Probabilistic squares and hexagons of opposition under coherence

TL;DR

This paper develops a probabilistic framework for the square and hexagon of opposition using coherence-based probability, introducing new constructions and applications with generalized quantifiers.

Contribution

It introduces a novel probabilistic interpretation of the square and hexagon of opposition based on coherence theory, including new constructions and applications.

Findings

Probabilistic versions of the square and hexagon of opposition are constructed.

The framework incorporates imprecise probability assessments and generalized quantifiers.

New relations of acceptability are analyzed within coherence-based probability theory.

Abstract

Various semantics for studying the square of opposition and the hexagon of opposition have been proposed recently. We interpret sentences by imprecise (set-valued) probability assessments on a finite sequence of conditional events. We introduce the acceptability of a sentence within coherence-based probability theory. We analyze the relations of the square and of the hexagon in terms of acceptability. Then, we show how to construct probabilistic versions of the square and of the hexagon of opposition by forming suitable tripartitions of the set of all coherent assessments. Finally, as an application, we present new versions of the square and of the hexagon involving generalized quantifiers.

Tables

Table 1. Table 1: Probabilistic interpretation of the sentence types A 𝐴 A , E 𝐸 E , I 𝐼 I , and O 𝑂 O involving generalized quantifiers Q 𝑄 Q defined by a threshold x 𝑥 x (with x ∈ ] 1 2 , 1 ] x\in]\frac{1}{2},1] ) on the subject S 𝑆 S and predicate P 𝑃 P and the respective imprecise probabilistic assessments ℐ A ​ ( x ) subscript ℐ 𝐴 𝑥 {\mathcal{I}}_{A(x)} , ℐ E ​ ( x ) subscript ℐ 𝐸 𝑥 \mathcal{I}_{E(x)} , ℐ I ​ ( x ) subscript ℐ 𝐼 𝑥 \mathcal{I}_{I(x)} , and ℐ O ​ ( x ) subscript ℐ 𝑂 𝑥 \mathcal{I}_{O(x)} on the conditional event P | S conditional 𝑃 𝑆 P|S (above). When x = 1 𝑥 1 x=1 , we obtain our probabilistic interpretation of the traditional sentence types A 𝐴 A , E 𝐸 E , I 𝐼 I , and O 𝑂 O (below).

Sentence		Probability constraints	Assessment on $P|S$
$A\left(x\right):$	$(Q_{\ge x}$ $S$ are $P$)	$p\left(P|S\right)\ge x$	${ℐ}_{A\left(x\right)}=[x,1]$
$E\left(x\right):$	$(Q_{\ge x}$ $S$ are not $P$)	$p\left(\overline{P}|S\right)\ge x$	${ℐ}_{E\left(x\right)}=[0,1-x]$
$I\left(x\right):$	($Q_{>1-x}$ $S$ are $P$)	$p\left(P|S\right)>1-x$	${ℐ}_{I\left(x\right)}=]1-x,1]$
$O\left(x\right):$	($Q_{>1-x}$ $S$ are not $P$)	$p\left(\overline{P}|S\right)>1-x$	${ℐ}_{O\left(x\right)}=[0,x[$
$A\left(1\right):$	(Every $S$ is $P$)	$p\left(P|S\right)=1$	${ℐ}_A=\left\{1\right\}$
$E\left(1\right):$	(No $S$ is $P$)	$p\left(\overline{P}|S\right)=1$	${ℐ}_E=\left\{0\right\}$
$I\left(1\right):$	(Some $S$ is $P$)	$p\left(P|S\right)>0$	${ℐ}_I=]0,1]$
$O\left(1\right):$	(Some $S$ is not $P$)	$p\left(\overline{P}|S\right)>0$	${ℐ}_O=[0,1[$

Table 2. Table 2: Probabilistic interpretation of the sentence types at the top ( U 𝑈 U ) and at the bottom ( Y 𝑌 Y ) of the hexagon of opposition involving generalized quantifiers Q 𝑄 Q defined by a threshold x 𝑥 x (with x ∈ ] 1 2 , 1 ] x\in]\frac{1}{2},1] ) on the subject S 𝑆 S and predicate P 𝑃 P and the respective imprecise probabilistic assessments ℐ U ​ ( x ) subscript ℐ 𝑈 𝑥 {\mathcal{I}}_{U(x)} , and ℐ Y ​ ( x ) subscript ℐ 𝑌 𝑥 \mathcal{I}_{Y(x)} on the conditional event P | S conditional 𝑃 𝑆 P|S (above). When x = 1 𝑥 1 x=1 , we obtain our probabilistic interpretation of the traditional sentence types U 𝑈 U , Y 𝑌 Y .

Sentence		Probability constr.	Assessment on $P|S$
$U\left(x\right):$	$A\left(x\right)\vee E\left(x\right)$	$\begin{array}{c}p\left(P|S\right)\ge x\\ \text{or}\\ p\left(\overline{P}|S\right)\ge x\end{array}$	${ℐ}_{U\left(x\right)}=[0,1-x]\cup [x,1]$
$Y\left(x\right):$	$I\left(x\right)\wedge O\left(x\right)$	$\{\begin{array}{c}p\left(P|S\right)>1-x\\ p\left(\overline{P}|S\right)>1-x\end{array}$	${ℐ}_{Y\left(x\right)}=]1-x,x[$
$U\left(1\right):$	$\begin{array}{c}\text{Every }S\text{ is }P\\ \text{or}\\ \text{No }S\text{ is }P\end{array}$	$\begin{array}{c}p\left(P|S\right)=1\\ \text{or}\\ p\left(\overline{P}|S\right)=1\end{array}$	${ℐ}_U=\left\{0\right\}\cup \left\{1\right\}$
$Y\left(1\right):$	$\begin{array}{c}\text{Some }S\text{ is }P\\ \text{and}\\ \text{Some }S\text{ is }\overline{P}\end{array}$	$\{\begin{array}{c}p\left(P|S\right)>0\\ p\left(\overline{P}|S\right)>0\end{array}$	${ℐ}_Y=]0,1[$

Equations

$$\begin{array}[]{ll}\mathcal{B}_{1}(x)=\{(p_{1},\ldots,p_{n})\in[0,1]^{n}:\sum_{i=1}^{n}\frac{p_{i}}{n}\geq x\},\\ \mathcal{B}_{2}(x)=\{(p_{1},\ldots,p_{n})\in[0,1]^{n}:\sum_{i=1}^{n}\frac{p_{i}}{n}\leq 1-x\},\\ \mathcal{B}_{3}(x)=\{(p_{1},\ldots,p_{n})\in[0,1]^{n}:1-x<\sum_{i=1}^{n}\frac{p_{i}}{n}<x\}.\\ \end{array}$$

$$\begin{array}[]{ll}\mathcal{I}_{A(x)}=\mathcal{B}_{1}(x),\;\mathcal{I}_{E(x)}=\mathcal{B}_{2}(x),\;\mathcal{I}_{Y(x)}=\mathcal{B}_{3}(x),\\ \mathcal{I}_{I(x)}=\mathcal{B}_{1}(x)\cup\mathcal{B}_{3}(x)=\{(p_{1},\ldots,p_{n})\in[0,1]^{n}:\sum_{i=1}^{n}\frac{p_{i}}{n}>1-x\},\\ \mathcal{I}_{O(x)}=\mathcal{B}_{2}(x)\cup\mathcal{B}_{3}(x)=\{(p_{1},\ldots,p_{n})\in[0,1]^{n}:\sum_{i=1}^{n}\frac{p_{i}}{n}<x\},\\ \mathcal{I}_{U(x)}=\mathcal{B}_{1}(x)\cup\mathcal{B}_{2}(x)=\{(p_{1},\ldots,p_{n})\in[0,1]^{n}:\sum_{i=1}^{n}\frac{p_{i}}{n}\geq x\text{ or }\sum_{i=1}^{n}\frac{p_{i}}{n}\leq 1-x\}.\end{array}$$

