositions G ∘ P and P ∘ G in (P3). Here and in the following $\mathcal P_{+}A$ denotes the set 2^A ∖{∅}.

2 Preliminaries

The following concepts are taken from [1, 2].

Consider a poset (A,≤). If a,b ∈ A and $\sup (a,b)$ exists then we will denote it by a ∨ b. If $\inf (a,b)$ exists, it will be denoted by a ∧ b.

A unary operation ^′ on A is called an antitone involution if a ≤ b implies $b^{\prime }\leq a^{\prime }$ and if $a^{\prime \prime }=a$ for each a,b ∈ A. If the poset (A,≤) has a bottom or top element, this element will be denoted by 0 or 1, respectively, and we will write (A,≤,0,1) in order to express the fact that (A,≤) is bounded.

A complementation on a bounded poset (A,≤,0,1) is a unary operation ^′ on A satisfying $a\vee a^{\prime }=1$ and $a\wedge a^{\prime }=0$ for each a ∈ A. For a bounded poset (A,≤,0,1) with an antitone involution ^′ that is a complementation we will write $(A,\leq ,{}^{\prime },0,1)$.

Definition 2.1

An orthomodular poset is a bounded poset $(A,\leq ,{}^{\prime },0,1)$ with an antitone involution ^′ that is a complementation satisfying the following two conditions:

• (i) If x,y ∈ A and $x\leq y^{\prime }$ then x ∨ y is defined,

• (ii) if x,y ∈ A and x ≤ y then $y=x\vee (y\wedge x^{\prime })$.

Condition (ii) is called the orthomodular law. Two elements a,b of A are called orthogonal to each other if $a\leq b^{\prime }$ (which is equivalent to $b\leq a^{\prime }$).

Let us note that due to De Morgan’s laws, (ii) of Definition 2.1 is equivalent to the condition

• (ii) If x,y ∈ A and x ≤ y then $x=y\wedge (x\vee y^{\prime })$.

Throughout the paper we will consider only finite orthomodular posets. An example of a finite orthomodular poset which is not a lattice is depicted in Fig. 1.

Fig. 1

A non-lattice orthomodular poset

This orthomodular poset is not a lattice since a ∨ b does not exist.

Consider a poset (A,≤) and a time frame (T,R) and let B,C be subsets of A and $x,y\in (\mathcal P_{+}A)^{T}$. We define

$$ \begin{array}{@{}rcl@{}} B\leq C & :\Leftrightarrow&b\leq c\text{ for all }b\in B\text{ and all }c\in C, \\ B\leq_{1}C & :\Leftrightarrow&\text{for every }b\in B\text{ there exists some }c\in C\text{ with }b\leq c, \\ B\leq_{2}C & :\Leftrightarrow&\text{for every }c\in C\text{ there exists some }b\in B\text{ with }b\leq c, \\ B\sqsubseteq C & :\Leftrightarrow&\text{there exists some }b\in B\text{ and some }c\in C\text{ with }b\leq c, \\ x\leq y & :\Leftrightarrow& x(t)\leq y(t)\text{ for all }t\in T, \\ x\leq_{1}y & :\Leftrightarrow& x(t)\leq_{1}y(t)\text{ for all }t\in T, \\ x\leq_{2}y & :\Leftrightarrow& x(t)\leq_{2}y(t)\text{ for all }t\in T, \\ x\sqsubseteq y & :\Leftrightarrow& x(t)\sqsubseteq y(t)\text{ for all }t\in T. \end{array} $$

For $B,C\in \mathcal P_{+}(A^{T})$ we define

$$ \begin{array}{@{}rcl@{}} B\leq C & :\Leftrightarrow& p\leq q\text{ for all }p\in B\text{ and all }q\in C, \\ B\leq_{1}C & :\Leftrightarrow& p\leq_{1}q\text{ for all }p\in B\text{ and all }q\in C, \\ B\leq_{2}C & :\Leftrightarrow& p\leq_{2} q\text{ for all }p\in B\text{ and all }q\in C, \\ B\sqsubseteq C & :\Leftrightarrow& p\sqsubseteq q\text{ for all }p\in B\text{ and all }q\in C. \end{array} $$

Hereby, for a ∈ A, a_t ∈ A for all t ∈ T and p ∈ A^T we identify {a} with a, the mapping assigning to each t∈ T the set {a_t} with the mapping assigning to each t∈ T the element a_t, {p} with p.

Let (A,≤) be a poset, b,c ∈ A and $B,C\subseteq A$. The sets

$$ \begin{array}{@{}rcl@{}} L(B) & :=&\{x\in A\mid x\leq B\}, \\ U(B) & :=&\{x\in A\mid B\leq x\} \end{array} $$

are called the lower cone and upper cone of B, respectively. Instead of L(B ∪ C), L(B ∪{c}), L({b,c}) and L(U(B)) we write L(B,C), L(B,c), L(b,c) and LU(B), respectively. Analogously, we proceed in similar cases. Moreover, we denote the set of all maximal (minimal) elements of B by MaxB (MinB). If A is finite and B≠∅ then MaxB,MinB≠∅. If b ∨ c exists for all c ∈ C then we will denote {b ∨ c∣c ∈ C} by b ∨ C. Analogously, we proceed in similar cases.

The following result was proved in [5]. For the convenience of the reader we repeat the proof.

Proposition 2.2

Let $(A,\leq ,{}^{\prime },0,1)$ be an orthomodular poset, a,b ∈ A and $B,C\subseteq A$. Then the following hold:

• (i) If a ≤ B then $B=a\vee (B\wedge a^{\prime })$,

• (ii) if C ≤ a then $C=a\wedge (C\vee a^{\prime })$,

• (iii) $\min U(a,b^{\prime })\wedge b$ and $a^{\prime }\vee \max L(a,b)$ are defined.

Here and in the following $\min U(a,b^{\prime })\wedge b$ means $(\min U(a,b^{\prime }))\wedge b$. Analogous notations are used in the sequel.

Proof Proof of Proposition 2.2

• (i) If a ≤ B then $B=\{b\mid b\in B\}=\{a\vee (b\wedge a^{\prime })\mid b\in B\}=a\vee (B\wedge a^{\prime })$.

• (ii) If C ≤ a then $C=\{c\mid c\in C\}=\{a\wedge (c\vee a^{\prime })\mid c\in C\}=a\wedge (C\vee a^{\prime })$.

• (iii) Because of $\max L(a^{\prime },b)\leq b^{\prime \prime }$, $\max L(a^{\prime },b)\vee b^{\prime }$ and hence $\min U(a,b^{\prime })\wedge b$ is defined. Because of $\max L(a,b)\leq a^{\prime \prime }$, $a^{\prime }\vee \max L(a,b)$ is defined.

□

Let us note that both ≤₁ and ≤₂ are extensions of ≤ and that they are reflexive and transitive, i.e. they are quasi-orders.

In [5] we investigated the so-called inexact connectives in the logic based on a finite orthomodular poset. These are denoted by ⊙ (conjunction) and → (implication) and defined by

• (13) $x\odot y:=\min U(x,y^{\prime })\wedge y\quad$ and $\quad x\rightarrow y:=x^{\prime }\vee \max L(x,y)$.

