Fitchean Ignorance and First-order Ignorance:
A Neighborhood Look

Jie Fan
Institute of Philosophy, Chinese Academy of Sciences;
School of Humanities, University of Chinese Academy of Sciences
[email protected]

Abstract

In a seminal work [15], Fine classifies several forms of ignorance, among which are Fitchean ignorance, first-order ignorance, Rumsfeld ignorance, and second-order ignorance. It is shown that there is interesting relationship among some of them, which includes that in ${\bf S4}$ , all higher-order ignorance are reduced to second-order ignorance. This is thought of as a bad consequence by some researchers. It is then natural to ask how to avoid this consequence. We deal with this issue in a much more general framework. In detail, we treat the forms of Fitchean ignorance and first-order ignorance as primitive modalities and study them as first-class citizens under neighborhood semantics, in which Rumsfeld ignorance and second-order ignorance are definable. The main contributions include model-theoretical results such as expressivity and frame definability, and axiomatizations. Last but not least, by updating the neighborhood models via the intersection semantics, we extend the results to the dynamic case of public announcements, which gives us some applications to successful formulas.

Keywords: Fitchean ignorance, first-order ignorance, contingency, accident, unknown truths, expressivity, frame definability, axiomatizations, intersection semantics, successful formulas

1 Introduction

Ignorance has been a hotly discussed theme in epistemology and many other fields since Socrates, who professed ignorance in e.g. the Apology [1]. Just as there has been no consensus on the definition of knowledge, there has been no consensus on the definition of ignorance. Instead, there has been at least three views in the literature: the standard view, the new view, and the logical view.¹¹1The terminology ‘the standard view’ is introduced in [25], ‘the new view’ is from [29], whereas the term ‘the logical view’ comes from [5]. The standard view thinks that ignorance is merely the negation of propositional knowledge, the new view thinks that ignorance is the lack of true belief,²²2For the discussion on the standard and new view, see [20] and references therein. whereas the logical view thinks that ignorance means neither knowing nor knowing not [34, 35, 33, 13, 14, 27].³³3To the best of our knowledge, the first to evidently investigate ignorance from the logical view is [34] — also see its extended journal version [35], which though includes an unsound transitive axiomatization, as shown in [14, pp. 102–103].

Recently there has been a flurry of research on ignorance. Various forms of ignorance are proposed in the literature, such as pluralistic ignorance [26, 2, 30], circumscriptive ignorance [18], chronological ignorance [32], factive ignorance [19], relative ignorance [16], disjunctive ignorance [10]. In a seminal paper [15], instead of discussing the definition of ignorance, Fine classifies several forms of ignorance, among which are ‘ignorance of (the fact that)’ (also called ‘Fitchean ignorance’ there), ‘first-order ignorance (whether)’, ‘Rumsfeld ignorance’ and ‘second-order ignorance’. One is ignorant of (the fact that) $\varphi$ , if $\varphi$ is the case but one does not know it. One is (first-order) ignorant whether $\varphi$ , if one neither knows $\varphi$ nor knows its negation. One is Rumsfeld ignorant of $\varphi$ , if one is ignorant of the fact that one is ignorant whether $\varphi$ . One is second-order ignorant whether $\varphi$ , if one is ignorant whether one is ignorant whether $\varphi$ .

As Fine [15] shows, there is interesting relationship among some of the forms. For instance, within the context of the system ${\bf S4}$ , second-order ignorance implies first-order ignorance; second-order ignorance implies Rumsfeld ignorance, and vice versa; one does not know one is Rumsfeld ignorant; one does not know one is second-order ignorant. However, all these results are based on the context of ${\bf S4}$ . It is then natural to ask what relationship among these forms there is in other contexts, based on the following reasons: firstly, although knowledge is usually based on ${\bf S4}$ (for instance in [17]), ignorance is not — it is argued on the new view that ignorance is not not-knowing (e.g. [29]); secondly, in the first explicitly logical studies on ignorance [34, 35], the semantic condition is arbitrary, without any restriction; moreover, in ${\bf S4}$ , all higher-order ignorance are reduced to second-order ignorance — this is called the black hole of ignorance in [15] and a quite problematic phenomenon in [3, p. 1060].

One may easily check that the latter two forms are definable with the former two ones. It is the former two forms that are our focus here.⁴⁴4Fitchean ignorance and first-order ignorance correspond to important metaphysical concepts — accident (or ‘accidental truths’) and contingency, respectively. For the history of the bimodal logic of contingency and accident and the importance of the two metaphysical concepts, we refer to [7] and the reference therein. It is important to distinguish these two forms. For instance, the Fitchean ignorance satisfies the so-called Factivity Principle (that is, if an agent is ignorant of $\varphi$ then $\varphi$ is true), but the first-order ignorance does not.⁵⁵5In a recent work [3, p. 7], the authors seem to think that ignorance has only one form, and say that ignorance should satisfy Factivity Principle since knowledge does. Moreover, since the operators of the two forms and their duals are not normal, the logic of Fitchean ignorance and first-order ignorance is not normal. As is well known, neighborhood semantics has been a standard semantics tool for non-normal modal logics since its introduction in 1970 [24, 31, 4, 28]. In the current paper, we will investigate the logical properties of the two forms of ignorance and their relationship under the neighborhood semantics. As we will show, there is interesting relationship among first-order ignorance, second-order ignorance, and Rumsfeld ignorance. For example, under any condition, Rumsfeld ignorance implies first-order ignorance, and second-order ignorance plus first-order ignorance implies Rumsfeld ignorance, whereas under the condition $(c)$ , Rumsfeld ignorance implies second-order ignorance, and thus Rumsfeld ignorance amounts to second-order ignorance plus first-order ignorance. However, similar to the case for relational semantics [7], the situation may become quite involved if we study the two notions in a unified framework under the neighborhood semantics. For instance, we will be confronted with a difficulty in axiomatizing the bimodal logic, since we have only one neighborhood function to deal with two modal operators uniformly, which makes it hard to find suitable interaction axioms.

The remainder of the paper is organized as follows. After briefly reviewing the syntax and the neighborhood semantics of the bimodal logic of Fitchean ignorance and first-order ignorance and also some related logics (Sec. 2), we compare the relative expressivity (Sec. 3) and investigate the frame definability of the bimodal logic (Sec. 4). We axiomatize the bimodal logic over various classes of neighborhood frames (Sec. 5). By updating the neighborhood models via the intersection semantics, we find suitable reduction axioms and thus reduce the public announcements operators to the bimodal logic, which gives us good applications to successful formulas (Sec. 6), where, as we shall show, any combination of $p$ , $\neg p$ , ${\neg\bullet}p$ , and ${\neg\nabla}p$ via conjunction (or, via disjunction) is successful under the intersection semantics. Finally, we conclude with some future work in Sec. 7.

2 Syntax and Neighborhood Semantics

This section introduces the languages and their neighborhood semantics involved in this paper.

Fix a nonempty set P of propositional variables, and let $p\in\textbf{P}$ . In what follows, $\mathcal{L}(\Box)$ is the language of standard epistemic logic, $\mathcal{L}(\nabla)$ is the language of the logic of (first-order) ignorance, $\mathcal{L}(\bullet)$ is the language of the logic of Fitchean ignorance⁶⁶6 $\mathcal{L}(\bullet)$ is also called ‘the logic of essence and accident’ or ‘the logic of unknown truths’, see e.g. [23, 33]., and $\mathcal{L}(\nabla,\bullet)$ is the language of the bimodal logic of Fitchean ignorance and first-order ignorance. We will mainly focus on $\mathcal{L}(\nabla,\bullet)$ . For the sake of simplicity, we only exhibit the single-agent languages, but all our results also apply to multi-agent cases.

Definition 1 (Languages).

\begin{array}[]{llll}\mathcal{L}(\Box)&\varphi&::=&p\mid\neg\varphi\mid\varphi\land\varphi\mid\Box\varphi\\ \mathcal{L}(\nabla)&\varphi&::=&p\mid\neg\varphi\mid\varphi\land\varphi\mid\nabla\varphi\\ \mathcal{L}(\bullet)&\varphi&::=&p\mid\neg\varphi\mid\varphi\land\varphi\mid\bullet\varphi\\ \mathcal{L}(\nabla,\bullet)&\varphi&::=&p\mid\neg\varphi\mid\varphi\land\varphi\mid\nabla\varphi\mid\bullet\varphi\\ \end{array}

$\Box\varphi$ is read “one knows that $\varphi$ ”, $\nabla\varphi$ is read “one is (first-order) ignorant whether $\varphi$ ”, and $\bullet\varphi$ is read “one is ignorant of (the fact that) $\varphi$ ”, or “ $\varphi$ is an unknown truth”. In the metaphysical setting, $\nabla\varphi$ and $\bullet\varphi$ are read, respectively, “it is contingent that $\varphi$ ” and “it is accidental that $\varphi$ ”. Among other connectives, $\Diamond\varphi$ , $\Delta\varphi$ , and $\circ\varphi$ abbreviate, respectively, $\neg\Box\neg\varphi$ , ${\neg\nabla}\varphi$ , and ${\neg\bullet}\varphi$ , read “it is epistemically possible that $\varphi$ ”, “ one knows whether $\varphi$ ”, and “one is non-ignorant of $\varphi$ ”.

Note that the forms of ‘Rumsfeld ignorance (of $\varphi$ )’ and ‘second-ignorance (whether $\varphi$ )’ can be defined as, respectively, $\bullet\nabla\varphi$ and $\nabla\nabla\varphi$ .

The above languages are interpreted over neighborhood models.

Definition 2 (Neighborhood structures).

A (neighborhood) model is a triple $\mathcal{M}=\langle S,N,V\rangle$ , where $S$ is a nonempty set of states (also called ‘points’ or ‘possible worlds’, $N$ is a neighborhood function from $S$ to $\mathcal{P}(\mathcal{P}(S))$ , and $V$ is a valuation function. A (neighborhood) frame is a model without a valuation; in this case, we say that the model is based on the frame. A pointed model is a pair of a model with a point in it. Given an $s\in S$ , an element of $N(s)$ is called ‘a neighborhood of $s$ ’.

The following list of neighborhood properties come from [12, Def. 3].

Definition 3 (Neighborhood properties).

Let $\mathcal{F}=\langle S,N\rangle$ be a frame, and $\mathcal{M}$ be a model based on $\mathcal{F}$ . Let $s\in S$ and $X,Y\subseteq S$ . We define various neighborhood properties as follows.

•

$(n)$ : $N(s)$ contains the unit, if $S\in N(s)$ .
•

$(r)$ : $N(s)$ contains its core, if $\bigcap N(s)\in N(s)$ .
•

$(i)$ : $N(s)$ is closed under intersections, if $X,Y\in N(s)$ implies $X\cap Y\in N(s)$ .
•

$(s)$ : $N(s)$ is supplemented, or closed under supersets, if $X\in N(s)$ and $X\subseteq Y\subseteq S$ implies $Y\in N(s)$ .
•

$(c)$ : $N(s)$ is closed under complements, if $X\in N(s)$ implies $S\backslash X\in N(s)$ .⁷⁷7The property $(c)$ provides a new perspective for $\mathcal{L}(\nabla)$ , see [6] for details.
•

$(d)$ : $X\in N(s)$ implies $S\backslash X\notin N(s)$ .
•

$(t)$ : $X\in N(s)$ implies $s\in X$ .
•

$(b)$ : $s\in X$ implies $\{u\in S\mid S\backslash X\notin N(u)\}\in N(s)$ .
•

$(4)$ : $X\in N(s)$ implies $\{u\in S\mid X\in N(u)\}\in N(s)$ .
•

$(5)$ : $X\notin N(s)$ implies $\{u\in S\mid X\notin N(u)\}\in N(s)$ .