Full text

1 Introduction

There is a long history of investigations on the square of opposition spanning over two millenia [4, 38]. A square of opposition represents logical key relations among basic sentence types in a diagrammatic way. The basic sentence types, traditionally denoted by $A$ (universal affirmative: “Every $S$ is $P$”), $E$ (universal negative: “No $S$ is $P$”), $I$ (particular affirmative: “Some $S$ are $P$”), and $O$ (particular negative: “Some $S$ are not $P$”), constitute the corners of the square. The diagonals and the sides of the square of opposition are formed by the following logical relations among the basic sentence types: $A$ and $E$ are contraries (i.e., they cannot both be true), $I$ and $O$ are subcontraries (i.e., they cannot both be false), $A$ and $O$ as well as $E$ and $I$ are contradictories (i.e., they cannot both be true and they cannot both be false), $I$ is a subaltern of $A$ and $O$ is a subaltern of $E$ (i.e., $A$ entails $I$ and $E$ entails $O$; for a visual representation see, e.g., Figure 3 below, and cover the probabilities for seeing the traditional square of opposition). In the early 1950ies, the square of opposition was expanded to the hexagon of opposition, by adding a sentence at the top and another one at the bottom of the square (see, e.g., Figure 5). Recently, the square of opposition as well as the hexagon of opposition and its extensions have been investigated from various semantic points of view (see, e.g., [3, 4, 11, 20, 21, 22, 28, 34, 35, 36]). In this paper we present a probabilistic analysis of the square of opposition under coherence, introduce the hexagon of opposition under coherence, and study the semantics of basic key relations among quantified statements.

After preliminary notions (Section 2), we introduce, based on g-coherence, a (probabilistic) notion of sentences and their acceptability and show how to construct squares of opposition under coherence from suitable tripartitions (Section 3). Then, we present an application of our square to the study of generalized quantifiers (Section 4). In Section 5 we introduce the hexagon of opposition under coherence. Section 6 concludes the paper by some remarks on future work.

2 Preliminary Notions

The coherence-based approach to probability and to other uncertain measures has been adopted by many authors (see, e.g., [5, 6, 9, 10, 14, 15, 16, 17, 18, 24, 26, 30, 31, 41, 42, 44]); we therefore recall only selected key features of coherence and its generalizations in this section.

An event $E$ is a two-valued logical entity which can be either true or false. The indicator of $E$ is a two-valued numerical quantity which is 1, or 0, according to whether the event $E$ is true, or false, respectively. We use the same symbols for events and their indicators. We denote by $\top$ the sure event (i.e., tautology or logical truth) and by $\bot$ the impossible event (i.e., contradiction or logical falsehood). Moreover, given two events $E$ and $H$, we denote by $E\land H$ (resp., $E\vee H$) conjunction (resp., disjunction). To simplify notation, we will use the product $EH$ to denote the conjunction $E\land H$, which also denotes the indicator of $E\wedge H$. We denote by $\widebar{E}$ the negation of $E$.

Given two events $E$ and $H$, with $H\neq\bot$, the conditional event $E|H$ is defined as a three-valued logical entity which is true if $EH$ (i.e., $E\wedge H$) is true, false if $\widebar{E}H$ is true, and indetermined (void) if $H$ is false ([19, p. 307]). In terms of the betting metaphor, if you assess $p(E|H)=p$, then you are willing to pay (resp., to receive) an amount $p$ and to receive (resp., to pay) 1, or 0, or $p$, according to whether $EH$ is true, or $\widebar{E}H$ is true, or $\widebar{H}$ is true (bet called off), respectively. For defining coherence, consider a real function $p:\;\mathcal{F}\,\rightarrow\,\mathcal{R}$, where $\mathcal{F}$ is an arbitrary family of conditional events. Consider a finite sub-family $\mathcal{F}_{n}=(E_{1}|H_{1},\ldots,E_{n}|H_{n})\subseteq\mathcal{F}$, and the vector $\mathcal{P}_{n}=(p_{1},\ldots,p_{n})$, where $p_{i}=p(E_{i}|H_{i})\,,\;\;i=1,\ldots,n$. We denote by $\mathcal{H}_{n}$ the disjunction $H_{1}\vee\cdots\vee H_{n}$. With the pair $(\mathcal{F}_{n},\mathcal{P}_{n}$) we associate the random gain ${\mathcal{G}}=\sum_{i=1}^{n}s_{i}H_{i}(E_{i}-p_{i})$, where $s_{1},\ldots,s_{n}$ are $n$ arbitrary real numbers. $\mathcal{G}$ represents the net gain of $n$ transactions, where for each transaction its meaning is specified by the sign of $s_{i}$ (plus for buying or minus for selling) and its scaling is specified by the magnitude of $s_{i}$. Denoting by $G_{\mathcal{H}_{n}}$ the set of values of $\mathcal{G}$ restricted to $\mathcal{H}_{n}$, we recall

Definition 1**.**

The function $p$ defined on $\mathcal{F}$ is called coherent if and only if, for every integer $n$, for every finite sub-family $\mathcal{F}_{n}$ $\subseteq\mathcal{F}$ and for every $s_{1},\ldots,s_{n}$, it holds that: $\min G_{\mathcal{H}_{n}}\leq 0\leq\max G_{\mathcal{H}_{n}}$.

We say that $p$ is incoherent if and only if $p$ is not coherent.

As shown by Definition 1, a probability assessment is coherent if and only if, in any finite combination of $n$ bets, it does not happen that the values in the set $G_{\mathcal{H}_{n}}$ are all positive, or all negative (no Dutch Book). Moreover, coherence of $p(E|H)$ requires that $p(E|H)\in[0,1]$ for every $E|H\in\mathcal{F}$. If $p$ on $\mathcal{F}$ is coherent, we call it a conditional probability on $\mathcal{F}$ (see, e.g., [1, 17, 47]). Notice that, if $p$ is coherent, then $p$ also satisfies all the well known properties of finitely additive conditional probability (while the converse does not hold; see, e.g., [17, Example 13] or [23, Example 8]).

In what follows $\mathcal{F}$ will denote finite sequence of conditional events. Let $\mathcal{F}=(E_{1}|H_{1},\ldots,E_{n}|H_{n})$. We denote by $\mathcal{P}$ a (precise) probability assessment $\mathcal{P}=(p_{1},\ldots,p_{n})$ on $\mathcal{F}$, where $p_{j}=p(E_{j}|H_{j})\in[0,1]$, $j=1,\dots,n$. Moreover, we denote by $\Pi$ the set of all coherent precise assessments on $\mathcal{F}$. We recall that when there are no logical relations among the events $E_{1},H_{1},\ldots,E_{n},H_{n}$ involved in $\mathcal{F}$, that is $E_{1},H_{1},\ldots,E_{n},H_{n}$ are logically independent, then the set $\Pi$ associated with $\mathcal{F}$ is the whole unit hypercube $[0,1]^{n}$. If there are logical relations, then the set $\Pi$ could be a strict subset of $[0,1]^{n}$. As it is well known $\Pi\neq\emptyset$; therefore, $\emptyset\neq\Pi\subseteq[0,1]^{n}$. If not stated otherwise, we do not make any assumptions concerning logical independence.

Definition 2**.**

An imprecise, or set-valued, assessment ${\mathcal{I}}$ on a family of conditional events $\mathcal{F}$ is a (possibly empty) set of precise assessments $\mathcal{P}$ on $\mathcal{F}$.

Definition 2 states that an imprecise (probability) assessment ${\mathcal{I}}$ on a sequence of $n$ conditional events $\mathcal{F}$ is just a (possibly empty) subset of $[0,1]^{n}$ ([25, 27, 28]). For instance, think about an agent (like Pythagoras) who considers only rational numbers to evaluate the probability of an event $E|H$. Pythagoras’ evaluation can be represented by the imprecise assessment $\mathcal{I}=[0,1]\cap\mathbb{Q}$ on $E|H$. Moreover, a constraint like $p(E|H)>0$ can be represented by the imprecise assessment $\mathcal{I}=]0,1]$ on $E|H$.

Given an imprecise assessment ${\mathcal{I}}$ we denote by $\widebar{\mathcal{I}}$ the complementary imprecise assessment of ${\mathcal{I}}$, i.e. $\widebar{\mathcal{I}}=[0,1]^{n}\setminus{\mathcal{I}}$. We now recall the notions of g-coherence and total coherence in the general case of imprecise (in the sense of set-valued) probability assessments [28].