Due to Proposition 2.2 these expressions are everywhere defined and ⊙ and → are binary operators on A, more precisely, mappings from A² to $\mathcal P_{+}A$. We extend them to $(\mathcal P_{+}A)^{2}$ by defining

$$ \begin{array}{@{}rcl@{}} B\odot C & :=\bigcup\{b\odot c\mid b\in B,c\in C\}, \\ B\rightarrow C & :=\bigcup\{b\rightarrow c\mid b\in B,c\in C\} \end{array} $$

for all $B,C\in \mathcal P_{+}A$. The extended operators ⊙ and → are now mappings from $(\mathcal P_{+}A)^{2}$ to $\mathcal P_{+}A$. In [5] we showed that $(\odot ,\rightarrow )$ form an adjoint pair, i.e.

$$ x\odot y\sqsubseteq z\text{ if and only if }x\sqsubseteq y\rightarrow z. $$

Moreover, we define

$$ \begin{array}{@{}rcl@{}} (x\odot y)(t) & :=x(t)\odot y(t), \\ (x\rightarrow y)(t) & :=x(t)\rightarrow y(t) \end{array} $$

for all $x,y\in (\mathcal P_{+}A)^{T}$ and all t ∈ T.

We recall from [5] the following definition and theorem:

Definition 2.3

An operator residuated structure is a six-tuple $\mathbf A=(A,\leq ,\odot ,\rightarrow ,0,1)$ satisfying the following conditions:

• (i) (A,≤,0,1) is a bounded poset,

• (ii) ⊙ and → are mappings from A² to $\mathcal P_{+}A$,

• (iii) $x\odot y\sqsubseteq z$ if and only if $x\sqsubseteq y\rightarrow z$,

• (iv) x ⊙ 1 ≈ 1 ⊙ x ≈ x,

• (v) $y\rightarrow 0\leq x$ implies x ⊙ y = x ∧ y,

• (vi) y ≤ x implies $x\rightarrow y=(x\rightarrow 0)\vee y$.

A is called

• idempotent if x ⊙ x ≈ x,

• divisible if $(x\rightarrow y)\odot x\approx \max L(x,y)$.

The expression x ⊙ 1 ≈ 1 ⊙ x ≈ x means that x ⊙ 1 = 1 ⊙ x = x holds for all x ∈ A. Similarly, the symbol ≈ is used in other places of the paper.

Theorem 2.4

Let $\mathbf A=(A,\leq ,{}^{\prime },0,1)$ be a finite orthomodular poset and define mappings $\odot ,\rightarrow \colon A^{2}\rightarrow \mathcal P_{+}A$ by

$$ \begin{array}{@{}rcl@{}} x\odot y & :=&\min U(x,y^{\prime})\wedge y, \\ x\rightarrow y & :=&x^{\prime}\vee\max L(x,y) \end{array} $$

for all x,y ∈ A. Then $(A,\leq ,\odot ,\rightarrow ,0,1)$ is an idempotent divisible operator residuated structure.

In the following we will investigate tense operators on finite orthomodular posets equipped with the operators ⊙ and → forming a certain propositional logic of quantum mechanics.

Lemma 2.5

Let $(A,\leq ,{}^{\prime },0,1)$ be a finite orthomodular poset and p,q ∈ A^T. Then

• (i) $p\leq q\rightarrow (p\odot q)$,

• (ii) $(p\rightarrow q)\odot p\leq q$.

Proof

• (i) Since $r\in \min U(p,q^{\prime })\wedge q$ implies r ≤ q and hence MaxL(q,r) = r and since $q^{\prime }\leq \min U(p,q^{\prime })$ we have

  $$ \begin{array}{@{}rcl@{}} q\rightarrow(p\odot q) & =&q\rightarrow\big(\min U(p,q^{\prime})\wedge q\big)=\bigcup\{q\rightarrow r\mid r\in\min U(p,q^{\prime})\wedge q\} \\ & =&\bigcup\{q^{\prime}\vee\max L(q,r)\mid r\in\min U(p,q^{\prime})\wedge q\} \\ & =&\bigcup\{q^{\prime}\vee r\mid r\in\min U(p,q^{\prime})\wedge q\}=q^{\prime}\vee\big(\min U(p,q^{\prime})\wedge q\big) \\ & =&\min U(p,q^{\prime})\geq p \end{array} $$

  according to Proposition 2.2 (i).

• (ii) Since $r\in p^{\prime }\vee \max L(p,q)$ implies $p^{\prime }\leq r$ and hence $\min U(r,p^{\prime })=r$ and since MaxL(p,q) ≤ p we have

  $$ \begin{array}{@{}rcl@{}} (p\rightarrow q)\odot p & =&(p^{\prime}\vee\max L(p,q))\odot p=\bigcup\{r\odot p\mid r\in p^{\prime}\vee\max L(p,q)\} \\ & =&\bigcup\{\min U(r,p^{\prime})\wedge p\mid r\in p^{\prime}\vee\max L(p,q)\} \\ & =&\bigcup\{r\wedge p\mid r\in p^{\prime}\vee\max L(p,q)\}=\big(p^{\prime}\vee\max L(p,q)\big)\wedge p \\ & =&\max L(p,q)\leq q \end{array} $$

  according to Proposition 2.2 (ii).

□

Lemma 2.6

Let $(A,\leq ,{}^{\prime },0,1)$ be a finite orthomodular poset, $x\in (\mathcal P_{+}A)^{T}$ and p ∈ A^T. Then

• (i) $x\leq _{1}p\rightarrow (x\odot p)$ and $x\leq _{2}p\rightarrow (x\odot p)$,

• (ii) $(p\rightarrow x)\odot p\leq _{1}x$ and $(p\rightarrow x)\odot p\leq _{2}x$.

Proof

• (i) We have

  $$ \begin{array}{@{}rcl@{}} p\rightarrow(x\odot p) & =&p\rightarrow\bigcup\{q\odot p\mid q\in x\}=p\rightarrow\bigcup\{\min U(q,p^{\prime})\wedge p\mid q\in x\} \\ & =&\bigcup\{p\rightarrow r\mid r\in\bigcup\{\min U(q,p^{\prime})\wedge p\mid q\in x\}\} \\ & =&\bigcup\{p^{\prime}\vee\max L(p,r)\mid r\in\bigcup\{\min U(q,p^{\prime})\wedge p\mid q\in x\}\} \\ & =&\bigcup\{p^{\prime}\vee r\mid r\in\bigcup\{\min U(q,p^{\prime})\wedge p\mid q\in x\}\} \\ & =&p^{\prime}\vee\bigcup\{\min U(q,p^{\prime})\wedge p\mid q\in x\} \\ & =&\bigcup\{p^{\prime}\vee\big(\min U(q,p^{\prime})\wedge p\big)\mid q\in x\} = \bigcup\{\min U(q,p^{\prime})\mid q\in x\}. \end{array} $$

  (The third and the fourth line are equal since

  $$ \max L(p,r)=r\text{ for all }r\in\bigcup\{\min U(q,p^{\prime})\wedge p\mid q\in x\}.) $$

  Let q ∈ x. Then $\min U(q,p^{\prime })\subseteq p\rightarrow (x\odot p)$. Let $b\in \min U(q,p^{\prime })$. Then q ≤ b. Now $b\in p\rightarrow (x\odot p)$ showing $x\leq _{1}p\rightarrow (x\odot p)$. Conversely, let $b\in p\rightarrow (x\odot p)$. Then there exists some q ∈ x with $b\in \min U(q,p^{\prime })$. Now q ∈ x and q ≤ b showing $x\leq _{2}p\rightarrow (x\odot p)$.