The function $N$ possesses such a property, if for all $s\in S$ , $N(s)$ has the property. $\mathcal{F}$ (and $\mathcal{M}$ ) has a property, if $N$ has. In particular, we say that $\mathcal{F}$ (and $\mathcal{M}$ ) is monotone, if $N$ has $(s)$ . $\mathcal{F}$ (and $\mathcal{M}$ ) is a quasi-filter, if $N$ has $(i)$ and $(s)$ ; $\mathcal{F}$ (and $\mathcal{M}$ ) is a filter, if $N$ has also $(n)$ .

Also, in what follows, we will use $\mathbb{C}_{n}$ to denote the class of $(n)$ -models, and similarly for $\mathbb{C}_{r}$ , etc. We use $\mathbb{C}_{\text{all}}$ for the class of all neighborhood models.

Definition 4 (Semantics).

Let $\mathcal{M}=\langle S,N,V\rangle$ be a model. Given a pointed model $(\mathcal{M},s)$ , the truth condition of formulas is defined recursively as follows:

\begin{array}[]{|lll|}\hline\cr\mathcal{M},s\vDash p&\iff&s\in V(p)\\ \mathcal{M},s\vDash\neg\varphi&\iff&\mathcal{M},s\nvDash\varphi\\ \mathcal{M},s\vDash\varphi\land\psi&\iff&\mathcal{M},s\vDash\varphi\text{ and }\mathcal{M},s\vDash\psi\\ \mathcal{M},s\vDash\Box\varphi&\iff&\varphi^{\mathcal{M}}\in N(s)\\ \mathcal{M},s\vDash\nabla\varphi&\iff&\varphi^{\mathcal{M}}\notin N(s)\text{ and }S\backslash\varphi^{\mathcal{M}}\notin N(s)\\ \mathcal{M},s\vDash\bullet\varphi&\iff&\mathcal{M},s\vDash\varphi\text{ and }\varphi^{\mathcal{M}}\notin N(s)\\ \hline\cr\end{array}

where $\varphi^{\mathcal{M}}$ denotes the truth set of $\varphi$ in $\mathcal{M}$ , in symbols, $\varphi^{\mathcal{M}}=\{s\in S\mid\mathcal{M},s\vDash\varphi\}$ ; given a set $X\subseteq S$ , $S\backslash X$ denotes the complement of $X$ with respect to $S$ .

We say that $\varphi$ is true in $(\mathcal{M},s)$ , if $\mathcal{M},s\vDash\varphi$ ; we say that $\varphi$ is valid on a model $\mathcal{M}$ , notation: $\mathcal{M}\vDash\varphi$ , if for all $s$ in $\mathcal{M}$ , we have $\mathcal{M},s\vDash\varphi$ ; we say that $\varphi$ is valid on a frame $\mathcal{F}$ , notation: $\mathcal{F}\vDash\varphi$ , if for all $\mathcal{M}$ based on $\mathcal{F}$ , we have $\mathcal{M}\vDash\varphi$ ; we say that $\varphi$ is valid over a class $\mathbb{F}$ of frames, notation: $\mathbb{F}\vDash\varphi$ , if for all $\mathcal{F}$ in $\mathbb{F}$ , we have $\mathcal{F}\vDash\varphi$ ; we say that $\varphi$ is satisfiable over the class $\mathbb{F}$ , if $\mathbb{F}\nvDash\neg\varphi$ . Similar notions go to a set of formulas.

For the sake of reference, we also list the semantics of the aforementioned defined modalities as follows:

\begin{array}[]{lll}\mathcal{M},s\vDash\Diamond\varphi&\iff&S\backslash\varphi^{\mathcal{M}}\notin N(s)\\ \mathcal{M},s\vDash\Delta\varphi&\iff&\varphi^{\mathcal{M}}\in N(s)\text{ or }S\backslash\varphi^{\mathcal{M}}\in N(s)\\ \mathcal{M},s\vDash\circ\varphi&\iff&\mathcal{M},s\vDash\varphi\text{ implies }\varphi^{\mathcal{M}}\in N(s).\\ &\end{array}

3 Expressivity

In this section, we compare the relative expressivity of $\mathcal{L}(\nabla,\bullet)$ and other languages introduced before, over various classes of neighborhood models. Some expressivity results over the class of relational models have been obtained in [7] and [9].

To make our presentation self-contained, we introduce some necessary technical terms.

Definition 5.

Let $\mathcal{L}_{1}$ and $\mathcal{L}_{2}$ be two languages that are interpreted on the same class of models $\mathbb{C}$ , where $\mathbb{C}$ ranges over classes of models which are models for $\mathcal{L}_{1}$ and for $\mathcal{L}_{2}$ .

•

$\mathcal{L}_{2}$ is at least as expressive as $\mathcal{L}_{1}$ over $\mathbb{C}$ , notation: $\mathcal{L}_{1}\preceq\mathcal{L}_{2}[\mathbb{C}]$ , if for all $\varphi\in\mathcal{L}_{1}$ , there exists $\psi\in\mathcal{L}_{2}$ such that for all $\mathcal{M}\in\mathbb{C}$ and all $s$ in $\mathcal{M}$ , we have that $\mathcal{M},s\vDash\varphi$ iff $\mathcal{M},s\vDash\psi$ .
•

$\mathcal{L}_{1}$ and $\mathcal{L}_{2}$ are equally expressive over $\mathbb{C}$ , notation: $\mathcal{L}_{1}\equiv\mathcal{L}_{2}[\mathbb{C}]$ , if $\mathcal{L}_{1}\preceq\mathcal{L}_{2}[\mathbb{C}]$ and $\mathcal{L}_{2}\preceq\mathcal{L}_{1}[\mathbb{C}]$ .
•

$\mathcal{L}_{1}$ is less expressive than $\mathcal{L}_{2}$ over $\mathbb{C}$ , notation: $\mathcal{L}_{1}\prec\mathcal{L}_{2}[\mathbb{C}]$ , if $\mathcal{L}_{1}\preceq\mathcal{L}_{2}[\mathbb{C}]$ but $\mathcal{L}_{2}\not\preceq\mathcal{L}_{1}[\mathbb{C}]$ .
•

$\mathcal{L}_{1}$ and $\mathcal{L}_{2}$ are incomparable in expressivity over $\mathbb{C}$ , notation: $\mathcal{L}_{1}\asymp\mathcal{L}_{2}[\mathbb{C}]$ , if $\mathcal{L}_{1}\not\preceq\mathcal{L}_{2}[\mathbb{C}]$ and $\mathcal{L}_{2}\not\preceq\mathcal{L}_{1}[\mathbb{C}]$ .

It turns out that over the class of $(c)$ -models and the class of $(t)$ -models, $\mathcal{L}(\nabla)$ is at least as expressive as $\mathcal{L}(\bullet)$ (Prop. 10 and Prop. 11), whereas $\mathcal{L}(\nabla)$ is not at least as expressive as $\mathcal{L}(\bullet)$ over the class of models possessing either of other eight neighborhood properties (Prop. 6-Prop. 8).

Proposition 6.

$\mathcal{L}(\bullet)\not\preceq\mathcal{L}(\nabla)[\mathbb{C}]$ , where $\mathbb{C}\in\{\mathbb{C}_{\text{all}},\mathbb{C}_{r},\mathbb{C}_{i},\mathbb{C}_{s},\mathbb{C}_{d}\}$ .

Proof.

Consider the following models, which comes from [12, Prop. 2]. An arrow from a state $x$ to a set $X$ means that $X$ is a neighborhood of $x$ (Idem for other arrows).

It has been shown in [12, Prop. 2] that both $\mathcal{M}$ and $\mathcal{M}^{\prime}$ satisfy $(r)$ , $(i)$ , $(s)$ and $(d)$ , and $(\mathcal{M},s)$ and $(\mathcal{M}^{\prime},s^{\prime})$ cannot be distinguished by $\mathcal{L}(\nabla)$ .

However, both pointed models can be distinguished by an $\mathcal{L}(\bullet)$ . To see this, note that $p^{\mathcal{M}}=\{s\}$ and $\{s\}\notin N(s)$ , and thus $\mathcal{M},s\vDash\bullet p$ , whereas $\mathcal{M}^{\prime},s^{\prime}\nvDash\bullet p$ , as $p^{\mathcal{M}^{\prime}}=\{s^{\prime},t^{\prime}\}\in N^{\prime}(s^{\prime})$ . ∎

Proposition 7.

$\mathcal{L}(\bullet)\not\preceq\mathcal{L}(\nabla)[\mathbb{C}]$ , where $\mathbb{C}\in\{\mathbb{C}_{n},\mathbb{C}_{b}\}$ .

Proof.

Consider the following models, which comes from [12, Prop. 3]:

It has been shown in [12, Prop. 3] that both $\mathcal{M}$ and $\mathcal{M}^{\prime}$ satisfy $(n)$ and $(b)$ , and $(\mathcal{M},s)$ and $(\mathcal{M}^{\prime},s^{\prime})$ cannot be distinguished by $\mathcal{L}(\nabla)$ .

Proposition 8.

$\mathcal{L}(\bullet)\not\preceq\mathcal{L}(\nabla)[\mathbb{C}]$ , where $\mathbb{C}\in\{\mathbb{C}_{4},\mathbb{C}_{5}\}$ .

Proof.

Consider the following models, which is a revision of the figures in [12, Prop. 4]:

Firstly, $\mathcal{M}$ and $\mathcal{M}^{\prime}$ satisfy $(4)$ and $(5)$ . In what follows we only show the claim for $\mathcal{M}$ ; the proof for the case $\mathcal{M}^{\prime}$ is analogous.

-

For $(4)$ : Suppose that $X\in N(s)$ . Then $X=\emptyset$ or $X=\{s\}$ . Notice that $\{u\mid X\in N(u)\}=\{s\}\in N(s)$ . Similarly, we can demonstrate that $(4)$ holds for $N(t)$ .
-

For $(5)$ : Assume that $X\notin N(s)$ . Then $X=\{t\}$ or $X=\{s,t\}$ . Notice that $\{u\mid X\notin N(u)\}=\{s\}\in N(s)$ . A similar argument goes for $N(t)$ .