Definition 3** (g-coherence).**

Given a sequence of $n$ conditional events $\mathcal{F}$. An imprecise assessment ${\mathcal{I}}\subseteq[0,1]^{n}$ on $\mathcal{F}$ is g-coherent iff there exists a coherent precise assessment $\mathcal{P}$ on $\mathcal{F}$ such that $\mathcal{P}\in{\mathcal{I}}$.

Definition 4** (t-coherence).**

An imprecise assessment ${\mathcal{I}}$ on $\mathcal{F}$ is totally coherent (t-coherent) iff the following two conditions are satisfied: (i) ${\mathcal{I}}$ is non-empty; (ii) if $\mathcal{P}\in{\mathcal{I}}$, then $\mathcal{P}$ is a coherent precise assessment on $\mathcal{F}$.

Definition 5** (t-coherent part).**

Given a sequence of $n$ conditional events $\mathcal{F}$. Let $\Pi$ be the set of all coherent assessments on $\mathcal{F}$. We denote by $\pi:\wp([0,1]^{n})\rightarrow\wp(\Pi)$ the function defined by $\pi(\mathcal{I})=\Pi\cap\mathcal{I}$, for any imprecise assessment $\mathcal{I}\in\wp([0,1]^{n})$. Moreover, for each subset $\mathcal{I}\in\wp([0,1]^{n})$ we call $\pi(\mathcal{I})$ the t-coherent part of $\mathcal{I}$.

Of course, if $\pi(\mathcal{I})\neq\emptyset$, then $\mathcal{I}$ is g-coherent and $\pi(\mathcal{I})$ is t-coherent.

3 From Imprecise Assessments to the Square of Opposition

In this section we consider imprecise assessments on a given sequence $\mathcal{F}$ of $n$ conditional events. In our approach, a sentence $s$ is a pair $(\mathcal{F},\mathcal{I})$, where $\mathcal{I}\subseteq[0,1]^{n}$ is an imprecise assessment on $\mathcal{F}$. We introduce the following equivalence relation under t-coherence:

Definition 6**.**

Given two sentences $s_{1}:(\mathcal{F},\mathcal{I}_{1})$ and $s_{2}:(\mathcal{F},\mathcal{I}_{2})$, $s_{1}$ and $s_{2}$ are equivalent (under t-coherence), denoted by $s_{1}\equiv s_{2}$, iff $\pi(\mathcal{I}_{1})=\pi(\mathcal{I}_{2})$.

Definition 7**.**

Given three sentences $s:(\mathcal{F},\mathcal{I})$, $s_{1}:(\mathcal{F},\mathcal{I}_{1})$, and $s_{2}:(\mathcal{F},\mathcal{I}_{2})$. We define: $s_{1}\wedge s_{2}:(\mathcal{F},\mathcal{I}_{1}\cap\mathcal{I}_{2})$ (conjunction); $s_{1}\vee s_{2}:(\mathcal{F},\mathcal{I}_{1}\cup\mathcal{I}_{2})$ (disjunction); $\widebar{s}:(\mathcal{F},\widebar{\mathcal{I}})$, where $\widebar{\mathcal{I}}=[0,1]^{n}\setminus\mathcal{I}$ (negation).

Remark 1**.**

As the basic operations among sentences are defined by set-theoretical operations, they inherit the corresponding properties (including associativity, commutativity, De Morgan’s law, etc.). Moreover, as $\pi(\mathcal{I}_{1}\cap\mathcal{I}_{2})=\pi(\mathcal{I}_{1})\cap\pi(\mathcal{I}_{2})$, by setting $s_{1}^{*}=(\mathcal{F},\pi(\mathcal{I}_{1}))$, $s_{2}^{*}=(\mathcal{F},\pi(\mathcal{I}_{2}))$ and $(s_{1}\wedge s_{2})^{*}:(\mathcal{F},\pi(\mathcal{I}_{1}\cap\mathcal{I}_{2}))$, it follows that $(s_{1}\wedge s_{2})\equiv(s_{1}\wedge s_{2})^{*}\equiv s_{1}^{*}\wedge s_{2}^{*}$. Likewise, $s_{1}\vee s_{2}\equiv(s_{1}\vee s_{2})^{*}\equiv s_{1}^{*}\vee s_{2}^{*}$.

As we interpret the basic sentence types involved in the square of opposition by imprecise probability assessments on sequences of conditional events, we will introduce the following notion of acceptability, which serves as a semantic bridge between basic sentence types and imprecise assessments:

Definition 8**.**

A sentence $s:(\mathcal{F},\mathcal{I})$ is (resp., is not) acceptable iff the assessment $\mathcal{I}$ on $\mathcal{F}$ is (resp., is not) g-coherent, i.e. $\pi(\mathcal{I})$ is not (resp., is) empty.

Remark 2**.**

If $s_{1}\wedge s_{2}$ is acceptable, then $s_{1}$ is acceptable and $s_{2}$ is acceptable. However, the converse does not hold, indeed $s_{1}:(E|H,\{1\})$ is acceptable and $s_{2}:(E|H),\{0\})$ is acceptable, but $s_{1}\wedge s_{2}:(E|H,\emptyset)$ is not acceptable (as $\pi(\emptyset)=\emptyset$).

Definition 9**.**

Given two sentences $s_{1}:(\mathcal{F},\mathcal{I}_{1})$ and $s_{2}:(\mathcal{F},\mathcal{I}_{2})$, we say, under coherence: $s_{1}$ and $s_{2}$ are contraries iff the sentence $s_{1}\wedge s_{2}$ is not acceptable;222 Some definitions of contrariety additionally require that “$s_{1}$ and $s_{2}$ can both be acceptable”. For reasons stated in [28], we omit this additional requirement. Similarly, mutatis mutandis, in our definition of subcontrariety. $s_{1}$ and $s_{2}$ are subcontraries iff $\widebar{s}_{1}\wedge\widebar{s}_{2}$ is not acceptable; $s_{1}$ and $s_{2}$ are contradictories iff $s_{1}$ and $s_{2}$ are both, contraries and subcontraries; $s_{2}$ is a subaltern of $s_{1}$ iff the sentence $s_{1}\wedge\widebar{s}_{2}$ is not acceptable.

Remark 3**.**

By Remark 1, we observe that two sentences $s_{1}$ and $s_{2}$ are contraries if and only if $\pi(\mathcal{I}_{1}\cap\mathcal{I}_{2})=\pi(\mathcal{I}_{1})\cap\pi(\mathcal{I}_{2})=\emptyset$. Moreover, two sentences $s_{1}$ and $s_{2}$ are subcontraries if and only if $\pi(\widebar{\mathcal{I}}_{1}\cap\widebar{\mathcal{I}}_{2})=\pi(\widebar{\mathcal{I}}_{1})\cap\pi(\widebar{\mathcal{I}}_{2})=\emptyset$, that is (by De Morgan’s law) if and only if $\pi(\mathcal{I}_{1})\cup\pi(\mathcal{I}_{2})=\Pi$. Then, two sentences $s_{1}$ and $s_{2}$ are contradictories if and only if $\pi(\mathcal{I}_{1})\cap\pi(\mathcal{I}_{2})=\emptyset$ and $\pi(\mathcal{I}_{1})\cup\pi(\mathcal{I}_{2})=\Pi$, that is if and only if $s_{2}=\widebar{s}_{1}$ (and, of course, $s_{1}=\widebar{s}_{2}$). Given two sentences $s_{1},s_{2}$ we also observe that $s_{2}$ is a subaltern of $s_{1}$ if and only if $\Pi\cap(\mathcal{I}_{1}\cap\widebar{\mathcal{I}}_{2})=\emptyset$, which also amounts to say that $\Pi\cap\mathcal{I}_{1}\subseteq\Pi\cap\mathcal{I}_{2}$, that is if and only if $\pi(\mathcal{I}_{1})\subseteq\pi(\mathcal{I}_{2})$. For instance, $s_{1}\vee s_{2}$ is a subaltern of $s_{1}$ and also of $s_{2}$; similarly, $s_{1}$ is a subaltern of $s_{1}\wedge s_{2}$, and $s_{2}$ is a subaltern of $s_{1}\wedge s_{2}$. Furthermore, if $s_{1}$ is not acceptable, that is $\pi(\mathcal{I}_{1})=\emptyset$, then any sentence $s_{2}$ is a subaltern of $s_{1}$. For example, the sentence $s_{1}:(E|\widebar{E},\{1\})$ is not acceptable because $\Pi=\{0\}$ and then any sentence $s_{2}:(E|\widebar{E},\mathcal{I})$, where $\mathcal{I}\subseteq[0,1]$, is a subaltern of $s_{1}$.