• (ii) We have

  $$ \begin{array}{@{}rcl@{}} (p\rightarrow x)\odot p & =&\bigcup\{p\rightarrow q\mid q\in x\}\odot p=\bigcup\{p^{\prime}\vee\max L(p,q)\mid q\in x\}\odot p \\ & =&\bigcup\{r\odot p\mid r\in\bigcup\{p^{\prime}\vee\max L(p,q)\mid q\in x\}\} \\ & =&\bigcup\{\min U(r,p^{\prime})\wedge p\mid r\in\bigcup\{p^{\prime}\vee\max L(p,q)\mid q\in x\}\} \\ & =&\bigcup\{r\wedge p\mid r\in\bigcup\{p^{\prime}\vee\max L(p,q)\mid q\in x\}\} \\ & =&\bigcup\{p^{\prime}\vee\max L(p,q)\mid q\in x\}\wedge p \\ & =&\bigcup\{\big(p^{\prime}\vee\max L(p,q)\big)\wedge p\mid q\in x\}=\bigcup\{\max L(p,q)\mid q\in x\}. \end{array} $$

  The third and the fourth line are equal since

  $$ \min U(r,p^{\prime})=r\text{ for all }r\in\bigcup\{p^{\prime}\vee\max L(p,q)\mid q\in x\}. $$

  Let $a\in (p\rightarrow x)\odot p$. Then there exists some q ∈ x with a ∈ MaxL(p,q). Now q ∈ x and a ≤ q showing $(p\rightarrow x)\odot p\leq _{1}x$. Conversely, let q ∈ x. Then $\max L(p,q)\subseteq (p\!\rightarrow \! x)\odot p$. Let a ∈ MaxL(p,q). Then a ≤ q. Now $a\in (p\rightarrow x)\odot p$ showing $(p\rightarrow x)\odot p\leq _{2}x$.

□

3 Dynamic pairs

As mentioned in the introduction, by a dynamic pair is meant a couple (P,G) of tense operators satisfying axioms (P1) – (P3). We firstly define tense operators on a finite orthomodular poset $(A,\leq ,{}^{\prime },0,1)$ as follows: Let a time frame (T,R) be given. Define $P,F,H,G\colon A^{T}\rightarrow (\mathcal P_{+}A)^{T}$ by

$$ \begin{array}{@{}rcl@{}} P(q)(s) & :=&\min U(\{q(t)\mid t\mathrel Rs\}), \\ F(q)(s) & :=&\min U(\{q(t)\mid s\mathrel Rt\}), \\ H(q)(s) & :=&\max L(\{q(t)\mid t\mathrel Rs\}), \\ G(q)(s) & :=&\max L(\{q(t)\mid s\mathrel Rt\}) \end{array} $$

for all q ∈ A^T and all s ∈ T.

It is elementary to show that if the orthomodular poset $(A,\leq ,{}^{\prime },0,1)$ is a complete lattice then these tense operators coincide with those mentioned in the introduction. However, in general P(q)(s) for q ∈ A^T and s ∈ T need not be a single element of A (i.e. a singleton), but may be a non-empty subset of A. Analogously for the remaining tense operators. Hence, these operators are again “inexact” in the sense that they reach maximal (for H, G) or minimal (for P, F) incomparable values and we cannot distinguished among them. On the other hand, the results of these operators belong to A contrary to the case when the poset $(A,\leq ,{}^{\prime },0,1)$ is embedded into a complete lattice, the method used in [6].

Since we want to compose our tense operators we need to solve two essential tasks:

• (i) Extend P,F,H,G from A^T to $\mathcal P_{+}(A^{T})$ (i.e. lifting of the operators).

• (ii) Define how to compose them.

Concerning the first task we define for $B\in \mathcal P_{+}(A^{T})$ and all s ∈ T

$$ \begin{array}{@{}rcl@{}} P(B)(s) & :=\min U(\{q(t)\mid q\in B\text{ and }t\mathrel Rs\}), \\ F(B)(s) & :=\min U(\{q(t)\mid q\in B\text{ and }s\mathrel Rt\}), \\ H(B)(s) & :=\max L(\{q(t)\mid q\in B\text{ and }t\mathrel Rs\}), \\ G(B)(s) & :=\max L(\{q(t)\mid q\in B\text{ and }s\mathrel Rt\}). \end{array} $$

In order to solve the second task, we should mention that our extended tense operators are not mappings from $\mathcal P_{+}(A^{T})$ to $\mathcal P_{+}(A^{T})$, but from $\mathcal P_{+}(A^{T})$ to $(\mathcal P_{+}A)^{T}$. Hence we introduce the so-called transformation function $\varphi \colon (\mathcal P_{+}A)^{T}\rightarrow \mathcal P_{+}(A^{T})$ as follows:

$$ \varphi(x):=\{q\in A^{T}\mid q(t)\in x(t)\text{ for all }t\in T\}\text{ for all }x\in(\mathcal P_{+}A)^{T}. $$

By means of the transformation function we can define the compositionG ∗ P of the tense operators G and P by

$$ G*P:=G\circ\varphi\circ P. $$

Lemma 3.1

Let (A,≤) be a poset, T≠∅, p ∈ A^T and $x,y\in (\mathcal P_{+}A)^{T}$. Then

• (i) φ is injective,

• (ii) if z(t) := {p(t)} for all t ∈ T then φ(z) = {p},

• (iii) we have x ≤ y if and only if φ(x) ≤ φ(y).

Proof

• (i) Assume x≠y. Then there exists some t ∈ T with x(t)≠y(t). Without loss of generality, assume x(t) ∖(y(t))≠∅. Let a ∈ x(t) ∖(y(t)) and p ∈ A^T with p(t) = a and p(s) ∈ x(s) for all s ∈ T ∖{t}. Then p ∈ φ(x) ∖(φ(y)) and hence φ(x)≠φ(y).

• (ii) If z(t) := {p(t)} for all t ∈ T then φ(z) = {q ∈ A^T∣q(t) ∈ z(t) for all t ∈ T} = {p}.