Secondly, $(\mathcal{M},s)$ and $(\mathcal{M}^{\prime},s^{\prime})$ cannot be distinguished by $\mathcal{L}(\nabla)$ , that is to say, for all $\varphi\in\mathcal{L}(\nabla)$ , we have that $\mathcal{M},s\vDash\varphi$ iff $\mathcal{M}^{\prime},s^{\prime}\vDash\varphi$ . The proof goes by induction on $\varphi$ , where the only nontrivial case is $\nabla\varphi$ . By semantics, we have the following equivalences:

\begin{array}[]{ll}&\mathcal{M},s\vDash\nabla\varphi\\ \iff&\varphi^{\mathcal{M}}\notin N(s)\text{ and }(\neg\varphi)^{\mathcal{M}}\notin N(s)\\ \iff&\varphi^{\mathcal{M}}\notin\{\emptyset,\{s\}\}\text{ and }(\neg\varphi)^{\mathcal{M}}\notin\{\emptyset,\{s\}\}\\ \iff&\varphi^{\mathcal{M}}\neq\emptyset\text{ and }\varphi^{\mathcal{M}}\neq\{s\}\text{ and }(\neg\varphi)^{\mathcal{M}}\neq\emptyset\text{ and }(\neg\varphi)^{\mathcal{M}}\neq\{s\}\\ \iff&\varphi^{\mathcal{M}}\neq\emptyset\text{ and }\varphi^{\mathcal{M}}\neq\{s\}\text{ and }\varphi^{\mathcal{M}}\neq\{s,t\}\text{ and }\varphi^{\mathcal{M}}\neq\{t\}\\ \iff&\text{false}\\ \end{array}

\begin{array}[]{ll}&\mathcal{M}^{\prime},s^{\prime}\vDash\nabla\varphi\\ \iff&\varphi^{\mathcal{M}^{\prime}}\notin N^{\prime}(s^{\prime})\text{ and }(\neg\varphi)^{\mathcal{M}^{\prime}}\notin N^{\prime}(s^{\prime})\\ \iff&\varphi^{\mathcal{M}^{\prime}}\notin\{\{s^{\prime},t^{\prime}\},\{s^{\prime}\}\}\text{ and }(\neg\varphi)^{\mathcal{M}^{\prime}}\notin\{\{s^{\prime},t^{\prime}\},\{s^{\prime}\}\}\\ \iff&\varphi^{\mathcal{M}^{\prime}}\neq\{s^{\prime},t^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{s^{\prime}\}\text{ and }(\neg\varphi)^{\mathcal{M}^{\prime}}\neq\{s^{\prime},t^{\prime}\}\text{ and }(\neg\varphi)^{\mathcal{M}^{\prime}}\neq\{s^{\prime}\}\\ \iff&\varphi^{\mathcal{M}^{\prime}}\neq\{s^{\prime},t^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{s^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\emptyset\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{t^{\prime}\}\\ \iff&\text{false}\\ \end{array}

In either case, the penultimate line of the proof merely states that $\varphi$ cannot be interpreted on the related model: its denotation is not one of all possible subsets of the domain. We conclude that $\mathcal{M},s\vDash\nabla\varphi$ iff $\mathcal{M}^{\prime},s^{\prime}\vDash\nabla\varphi$ .

Finally, we show that $(\mathcal{M},s)$ and $(\mathcal{M}^{\prime},s^{\prime})$ can be distinguished by $\mathcal{L}(\bullet)$ . To see this, note that $(\neg p)^{\mathcal{M}}=\{s,t\}\notin N(s)$ , and thus $\mathcal{M},s\vDash\bullet\neg p$ . However, since $(\neg p)^{\mathcal{M}^{\prime}}=\{s^{\prime},t^{\prime}\}\in N^{\prime}(s^{\prime})$ , we have $\mathcal{M},s\nvDash\bullet\neg p$ . ∎

Remark 9.

The reader may ask whether the figure in [12, Prop. 4] (as below) applies to the above proposition.

The answer is negative. This is because the pointed models $(\mathcal{M},s)$ and $(\mathcal{M}^{\prime},s^{\prime})$ in this figure cannot be distinguished by $\mathcal{L}(\bullet)$ either. To see this, note that $\mathcal{M},s\vDash\bullet\varphi$ iff $\mathcal{M},s\vDash\varphi$ and $\varphi^{\mathcal{M}}\notin N(s)$ , which by the construction of $N(s)$ implies that $s\in\varphi^{\mathcal{M}}$ and $\varphi^{\mathcal{M}}\neq\{s\}$ and $\varphi^{\mathcal{M}}\neq\{s,t\}$ , which is impossible. It then follows that $\mathcal{M},s\nvDash\bullet\varphi$ . A similar argument can show that $\mathcal{M}^{\prime},s^{\prime}\nvDash\bullet\varphi$ . Therefore, $\mathcal{M},s\vDash\bullet\varphi$ iff $\mathcal{M}^{\prime},s^{\prime}\vDash\bullet\varphi$ .

Proposition 10.

$\mathcal{L}(\bullet)\preceq\mathcal{L}(\nabla)[\mathbb{C}_{c}]$ .

Proof.

It suffices to show that $\bullet\varphi\leftrightarrow(\varphi\land\nabla\varphi)$ is valid over the class of $(c)$ -models. Let $\mathcal{M}=\langle S,N,V\rangle$ be a $(c)$ -model and $s\in S$ . Suppose that $\mathcal{M},s\vDash\bullet\varphi$ , it remains only to prove that $\mathcal{M},s\vDash\varphi\land\nabla\varphi$ . By supposition, we have $\mathcal{M},s\vDash\varphi$ and $\varphi^{\mathcal{M}}\notin N(s)$ . We have also $S\backslash\varphi^{\mathcal{M}}\notin N(s)$ : otherwise, by $(c)$ , $S\backslash(S\backslash\varphi^{\mathcal{M}})\in N(s)$ , that is, $\varphi^{\mathcal{M}}\in N(s)$ : a contradiction. Thus $\mathcal{M},s\vDash\nabla\varphi$ , and therefore $\mathcal{M},s\vDash\varphi\land\nabla\varphi$ . The converse is clear from the semantics. ∎

Proposition 11.

$\mathcal{L}(\bullet)\preceq\mathcal{L}(\nabla)[\mathbb{C}_{t}]$ .

Proof.

It suffices to show that $\bullet\varphi\leftrightarrow(\varphi\land\nabla\varphi)$ over the class of $(t)$ -models. The proof is almost the same as that in Prop. 10, except that $S\backslash\varphi^{\mathcal{M}}\notin N(s)$ (that is, $(\neg\varphi)^{\mathcal{M}}\notin N(s)$ ) is obtained from $\mathcal{M},s\vDash\varphi$ and the property $(t)$ . ∎

Conversely, on the class of $(c)$ -models and the class of $(t)$ -models, $\mathcal{L}(\bullet)$ is at least as expressive as $\mathcal{L}(\nabla)$ (Prop. 15 and Prop. 16), whereas on the class of models possessing either of other eight neighborhood properties, $\mathcal{L}(\bullet)$ is not at least as expressive as $\mathcal{L}(\nabla)$ (Prop. 12-Prop. 14). As a corollary, on the class of $(c)$ -models and the class of $(t)$ -models, $\mathcal{L}(\nabla)$ , $\mathcal{L}(\bullet)$ , and $\mathcal{L}(\nabla,\bullet)$ are equally expressive, whereas over the class of models possessing the eight neighborhood properties in question, $\mathcal{L}(\nabla)$ and $\mathcal{L}(\bullet)$ are both less expressive than $\mathcal{L}(\nabla,\bullet)$ (Coro. 17).

Proposition 12.

$\mathcal{L}(\nabla)\not\preceq\mathcal{L}(\bullet)[\mathbb{C}]$ , where $\mathbb{C}\in\{\mathbb{C}_{\text{all}},\mathbb{C}_{n},\mathbb{C}_{r},\mathbb{C}_{i},\mathbb{C}_{s},\mathbb{C}_{d},\mathbb{C}_{b}\}$ .

Proof.

Consider the following models:

It is straightforward to check that both $\mathcal{M}$ and $\mathcal{M}^{\prime}$ satisfy $(n)$ , $(r)$ , $(i)$ , $(s)$ , and $(d)$ . In what follows, we show that $\mathcal{M}$ and $\mathcal{M}^{\prime}$ both have the property $(b)$ .

-

For $\mathcal{M}$ : suppose that $s\in X$ . Then $X=\{s\}$ or $X=\{s,t\}$ . This implies that $\{u\mid S\backslash X\notin N(u)\}=\{s,t\}\in N(s)$ . Similarly, we can show that $(b)$ holds for $N(t)$ .
-

For $\mathcal{M}^{\prime}$ : assume that $s^{\prime}\in X$ . Then $X=\{s^{\prime}\}$ or $X=\{s^{\prime},t^{\prime}\}$ . If $X=\{s^{\prime}\}$ , then $\{u\mid S^{\prime}\backslash X\notin N^{\prime}(u)\}=\{t^{\prime}\}\in N^{\prime}(s^{\prime})$ ; if $X=\{s^{\prime},t^{\prime}\}$ , then $\{u\mid S^{\prime}\backslash X\notin N^{\prime}(u)\}=\{s^{\prime},t^{\prime}\}\in N^{\prime}(s^{\prime})$ . Now assume that $t^{\prime}\in X$ . Then $X=\{t^{\prime}\}$ or $X=\{s^{\prime},t^{\prime}\}$ . If $X=\{t^{\prime}\}$ , then $\{u\mid S^{\prime}\backslash X\notin N^{\prime}(u)\}=\{s^{\prime},t^{\prime}\}\in N^{\prime}(t^{\prime})$ ; if $X=\{s^{\prime},t^{\prime}\}$ , we can also show that $\{u\mid S^{\prime}\backslash X\notin N^{\prime}(u)\}=\{s^{\prime},t^{\prime}\}\in N^{\prime}(t^{\prime})$ .

Moreover, $(\mathcal{M},s)$ and $(\mathcal{M}^{\prime},s^{\prime})$ cannot be distinguished by $\mathcal{L}(\bullet)$ . Here we use the notion of $\bullet$ -morphisms introduced in [11, Def. 4.1].⁸⁸8Recall that the notion of $\bullet$ -morphisms is defined as follows. Let $\mathcal{M}=\langle S,N,V\rangle$ and $\mathcal{M}^{\prime}=\langle S^{\prime},N^{\prime},V^{\prime}\rangle$ be neighborhood models. A function $f:S\to S^{\prime}$ is a $\bullet$ -morphism from $\mathcal{M}$ to $\mathcal{M}^{\prime}$ , if for all $s\in S$ , (Var) $s\in V(p)$ iff $f(s)\in V^{\prime}(p)$ for all $p\in\textbf{P}$ , ( $\bullet$ -Mor) for all $X^{\prime}\subseteq S^{\prime}$ , $[s\in f^{-1}[X^{\prime}]\text{ and }f^{-1}[X^{\prime}]\notin N(s)]\Longleftrightarrow[f(s)\in X^{\prime}\text{ and }X^{\prime}\notin N^{\prime}(f(s))]$ . It is then demonstrated in [11, Prop. 4.1] that the formulas of $\mathcal{L}(\bullet)$ are invariant under $\bullet$ -morphisms. In details, let $\mathcal{M}$ and $\mathcal{M}^{\prime}$ be neighborhood models, and let $f$ be a $\bullet$ -morphism from $\mathcal{M}$ to $\mathcal{M}^{\prime}$ . Then for all $s\in S$ , for all $\varphi\in\mathcal{L}(\bullet)$ , we have that $\mathcal{M},s\vDash\varphi$ iff $\mathcal{M}^{\prime},f(s)\vDash\varphi$ . Define a function $f:S\to S^{\prime}$ such that $f(s)=s^{\prime}$ and $f(t)=t^{\prime}$ . We prove that $f$ is a $\bullet$ -morphism from $\mathcal{M}$ to $\mathcal{M}^{\prime}$ . The condition (Var) follows directly from the valuations. For the condition ( $\bullet$ -Mor), we first prove that it holds for $s$ : assume that $s\in f^{-1}[X^{\prime}]$ and $f^{-1}[X^{\prime}]\notin N(s)$ , then it must be that $X^{\prime}=\{s^{\prime}\}$ . Then we have $f(s)=s^{\prime}\in X^{\prime}$ and $X^{\prime}\notin N^{\prime}(f(s))$ . The converse is similar. In a similar way, we can show that ( $\bullet$ -Mor) also holds for $t$ . Then by [11, Prop. 4.1] (see also fn. 8), we have $\mathcal{M},s\vDash\varphi$ iff $\mathcal{M}^{\prime},s^{\prime}\vDash\varphi$ for all $\varphi\in\mathcal{L}(\bullet)$ .