Based on the relations given in Definition 9 we define a square of opposition as follows.

Definition 10**.**

Let $s_{k}:(\mathcal{F},\mathcal{I}_{k})$, $k=1,2,3,4$, be four sentences. We call the ordered quadruple $(s_{1},s_{2},s_{3},s_{4})$ a square of opposition (under coherence), iff the following relations among the four sentences hold:

1. (a)

$s_{1}$* and $s_{2}$ are contraries, i.e., $\pi(\mathcal{I}_{1})\cap\pi(\mathcal{I}_{2})=\emptyset$;* 2. (b)

$s_{3}$* and $s_{4}$ are subcontraries, i.e., $\pi(\mathcal{I}_{3})\cup\pi(\mathcal{I}_{4})=\Pi$;* 3. (c)

$s_{1}$* and $s_{4}$ are contradictories, i.e., $\pi(\mathcal{I}_{1})\cap\pi(\mathcal{I}_{4})=\emptyset$ and ${\pi(\mathcal{I}_{1})\cup\pi(\mathcal{I}_{4})=\Pi}$; *

$s_{2}$* and $s_{3}$ are contradictories, i.e., $\pi(\mathcal{I}_{2})\cap\pi(\mathcal{I}_{3})=\emptyset$ and ${\pi(\mathcal{I}_{2})\cup\pi(\mathcal{I}_{3})=\Pi}$;* 4. (d)

$s_{3}$* is a subaltern of $s_{1}$, i.e., $\pi(\mathcal{I}_{1})\subseteq\pi(\mathcal{I}_{3})$;*

$s_{4}$* is a subaltern of $s_{2}$, i.e., $\pi(\mathcal{I}_{2})\subseteq\pi(\mathcal{I}_{4})$.*

Figure 1 shows the square of opposition based on Definition 10.

Remark 4**.**

Based on Definition 10, we observe that in order to verify if a quadruple of sentences $(s_{1},s_{2},s_{3},s_{4})$, where $s_{k}:(\mathcal{F},\mathcal{I}_{k})$ and $k=1,2,3,4$, is a square of opposition, it is necessary and sufficient to check that the quadruple $(s_{1}^{\prime},s_{2}^{\prime},s_{3}^{\prime},s_{4}^{\prime})$, where $s_{k}^{\prime}=(\mathcal{F},\mathcal{I}_{k}^{\prime})$, $\mathcal{I}_{k}^{\prime}=\pi(\mathcal{I}_{k})$, is a square of opposition. Then, we say that two squares $(s_{1},s_{2},s_{3},s_{4})$ and $(s_{1}^{\prime},s_{2}^{\prime},s_{3}^{\prime},s_{4}^{\prime})$ coincide iff $\pi(\mathcal{I}_{k})=\pi(\mathcal{I}_{k}^{\prime})$ for each $k$. Moreover, based on Definition 10, we observe that $(s_{1},s_{2},s_{3},s_{4})$ is a square of opposition iff $(s_{2},s_{1},s_{4},s_{3})$ is a square of opposition.

Definition 11**.**

An (ordered) tripartition of a set $\mathfrak{S}$ is a triple $(\mathcal{D}_{1},\mathcal{D}_{2},\mathcal{D}_{3})$, where $\mathcal{D}_{1}$, $\mathcal{D}_{2}$, and $\mathcal{D}_{3}$ are subsets of $\mathfrak{S}$, such that the following conditions are satisfied: (i) $\mathcal{D}_{i}\cap\mathcal{D}_{j}=\emptyset$, $i\neq j$ for all $i,j=1,2,3$; (ii); $\mathcal{D}_{1}\cup\mathcal{D}_{2}\cup\mathcal{D}_{3}=\mathfrak{S}$.

Theorem 1**.**

Given any sequence of $n$ conditional events $\mathcal{F}$ and a quadruple $(s_{1},s_{2},s_{3},s_{4})$ of sentences, with $s_{k}:(\mathcal{F},\mathcal{I}_{k})$, $k=1,2,3,4$. Define $\mathcal{D}_{1}=\pi(\mathcal{I}_{1})$, $\mathcal{D}_{2}=\pi(\mathcal{I}_{2})$, and $\mathcal{D}_{3}=\pi(\mathcal{I}_{3})\cap\pi(\mathcal{I}_{4})$. Then, the quadruple $(s_{1},s_{2},s_{3},s_{4})$ is a square of opposition if and only if $(\mathcal{D}_{1},\mathcal{D}_{2},\mathcal{D}_{3})$ is a tripartition of (the non-empty set) $\Pi$ such that: $\pi(\mathcal{I}_{3})=\mathcal{D}_{1}\cup\mathcal{D}_{3}$, $\pi(\mathcal{I}_{4})=\mathcal{D}_{2}\cup\mathcal{D}_{3}$.

Proof.

$(\Rightarrow)$. We assume that $\mathcal{D}_{1}=\pi(\mathcal{I}_{1})$, $\mathcal{D}_{2}=\pi(\mathcal{I}_{2})$, and $\mathcal{D}_{3}=\pi(\mathcal{I}_{3})\cap\pi(\mathcal{I}_{4})$. Of course, $\mathcal{D}_{i}\subseteq\Pi$, $i=1,2,3$. We now prove that: $(i)$ $\mathcal{D}_{1}\cap\mathcal{D}_{2}=\emptyset$; $(ii)$ $\mathcal{D}_{3}=\Pi\setminus(\mathcal{D}_{1}\cup\mathcal{D}_{2})$. $(i)$ From condition (a) in Definition 10, as $s_{1}$ and $s_{2}$ are contraries, it follows that $\mathcal{D}_{1}\cap\mathcal{D}_{2}=\emptyset$. $(ii)$ We first prove that $\mathcal{D}_{3}\subseteq\Pi\setminus(\mathcal{D}_{1}\cup\mathcal{D}_{2})$. This trivially follows when $\mathcal{D}_{3}=\emptyset$. If $\mathcal{D}_{3}\neq\emptyset$, then let $x\in\mathcal{D}_{3}=\pi(\mathcal{I}_{3})\cap\pi(\mathcal{I}_{4})$. As $x\in\pi(\mathcal{I}_{3})$, from condition (c) in Definition 10, we obtain $x\notin\pi(\mathcal{I}_{2})$. Likewise, as $x\in\pi(\mathcal{I}_{4})$, from condition (c) in Definition 10, we obtain $x\notin\pi(\mathcal{I}_{1})$. Then, $x\in\Pi$ and $x\notin(\pi(\mathcal{I}_{1})\cup\pi(\mathcal{I}_{2}))$, that is $x\in\Pi\setminus(\mathcal{D}_{1}\cup\mathcal{D}_{2})$. We now prove that $\Pi\setminus(\mathcal{D}_{1}\cup\mathcal{D}_{2})\subseteq\mathcal{D}_{3}$. This trivially follows when $\Pi\setminus(\mathcal{D}_{1}\cup\mathcal{D}_{2})=\emptyset$. If $\Pi\setminus(\mathcal{D}_{1}\cup\mathcal{D}_{2})\neq\emptyset$, let $x\in\Pi\setminus(\pi(\mathcal{I}_{1})\cup\pi(\mathcal{I}_{2}))$. As $x\in\Pi\setminus\pi(\mathcal{I}_{1})$, from condition (c) in Definition 10, we obtain $x\in\pi(\mathcal{I}_{4})$. Likewise, as $x\in\Pi\setminus\pi(\mathcal{I}_{2})$ from condition (c) in Definition 10, we obtain $x\in\pi(\mathcal{I}_{3})$. Then, $x\in(\pi(\mathcal{I}_{3})\cap\pi(\mathcal{I}_{4}))=\mathcal{D}_{3}$. Therefore $(\mathcal{D}_{1},\mathcal{D}_{2},\mathcal{D}_{3})$ is a tripartition of $\Pi$. By our assumption, $\pi(\mathcal{I}_{1})=\mathcal{D}_{1}$ and $\pi(\mathcal{I}_{2})=\mathcal{D}_{2}$. We observe that $\pi(\mathcal{I}_{3})\cap\mathcal{D}_{3}=\mathcal{D}_{3}$; moreover, from conditions (c) and (d), we obtain $\pi(\mathcal{I}_{3})\cap\mathcal{D}_{2}=\pi(\mathcal{I}_{3})\cap\pi(\mathcal{I}_{2})=\emptyset$ and $\pi(\mathcal{I}_{3})\cap\mathcal{D}_{1}=\pi(\mathcal{I}_{1})\cap\pi(\mathcal{I}_{3})=\pi(\mathcal{I}_{1})=\mathcal{D}_{1}$; then $\pi(\mathcal{I}_{3})=\pi(\mathcal{I}_{3})\cap(\mathcal{D}_{1}\cup\mathcal{D}_{2}\cup\mathcal{D}_{3})=\mathcal{D}_{1}\cup\mathcal{D}_{3}$. Likewise, we observe that $\pi(\mathcal{I}_{4})\cap\mathcal{D}_{3}=\mathcal{D}_{3}$; moreover, from conditions (c),(d) in Definition 10, we obtain $\mathcal{D}_{1}\cap\pi(\mathcal{I}_{4})=\pi(\mathcal{I}_{1})\cap\pi(\mathcal{I}_{4})=\emptyset$ and $\mathcal{D}_{2}\cap\pi(\mathcal{I}_{4})=\pi(\mathcal{I}_{2})\cap\pi(\mathcal{I}_{4})=\pi(\mathcal{I}_{2})=\mathcal{D}_{2}$; then $\pi(\mathcal{I}_{4})=\pi(\mathcal{I}_{4})\cap(\mathcal{D}_{1}\cup\mathcal{D}_{2}\cup\mathcal{D}_{3})=\mathcal{D}_{2}\cup\mathcal{D}_{3}$.