• (iii) Assume x ≤ y. Let p ∈ φ(x) and q ∈ φ(y). Then p(s) ∈ x(s) and q(s) ∈ y(s) for all s ∈ T. Hence p(s) ≤ q(s) for all s ∈ T, i.e. p ≤ q. This shows φ(x) ≤ φ(y). Conversely, assume φ(x) ≤ φ(y). Let t ∈ T, a ∈ x(t) and b ∈ y(t). Take p,q ∈ A^T with p(t) = a, p(s) ∈ x(s) for all s ∈ T ∖{t}, q(t) = b and q(s) ∈ y(s) for all s ∈ T ∖{t}. Then p(s) ∈ x(s) and q(s) ∈ y(s) for all s ∈ T. Hence p ∈ φ(x) and q ∈ φ(y) which implies p ≤ q whence a = p(t) ≤ q(t) ≤ b. This shows x(t) ≤ y(t). Since t was an arbitrary element of T we conclude x ≤ y.

□

Now we are ready to prove our result. But first we make the following more or less evident observation.

Observation 3.2

Let (A,≤) be a finite poset and B,C subsets of A with C≠∅. Then $B\subseteq C$ implies B ≤₁ MaxC and MinC ≤₂B.

Theorem 3.3

Let $\mathbf A=(A,\leq ,{}^{\prime },0,1)$ be a finite orthomodular poset, (T,R) a time frame and P and G tense operators as defined above. Then the couple (P,G) forms a dynamic pair, i.e.

• (P1) G(1) = 1 and P(0) = 0,

• (P2) if p,q ∈ A^T and p ≤ q then G(p) ≤₁G(q) and P(p) ≤₂P(q),

• (P3) if q ∈ A^T then q ≤₁(G ∗ P)(q) and (P ∗ G)(q) ≤₂q.

Proof

Let p,q ∈ A^T and s ∈ T.

• (P1) G(1)(s) = MaxL(1) = MaxA = 1 and P(0)(s) = MinU(0) = MinA = 0

• (P2) Assume p ≤ q. Then

  $$ \max L(\{p(t)\mid s\mathrel Rt\})\subseteq L(\{p(t)\mid s\mathrel Rt\})\subseteq L(\{q(t)\mid s\mathrel Rt\}) $$

  and hence

  $$ G(p)(s)=\max L(\{p(t)\mid s\mathrel Rt\})\leq_{1}\max L(\{q(t)\mid s\mathrel Rt\})=G(q)(s) $$

  according to Observation 3.2. Similarly,

  $$ \min U(\{q(t)\mid t\mathrel Rs\})\subseteq U(\{q(t)\mid t\mathrel Rs\})\subseteq U(\{p(t)\mid t\mathrel Rs\}) $$

  and hence

  $$ P(p)(s)=\min U(\{p(t)\mid t\mathrel Rs\})\leq_{2}\min U(\{q(t)\mid t\mathrel Rs\})=P(q)(s) $$

  according to Observation 3.2.

• (P3) We have

  $$ \begin{array}{@{}rcl@{}} P(q)(s) & =&\min U(\{q(t)\mid t\mathrel Rs\}), \\ \varphi\big(P(q)\big) & =&\{r\in A^{T}\mid r(u)\in P(q)(u)\text{ for all }u\in T\} \\ & =&\{r\in A^{T}\mid r(u)\in \min U(\{q(t)\mid t\mathrel Ru\})\text{ for all }u\in T\}, \\ (G*P)(q)(s) & =&G\Big(\varphi\big(P(q)\big)\Big)(s)=\max L(\{r(v)\mid r\in\varphi\big(P(q)\big)\text{ and }s\mathrel Rv\}) \\ & =&\max L\big(\bigcup\{\min U(\{q(t)\mid t\mathrel Rv\})\mid s\mathrel Rv\}\big), \\ q(s) & \in& L\big(\bigcup\{U(\{q(t)\mid t\mathrel Rv\})\mid s\mathrel Rv\}\big) \\ & \subseteq& L\big(\bigcup\{\min U(\{q(t)\mid t\mathrel Rv\})\mid s\mathrel Rv\}\big) \end{array} $$

  and hence q(s) ≤₁(G ∗ P)(q)(s) according to Observation 3.2. The last but one line can be seen as follows: If $a\in \bigcup \{U(\{q(t)\mid t\mathrel Rv\})\mid s\mathrel Rv\}$ then there exists some v ∈ T with s R v and a ∈ U({q(t)∣t R v}). But then q(s) ≤ a. Analogously, we have

  $$ \begin{array}{@{}rcl@{}} G(q)(s) & =&\max L(\{q(t)\mid s\mathrel Rt\}), \\ \varphi\big(G(q)\big) & =&\{r\in A^{T}\mid r(u)\in G(q)(u)\text{ for all }u\in T\} \\ & =&\{r\in A^{T}\mid r(u)\in\max L(\{q(t)\mid u\mathrel Rt\})\text{ for all }u\in T\}, \\ (P*G)(q)(s) & =&P\Big(\varphi\big(G(q)\big)\Big)(s)=\min U(\{r(v)\mid r\in\varphi\big(G(q)\big)\text{ and }v\mathrel Rs\}) \\ & =&\min U\big(\bigcup\{\max L\big(\{q(t)\mid v\mathrel Rt\})\mid v\mathrel Rs\}\big), \\ q(s) & \in &U\big(\bigcup\{L(\{q(t)\mid v\mathrel Rt\})\mid v\mathrel Rs\}\big) \\ & \subseteq& U\big(\bigcup\{\max L(\{q(t)\mid v\mathrel Rt\})\mid v\mathrel Rs\}\big) \end{array} $$

  and hence (P ∗ G)(q)(s) ≤₂q(s) according to Observation 3.2. The last but one line can be seen as follows: If $a\in \bigcup \{L(\{q(t)\mid v\mathrel Rt\})\mid v\mathrel Rs\}$ then there exists some v ∈ T with v R s and a ∈ L({q(t)∣v R t}). But then a ≤ q(s).

□

Similarly one can show that also the couple (F,H) forms a dynamic pair.

4 Properties of tense operators

As mentioned in our previous section, our logic based on a finite orthomodular poset is equipped with logical connectives ⊙ and → forming an adjoint pair. The aim of this section is to show that our tense operators satisfy properties considered in classical or non-classical tense logics as collected in [7] although they are defined in an inexact way.

Before formulating of our results, let us illuminate our concepts by the following example.

Example 4.1

We consider the orthomodular poset $(A,\leq ,{}^{\prime },0,1)$ from Fig. 1 and the time frame ({1,2,3},≤). Define p,q ∈ A^T as follows:

We obtain

This shows

$$ \begin{array}{@{}rcl@{}} H(\varphi(p\odot q)) & \leq& H(p)\odot H(q), \\ H(\varphi(p\rightarrow q)) & \leq&{~}_{2}H(p)\rightarrow H(q), \\ G(\varphi(p\odot q)) & \leq& G(p)\odot G(q), \\ G(\varphi(p\rightarrow q)) & \leq&{~}_{1}G(p)\rightarrow G(q) \end{array} $$

and these inequalities are proper.