However, these pointed models can be distinguished by $\mathcal{L}(\nabla)$ . This is because $\mathcal{M},s\vDash\nabla p$ (as $p^{\mathcal{M}}=\{t\}\notin N(s)$ and $(\neg p)^{\mathcal{M}}=\{s\}\notin N(s)$ ) and $\mathcal{M}^{\prime},s^{\prime}\nvDash\nabla p$ (as $p^{\mathcal{M}^{\prime}}=\{t^{\prime}\}\in N^{\prime}(s^{\prime})$ ). ∎

Proposition 13.

$\mathcal{L}(\nabla)\not\preceq\mathcal{L}(\bullet)[\mathbb{C}_{4}]$ .

Proof.

Consider the following models:

Firstly, both $\mathcal{M}$ and $\mathcal{M}^{\prime}$ have $(4)$ .

-

For $\mathcal{M}$ : Suppose that $X\in N(s)$ . Then $X=\{s,t\}$ , and so $\{u\mid X\in N(u)\}=\{s,t\}\in N(s)$ . Now assume that $X\in N(t)$ . Then $X=\{t\}$ or $X=\{s,t\}$ . If $X=\{t\}$ , then $\{u\mid X\in N(u)\}=\{t\}\in N(t)$ ; if $X=\{s,t\}$ , then $\{u\mid X\in N(u)\}=\{s,t\}\in N(t)$ .
-

For $\mathcal{M}^{\prime}$ : Suppose that $X\in N^{\prime}(s^{\prime})$ . Then $X=\{t^{\prime}\}$ or $X=\{s^{\prime},t^{\prime}\}$ . Either case implies that $\{u\mid X\in N^{\prime}(u)\}=\{s^{\prime},t^{\prime}\}\in N^{\prime}(s^{\prime})$ . Now assume that $X\in N^{\prime}(t^{\prime})$ . Then $X=\{t^{\prime}\}$ or $X=\{s^{\prime},t^{\prime}\}$ . Again, either case implies that $\{u\mid X\in N^{\prime}(u)\}=\{s^{\prime},t^{\prime}\}\in N^{\prime}(t^{\prime})$ .

Secondly, similar to the proof of the corresponding part in Prop. 12, we can show that $(\mathcal{M},s)$ and $(\mathcal{M}^{\prime},s^{\prime})$ cannot be distinguished by $\mathcal{L}(\bullet)$ .

It remains only to show that $(\mathcal{M},s)$ and $(\mathcal{M}^{\prime},s^{\prime})$ can be distinguished by $\mathcal{L}(\nabla)$ . The proof for this is analogous to that in Prop. 12. ∎

Proposition 14.

$\mathcal{L}(\nabla)\not\preceq\mathcal{L}(\bullet)[\mathbb{C}_{5}]$ .

Proof.

Consider the following models:

Firstly, both $\mathcal{M}$ and $\mathcal{M}^{\prime}$ possess the property $(5)$ . Since for all $X\subseteq S=\{s,t\}$ , $X\in N(t)$ , the property $(5)$ is possessed vacuously by $N(t)$ Similarly, $(5)$ is also possessed vacuously by $N^{\prime}(t^{\prime})$ . It remains only to show that both $N(s)$ and $N^{\prime}(s^{\prime})$ have $(5)$ .

-

For $N(s)$ : suppose that $X\notin N(s)$ , then $X=\emptyset$ or $X=\{s,t\}$ . Either case implies that $\{u\in S\mid X\notin N(u)\}=\{s\}\in N(s)$ .
-

For $N^{\prime}(s^{\prime})$ : assume that $X\notin N^{\prime}(s^{\prime})$ , then $X=\{s^{\prime},t^{\prime}\}$ . This follows that $\{u\in S^{\prime}\mid X\notin N^{\prime}(u)\}=\{s^{\prime}\}\in N^{\prime}(s^{\prime})$ .

Secondly, $(\mathcal{M},s)$ and $(\mathcal{M}^{\prime},s)$ cannot be distinguished by $\mathcal{L}(\bullet)$ . Again, this can be shown as the corresponding part in Prop. 12.

Finally, $(\mathcal{M},s)$ and $(\mathcal{M}^{\prime},s^{\prime})$ can be distinguished by $\mathcal{L}(\nabla)$ . On one hand, $p^{\mathcal{M}}=\{s,t\}\notin N(s)$ and $S\backslash p^{\mathcal{M}}=\emptyset\notin N(s)$ , thus $\mathcal{M},s\vDash\nabla p$ . On the other hand, $S^{\prime}\backslash p^{\mathcal{M}^{\prime}}=\emptyset\in N^{\prime}(s^{\prime})$ , thus $\mathcal{M}^{\prime},s^{\prime}\nvDash\nabla p$ . ∎

Proposition 15.

$\mathcal{L}(\nabla)\preceq\mathcal{L}(\bullet)[\mathbb{C}_{c}]$ .

Proof.

We claim that over the class of $(c)$ -models, $\vDash\nabla\varphi\leftrightarrow\bullet\varphi\vee{\bullet\neg}\varphi$ .

First, we show that $\vDash\nabla\varphi\to\bullet\varphi\vee{\bullet\neg}\varphi$ .

Let $\mathcal{M}=\langle S,N,V\rangle$ be any model and $s\in S$ . Suppose that $\mathcal{M},s\vDash\nabla\varphi$ . Then $\varphi^{\mathcal{M}}\notin N(s)$ and $S\backslash\varphi^{\mathcal{M}}\notin N(s)$ , that is, $(\neg\varphi)^{\mathcal{M}}\notin N(s)$ . We have either $\mathcal{M},s\vDash\varphi$ or $\mathcal{M},s\vDash\neg\varphi$ . If $\mathcal{M},s\vDash\varphi$ , since $\varphi^{\mathcal{M}}\notin N(s)$ , we infer that $\mathcal{M},s\vDash\bullet\varphi$ ; if $\mathcal{M},s\vDash\neg\varphi$ , since $(\neg\varphi)^{\mathcal{M}}\notin N(s)$ , we derive that $\mathcal{M},s\vDash{\bullet\neg}\varphi$ . Therefore, $\mathcal{M},s\vDash\bullet\varphi\vee{\bullet\neg}\varphi$ . Since $(\mathcal{M},s)$ is arbitrary, this establishes the validity of $\nabla\varphi\to\bullet\varphi\vee{\bullet\neg}\varphi$ .

Conversely, we prove that over the class of $(c)$ -models, $\vDash\bullet\varphi\vee{\bullet\neg}\varphi\to\nabla\varphi$ .

Suppose that $\mathcal{M}=\langle S,N,V\rangle$ be a $(c)$ -model and $s\in S$ . Assume that $\mathcal{M},s\vDash\bullet\varphi\vee{\bullet\neg}\varphi$ . Then $\mathcal{M},s\vDash\bullet\varphi$ or $\mathcal{M},s\vDash{\bullet\neg}\varphi$ . If $\mathcal{M},s\vDash\bullet\varphi$ , then $\varphi^{\mathcal{M}}\notin N(s)$ . By $(c)$ , we have $S\backslash\varphi^{\mathcal{M}}\notin N(s)$ , so $\mathcal{M},s\vDash\nabla\varphi$ . If $\mathcal{M},s\vDash{\bullet\neg}\varphi$ , then $(\neg\varphi)^{\mathcal{M}}\notin N(s)$ , namely $S\backslash\varphi^{\mathcal{M}}\notin N(s)$ . By $(c)$ again, we obtain $\varphi^{\mathcal{M}}\notin N(s)$ , and thus $\mathcal{M},s\vDash\nabla\varphi$ . Therefore, $\mathcal{M},s\vDash\nabla\varphi$ . ∎

Proposition 16.

$\mathcal{L}(\nabla)\preceq\mathcal{L}(\bullet)[\mathbb{C}_{t}]$ .

Proof.

We claim that over the class of $(t)$ -models, $\vDash\nabla\varphi\leftrightarrow\bullet\varphi\vee\bullet\neg\varphi$ . The proof is almost the same as that in Prop. 15, except that in the proof of the validity of $\bullet\varphi\vee\bullet\neg\varphi\to\nabla\varphi$ , $\mathcal{M},s\vDash\nabla\varphi$ is obtained as follows: if $\mathcal{M},s\vDash\bullet\varphi$ , then $\mathcal{M},s\vDash\varphi$ and $\varphi^{\mathcal{M}}\notin N(s)$ , thus $\mathcal{M},s\nvDash\neg\varphi$ , namely $s\notin(\neg\varphi)^{\mathcal{M}}$ , and then by $(t)$ , we infer that $(\neg\varphi)^{\mathcal{M}}\notin N(s)$ , namely $S\backslash\varphi^{\mathcal{M}}\notin N(s)$ , so $\mathcal{M},s\vDash\nabla\varphi$ ; similarly, we can show that if $\mathcal{M},s\vDash\bullet\neg\varphi$ then $\mathcal{M},s\vDash\nabla\varphi$ . ∎

With the above results in mind, we have the following result, which extends the expressivity results over Kripke models in [7].

Corollary 17.

Where $\mathbb{C}\in\{\mathbb{C}_{\text{all}},\mathbb{C}_{n},\mathbb{C}_{r},\mathbb{C}_{i},\mathbb{C}_{s},\mathbb{C}_{d},\mathbb{C}_{b},\mathbb{C}_{4},\mathbb{C}_{5}\}$ , $\mathcal{L}(\nabla)\asymp\mathcal{L}(\bullet)[\mathbb{C}]$ , and $\mathcal{L}\prec\mathbb{L}(\nabla,\bullet)[\mathbb{C}]$ , where $\mathcal{L}\in\{\mathcal{L}(\nabla),\mathcal{L}(\bullet)\}$ . Where $\mathbb{C}\in\{\mathbb{C}_{c},\mathbb{C}_{t}\}$ , $\mathcal{L}_{1}\equiv\mathcal{L}_{2}[\mathbb{C}]$ , where $\mathcal{L}_{1},\mathcal{L}_{2}\in\{\mathcal{L}(\nabla),\mathcal{L}(\bullet),\mathcal{L}(\nabla,\bullet)\}$ .

Moreover, over the class of $(c)$ -models and the class of $(t)$ -models, $\mathcal{L}(\nabla,\bullet)$ and $\mathcal{L}(\Box)$ are equally expressive (Prop. 20), whereas over the class of models possessing either of other eight neighborhood properties except for $(d)$ , $\mathcal{L}(\nabla,\bullet)$ is less expressive than $\mathcal{L}(\Box)$ (Prop. 18 and Prop. 19).

Proposition 18.

$\mathcal{L}(\nabla,\bullet)\prec\mathcal{L}(\Box)[\mathbb{C}]$ , where $\mathbb{C}\in\{\mathbb{C}_{\text{all}},\mathbb{C}_{r},\mathbb{C}_{i},\mathbb{C}_{4},\mathbb{C}_{5}\}$ .

Proof.

Use Remark 9 and [12, Prop. 4]. ∎

Proposition 19.

$\mathcal{L}(\nabla,\bullet)\prec\mathcal{L}(\Box)[\mathbb{C}]$ , where $\mathbb{C}\in\{\mathbb{C}_{n},\mathbb{C}_{s},\mathbb{C}_{b}\}$ .

Proof.

Consider the following models $\mathcal{M}=\langle S,N,V\rangle$ and $\mathcal{M}^{\prime}=\langle S^{\prime},N^{\prime},V^{\prime}\rangle$ , where $S=\{s,t\}$ and $S^{\prime}=\{s^{\prime},t^{\prime}\}$ .