$(\Leftarrow)$ Assume that $(\mathcal{D}_{1},\mathcal{D}_{2},\mathcal{D}_{3})$, where $\mathcal{D}_{1}=\pi(\mathcal{I}_{1})$, $\mathcal{D}_{2}=\pi(\mathcal{I}_{2})$, $\mathcal{D}_{3}=\pi(\mathcal{I}_{3})\cap\pi(\mathcal{I}_{4})$, is a tripartition of $\Pi$ such that $\mathcal{D}_{1}\cup\mathcal{D}_{3}=\pi(\mathcal{I}_{3})$ and $\mathcal{D}_{2}\cup\mathcal{D}_{3}=\pi(\mathcal{I}_{4})$, we prove that the quadruple $(s_{1},s_{2},s_{3},s_{4})$ satisfies conditions (a), (b), (c), and (d) in Definition 10. We observe that $\pi(\mathcal{I}_{1})\cap\pi(\mathcal{I}_{2})=\mathcal{D}_{1}\cap\mathcal{D}_{2}=\emptyset$, which coincides with (a). Condition (b) is satisfied because $\pi(\mathcal{I}_{3})\cup\pi(\mathcal{I}_{4})=\mathcal{D}_{1}\cup\mathcal{D}_{3}\cup\mathcal{D}_{2}\cup\mathcal{D}_{3}=\Pi$. Moreover, $\pi(\mathcal{I}_{1})\cap\pi(\mathcal{I}_{4})=\mathcal{D}_{1}\cap(\mathcal{D}_{2}\cup\mathcal{D}_{3})=\emptyset$ and $\pi(\mathcal{I}_{1})\cup\pi(\mathcal{I}_{4})=\mathcal{D}_{1}\cup(\mathcal{D}_{2}\cup\mathcal{D}_{3})=\Pi$; likewise, $\pi(\mathcal{I}_{2})\cap\pi(\mathcal{I}_{3})=\mathcal{D}_{2}\cap(\mathcal{D}_{1}\cup\mathcal{D}_{3})=\emptyset$ and $\pi(\mathcal{I}_{2})\cup\pi(\mathcal{I}_{3})=\mathcal{D}_{2}\cup(\mathcal{D}_{1}\cup\mathcal{D}_{3})=\Pi$. Thus, the conditions in (c) are satisfied. Finally, $\pi(\mathcal{I}_{1})=\mathcal{D}_{1}\subseteq\mathcal{D}_{1}\cup\mathcal{D}_{3}=\pi(\mathcal{I}_{3})$ and $\pi(\mathcal{I}_{2})=\mathcal{D}_{2}\subseteq\mathcal{D}_{2}\cup\mathcal{D}_{3}=\pi(\mathcal{I}_{4})$ which satisfy conditions in (d). ∎

A method to construct a square of opposition by starting from a tripartition of $\Pi$ is given in the following result (see also [20]).

Corollary 1**.**

Given any sequence of $n$ conditional events $\mathcal{F}$ and a tripartition $(\mathcal{D}_{1},\mathcal{D}_{2},\mathcal{D}_{3})$ of $\Pi$, then the quadruple $(s_{1},s_{2},s_{3},s_{4})$, with $s_{k}:(\mathcal{F},\mathcal{I}_{k})$, $k=1,2,3,4$ and $\pi(\mathcal{I}_{1})=\mathcal{D}_{1}$, $\pi(\mathcal{I}_{2})=\mathcal{D}_{2}$, $\pi(\mathcal{I}_{3})=\mathcal{D}_{1}\cup\mathcal{D}_{3}$, $\pi(\mathcal{I}_{4})=\mathcal{D}_{2}\cup\mathcal{D}_{3}$ is a square of opposition.

Proof.

The proof immediately follows by observing $\pi(\mathcal{I}_{3})\cap\pi(\mathcal{I}_{4})=\mathcal{D}_{3}$ and by the ($\Leftarrow$) side proof of Theorem 1. ∎

The following result allows to construct a square of opposition by starting from a tripartition of the whole set $[0,1]^{n}$:

Corollary 2**.**

Given a tripartition $(\mathcal{B}_{1},\mathcal{B}_{2},\mathcal{B}_{3})$ of $[0,1]^{n}$, let $\mathcal{I}_{1}=\mathcal{B}_{1}$, $\mathcal{I}_{2}=\mathcal{B}_{2}$, $\mathcal{I}_{3}=\mathcal{B}_{1}\cup\mathcal{B}_{3}$, and $\mathcal{I}_{4}=\mathcal{B}_{2}\cup\mathcal{B}_{3}$. For any sequence of $n$ conditional events $\mathcal{F}$, the quadruple $(s_{1},s_{2},s_{3},s_{4})$, where $s_{k}:(\mathcal{F},\mathcal{I}_{k})$, $k=1,2,3,4$, is a square of opposition.

Proof.

Let $\mathcal{F}$ be any sequence of $n$ conditional events and $\Pi$ be the associated set of all coherent precise assessments. We set $\mathcal{D}_{i}=\pi(\mathcal{B}_{i})$, $i=1,2,3$. Of course, $(\pi(\mathcal{B}_{1}),\pi(\mathcal{B}_{2}),\pi(\mathcal{B}_{3}))$ is a tripartition of $\Pi$. Moreover, $\pi(\mathcal{I}_{1})=\mathcal{D}_{1}$, $\pi(\mathcal{I}_{2})=\mathcal{D}_{2}$, $\pi(\mathcal{I}_{3})=\mathcal{D}_{1}\cup\mathcal{D}_{3}$, $\pi(\mathcal{I}_{4})=\mathcal{D}_{2}\cup\mathcal{D}_{3}$. Then, by applying Corollary 1 we obtain that $(s_{1},s_{2},s_{3},s_{4})$ is a square of opposition. ∎

Traditionally the square of opposition can be constructed based on the fragmented square of opposition which requires only the contrariety and contradiction relations (which goes back to Aristotle’s De Interpretatione 6–7, 17b.17–26, see [38, Section 2]). This result also holds in our framework:

Theorem 2**.**

The quadruple $(s_{1},s_{2},s_{3},s_{4})$ of sentences, with $s_{k}:(\mathcal{F},\mathcal{I}_{k})$, $k=1,2,3,4$, is a square of opposition iff relations (a) and (c) in Definition 10 are satisfied.

Proof.