We introduce the following notations:

For p ∈ A^T, $x\in (\mathcal P_{+}A)^{T}$ and $B\in \mathcal P_{+}(A^{T})$ we define $p^{\prime }\in A^{T}$, $x^{\prime }\in (\mathcal P_{+}A)^{T}$ and $B^{\prime }\in \mathcal P_{+}(A^{T})$ as follows:

$$ \begin{array}{@{}rcl@{}} p^{\prime}(t) & :=&\big(p(t)\big)'\text{ for all }t\in T, \\ x^{\prime}(t) & :=&\{a^{\prime}\mid a\in x(t)\}\text{ for all }t\in T, \\ B^{\prime} & :=&\{q^{\prime}\mid q\in B\}. \end{array} $$

We have $\big (\varphi (x)\big )'=\varphi (x^{\prime })$ since

$$ \begin{array}{@{}rcl@{}} \big(\varphi(x)\big)' & =&\{q^{\prime}\mid q\in\varphi(x)\}=\{q\in A^{T}\mid q^{\prime}\in\varphi(x)\}= \\ & =&\{q\in A^{T}\mid\big(q(t)\big)'\in x(t)\text{ for all }t\in T\}=\{q\in A^{T}\mid q(t)\in x^{\prime}(t)\text{ for all }t\in T\} \\ & =&\varphi(x^{\prime}). \end{array} $$

When combining operators, both tense ones and logical connectives, we must use the transformation function φ because of the reasons explained in the previous section. We can state and prove the following propositions.

Proposition 4.2

Let (A,≤,0,1) be a finite bounded poset, (T,R) a time frame with reflexive R and q ∈ A^T. Then

$$ \begin{array}{@{}rcl@{}} q & \leq&(\varphi\circ P)(q), \\ q & \leq&(\varphi\circ F)(q), \\ (\varphi\circ H)(q) & \leq& q, \\ (\varphi\circ G)(q) & \leq& q. \end{array} $$

Proof

Let s ∈ T. We have

$$ \begin{array}{@{}rcl@{}} P(q)(s) & =&\min U(\{q(t)\mid t\mathrel Rs\}), \\ (\varphi\circ P)(q) & =&\{p\in A^{T}\mid p(t)\in\min U(\{q(u)\mid u\mathrel Rt\})\text{ for all }t\in T\}, \\ q & \leq& p\text{ for all }p\in(\varphi\circ P)(q), \\ q & \leq&(\varphi\circ P)(q). \end{array} $$

(In the last but one line reflexivity of R is used. Namely, if p ∈ (φ ∘ P)(q) and t ∈ T then because of t R t we have q(t) ≤ p(t).) The result for F is analogous. Now

$$ \begin{array}{@{}rcl@{}} H(q)(s) & =&\max L(\{q(t)\mid t\mathrel Rs\}), \\ (\varphi\circ H)(q) & =&\{p\in A^{T}\mid p(t)\in\max L(\{q(u)\mid u\mathrel Rt\})\text{ for all }t\in T\}, \\ p & \leq& q\text{ for all }p\in(\varphi\circ H)(q), \\ (\varphi\circ H)(q) & \leq& q. \end{array} $$

(In the last but one line reflexivity of R is used. Namely, if p ∈ (φ ∘ H)(q) and t ∈ T then t R t and hence p(t) ≤ q(t).) The result for G is analogous. □

Proposition 4.3

Let $(A,\leq ,{}^{\prime },0,1)$ be a finite orthomodular poset, (T,R) a time frame, $x\in (\mathcal P_{+}A)^{T}$, $B,C\in \mathcal P_{+}(A^{T})$ with B ≤ C and s ∈ T and denote by φ the transformation function. Then the following holds:

• (i)
  $$ \begin{array}{@{}rcl@{}} P\big(\varphi(x)\big)(s) & =\min U\big(\bigcup\{x(t)\mid t\mathrel Rs\}\big), \\ F\big(\varphi(x)\big)(s) & =\min U\big(\bigcup\{x(t)\mid s\mathrel Rt\}\big), \\ H\big(\varphi(x)\big)(s) & =\max L\big(\bigcup\{x(t)\mid t\mathrel Rs\}\big), \\ G\big(\varphi(x)\big)(s) & =\max L\big(\bigcup\{x(t)\mid s\mathrel Rt\}\big), \end{array} $$

• (ii) $H(B)=P(B^{\prime })'$ and $G(B)=F(B^{\prime })'$,

• (iii) P(B) ≤₂P(C), F(B) ≤₂F(C), H(B) ≤₁H(C) and G(B) ≤₁G(C),

• (iv) H(B) ≤ P(B) and G(B) ≤ F(B).

Proof

• (i) We have

  $$ P\big(\varphi(x)\big)(s)=\min U(\{p(t)\mid p\in\varphi(x)\text{ and }t\mathrel Rs\})=\min U\big(\bigcup\{x(t)\mid t\mathrel Rs\}\big). $$

  The proof for F, H and G is analogous.

• (ii) We have

  $$ \begin{array}{@{}rcl@{}} P(B)(s) & =&\min U(\{q(t)\mid q\in B\text{ and }t\mathrel Rs\})=\min U(\{q^{\prime}(t)\mid q\in B^{\prime}\text{ and }t\mathrel Rs\}) \\ & =&\min U(\{\big(q(t)\big)'\mid q\in B^{\prime}\text{ and }t\mathrel Rs\}) \\ & =&\big(\max L(\{q(t)\mid q\in B^{\prime}\text{ and }t\mathrel Rs\}\big)'=H(B^{\prime})'(s). \end{array} $$

  The second assertion can be proved in an analogous way.

• (iii) We have

  $$ \begin{array}{@{}rcl@{}} P(C)(s) & =&\min U(\{q(t)\mid q\in C\text{ and }t\mathrel Rs\})\subseteq U(\{q(t)\mid q\in C\text{ and }t\mathrel Rs\}) \\ & \subseteq& U(\{p(t)\mid p\in B\text{ and }t\mathrel Rs\}) \end{array} $$

  and hence

  $$ P(B)(s)=\min U(\{p(t)\mid p\in B\text{ and }t\mathrel Rs\})\leq_{2} P(C)(s). $$

  Analogously, we obtain F(B) ≤₂F(C). Now

  $$ \begin{array}{@{}rcl@{}} H(B)(s) & =&\max L(\{p(t)\mid p\in B\text{ and }t\mathrel Rs\})\subseteq L(\{p(t)\mid p\in B\text{ and }t\mathrel Rs\}) \\ & \subseteq& L(\{q(t)\mid q\in C\text{ and }t\mathrel Rs\}) \end{array} $$

  and hence

  $$ H(B)(s)\leq_{1}\max L(\{q(t)\mid q\in C\text{ and }t\mathrel Rs\})=H(C)(s). $$

  Analogously, we obtain G(B) ≤₁G(C).

• (iv) Since R is serial there exists some u ∈ T with u R s. We have

  $$ \begin{array}{@{}rcl@{}} H(B)(s) & =&\max L(\{q(t)\mid q\in B\text{ and }t\mathrel Rs\})\leq\{q(u)\mid q\in B\} \\ & \leq&\min U(\{q(t)\mid q\in B\text{ and }t\mathrel Rs\})=P(B)(s). \end{array} $$

  The second assertion follows analogously.