It should be straightforward to check that both $\mathcal{M}$ and $\mathcal{M}^{\prime}$ possess $(n)$ and $(s)$ . Moreover, both models also possess $(b)$ , shown as follows. Since for all $X\subseteq S$ , we have $X\in N(s)$ , one may easily see that $N(s)$ has $(b)$ . Similarly, we can show that $N^{\prime}(t^{\prime})$ has $(b)$ . Besides,

-

$N(t)$ has $(b)$ : suppose that $t\in X$ , then $X=\{t\}$ or $X=\{s,t\}$ . The first case implies $\{u\in S\mid S\backslash X\notin N(u)\}=\{u\in S\mid\{s\}\notin N(u)\}=\{t\}\in N(t)$ , and the second case implies that $\{u\in S\mid S\backslash X\notin N(u)\}=\{u\in S\mid\emptyset\notin N(u)\}=\{t\}\in N(t)$ , as desired.
-

$N^{\prime}(s^{\prime})$ has $(b)$ : assume that $s^{\prime}\in X$ , then $X=\{s^{\prime}\}$ or $X=\{s^{\prime},t^{\prime}\}$ . The first case entails that $\{u\in S^{\prime}\mid S^{\prime}\backslash X\notin N^{\prime}(u)\}=\{u\in S^{\prime}\mid\{t^{\prime}\}\notin N^{\prime}(u)\}=\{s^{\prime}\}\in N^{\prime}(s^{\prime})$ , and the second case entails that $\{u\in S^{\prime}\mid S^{\prime}\backslash X\notin N^{\prime}(u)\}=\{u\in S^{\prime}\mid\emptyset\notin N^{\prime}(u)\}=\{s^{\prime}\}\in N^{\prime}(s^{\prime})$ , as desired.

Next, we show $(\mathcal{M},s)$ and $(\mathcal{M}^{\prime},s^{\prime})$ cannot be distinguished by $\mathcal{L}(\nabla,\bullet)$ . That is, for all $\varphi\in\mathcal{L}(\nabla,\bullet)$ , we have $\mathcal{M},s\vDash\varphi$ iff $\mathcal{M}^{\prime},s^{\prime}\vDash\varphi$ . The proof proceeds by induction on $\varphi$ , where the nontrivial cases are $\nabla\varphi$ and $\bullet\varphi$ . The proof for the case $\bullet\varphi$ is shown as in Remark 9. For the case $\nabla\varphi$ , we have the following equivalences.

\begin{array}[]{ll}&\mathcal{M},s\vDash\nabla\varphi\\ \iff&\varphi^{\mathcal{M}}\notin N(s)\text{ and }S\backslash\varphi^{\mathcal{M}}\notin N(s)\\ \iff&\varphi^{\mathcal{M}}\neq\emptyset\text{ and }\varphi^{\mathcal{M}}\neq\{s,t\}\text{ and }\varphi^{\mathcal{M}}\neq\{s\}\text{ and }\varphi^{\mathcal{M}}\neq\{t\}\\ \iff&\text{false}\\ \end{array}

\begin{array}[]{ll}&\mathcal{M}^{\prime},s^{\prime}\vDash\nabla\varphi\\ \iff&\varphi^{\mathcal{M}^{\prime}}\notin N^{\prime}(s^{\prime})\text{ and }S^{\prime}\backslash\varphi^{\mathcal{M}^{\prime}}\notin N^{\prime}(s^{\prime})\\ \iff&\varphi^{\mathcal{M}^{\prime}}\notin\{\{s^{\prime},t^{\prime}\},\{s^{\prime}\}\}\text{ and }S\backslash\varphi^{\mathcal{M}^{\prime}}\notin\{\{s^{\prime},t^{\prime}\},\{s^{\prime}\}\}\\ \iff&\varphi^{\mathcal{M}^{\prime}}\neq\{s^{\prime},t^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{s^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\emptyset\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{t^{\prime}\}\\ \iff&\text{false}\\ \end{array}

In either case, the penultimate line of the equivalences states that $\varphi$ cannot be interpreted on the related model: its denotation is not one of all possible subsets of the domain. We therefore conclude that $\mathcal{M},s\vDash\nabla\varphi$ iff $\mathcal{M}^{\prime},s^{\prime}\vDash\nabla\varphi$ .

Finally, $(\mathcal{M},s)$ and $(\mathcal{M}^{\prime},s^{\prime})$ can be distinguished by $\mathcal{L}(\Diamond)$ . To see this, note that $p^{\mathcal{M}}=\emptyset\in N(s)$ , thus $\mathcal{M},s\vDash\Box p$ ; however, $p^{\mathcal{M}^{\prime}}=\emptyset\notin N^{\prime}(s^{\prime})$ , and thus $\mathcal{M}^{\prime},s^{\prime}\nvDash\Box p$ . ∎

Proposition 20.

$\mathcal{L}(\nabla,\bullet)\equiv\mathcal{L}(\Box)[\mathbb{C}]$ , where $\mathbb{C}\in\{\mathbb{C}_{c},\mathbb{C}_{t}\}$ .

Proof.

Straightforward from $\mathcal{L}(\nabla,\bullet)\equiv\mathbb{L}(\nabla)[\mathbb{C}]$ (see Coro. 17) and $\mathcal{L}(\nabla)\equiv\mathcal{L}(\Box)[\mathbb{C}]$ [12, Prop. 5, Prop. 6], where $\mathbb{C}\in\{\mathbb{C}_{c},\mathbb{C}_{t}\}$ . ∎

We do not know whether $\mathcal{L}(\nabla,\bullet)$ is less expressive than $\mathcal{L}(\Box)$ over the class of $(d)$ -models. We conjecture the answer is positive. We leave it for future work.

We summarize the results in this section as follows.

\begin{array}[]{ll}\mathcal{L}(\nabla)\asymp\mathcal{L}(\bullet)[\mathbb{C}],\text{ where }\mathbb{C}\in\{\mathbb{C}_{\text{all}},\mathbb{C}_{n},\mathbb{C}_{r},\mathbb{C}_{i},\mathbb{C}_{s},\mathbb{C}_{d},\mathbb{C}_{b},\mathbb{C}_{4},\mathbb{C}_{5}\}&(\text{Coro.}\leavevmode\nobreak\ \ref{coro.exp-nabla-bullet-nablabullet})\\ \mathcal{L}(\nabla)\equiv\mathcal{L}(\bullet)[\mathbb{C}],\text{ where }\mathbb{C}\in\{\mathbb{C}_{c},\mathbb{C}_{t}\}&(\text{Coro.}\leavevmode\nobreak\ \ref{coro.exp-nabla-bullet-nablabullet})\\ \mathcal{L}(\nabla)\prec\mathcal{L}(\nabla,\bullet)[\mathbb{C}],\text{ where }\mathbb{C}\in\{\mathbb{C}_{\text{all}},\mathbb{C}_{n},\mathbb{C}_{r},\mathbb{C}_{i},\mathbb{C}_{s},\mathbb{C}_{d},\mathbb{C}_{b},\mathbb{C}_{4},\mathbb{C}_{5}\}&(\text{Coro.}\leavevmode\nobreak\ \ref{coro.exp-nabla-bullet-nablabullet})\\ \mathcal{L}(\bullet)\prec\mathcal{L}(\nabla,\bullet)[\mathbb{C}],\text{ where }\mathbb{C}\in\{\mathbb{C}_{\text{all}},\mathbb{C}_{n},\mathbb{C}_{r},\mathbb{C}_{i},\mathbb{C}_{s},\mathbb{C}_{d},\mathbb{C}_{b},\mathbb{C}_{4},\mathbb{C}_{5}\}&(\text{Coro.}\leavevmode\nobreak\ \ref{coro.exp-nabla-bullet-nablabullet})\\ \mathcal{L}(\nabla)\equiv\mathcal{L}(\nabla,\bullet)[\mathbb{C}],\text{ where }\mathbb{C}\in\{\mathbb{C}_{c},\mathbb{C}_{t}\}&(\text{Coro.}\leavevmode\nobreak\ \ref{coro.exp-nabla-bullet-nablabullet})\\ \mathcal{L}(\bullet)\equiv\mathcal{L}(\nabla,\bullet)[\mathbb{C}],\text{ where }\mathbb{C}\in\{\mathbb{C}_{c},\mathbb{C}_{t}\}&(\text{Coro.}\leavevmode\nobreak\ \ref{coro.exp-nabla-bullet-nablabullet})\\ \mathcal{L}(\nabla,\bullet)\prec\mathcal{L}(\Box)[\mathbb{C}],\text{ where }\mathbb{C}\in\{\mathbb{C}_{\text{all}},\mathbb{C}_{n},\mathbb{C}_{r},\mathbb{C}_{i},\mathbb{C}_{s},\mathbb{C}_{b},\mathbb{C}_{4},\mathbb{C}_{5}\}&(\text{Props.}\leavevmode\nobreak\ \ref{prop.exp-nablabullet-diamond-ri45},\ref{prop.exp-nablabullet-diamond-nsb})\\ \mathcal{L}(\nabla,\bullet)\equiv\mathcal{L}(\Box)[\mathbb{C}],\text{ where }\mathbb{C}\in\{\mathbb{C}_{c},\mathbb{C}_{t}\}&(\text{Coro.}\leavevmode\nobreak\ \ref{prop.nablabullet-diamond-ct})\\ \end{array}

4 Frame Definability

We have shown in the previous section that $\mathcal{L}(\nabla,\bullet)$ is more expressive than $\mathcal{L}(\nabla)$ and $\mathcal{L}(\bullet)$ (at the level of models). It may then be natural to ask whether a similar situation holds at the level of frames. Recall that it is shown in [12, Prop. 7] that all frame properties in Def. 3, in particular $(n)$ , are undefinable in $\mathcal{L}(\nabla)$ . In what follows, we shall show that all frame properties in question except for $(n)$ are undefinable in $\mathcal{L}(\nabla,\bullet)$ , thus $\mathcal{L}(\nabla,\bullet)$ is also more expressive than $\mathcal{L}(\nabla)$ and $\mathcal{L}(\bullet)$ at the level of frames. First, we need some related notion.

Definition 21.

Let $\Gamma$ be a set of $\mathcal{L}(\nabla,\bullet)$ -formulas, and $P$ a neighborhood property. We say that $\Gamma$ defines $P$ , if for all frames $\mathcal{F}$ , $\mathcal{F}$ has $P$ if and only if $\mathcal{F}\vDash\Gamma$ . If $\Gamma$ is a singleton, say $\{\varphi\}$ , we will write $\mathcal{F}\vDash\varphi$ rather than $\mathcal{F}\vDash\{\varphi\}$ . We say that $P$ is definable in $\mathcal{L}(\nabla,\bullet)$ , if there exists a set of $\mathcal{L}(\nabla,\bullet)$ -formulas that defines it.

Proposition 22.

The frame property $(n)$ is definable in $\mathcal{L}(\nabla,\bullet)$ .

Proof.

[11] has shown that $(n)$ is defined in $\mathcal{L}(\bullet)$ , by $\circ\top$ . Therefore, $(n)$ is also definable in $\mathcal{L}(\nabla,\bullet)$ , by $\circ\top$ . ∎

Proposition 23.

The frame properties $(r)$ , $(i)$ , $(c)$ , $(d)$ , $(t)$ and $(b)$ are undefinable in $\mathcal{L}(\nabla,\bullet)$ .

Proof.