$(\Rightarrow)$ It follows directly from Definition 10. $(\Leftarrow)$ We prove that (d) and (b) in Definition 10 follow from (a) and (c). If $\pi(\mathcal{I}_{1})=\emptyset$, then of course $\pi(\mathcal{I}_{1})\subseteq\pi(\mathcal{I}_{3})$. If $\pi(\mathcal{I}_{1})\neq\emptyset$, let $x\in\pi(\mathcal{I}_{1})\subseteq\Pi$, from (a) it follows that $x\notin\pi(\mathcal{I}_{2})$, and since (c) requires $\pi(\mathcal{I}_{2})\cup\pi(\mathcal{I}_{3})=\Pi$, we obtain $x\in\pi(\mathcal{I}_{3})$. Thus, $\pi(\mathcal{I}_{1})\subseteq\pi(\mathcal{I}_{3})$; likewise, $\pi(\mathcal{I}_{2})\subseteq\pi(\mathcal{I}_{4})$. Therefore, (d) is satisfied. Now we prove that (b) is satisfied, i.e. $\pi(\mathcal{I}_{3})\cup\pi(\mathcal{I}_{4})=\Pi$. Of course, $\pi(\mathcal{I}_{3})\cup\pi(\mathcal{I}_{4})\subseteq\Pi$. Let $x\in\Pi$. If $x\notin\pi(\mathcal{I}_{3})$, then, $x\in\pi(\mathcal{I}_{2})$ from (c). Moreover, from (d), $x\in\pi(\mathcal{I}_{4})$. Then, $\Pi\subseteq\pi(\mathcal{I}_{3})\cup\pi(\mathcal{I}_{4})$. Therefore, (b) is satisfied. ∎

Corollary 3**.**

The quadruple $(s_{1},s_{2},s_{3},s_{4})$ of sentences, with $s_{k}:(\mathcal{F},\mathcal{I}_{k})$, $k=1,2,3,4$, is a square of opposition if and only if $(s_{1},s_{2},s_{3},s_{4})=(s_{1},s_{2},\widebar{s}_{2},\widebar{s}_{1})$ with $s_{1}$ and $s_{2}$ being contraries.

Proof.

Of course, if $(s_{1},s_{2},s_{3},s_{4})$ is a square of opposition, then $s_{1}$ and $s_{2}$ are contraries. Moreover, $s_{1}$ and $s_{4}$ are contradictories, that is: $\pi(\mathcal{I}_{1})\cap\pi(\mathcal{I}_{4})=\emptyset$ and $\pi(\mathcal{I}_{1})\cup\pi(\mathcal{I}_{4})=\Pi$. Therefore, $\Pi\setminus\pi(\mathcal{I}_{4})=\pi(\mathcal{I}_{1})$, which amounts to $s_{4}=\widebar{s}_{1}$. Similary, as $s_{2}$ and $s_{3}$ are contradictories, it holds that $s_{3}=\widebar{s}_{2}$. Conversely, assume that $s_{1}$ and $s_{2}$ are contraries. By instantiating Theorem 2 with $s_{3}=\widebar{s}_{2}$ and with $s_{4}=\widebar{s}_{1}$, it follows that the quadruple $(s_{1},s_{2},\widebar{s}_{2},\widebar{s}_{1})$ is a square of opposition. ∎

4 Square of Opposition and Generalized Quantifiers

Let $\mathcal{F}$ be a conditional event $P|S$ (where $S\neq\bot$) and $(\mathcal{B}_{1}(x),\mathcal{B}_{2}(x),\mathcal{B}_{3}(x))$ be a tripartition of $[0,1]$, where $\mathcal{B}_{1}(x)=[x,1]$, $\mathcal{B}_{2}(x)=[0,1-x]$, $\mathcal{B}_{3}(x)=]1-x,x[$ and $x\in]\frac{1}{2},1]$ (see Figure 2).

Consider the quadruple of sentences $(A(x),E(x),I(x),O(x))$, with $A(x):(P|S,\mathcal{I}_{A(x)})$, $E(x):(P|S,\mathcal{I}_{E(x)})$, $I(x):(P|S,\mathcal{I}_{I(x)})$, $O(x):(P|S,\mathcal{I}_{O(x)})$, where $\mathcal{I}_{A(x)}=\mathcal{B}_{1}(x)=[x,1]$, $\mathcal{I}_{E(x)}=\mathcal{B}_{2}(x)=[0,1-x]$, $\mathcal{I}_{I(x)}=\mathcal{B}_{1}(x)\cup\mathcal{B}_{3}(x)=]1-x,1]$, and $\mathcal{I}_{O(x)}=\mathcal{B}_{2}(x)\cup\mathcal{B}_{3}(x)=[0,x[$. By applying Corollary 2 with $(s_{1},s_{2},s_{3},s_{4})=(A(x),E(x),I(x),O(x))$, it follows that $(A(x),E(x),I(x),O(x))$ is a square of opposition for any $x\in]\frac{1}{2},1]$ (see Figure 3).

We recall that in presence of some logical relations between $P$ and $S$ the set $\Pi$ could be a strict subset of $[0,1]$. In particular, we have the following three cases (see, [29, 30]): (i) if $P\wedge S\neq\bot$ and $P\wedge S\neq S$, then $\Pi=[0,1]$; (ii) if $P\wedge S=S$, then $\Pi=\{1\}$; (iii) if $P\wedge S=\bot$, then $\Pi=\{0\}$. The quadruple $(A(x),E(x),I(x),O(x))$, with the threshold $\frac{1}{2}<x\leq 1$, is a square of opposition in each of the three cases. In particular we obtain: case (i) $\pi(\mathcal{I}_{A(x)})=\mathcal{I}_{A(x)}$, $\pi(\mathcal{I}_{E(x)})=\mathcal{I}_{E(x)}$,$\pi(\mathcal{I}_{I(x)})=\mathcal{I}_{I(x)}$, and $\pi(\mathcal{I}_{O(x)})=\mathcal{I}_{O(x)}$; case (ii): $\pi(\mathcal{I}_{A(x)})=\{1\}$, $\pi(\mathcal{I}_{E(x)})=\emptyset$,$\pi(\mathcal{I}_{I(x)})=\{1\}$, and $\pi(\mathcal{I}_{O(x)})=\emptyset$; case (iii): $\pi(\mathcal{I}_{A(x)})=\emptyset$, $\pi(\mathcal{I}_{E(x)})=\{1\}$,$\pi(\mathcal{I}_{I(x)})=\emptyset$, and $\pi(\mathcal{I}_{O(x)})=\{1\}$. We note that in cases (ii) and (iii) we obtain degenerated squares each, where—apart from the contradictory relations—all relations are strengthened. Specifically, both contrary and the subcontrary become contradictory relations. Moreover, both subalternation relations become symmetric. As by coherence $p(P|S)+p(\widebar{P}|S)=1$, a sentence $s:(P|S,\mathcal{I})$ is equivalent to the sentence $s^{\prime}:(\widebar{P}|S,\widebar{\mathcal{I}})$, where $\widebar{\mathcal{I}}=[0,1]\setminus\mathcal{I}$. Table 1

presents generalization of basic sentence types $A(x)$, $E(x)$, $I(x)$, and $O(x)$ involving generalized quantifiers $Q$. The generalized quantifiers are defined on a threshold $x>\frac{1}{2}$. The value of the threshold may be context dependent and provides lots of flexibility for modeling various instances of generalized quantifiers (like “most”, “almost all”).

Given two thresholds $x_{1}$ and $x_{2}$, with $\frac{1}{2}<x_{2}<x_{1}\leq 1$, we analyze the relations among the same sentence types in the two squares of opposition $\mathbf{S}(x_{1})$ and $\mathbf{S}(x_{2})$, with $\mathbf{S}(x_{i})=(A(x_{i}),E(x_{i}),I(x_{i}),O(x_{i}))$, $i=1,2$. It can be easily proved that: $A(x_{2})$ is a subaltern of $A(x_{1})$, $E(x_{2})$ is a subaltern of $E(x_{1})$, $I(x_{1})$ is a subaltern of $I(x_{2})$, and $O(x_{1})$ is a subaltern of $O(x_{2})$. In the extreme case $x=1$ we obtain the probabilistic interpretation under coherence of the basic sentence types involved in the traditional square of opposition $(A,E,I,O)$ (see [27, 28] for the default square of opposition, involving defaults and negated defaults).