□

By (ii) of Proposition 4.3 we see that the tense operators P and F are fully determined by means of H and G, respectively. Condition (iii) of Proposition 4.3 shows monotonicity of all tense operators in a finite orthomodular poset.

In the next theorem we verify that the tense operators P, F, H and G as defined in Section 2 satisfy the composition laws in accordance with known sources, see e.g. [4, 7] or [11].

Theorem 4.4

Let (A,≤,0,1) be a finite bounded poset and (T,R) a time frame with reflexive R. Then

$$ \begin{array}{llll} P\leq_{2}P*F, & F\leq_{2}F*P, & H\leq_{1}H*P, & G\leq_{1}G*P, \\ P*H\leq_{2}P, & F*H\leq_{2}F, & H\leq_{1}H*F, & G\leq_{1}G*F, \\ P*G\leq_{2}P, & F*G\leq_{2}F, & H*G\leq_{1}H, & G*H\leq_{1}G. \end{array} $$

Proof

Let q ∈ A^T. According to Proposition 4.2 we have q ≤ (φ ∘ F)(q). Hence by (iii) of Proposition 4.3 we obtain

$$ P(q)\leq_{2}P\big((\varphi\circ F)(q)\big)=(P\circ\varphi\circ F)(q)=(P*F)(q). $$

This shows P ≤₂P ∗ F. The remaining inequalities can be proved analogously. □

However, in some cases we can prove results which are stronger than those of Theorem 4.4.

Theorem 4.5

Let (A,≤,0,1) be a finite bounded poset, (T,R) a time frame with reflexive R and X ∈{P,F,H,G}. Then

$$ H*X,G*X\leq X\leq P*X,F*X, $$

especially

$$ P\leq P*P, F\leq F*F, H*H\leq H, G*G\leq G. $$

Proof

If q ∈ A^T and s ∈ T then

$$ (\varphi\circ X)(q)=\{r\in A^{T}\mid r(u)\in X(q)(u)\text{ for all }u\in T\} $$

and hence

$$ \begin{array}{@{}rcl@{}} (P*X)(q)(s) & =&\min U(\bigcup\{X(q)(t)\mid t\mathrel Rs\})\geq X(q)(s), \\ (F*X)(q)(s) & =&\min U(\bigcup\{X(q)(t)\mid s\mathrel Rt\})\geq X(q)(s), \\ (H*X)(q)(s) & =&\max L(\bigcup\{X(q)(t)\mid t\mathrel Rs\})\leq X(q)(s), \\ (G*X)(q)(s) & =&\max L(\bigcup\{X(q)(t)\mid s\mathrel Rt\})\leq X(q)(s). \end{array} $$

□

That P ∗ P = P does not hold in general can be seen from the following example.

Example 4.6

Consider the orthomodular poset $(A,\leq ,{}^{\prime },0,1)$ from Fig. 1 and the time frame ({1,2,3},≤) as in Example 4.1. Define r ∈ A^T as follows:

Then

$$ \begin{array}{@{}rcl@{}} P(r)(1) & =&\min U(a)=a, \\ P(r)(2) & =&\min U(a,b)=\{f^{\prime},i^{\prime}\}, \\ P(r)(3) & =&\min U(a,b)=\{f^{\prime},i^{\prime}\}, \\ (P*P)(r)(1) & =&\min U(a)=a, \\ (P*P)(r)(2) & =&\min U(a,f^{\prime},i^{\prime})=1, \\ (P*P)(r)(3) & =&\min U(a, ,i^{\prime})=1. \end{array} $$

and hence

In the following we show that also preserving of the connective → with respect to H or G can be derived by the corresponding property for ⊙.

Lemma 4.7

Let $(A,\leq ,{}^{\prime },0,1)$ be a finite orthomodular poset, ⊙ and → defined by (13) and $B,C,D\in \mathcal P_{+}A$. Then $B\odot C\sqsubseteq D$ is equivalent to $B\sqsubseteq C\rightarrow D$.

Proof

First assume $B\odot C\sqsubseteq D$. Then there exist b ∈ B, c ∈ C, d ∈ D and a ∈ b ⊙ c with a ≤ d. Hence $b\odot c\sqsubseteq d$. By operator left adjointness this implies $b\sqsubseteq c\rightarrow d$, i.e. there exists some $e\in c\rightarrow d$ with b ≤ e. Since b ∈ B and $e\in C\rightarrow D$ we conclude $B\sqsubseteq C\rightarrow D$. Conversely, assume $B\sqsubseteq C\rightarrow D$. Then there exist b ∈ B, c ∈ C, d ∈ D and $e\in c\rightarrow d$ with b ≤ e. Hence $b\sqsubseteq c\rightarrow d$. By operator left adjointness this implies $b\odot c\sqsubseteq d$, i.e. there exists some f ∈ b ⊙ c with f ≤ d. Since f ∈ B ⊙ C and d ∈ D we conclude $B\odot C\sqsubseteq D$. □

Theorem 4.8

Let $(A,\leq ,{}^{\prime },0,1)$ be a finite orthomodular poset, (T,R) a time frame and X,Y,Z ∈{P,F,H,G}. Then the following hold:

• (i) Put i := 1 if Z ∈{H,G} and i := 2 otherwise and assume

  $$ X\big(\varphi(x)\big)\odot Y(q)\leq_{i}Z\big(\varphi(x\odot q)\big)\text{ for all }x\in(\mathcal P_{+}A)^{T}\text{ and all }q\in A^{T}. $$

  Then

  $$ X\big(\varphi(p\rightarrow q)\big)\sqsubseteq Y(p)\rightarrow Z(q)\text{ for all }p,q\in A^{T}. $$

• (ii) Put i := 1 if X ∈{H,G} and i := 2 otherwise and assume

  $$ X\big(\varphi(p\rightarrow x)\big)\leq_{i}Y(p)\rightarrow Z\big(\varphi(x)\big)\text{ for all }p\in A^{T}\text{ and all }x\in(\mathcal P_{+}A)^{T}. $$

  Then

  $$ X(p)\odot Y(q)\sqsubseteq Z\big(\varphi(p\odot q)\big)\text{ for all }p,q\in A^{T}. $$

Proof

Let p,q ∈ A and $x\in \mathcal P_{+}A$.