Consider the following frames $\mathcal{F}_{1}=\langle S_{1},N_{1}\rangle$ , $\mathcal{F}_{2}=\langle S_{2},N_{2}\rangle$ , and $\mathcal{F}_{3}=\langle S_{3},N_{3}\rangle$ ⁹⁹9This come from [12, Prop. 7].:

It has been observed in [12, Prop. 7] that $\mathcal{F}_{1}$ satisfies $(d)$ and $(t)$ but $\mathcal{F}_{2}$ does not. Also, it is straightforward to check that $\mathcal{F}_{2}$ satisfies $(c)$ but $\mathcal{F}_{1}$ does not. Moreover, $\mathcal{F}_{2}$ satisfies $(r)$ , $(i)$ and $(b)$ , whereas $\mathcal{F}_{3}$ does not. To see $\mathcal{F}_{3}$ does not satisfy $(b)$ , note that $s_{3}\in\{s_{3}\}$ but $\{u\in S_{3}\mid\{t_{3}\}\notin N_{3}(u)\}=\emptyset\notin N_{3}(s_{3})$ . In what follows, we show that for all $\varphi\in\mathcal{L}(\nabla,\bullet)$ , $\mathcal{F}_{1}\vDash\varphi$ iff $\mathcal{F}_{2}\vDash\varphi$ iff $\mathcal{F}_{3}\vDash\varphi$ .

Suppose that $\mathcal{F}_{1}\nvDash\varphi$ . Then there exists $\mathcal{M}_{1}=\langle\mathcal{F}_{1},V_{1}\rangle$ such that $\mathcal{M}_{1},s_{1}\nvDash\varphi$ . Define a valuation $V_{2}$ on $\mathcal{F}_{2}$ as $s_{2}\in V_{2}(p)$ iff $s_{1}\in V_{1}(p)$ for all $p\in\textbf{P}$ . By induction on $\varphi$ , we show that $(\ast)$ : $\mathcal{M}_{1},s_{1}\vDash\varphi$ iff $\mathcal{M}_{2},s_{2}\vDash\varphi$ , where $\mathcal{M}_{2}=\langle\mathcal{F}_{2},V_{2}\rangle$ . The nontrivial cases are $\nabla\varphi$ and $\bullet\varphi$ . The case $\nabla\varphi$ can be shown as in [12, Prop. 7]. For the case $\bullet\varphi$ , notice that $\mathcal{M}_{1},s_{1}\vDash\bullet\varphi$ iff ( $\mathcal{M}_{1},s_{1}\vDash\varphi$ and $\varphi^{\mathcal{M}_{1}}\notin N_{1}(s_{1})$ ) iff ( $\mathcal{M}_{1},s_{1}\vDash\varphi$ and $\varphi^{\mathcal{M}_{1}}\neq\{s_{1}\}$ ), where the last one is a contradiction, and thus $\mathcal{M}_{1},s_{1}\nvDash\bullet\varphi$ ; a similar argument gives us $\mathcal{M}_{2},s_{2}\nvDash\bullet\varphi$ . We have thus proved $(\ast)$ . This entails that $\mathcal{M}_{2},s_{2}\nvDash\varphi$ , and thus $\mathcal{F}_{2}\nvDash\varphi$ . The converse is similar. Therefore, $\mathcal{F}_{1}\vDash\varphi$ iff $\mathcal{F}_{2}\vDash\varphi$ .

It remains only to show that $\mathcal{F}_{2}\vDash\varphi$ iff $\mathcal{F}_{3}\vDash\varphi$ . Assume that $\mathcal{F}_{2}\nvDash\varphi$ . Then there exists $\mathcal{M}_{2}=\langle\mathcal{F}_{2},V_{2}\rangle$ such that $\mathcal{M}_{2},s_{2}\nvDash\varphi$ . Define a valuation $V_{3}$ on $\mathcal{F}_{3}$ such that $s_{3}\in V_{3}(p)$ iff $s_{2}\in V_{2}(p)$ for all $p\in\textbf{P}$ . By induction on $\varphi\in\mathcal{L}(\nabla,\bullet)$ , we show that $(\ast\ast)$ : $\mathcal{M}_{2},s_{2}\vDash\varphi$ iff $\mathcal{M}_{3},s_{3}\vDash\varphi$ , where $\mathcal{M}_{3}=\langle\mathcal{F}_{3},V_{3}\rangle$ . The nontrivial cases are $\nabla\varphi$ and $\bullet\varphi$ . Again, the case $\nabla\varphi$ can be shown as in [12, Prop. 7]. For the case $\bullet\varphi$ , just note that $\mathcal{M}_{3},s_{3}\vDash\bullet\varphi$ iff ( $\mathcal{M}_{3},s_{3}\vDash\varphi$ and $\varphi^{\mathcal{M}_{3}}\notin N_{3}(s_{3})$ ) iff ( $\mathcal{M}_{3},s_{3}\vDash\varphi$ and $\varphi^{\mathcal{M}_{3}}\neq\{s_{3}\}$ and $\varphi^{\mathcal{M}_{3}}\neq\{t_{3}\}$ and $\varphi^{\mathcal{M}_{3}}\neq\{s_{3},t_{3}\}$ ) iff false. Thus $(\ast\ast)$ holds. This implies that $\mathcal{M}_{3},s_{3}\nvDash\varphi$ , and then $\mathcal{F}_{3}\nvDash\varphi$ . The converse is analogous. Therefore, $\mathcal{F}_{2}\vDash\varphi$ iff $\mathcal{F}_{3}\vDash\varphi$ .

If $(r)$ were to be defined by a set of $\mathcal{L}(\nabla,\bullet)$ -formulas, say $\Sigma$ , then as $\mathcal{F}_{2}$ satisfies $(r)$ , we have $\mathcal{F}_{2}\vDash\Sigma$ . Then we should also have $\mathcal{F}_{3}\vDash\Sigma$ , which means that $\mathcal{F}_{3}$ has $(r)$ : a contradiction. Therefore, $(r)$ is undefinable in $\mathcal{L}(\nabla,\bullet)$ . Similarly, we can show other frame properties in question are undefinable in $\mathcal{L}(\nabla,\bullet)$ . ∎

Proposition 24.

The frame properties $(s)$ and $(4)$ are undefinable in $\mathcal{L}(\nabla,\bullet)$ .

Proof.

Consider the following frames $\mathcal{F}=\langle S,N\rangle$ and $\mathcal{F}^{\prime}=\langle S^{\prime},N^{\prime}\rangle$ , where $S=\{s,t\}$ and $S^{\prime}=\{s^{\prime},t^{\prime}\}$ :

Firstly, one may easily see that $\mathcal{F}$ has $(s)$ . Also, $\mathcal{F}$ has $(4)$ . Suppose that $X\in N(s)$ , to show that $\{u\in S\mid X\in N(u)\}\in N(s)$ . By supposition, $X=\{s\}$ or $X=\{s,t\}$ . Either case implies that $\{u\in S\mid X\in N(u)\}=\{s,t\}\in N(s)$ . Thus $N(s)$ has $(4)$ . A similar argument applies to showing that $N(t)$ has $(4)$ .

Secondly, $\mathcal{F}^{\prime}$ does not have $(s)$ , since $\emptyset\in N^{\prime}(t^{\prime})$ and $\emptyset\subseteq\{t^{\prime}\}$ but $\{t^{\prime}\}\notin N^{\prime}(t^{\prime})$ . Moreover, $\mathcal{F}^{\prime}$ does not have $(4)$ . This is because, for instance, $\emptyset\in N^{\prime}(t^{\prime})$ but $\{u\in S^{\prime}\mid\emptyset\in N^{\prime}(u)\}=\{t^{\prime}\}\notin N^{\prime}(t^{\prime})$ .

Thirdly, for all $\psi\in\mathcal{L}(\nabla,\bullet)$ , we have that $\mathcal{F}\vDash\psi$ iff $\mathcal{F}^{\prime}\vDash\psi$ . Suppose that $\mathcal{F}\nvDash\psi$ . Then there exists $\mathcal{M}=\langle\mathcal{F},V\rangle$ and $x\in S$ such that $\mathcal{M},x\nvDash\psi$ . Define $V^{\prime}$ to be a valuation on $\mathcal{F}^{\prime}$ such that $s\in V(p)$ iff $s^{\prime}\in V^{\prime}(p)$ , and $t\in V(p)$ iff $t^{\prime}\in V^{\prime}(p)$ . In what follows, we show $(\ast)$ : for all $\varphi\in\mathcal{L}(\nabla,\bullet)$ , $\mathcal{M},s\vDash\varphi$ iff $\mathcal{M}^{\prime},s^{\prime}\vDash\varphi$ , and $\mathcal{M},t\vDash\varphi$ iff $\mathcal{M}^{\prime},t^{\prime}\vDash\varphi$ , where $\mathcal{M}^{\prime}=\langle\mathcal{F}^{\prime},N^{\prime}\rangle$ . We proceed by induction on $\varphi$ , where the nontrivial cases are $\nabla\varphi$ and $\bullet\varphi$ .

For the case $\nabla\varphi$ , we have the following equivalences.

\begin{array}[]{ll}&\mathcal{M},s\vDash\nabla\varphi\\ \iff&\varphi^{\mathcal{M}}\notin N(s)\text{ and }S\backslash\varphi^{\mathcal{M}}\notin N(s)\\ \iff&\varphi^{\mathcal{M}}\notin\{\{s\},\{s,t\}\}\text{ and }S\backslash\varphi^{\mathcal{M}}\notin\{\{s\},\{s,t\}\}\\ \iff&\varphi^{\mathcal{M}}\neq\{s\}\text{ and }\varphi^{\mathcal{M}}\neq\{s,t\}\text{ and }\varphi^{\mathcal{M}}\neq\{t\}\text{ and }\varphi^{\mathcal{M}}\neq\emptyset\\ \end{array}

\begin{array}[]{ll}&\mathcal{M}^{\prime},s^{\prime}\vDash\nabla\varphi\\ \iff&\varphi^{\mathcal{M}^{\prime}}\notin N^{\prime}(s^{\prime})\text{ and }S^{\prime}\backslash\varphi^{\mathcal{M}^{\prime}}\notin N^{\prime}(s^{\prime})\\ \iff&\varphi^{\mathcal{M}^{\prime}}\notin\{\{s^{\prime}\},\{s^{\prime},t^{\prime}\}\}\text{ and }S^{\prime}\backslash\varphi^{\mathcal{M}^{\prime}}\notin\{\{s^{\prime}\},\{s^{\prime},t^{\prime}\}\}\\ \iff&\varphi^{\mathcal{M}}\neq\{s^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{s^{\prime},t^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{t^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\emptyset\\ \end{array}

In each case, the last line of the above proofs states that $\varphi$ cannot be interpreted on the related models, which is impossible. Thus $\mathcal{M},s\nvDash\nabla\varphi$ and $\mathcal{M}^{\prime},s^{\prime}\nvDash\nabla\varphi$ . Analogously, we can show that $\mathcal{M},t\nvDash\nabla\varphi$ and $\mathcal{M}^{\prime},t^{\prime}\nvDash\nabla\varphi$ .