In agreement with De Morgan (as pointed out by [20]) by the quadruple $(a,e,i,o)$ we denotes the square of opposition obtained from $(A,E,I,O)$ when the events $P$ and $S$ are replaced by $\widebar{P}$ and $\widebar{S}$, respectively. Specifically, $a:(\widebar{P}|\widebar{S},\{1\})$, $e:(\widebar{P}|\widebar{S},\{0\})$, $i:(\widebar{P}|\widebar{S},]0,1])$, and $o:(\widebar{P}|\widebar{S},[0,1[)$.

In the general case when $P$ and $S$ are logically independent it can be proved that the set of all coherent assessments on $(P|S,\widebar{P}|\widebar{S})$ is the square $[0,1]^{2}$ (see e.g. [17]; see also [14, Proposition 1] [15, Theorem 4]). Thus, in the general case there are no relations between any two sentences $s_{1}$ and $s_{2}$, where $s_{1}\in\{A,E,I,O\}$ and $s_{2}\in\{a,e,i,o\}$. Therefore, the two squares $(A,E,I,O)$ and $(a,e,i,o)$ do not form a cube of opposition (with these two squares as opposite facing sides).

5 Hexagon of Opposition

Compared to the millennia long history of investigations on the square of opposition, the hexagon of opposition was discovered fairly recently, namely in the 1950ies. The hexagon generalizes the square by adding the disjunction of the top vertices of the square to build a new vertex at the top and by adding the conjunction of the bottom vertices of the square to build a new vertex at the bottom. According to Béziau ([2]), the hexagon of opposition was introduced by the French priest and logician Augustin Sesmat ([48]) and by the philosopher Robert Blanché ([7]), who worked out the full structure of the hexagon of opposition (for his main work on the hexagon of opposition see [8]). Jaspers and Seuren ([33]) trace the history of the hexagon back also to the American philosopher Paul Jacoby ([32], see also [20]). In this section we will use the tools developed in Section 3, to construct a hexagon of opposition by starting from a square of opposition. More precisely, given a traditional square of opposition $(A,E,I,O)$, by setting $U=A\lor E$, $Y=I\land O$, the tuple $(A,E,I,O,U,Y)$ defines a hexagon of opposition. Accordingly, we define the (probabilistic) hexagon of opposition in our approach as follows:

Definition 12** (Hexagon of opposition).**

Let $s_{k}:(\mathcal{F},\mathcal{I}_{k})$, $k=1,2,3,4,5,6$, be six sentences. We call the ordered tuple $(s_{1},s_{2},s_{3},s_{4},s_{5},s_{6})$ a hexagon of opposition (under coherence), if and only if the following relations among the six sentences hold:

1. (i)

$(s_{1},s_{2},s_{3},s_{4})$* is a square of opposition;* 2. (ii)

$s_{5}=s_{1}\vee s_{2}$; 3. (iii)

$s_{6}=s_{3}\wedge s_{4}$.

Figure 4 shows the probabilistic hexagon of opposition as given by Definition 12.

Theorem 3**.**

Let $s_{k}:(\mathcal{F},\mathcal{I}_{k})$, $k=1,2,3,4,5,6$, be six sentences. The tuple $(s_{1},s_{2},s_{3},s_{4},s_{5},s_{6})$ is a hexagon of opposition, if and only if $(s_{1},s_{2},s_{3},s_{4},s_{5},s_{6})=(s_{1},s_{2},\widebar{s}_{2},\widebar{s}_{1},s_{1}\vee s_{2},\widebar{s}_{1}\wedge\widebar{s}_{2})$, with $s_{1}$ and $s_{2}$ being contraries.

Proof.

$(\Rightarrow)$. Let $(s_{1},s_{2},s_{3},s_{4},s_{5},s_{6})$ be a hexagon of opposition. Then, as $(s_{1},s_{2},s_{3},s_{4})$ is a square of opposition, $s_{1}$ and $s_{2}$ are contraries. Moreover, by Corollary 3, it follows that $(s_{1},s_{2},s_{3},s_{4})=(s_{1},s_{2},\widebar{s}_{2},\widebar{s}_{1})$. Then, by Definition 12, $s_{5}=s_{1}\vee s_{2}$ and $s_{6}=s_{3}\wedge s_{4}=\widebar{s}_{1}\wedge\widebar{s}_{2}$. Therefore, $(s_{1},s_{2},s_{3},s_{4},s_{5},s_{6})=(s_{1},s_{2},\widebar{s}_{2},\widebar{s}_{1},s_{1}\vee s_{2},\widebar{s}_{1}\wedge\widebar{s}_{2})$.

$(\Leftarrow)$. Let $(s_{1},s_{2},s_{3},s_{4},s_{5},s_{6})=(s_{1},s_{2},\widebar{s}_{2},\widebar{s}_{1},s_{1}\vee s_{2},\widebar{s}_{1}\wedge\widebar{s}_{2})$, with $s_{1}$ and $s_{2}$ being contraries. From Corollary 3, it follows that $(s_{1},s_{2},s_{3},s_{4})$ is a square of opposition. Then, by relations $(ii)$ and $(iii)$ in Definition 12, it follows that $(s_{1},s_{2},s_{3},s_{4},s_{5},s_{6})$ is a hexagon of opposition.

∎

Remark 5**.**

Assume that $s_{1}$ and $s_{2}$ are contraries. Then, by Corollary 3, the quadruple $(s_{1},s_{2},\widebar{s}_{2},\widebar{s}_{1})$ is a square of opposition, and by Definition 12, the tuple $(s_{1},s_{2},\widebar{s}_{2},\widebar{s}_{1},s_{1}\vee s_{2},\widebar{s}_{1}\wedge\widebar{s}_{2})$ is a hexagon of opposition.

We now consider relations among a tripartition of the set of all coherent assessments $\Pi$ and a hexagon of opposition.

Remark 6**.**

Given a hexagon of opposition $(s_{1},s_{2},s_{3},s_{4},s_{5},s_{6})$, we observe that the sentence $s_{6}=s_{3}\wedge s_{4}$ represents the pair $(\mathcal{F},\mathcal{I}_{6})$, where $\mathcal{I}_{6}=\mathcal{I}_{3}\cap\mathcal{I}_{4}$. Moreover, by Remark 1, $\pi(\mathcal{I}_{6})=\pi(\mathcal{I}_{3}\cap\mathcal{I}_{4})=\pi(\mathcal{I}_{3})\cap\pi(\mathcal{I}_{4})$. Therefore, based on Theorem 1, the triple $(\mathcal{D}_{1},\mathcal{D}_{2},\mathcal{D}_{3})$, where $\mathcal{D}_{1}=\pi(\mathcal{I}_{1})$, $\mathcal{D}_{2}=\pi(\mathcal{I}_{2})$, and $\mathcal{D}_{3}=\pi(\mathcal{I}_{6})$, is a tripartition of $\Pi$. Conversely, based on Corollary 1, given a tripartition $(\mathcal{D}_{1},\mathcal{D}_{2},\mathcal{D}_{3})$ of $\Pi$, the sequence $(s_{1},s_{2},s_{3},s_{4},s_{5},s_{6})$ where ${s_{k}:(\mathcal{F},\mathcal{I}_{k})}$, $k=1,\ldots,6$, with $\pi(\mathcal{I}_{1})=\mathcal{D}_{1}$, $\pi(\mathcal{I}_{2})=\mathcal{D}_{2}$, $\pi(\mathcal{I}_{3})=\mathcal{D}_{1}\cup\mathcal{D}_{3}$, $\pi(\mathcal{I}_{4})=\mathcal{D}_{2}\cup\mathcal{D}_{3}$, $\pi(\mathcal{I}_{5})=\mathcal{D}_{1}\cup\mathcal{D}_{2}$, and $\pi(\mathcal{I}_{6})=\mathcal{D}_{3}$, is a hexagon of opposition (see also [11, 20, 21]).

Next, we consider relations among a tripartition of $[0,1]^{n}$ and a hexagon of opposition.