• (i) Because of divisibility of the operator residuated structure $(A,\leq ,\odot ,\rightarrow ,0,1)$ we have

  $$ X\big(\varphi(p\rightarrow q)\big)\odot Y(p)\leq_{i}Z\Big(\varphi\big((p\rightarrow q)\odot p\big)\Big)=Z\Big(\varphi\big(\max L(p,q)\big)\Big). $$

  Now MaxL(p,q) ≤ q implies φ(MaxL(p,q)) ≤ q according to Lemma 3.1 whence

  $$ X\big(\varphi(p\rightarrow q)\big)\odot Y(p)\leq_{i}Z\Big(\varphi\big(\max L(p,q)\big)\Big)\leq_{i}Z(q) $$

  according to (iii) of Proposition 4.3. Hence $X\big (\varphi (p\rightarrow q)\big )\odot Y(p)\sqsubseteq Z(q)$. Operator left adjointness (cf. [5]) and Lemma 4.7 yields

  $$ X\big(\varphi(p\rightarrow q)\big)\sqsubseteq Y(p)\rightarrow Z(q). $$

• (ii) Because of (i) of Lemma 2.5 and (iii) of Proposition 4.3 we obtain

  $$ X(p)\leq_{i}X\Big(\varphi\big(q\rightarrow(p\odot q)\big)\Big)\leq_{i}Y(q)\rightarrow Z\big(\varphi(p\odot q)\big) $$

  and therefore

  $$ X(p)\sqsubseteq Y(q)\rightarrow Z\big(\varphi(p\odot q)\big). $$

  Operator left adjointness (cf. [5]) and Lemma 4.7 yields

  $$ X(p)\odot Y(q)\sqsubseteq Z\big(\varphi(p\odot q)\big). $$

□

Remark 4.9

Theorem 4.8 implies that the assertions (1) – (6) and (7) – (12), respectively, are in some sense equivalent.

5 How to construct a preference relation

In this section we will present a construction of a binary relation R on T for a given time set T and A equipped with the tense operators P, F, H and G.

Let A = (A,≤,,0,1) be a finite orthomodular poset, T a time set, (T,R) the time frame. We say that the tense operators P, F, H and G are constructed by (T,R) if for each q ∈ A^T and each s ∈ T

$$ \begin{array}{@{}rcl@{}} P(q)(s) & :=&\min U(\{q(t)\mid t\mathrel Rs\}), \\ F(q)(s) & :=&\min U(\{q(t)\mid s\mathrel Rt\}), \\ H(q)(s) & :=&\max L(\{q(t)\mid t\mathrel Rs\}), \\ G(q)(s) & :=&\max L(\{q(t)\mid s\mathrel Rt\}). \end{array} $$

Since (P,G) as well as (F,H) form a dynamic pair, the quintuple (A,P,F,H,G) is called a dynamic orthomodular poset.

Conversely, let (A,P,F,H,G) be a dynamic orthomodular poset and T a given time set. The question is how to construct a binary relation R^∗ on T such that P, F, H and G are constructed by (T,R^∗).

Let (A,≤) be a poset. On $\mathcal P_{+}A$ we define two binary relations ≈₁ and ≈₂ as follows:

$$ \begin{array}{@{}rcl@{}} B\approx_{1}C & \text{ if }B\leq_{1}C\text{ and }C\leq_{1}B, \\ B\approx_{2}C & \text{ if }B\leq_{2}C\text{ and }C\leq_{2}B \end{array} $$

($B,C\in \mathcal P_{+}A$). Since ≤₁ and ≤₂ are quasi-orders on $\mathcal P_{+}A$, ≈₁ and ≈₂ are the corresponding equivalence relations on $\mathcal P_{+}A$.

Define a binary relation R^∗ on T as follows:

• (14) s R^∗ t if both G(q)(s) ≤ q(t) ≤ F(q)(s) and H(q)(t) ≤ q(s) ≤ P(q)(t) hold for each q ∈ A^T.

Theorem 5.1

Let $(A,\leq ,{}^{\prime },0,1)$ be a finite orthomodular poset, (T,R) a time frame and P, F, H and G the tense operators constructed by (T,R). Let R^∗ be the relation defined by (14). Then $R\subseteq R^{*}$. Let X ∈{P,F,H,G}, q ∈ A^T and s ∈ T, let X^∗ denote the tense operator constructed by (T,R^∗) corresponding to X and put i := 1 if X ∈{H,G} and i := 2 otherwise. Then

$$ X(q)(s)\approx_{i}X^{*}(q)(s). $$

Proof

Let t ∈ T. If s R t then

$$ \begin{array}{@{}rcl@{}} P(q)(t) & =&\min U(\{q(u)\mid u\mathrel Rt\})\geq q(s), \\ F(q)(s) & =&\min U(\{q(u)\mid s\mathrel Ru\})\geq q(t), \\ H(q)(t) & =&\max L(\{q(u)\mid u\mathrel Rt\})\leq q(s), \\ G(q)(s) & =&\max L(\{q(u)\mid s\mathrel Ru\})\leq q(t). \end{array} $$

This shows $R\subseteq R^{*}$. Now

$$ \begin{array}{@{}rcl@{}} P^{*}(q)(s) & =&\min U(\{q(t)\mid t\mathrel{R^{*}}s\})\subseteq U(\{q(t)\mid t\mathrel{R^{*}}s\})\subseteq U(\{q(t)\mid t\mathrel Rs\}), \\ F^{*}(q)(s) & =&\min U(\{q(t)\mid s\mathrel{R^{*}}t\})\subseteq U(\{q(t)\mid s\mathrel{R^{*}}t\})\subseteq U(\{q(t)\mid s\mathrel Rt\}), \\ H^{*}(q)(s) & =&\max L(\{q(t)\mid t\mathrel{R^{*}}s\})\subseteq L(\{q(t)\mid t\mathrel{R^{*}}s\})\subseteq L(\{q(t)\mid t\mathrel Rs\}), \\ G^{*}(q)(s) & =&\max L(\{q(t)\mid s\mathrel{R^{*}}t\})\subseteq L(\{q(t)\mid s\mathrel{R^{*}}t\})\subseteq L(\{q(t)\mid s\mathrel Rt\}) \end{array} $$

and hence

$$ \begin{array}{@{}rcl@{}} P(q)(s) & =&\min U(\{q(t)\mid t\mathrel Rs\})\leq_{2}P^{*}(q)(s), \\ F(q)(s) & =&\min U(\{q(t)\mid s\mathrel Rt\})\leq_{2}F^{*}(q)(s), \\ H^{*}(q)(s) & \leq_{1}&\max L(\{q(t)\mid t\mathrel Rs\})=H(q)(s), \\ G^{*}(q)(s) & \leq_{1}&\max L(\{q(t)\mid s\mathrel Rt\})=G(q)(s) \end{array} $$

according to Observation 3.2. Moreover, we have

$$ \begin{array}{@{}rcl@{}} q(t) & \leq& P(q)(s)\text{ for all }t\text{ with }t\mathrel{R^{*}}s, \\ q(t) & \leq& F(q)(s)\text{ for all }t\text{ with }s\mathrel{R^{*}}t, \\ H(q)(s) & \leq& q(t)\text{ for all }t\text{ with }t\mathrel{R^{*}}s, \\ G(q)(s) & \leq& q(t)\text{ for all }t\text{ with }s\mathrel{R^{*}}t \end{array} $$

and hence

$$ \begin{array}{@{}rcl@{}} P(q)(s) & \in& U(\{q(t)\mid t\mathrel{R^{*}}s\}), \\ F(q)(s) & \in &U(\{q(t)\mid s\mathrel{R^{*}}t\}), \\ H(q)(s) & \in& L(\{q(t)\mid t\mathrel{R^{*}}s\}), \\ G(q)(s) & \in& L(\{q(t)\mid s\mathrel{R^{*}}t\}) \end{array} $$

whence

$$ \begin{array}{@{}rcl@{}} P^{*}(q)(s) & =&\min U(\{q(t)\mid t\mathrel{R^{*}}s\})\leq_{2}P(q)(s), \\ F^{*}(q)(s) & =&\min U(\{q(t)\mid s\mathrel{R^{*}}t\})\leq_{2}F(q)(s), \\ H(q)(s) & \leq_{1}&\max L(\{q(t)\mid t\mathrel{R^{*}}s\})=H^{*}(q)(s), \\ G(q)(s) & \leq_{1}&\max L(\{q(t)\mid s\mathrel{R^{*}}t\})=G^{*}(q)(s) \end{array} $$

according to Observation 3.2. □

Now we present another construction of a time preference relation on the time set.