For the case $\bullet\varphi$ , we have the following equivalences.

\begin{array}[]{ll}&\mathcal{M},s\vDash\bullet\varphi\\ \iff&\mathcal{M},s\vDash\varphi\text{ and }\varphi^{\mathcal{M}}\notin N(s)\\ \iff&\mathcal{M},s\vDash\varphi\text{ and }\varphi^{\mathcal{M}}\neq\{s\}\text{ and }\varphi^{\mathcal{M}}\neq\{s,t\}\\ \iff&\text{false}\\ \end{array}

\begin{array}[]{ll}&\mathcal{M}^{\prime},s^{\prime}\vDash\bullet\varphi\\ \iff&\mathcal{M}^{\prime},s^{\prime}\vDash\varphi\text{ and }\varphi^{\mathcal{M}^{\prime}}\notin N^{\prime}(s^{\prime})\\ \iff&\mathcal{M}^{\prime},s^{\prime}\vDash\varphi\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{s^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{s^{\prime},t^{\prime}\}\\ \iff&\text{false}\\ \end{array}

This shows that $\mathcal{M},s\vDash\bullet\varphi$ iff $\mathcal{M}^{\prime},s^{\prime}\vDash\bullet\varphi$ . Therefore, $\mathcal{M},s\vDash\varphi$ iff $\mathcal{M}^{\prime},s^{\prime}\vDash\varphi$ for all $\varphi\in\mathcal{L}(\nabla,\bullet)$ .

\begin{array}[]{ll}&\mathcal{M},t\vDash\bullet\varphi\\ \iff&\mathcal{M},t\vDash\varphi\text{ and }\varphi^{\mathcal{M}}\notin N(t)\\ \iff&\mathcal{M},t\vDash\varphi\text{ and }\varphi^{\mathcal{M}}\neq\{s\}\text{ and }\varphi^{\mathcal{M}}\neq\{s,t\}\\ \iff&\mathcal{M},t\vDash\varphi\text{ and }\mathcal{M},s\nvDash\varphi\\ \stackrel{{\scriptstyle\text{IH}}}{{\iff}}&\mathcal{M}^{\prime},t^{\prime}\vDash\varphi\text{ and }\mathcal{M}^{\prime},s^{\prime}\nvDash\varphi\\ \iff&\mathcal{M}^{\prime},t^{\prime}\vDash\varphi\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{s^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{s^{\prime},t^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\emptyset\\ \iff&\mathcal{M}^{\prime},t^{\prime}\vDash\varphi\text{ and }\varphi^{\mathcal{M}^{\prime}}\notin N^{\prime}(t^{\prime})\\ \iff&\mathcal{M}^{\prime},t^{\prime}\vDash\bullet\varphi\\ \end{array}

This gives us that $\mathcal{M},t\vDash\varphi$ iff $\mathcal{M}^{\prime},t^{\prime}\vDash\varphi$ for all $\varphi\in\mathcal{L}(\nabla,\bullet)$ . We have now completed the proof of $(\ast)$ .

If $(s)$ were to be defined by a set of $\mathcal{L}(\nabla,\bullet)$ -formulas, say $\Gamma$ , then as $\mathcal{F}$ has $(s)$ , we would have $\mathcal{F}\vDash\Gamma$ , thus we should also have $\mathcal{F}^{\prime}\vDash\Gamma$ , that is, $\mathcal{F}^{\prime}$ has $(s)$ : a contradiction. Therefore, $(s)$ is not definable in $\mathcal{L}(\nabla,\bullet)$ . Similarly, we can obtain the undefinability of $(4)$ in $\mathcal{L}(\nabla,\bullet)$ . ∎

Proposition 25.

The frame property $(5)$ is undefinable in $\mathcal{L}(\nabla,\bullet)$ .

Proof.

Consider the following frames $\mathcal{F}=\langle S,N\rangle$ and $\mathcal{F}^{\prime}=\langle S^{\prime},N^{\prime}\rangle$ , where $S=\{s,t\}$ and $S^{\prime}=\{s^{\prime},t^{\prime}\}$ :

Firstly, $\mathcal{F}$ has $(5)$ . Suppose that $X\notin N(s)$ , to prove that $\{u\in S\mid X\notin N(u)\}\in N(s)$ . By supposition, $X=\emptyset$ or $X=\{s\}$ . Either case implies that $\{u\in S\mid X\notin N(u)\}=\{s,t\}\in N(s)$ . Thus $N(s)$ has $(5)$ . A similar argument shows that $N(t)$ has $(5)$ .

Secondly, $\mathcal{F}^{\prime}$ does not $(5)$ . For instance, $\emptyset\notin N^{\prime}(s^{\prime})$ and $\{u\in S^{\prime}\mid\emptyset\notin N^{\prime}(u)\}=\{s^{\prime}\}\notin N^{\prime}(s^{\prime})$ .

Thirdly, for all $\psi\in\mathcal{L}(\nabla,\bullet)$ , we have that $\mathcal{F}\vDash\psi$ iff $\mathcal{F}^{\prime}\vDash\psi$ . Suppose that $\mathcal{F}\nvDash\psi$ . Then there exists $\mathcal{M}=\langle\mathcal{F},V\rangle$ and $x\in S$ such that $\mathcal{M},x\nvDash\psi$ . Define $V^{\prime}$ to be a valuation on $\mathcal{F}^{\prime}$ such that $s\in V(p)$ iff $s^{\prime}\in V^{\prime}(p)$ , and $t\in V(p)$ iff $t^{\prime}\in V^{\prime}(p)$ . In what follows, we show $(\ast\ast)$ : for all $\varphi\in\mathcal{L}(\nabla,\bullet)$ , $\mathcal{M},s\vDash\varphi$ iff $\mathcal{M}^{\prime},s^{\prime}\vDash\varphi$ , and $\mathcal{M},t\vDash\varphi$ iff $\mathcal{M}^{\prime},t^{\prime}\vDash\varphi$ , where $\mathcal{M}^{\prime}=\langle\mathcal{F}^{\prime},N^{\prime}\rangle$ . We proceed by induction on $\varphi$ , where the nontrivial cases are $\nabla\varphi$ and $\bullet\varphi$ .

For the case $\nabla\varphi$ , we have the following equivalences.

\begin{array}[]{ll}&\mathcal{M},s\vDash\nabla\varphi\\ \iff&\varphi^{\mathcal{M}}\notin N(s)\text{ and }S\backslash\varphi^{\mathcal{M}}\notin N(s)\\ \iff&\varphi^{\mathcal{M}}\notin\{\{t\},\{s,t\}\}\text{ and }S\backslash\varphi^{\mathcal{M}}\notin\{\{t\},\{s,t\}\}\\ \iff&\varphi^{\mathcal{M}}\neq\{t\}\text{ and }\varphi^{\mathcal{M}}\neq\{s,t\}\text{ and }\varphi^{\mathcal{M}}\neq\{s\}\text{ and }\varphi^{\mathcal{M}}\neq\emptyset\\ \end{array}

\begin{array}[]{ll}&\mathcal{M}^{\prime},s^{\prime}\vDash\nabla\varphi\\ \iff&\varphi^{\mathcal{M}^{\prime}}\notin N^{\prime}(s^{\prime})\text{ and }S^{\prime}\backslash\varphi^{\mathcal{M}^{\prime}}\notin N^{\prime}(s^{\prime})\\ \iff&\varphi^{\mathcal{M}^{\prime}}\notin\{\{t^{\prime}\},\{s^{\prime},t^{\prime}\}\}\text{ and }S^{\prime}\backslash\varphi^{\mathcal{M}^{\prime}}\notin\{\{t^{\prime}\},\{s^{\prime},t^{\prime}\}\}\\ \iff&\varphi^{\mathcal{M}^{\prime}}\neq\{t^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{s^{\prime},t^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{s^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\emptyset\\ \end{array}

For the case $\bullet\varphi$ , we have the following equivalences.

\begin{array}[]{ll}&\mathcal{M},t\vDash\bullet\varphi\\ \iff&\mathcal{M},t\vDash\varphi\text{ and }\varphi^{\mathcal{M}}\notin N(t)\\ \iff&\mathcal{M},t\vDash\varphi\text{ and }\varphi^{\mathcal{M}}\neq\{t\}\text{ and }\varphi^{\mathcal{M}}\neq\{s,t\}\\ \iff&\text{false}\\ \end{array}

\begin{array}[]{ll}&\mathcal{M}^{\prime},t^{\prime}\vDash\bullet\varphi\\ \iff&\mathcal{M}^{\prime},t^{\prime}\vDash\varphi\text{ and }\varphi^{\mathcal{M}^{\prime}}\notin N^{\prime}(t^{\prime})\\ \iff&\mathcal{M}^{\prime},t^{\prime}\vDash\varphi\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\emptyset\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{s^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{t^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{s^{\prime},t^{\prime}\}\\ \iff&\text{false}\\ \end{array}

Thus $\mathcal{M},t\vDash\bullet\varphi$ iff $\mathcal{M}^{\prime},t^{\prime}\vDash\bullet\varphi$ . Therefore, $\mathcal{M},t\vDash\varphi$ iff $\mathcal{M}^{\prime},t^{\prime}\vDash\varphi$ for all $\varphi\in\mathcal{L}(\nabla,\bullet)$ .

\begin{array}[]{ll}&\mathcal{M},s\vDash\bullet\varphi\\ \iff&\mathcal{M},s\vDash\varphi\text{ and }\varphi^{\mathcal{M}}\notin N(s)\\ \iff&\mathcal{M},s\vDash\varphi\text{ and }\varphi^{\mathcal{M}}\neq\{t\}\text{ and }\varphi^{\mathcal{M}}\neq\{s,t\}\\ \iff&\mathcal{M},s\vDash\varphi\text{ and }\mathcal{M},t\nvDash\varphi\\ \stackrel{{\scriptstyle\text{IH}}}{{\iff}}&\mathcal{M}^{\prime},s^{\prime}\vDash\varphi\text{ and }\mathcal{M}^{\prime},t^{\prime}\nvDash\varphi\\ \iff&\mathcal{M}^{\prime},s^{\prime}\vDash\varphi\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{t^{\prime}\}\text{ and }\varphi^{\mathcal{M}^{\prime}}\neq\{s^{\prime},t^{\prime}\}\\ \iff&\mathcal{M}^{\prime},s^{\prime}\vDash\varphi\text{ and }\varphi^{\mathcal{M}^{\prime}}\notin N^{\prime}(s^{\prime})\\ \iff&\mathcal{M}^{\prime},s^{\prime}\vDash\bullet\varphi\\ \end{array}

Therefore, $\mathcal{M},s\vDash\varphi$ iff $\mathcal{M}^{\prime},s^{\prime}\vDash\varphi$ for all $\varphi\in\mathcal{L}(\nabla,\bullet)$ . This completes the proof of $(\ast\ast)$ .

If $(5)$ were to be defined by a set of $\mathcal{L}(\nabla,\bullet)$ -formulas, say $\Gamma$ , then as $\mathcal{F}$ has $(5)$ , we would have $\mathcal{F}\vDash\Gamma$ , thus we should also have $\mathcal{F}^{\prime}\vDash\Gamma$ , that is, $\mathcal{F}^{\prime}$ has $(5)$ : a contradiction. Therefore, $(5)$ is not definable in $\mathcal{L}(\nabla,\bullet)$ . ∎

5 Axiomatizations

In this section, we axiomatize $\mathcal{L}(\nabla,\bullet)$ over various classes of neighborhood frames.

5.1 Classical logic

5.1.1 Proof system and Soundness

Definition 26.

The classical logic of $\mathcal{L}(\nabla,\bullet)$ , denoted ${\bf E^{\nabla\bullet}}$ , consists of the following axioms and inference rules:

\begin{array}[]{ll}\text{TAUT}&\text{all instances of tautologies}\\ \text{E1}&\nabla\varphi\leftrightarrow\nabla\neg\varphi\\ \text{E2}&\bullet\varphi\to\varphi\\ \text{E3}&\nabla\varphi\to\bullet\varphi\vee\bullet\neg\varphi\\ \text{MP}&\dfrac{\varphi,\varphi\to\psi}{\psi}\\ \text{RE}\nabla&\dfrac{\varphi\leftrightarrow\psi}{\nabla\varphi\leftrightarrow\nabla\psi}\\ \text{RE}\bullet&\dfrac{\varphi\leftrightarrow\psi}{\bullet\varphi\leftrightarrow\bullet\psi}\\ \end{array}

Intuitively, E1 says that one is (first-order) ignorant whether a proposition holds if and only if one is ignorant whether its negation holds; E2 says that one is (Fitchean) ignorant of the fact that $\varphi$ only if it is the case that $\varphi$ ; E3 describes the relationship between Fitchean ignorance and first-order ignorance: if one is ignorant whether $\varphi$ , then either one is ignorant of the fact that $\varphi$ or one is ignorant of the fact that $\varphi$ is not the case; $\text{RE}\nabla$ and $\text{RE}\bullet$ concerns the replacement of equivalences for first-order ignorance and Fitchean ignorance, respectively.