Remark 7**.**

Based on Corollary 2, we can also construct a hexagon of opposition by starting from a tripartition of the whole set $[0,1]^{n}$. Specifically, given a tripartition $(\mathcal{B}_{1},\mathcal{B}_{2},\mathcal{B}_{3})$ of $[0,1]^{n}$, let $\mathcal{I}_{1}=\mathcal{B}_{1}$, $\mathcal{I}_{2}=\mathcal{B}_{2}$, $\mathcal{I}_{3}=\mathcal{B}_{1}\cup\mathcal{B}_{3}$, $\mathcal{I}_{4}=\mathcal{B}_{2}\cup\mathcal{B}_{3}$, $\mathcal{I}_{5}=\mathcal{B}_{1}\cup\mathcal{B}_{2}$, and $\mathcal{I}_{6}=\mathcal{B}_{3}$. For any sequence of $n$ conditional events $\mathcal{F}$, the tuple $(s_{1},s_{2},s_{3},s_{4},s_{5},s_{6})$, where $s_{k}:(\mathcal{F},\mathcal{I}_{k})$, $k=1,\ldots,6$, is a hexagon of opposition.

Theorem 4**.**

Given a hexagon of opposition $(s_{1},s_{2},s_{3},s_{4},s_{5},s_{6})$, by Definition 12 all relations among the basic sentence types in the square $(s_{1},s_{2},s_{3},s_{4})$ hold. Moreover, by Theorem 3 (and also by Remark 3), the following relations hold:

1. (i)

$s_{1}$* and $s_{6}$ are contraries (since $s_{6}=\widebar{s}_{2}\wedge\widebar{s}_{1}$ and $\pi(\mathcal{I}_{1}\cap\widebar{\mathcal{I}}_{2}\cap\widebar{\mathcal{I}}_{1})=\emptyset$);* 2. (ii)

$s_{2}$* and $s_{6}$ are contraries (since $s_{6}=\widebar{s}_{2}\wedge\widebar{s}_{1}$ and $\pi(\mathcal{I}_{2}\cap\widebar{\mathcal{I}}_{2}\cap\widebar{\mathcal{I}}_{1})=\emptyset$);* 3. (iii)

$s_{3}$* is a subaltern of $s_{6}$ (since $s_{6}=s_{3}\wedge s_{4}$);* 4. (iv)

$s_{4}$* is a subaltern of $s_{6}$ (since $s_{6}=s_{3}\wedge s_{4}$);* 5. (v)

$s_{5}$* is a subaltern of $s_{1}$ (since $s_{5}=s_{1}\vee s_{2}$);* 6. (vi)

$s_{5}$* is a subaltern of $s_{2}$ (since $s_{5}=s_{1}\vee s_{2}$);* 7. (vii)

$s_{5}$* and $s_{3}$ are subcontraries (as $s_{5}=s_{1}\vee s_{2}$ and $s_{3}=\widebar{s}_{2}$, hence $\pi(\mathcal{I}_{1}\cup\mathcal{I}_{2})\cup\widebar{\mathcal{I}}_{2})=\Pi$);* 8. (viii)

$s_{5}$* and $s_{4}$ are subcontraries (as $s_{5}=s_{1}\vee s_{2}$ and $s_{4}=\widebar{s}_{1}$, hence $\pi(\mathcal{I}_{1}\cup\mathcal{I}_{2})\cup\widebar{\mathcal{I}}_{1})=\Pi$);* 9. (ix)

$s_{5}$* and $s_{6}$ are contradictories (as $s_{5}=s_{1}\vee s_{2}$, $s_{6}=s_{3}\wedge s_{4}=\widebar{s}_{2}\wedge\widebar{s}_{1}$, hence $\pi((\mathcal{I}_{1}\cup\mathcal{I}_{2})\cap(\widebar{\mathcal{I}}_{1}\cap\widebar{\mathcal{I}}_{2}))=\emptyset$) and $\pi((\mathcal{I}_{1}\cup\mathcal{I}_{2})\cup(\widebar{\mathcal{I}}_{1}\cap\widebar{\mathcal{I}}_{2}))=\Pi$).*

Figure 4 illustrates all the relations in the hexagon of opposition described in Theorem 4. This figure also shows the two triangles $T_{1}:(s_{1},s_{2},s_{6})$ and $T_{2}:(s_{3},s_{4},s_{5})$. We note that the sides of $T_{1}$ consist of contrary relations, whereas the sides of $T_{2}$ consist of subcontrary relations. Moreover, the coherent part of the imprecise assessments defined by sentences in $T_{1}$ (i.e., $\mathcal{D}_{1}=\pi(\mathcal{I}_{1})$, $\mathcal{D}_{2}=\pi(\mathcal{I}_{2})$ and $\mathcal{D}_{3}=\pi(\mathcal{I}_{6})$) forms a tripartition $(\mathcal{D}_{1},\mathcal{D}_{2},\mathcal{D}_{3})$ of $\Pi$. Whereas, the imprecise assessments defined by sentences in $T_{2}$ are such that $\pi(\mathcal{I}_{3})=\mathcal{D}_{1}\cup\mathcal{D}_{3}$, $\pi(\mathcal{I}_{4})=\mathcal{D}_{2}\cup\mathcal{D}_{3}$, and $\pi(\mathcal{I}_{5})=\mathcal{D}_{1}\cup\mathcal{D}_{2}$.

By basing the hexagon of opposition on the square of opposition $(A(x),E(x),I(x),O(x))$ (as introduced in Section 4) we obtain the following hexagon of opposition: $(A(x),E(x),I(x),O(x),U(x),Y(x))$ with $x\in]1/2,1]$, where $U(x)$ denotes $A(x)\vee E(x)$ and $Y(x)$ denotes $I(x)\wedge O(x)$ (see Table 2). Figure 5 illustrates the hexagon $(A(x),E(x),I(x),O(x),U(x),Y(x))$ with $x\in]1/2,1]$.

We now consider a generalization of the hexagon of opposition $(A(x),E(x),I(x),O(x),U(x),Y(x))$ by considering $n$ conditional events. In particular, let $\mathcal{F}=(P_{1}|S_{1},\ldots,P_{n}|S_{n})$ be a sequence of $n$ conditional events. Exploiting Remark 7, we construct a hexagon of opposition by considering the following tripartition of $[0,1]^{n}$: $(\mathcal{B}_{1}(x),\mathcal{B}_{2}(x),\mathcal{B}_{3}(x))$, with $x\in]1/2,1]$, where

[TABLE]

We obtain the following (generalized) hexagon of opposition $(A(x),E(x),I(x),O(x),U(x),Y(x))$, with the quantified statements $A(x):(\mathcal{F},\mathcal{I}_{A(x)})$, $E(x):(\mathcal{F},\mathcal{I}_{E(x)})$, $I(x):(\mathcal{F},\mathcal{I}_{I(x)})$, $O(x):(\mathcal{F},\mathcal{I}_{O(x)})$, $U(x):(\mathcal{F},\mathcal{I}_{U(x)})$, $Y(x):(\mathcal{F},\mathcal{I}_{Y(x)})$, where

[TABLE]

6 Concluding Remarks

Finally, we note that conditional probability interpretations of quantified statements were also proposed in psychology (see, e.g., [12, 13, 37, 39, 41, 43, 46]), since generalized quantifiers are psychologically much more plausible compared to the traditional logical quantifiers, as the latter are either too strict ($\forall$ does not allow for exceptions) or too weak ($\exists$ quantifies over at least one object) for formalizing everyday life sentences. Recent experimental data suggests that people negate conditionals and quantified statements mainly by building contraries (in the sense of inferring $p(\neg C|A)=1-x$ from the negated $p(C|A)=x$) but hardly ever by building contradictories (in the sense of inferring $p(C|A)<x$ from the negated $p(C|A)=x$; see [40, 46]). However, this empirical result calls for further experiments. The square presented in Section 4 and the hexagon presented in Section 5 can serve as a new rationality framework for formal-normative and psychological investigations of basic relations among quantified statements.

Acknowledgement

We thank Deutsche Forschungsgemeinschaft (DFG), Fondation Maison des Sciences de l’Homme (FMSH), and Villa Vigoni for supporting joint meetings at Villa Vigoni where parts of this work originated (Project: “Human Rationality: Probabilistic Points of View”). Niki Pfeifer is supported by his DFG project PF 740/2-2 (within the SPP1516 “New Frameworks of Rationality”). Giuseppe Sanfilippo is supported by the INdAM–GNAMPA Project (2016 Grant U 2016/000391).