Define a binary relation $\bar R$ on T as follows:

• (15) $s\mathrel {\bar R}t$ if both (G ∗ X)(q)(s) ≤ X(q)(t) ≤ (F ∗ X)(q)(s) and (H ∗ X)(q)(t) ≤ X(q)(s) ≤ (P ∗ X)(q)(t) hold for each X ∈{P,F,H,G} and each q ∈ A^T.

Theorem 5.2

Let $(A,\leq ,{}^{\prime },0,1)$ be a finite orthomodular poset, (T,R) a time frame and P, F, H and G the tense operators constructed by (T,R). Let $\bar R$ be the relation defined by (15). Then $R\subseteq \bar R$ and for each X,Y ∈{P,F,H,G}, each q ∈ A^T and each s ∈ T we have

$$ (Y*X)(q)(s)\approx_{i}(\bar Y*X)(q)(s) $$

where i = 1 if Y ∈{H,G} and i = 2 otherwise, and where $\bar Y$ denotes the corresponding tense operator constructed by $(T,\bar R)$.

Proof

Let X ∈{P,F,H,G}, q ∈ A^T and s,t ∈ T. If s R t then

$$ \begin{array}{@{}rcl@{}} (P*X)(q)(t) & =&\min U(\bigcup\{X(q)(u)\mid u\mathrel Rt\})\geq X(q)(s), \\ (F*X)(q)(s) & =&\min U(\bigcup\{X(q)(u)\mid s\mathrel Ru\})\geq X(q)(t), \\ (H*X)(q)(t) & =&\max L(\bigcup\{X(q)(u)\mid u\mathrel Rt\})\leq X(q)(s), \\ (G*X)(q)(s) & =&\max L(\bigcup\{X(q)(u)\mid s\mathrel Ru\})\leq X(q)(t). \end{array} $$

This shows $R\subseteq \bar R$. Now

$$ \begin{array}{@{}rcl@{}} (\bar P*X)(q)(s) & =&\min U(\bigcup\{X(q)(t)\mid t\mathrel{\bar R}s\})\subseteq U(\bigcup\{X(q)(t)\mid t\mathrel{\bar R}s\}) \\ & \subseteq& U(\bigcup\{X(q)(t)\mid t\mathrel Rs\}), \\ (\bar F*X)(q)(s) & =&\min U(\bigcup\{X(q)(t)\mid s\mathrel{\bar R}t\})\subseteq U(\bigcup\{X(q)(t)\mid s\mathrel{\bar R}t\}) \\ & \subseteq& U(\bigcup\{X(q)(t)\mid s\mathrel Rt\}), \\ (\bar H*X)(q)(s) & =&\max L(\bigcup\{X(q)(t)\mid t\mathrel{\bar R}s\})\subseteq L(\bigcup\{X(q)(t)\mid t\mathrel{\bar R}s\}) \\ & \subseteq& L(\bigcup\{X(q)(t)\mid t\mathrel Rs\}), \\ (\bar G*X)(q)(s) & =&\max L(\bigcup\{X(q)(t)\mid s\mathrel{\bar R}t\})\subseteq L(\bigcup\{X(q)(t)\mid s\mathrel{\bar R}t\}) \\ & \subseteq& L(\bigcup\{X(q)(t)\mid s\mathrel Rt\}) \end{array} $$

and hence

$$ \begin{array}{@{}rcl@{}} (P*X)(q)(s) & =&\min U(\bigcup\{X(q)(t)\mid t\mathrel Rs\})\leq_{2}(\bar P*X)(q)(s), \\ (F*X)(q)(s) & =&\min U(\bigcup\{X(q)(t)\mid s\mathrel Rt\})\leq_{2}(\bar F*X)(q)(s), \\ (\bar H*X)(q)(s) & \leq_{1}&\max L(\bigcup\{X(q)(t)\mid t\mathrel Rs\})=(H*X)(q)(s), \\ (\bar G*X)(q)(s) & \leq_{1}&\max L(\bigcup\{X(q)(t)\mid s\mathrel Rt\})=(G*X)(q)(s) \end{array} $$

according to Observation 3.2. Moreover, we have

$$ \begin{array}{@{}rcl@{}} X(q)(t) & \leq&(P*X)(q)(s)\text{ for all }t\text{ with }t\mathrel{\bar R}s, \\ X(q)(t) & \leq&(F*X)(q)(s)\text{ for all }t\text{ with }s\mathrel{\bar R}t, \\ (H*X)(q)(s) & \leq& X(q)(t)\text{ for all }t\text{ with }t\mathrel{\bar R}s, \\ (G*X)(q)(s) & \leq& X(q)(t)\text{ for all }t\text{ with }s\mathrel{\bar R}t \end{array} $$

and hence

$$ \begin{array}{@{}rcl@{}} (P*X)(q)(s) & \in& U(\bigcup\{X(q)(t)\mid t\mathrel{\bar R}s\}), \\ (F*X)(q)(s) & \in& U(\bigcup\{X(q)(t)\mid s\mathrel{\bar R}t\}), \\ (H*X)(q)(s) & \in& L(\bigcup\{X(q)(t)\mid t\mathrel{\bar R}s\}), \\ (G*X)(q)(s) & \in& L(\bigcup\{X(q)(t)\mid s\mathrel{\bar R}t\}) \end{array} $$

whence

$$ \begin{array}{@{}rcl@{}} (\bar P*X)(q)(s) & =&\min U(\bigcup\{X(q)(t)\mid t\mathrel{\bar R}s\})\leq_{2}(P*X)(q)(s), \\ (\bar F*X)(q)(s) & =&\min U(\bigcup\{X(q)(t)\mid s\mathrel{\bar R}t\})\leq_{2}(F*X)(q)(s), \\ (H*X)(q)(s) & \leq_{1}&\max L(\bigcup\{X(q)(t)\mid t\mathrel{\bar R}s\})=(\bar H*X)(q)(s), \\ (G*X)(q)(s) & \leq_{1}&\max L(\bigcup\{X(q)(t)\mid s\mathrel{\bar R}t\})=(\bar G*X)(q)(s) \end{array} $$

according to Observation 3.2. □