It is straightforward by axiom E2 that $\bullet\nabla\varphi\to\nabla\varphi$ is provable in ${\bf E^{\nabla\bullet}}$ , which says that under any condition, Rumsfeld ignorance implies first-order ignorance.

The following result states how to derive Fitchean ignorance, which means that if one is ignorant whether a true proposition holds, then one is ignorant of the proposition. It will be used in several places below (for instance, Lemma 31, Lemma 34, and Prop. 40).

Proposition 27.

$\vdash\nabla\varphi\land\varphi\to\bullet\varphi$ . Equivalently, $\vdash\circ\varphi\land\varphi\to\Delta\varphi$ .

Proof.

We have the following proof sequence:

\begin{array}[]{lll}(i)&\nabla\varphi\to\bullet\varphi\vee\bullet\neg\varphi&\text{E3}\\ (ii)&\bullet\neg\varphi\to\neg\varphi&\text{E2}\\ (iii)&\nabla\varphi\to\bullet\varphi\vee\neg\varphi&(i),(ii)\\ (iv)&\nabla\varphi\land\varphi\to\bullet\varphi&(iii)\\ \end{array}

∎

As a corollary, we have $\vdash\nabla\nabla\varphi\land\nabla\varphi\to\bullet\nabla\varphi$ , and thus $\vdash\nabla\nabla\varphi\land\nabla\varphi\to\bullet\nabla\varphi$ . This means, in Fine [15] terms, second-order ignorance plus first-order ignorance implies Rumsfeld ignorance. On one hand, this is not noticed in Fine [15]; on the other hand, this plus the transitivity entails that second-order ignorance implies Rumsfeld ignorance, a result in the paper in question.

The following result indicates how to derive a proposition from Fitchean ignorance.

Proposition 28.

$\vdash\bullet(\circ\varphi\vee\psi\to\varphi)\to\varphi$

Proof.

We have the following proof sequence.

\begin{array}[]{lll}(1)&\bullet(\circ\varphi\vee\psi\to\varphi)\to(\circ\varphi\vee\psi\to\varphi)&\text{E2}\\ (2)&(\circ\varphi\vee\psi\to\varphi)\to(\circ\varphi\to\varphi)&\text{TAUT}\\ (3)&\bullet(\circ\varphi\vee\psi\to\varphi)\to(\circ\varphi\to\varphi)&(1),(2)\\ (4)&\bullet\varphi\to\varphi&\text{E2}\\ (5)&\bullet(\circ\varphi\vee\psi\to\varphi)\to\varphi&(3),(4)\\ \end{array}

∎

Proposition 29.

${\bf E^{\nabla\bullet}}$ is sound with respect to the class of all (neighborhood) frames.

Proof.

It suffices to show the validity of axiom E3. This has been shown in the proof of Prop. 15. ∎

5.1.2 Completeness

Definition 30.

The canonical model for ${\bf E^{\nabla\bullet}}$ is $\mathcal{M}^{c}=\langle S^{c},N^{c},V^{c}\rangle$ , where

-

$S^{c}=\{s\mid s\text{ is a maximal consistent set for }{\bf E^{\nabla\bullet}}\}$ ,
-

$N^{c}(s)=\{|\varphi|\mid\circ\varphi\land\Delta\varphi\in s\}$ ,
-

$V^{c}(p)=\{s\in S^{c}\mid p\in s\}$ .

Lemma 31.

For all $\varphi\in\mathcal{L}(\nabla,\bullet)$ , for all $s\in S^{c}$ , we have

\mathcal{M}^{c},s\vDash\varphi\iff\varphi\in s.

That is, $|\varphi|=\varphi^{\mathcal{M}^{c}}$ .

Proof.

By induction on $\varphi$ . The nontrivial cases are $\nabla\varphi$ and $\bullet\varphi$ .

For case $\nabla\varphi$ :

First, suppose that $\nabla\varphi\in s$ , to show that $\mathcal{M}^{c},s\vDash\nabla\varphi$ . By supposition, we have $\Delta\varphi\notin s$ . Then by definition of $N^{c}$ , we infer that $|\varphi|\notin N^{c}(s)$ . By supposition again and axiom E1, we have $\nabla\neg\varphi\in s$ , and thus $\Delta\neg\varphi\notin s$ , and hence $|\neg\varphi|\notin N^{c}(s)$ , that is, $S^{c}\backslash|\varphi|\notin N^{c}(s)$ . By induction hypothesis, we have $\varphi^{\mathcal{M}^{c}}\notin N^{c}(s)$ and $S^{c}\backslash\varphi^{\mathcal{M}^{c}}\notin N^{c}(s)$ . Therefore, $\mathcal{M}^{c},s\vDash\nabla\varphi$ .

Conversely, assume that $\nabla\varphi\notin s$ (that is, $\nabla\neg\varphi\notin s$ ), to show that $\mathcal{M}^{c},s\nvDash\nabla\varphi$ . By assumption, $\Delta\varphi\in s$ and $\Delta\neg\varphi\in s$ . Since $s\in S^{c}$ , we have either $\varphi\in s$ or $\neg\varphi\in s$ . If $\varphi\in s$ , then by axiom E2, $\circ\neg\varphi\in s$ , and then $|\neg\varphi|\in N^{c}(s)$ , viz. $S^{c}\backslash|\varphi|\in N^{c}(s)$ , which by induction hypothesis implies that $S^{c}\backslash\varphi^{\mathcal{M}^{c}}\in N^{c}(s)$ . If $\neg\varphi\in s$ , then again by axiom E2, $\circ\varphi\in s$ , thus $|\varphi|\in N^{c}(s)$ , which by induction hypothesis entails that $\varphi^{\mathcal{M}^{c}}\in N^{c}(s)$ . We have now shown that either $\varphi^{\mathcal{M}^{c}}\in N^{c}(s)$ or $S^{c}\backslash\varphi^{\mathcal{M}^{c}}\in N^{c}(s)$ , and we therefore conclude that $\mathcal{M}^{c},s\nvDash\nabla\varphi$ .

For case $\bullet\varphi$ :

First, suppose that $\bullet\varphi\in s$ , to show that $\mathcal{M}^{c},s\vDash\bullet\varphi$ . By supposition and axiom E2, we have $\varphi\in s$ . By induction hypothesis, $\mathcal{M}^{c},s\vDash\varphi$ . By supposition and definition of $N^{c}$ , we infer that $|\varphi|\notin N^{c}(s)$ , which by induction means that $\varphi^{\mathcal{M}^{c}}\notin N^{c}(s)$ . Therefore, $\mathcal{M}^{c},s\vDash\bullet\varphi$ .

Conversely, assume that $\bullet\varphi\notin s$ , to demonstrate that $\mathcal{M}^{c},s\nvDash\bullet\varphi$ . By assumption, $\circ\varphi\in s$ . If $\mathcal{M}^{c},s\nvDash\varphi$ , it is obvious that $\mathcal{M}^{c},s\nvDash\bullet\varphi$ . Otherwise, by induction hypothesis, we have $\varphi\in s$ , then $\circ\varphi\land\varphi\in s$ . By Prop. 27, $\Delta\varphi\in s$ , and thus $|\varphi|\in N^{c}(s)$ , by induction we obtain $\varphi^{\mathcal{M}^{c}}\in N^{c}(s)$ , and therefore we have also $\mathcal{M}^{c},s\nvDash\bullet\varphi$ . ∎

It is then a standard exercise to show the following.

Theorem 32.

${\bf E^{\nabla\bullet}}$ is sound and strongly complete with respect to the class of all neighborhood frames.

5.2 Extensions

5.2.1 ${\bf E_{c}^{\nabla\bullet}}$

Define ${\bf E_{c}^{\nabla\bullet}}$ be the smallest extension of ${\bf E^{\nabla\bullet}}$ with the following axiom, denoted E4:

\bullet\varphi\to\nabla\varphi.

Intuitively, E4 says that Fitchean ignorance implies first-order ignorance.

From E4 we can easily prove $\Delta\varphi\to\circ\varphi$ . This turns the canonical model for ${\bf E^{\nabla\bullet}}$ (Def. 30) into the following simpler one.

Definition 33.

The canonical model for ${\bf E_{c}^{\nabla\bullet}}$ is a triple $\mathcal{M}^{c}=\langle S^{c},N^{c},V^{c}\rangle$ , where

•

$S^{c}=\{s\mid s\text{ is a maximal consistent set for }{\bf E_{c}^{\nabla\bullet}}\}$
•

$N^{c}(s)=\{|\varphi|\mid\Delta\varphi\in s\}$
•

$V^{c}(p)=\{s\in S^{c}\mid p\in s\}$ .

Lemma 34.

For all $\varphi\in\mathcal{L}(\nabla,\bullet)$ , for all $s\in S^{c}$ , we have

\mathcal{M}^{c},s\vDash\varphi\iff\varphi\in s.

That is, $|\varphi|=\varphi^{\mathcal{M}^{c}}$ .

Proof.

By induction on $\varphi$ . The nontrivial cases are $\nabla\varphi$ and $\bullet\varphi$ . The case $\nabla\varphi$ has been shown in [12, Lemma 1]. It suffices to show the case $\bullet\varphi$ .

Suppose that $\bullet\varphi\in s$ . Then by axiom E2, we have $\varphi\in s$ ; by axiom E4, we derive that $\nabla\varphi\in s$ , and thus $\Delta\varphi\notin s$ . This follows that $|\varphi|\notin N^{c}(s)$ . By induction hypothesis, $\mathcal{M}^{c},s\vDash\varphi$ and $\varphi^{\mathcal{M}^{c}}\notin N^{c}(s)$ . Therefore, $\mathcal{M}^{c},s\vDash\bullet\varphi$ .

Conversely, assume that $\bullet\varphi\notin s$ , to show that $\mathcal{M}^{c},s\nvDash\bullet\varphi$ , which by induction hypothesis amounts to showing that $\varphi\notin s$ or $|\varphi|\in N^{c}(s)$ . For this, suppose that $\varphi\in s$ , this plus the assumption implies that $\circ\varphi\land\varphi\in s$ . Then by Prop. 27, $\Delta\varphi\in s$ , and therefore $|\varphi|\in N^{c}(s)$ . ∎

Proposition 35.

$\mathcal{M}^{c}$ possesses the property $(c)$ .

Proof.

Refer to [12, Thm. 2]. ∎

Now it is a standard exercise to show the following.

Theorem 36.

${\bf E_{c}^{\nabla\bullet}}$ is sound and strongly complete with respect to the class of $(c)$ -frames.

In the neighborhood context $(c)$ , there is some relationship between Rumsfeld ignorance, second-order ignorance and first-order ignorance. The following is immediate from the axiom E4.

Proposition 37.

$\bullet\nabla\varphi\to\nabla\nabla\varphi$ is provable in ${\bf E^{\nabla\bullet}_{c}}$ .

This says that under the condition $(c)$ , Rumsfeld ignorance implies second-order ignorance.

Combined with an instance of the axiom E2 ( $\bullet\nabla\varphi\to\nabla\varphi$ ) and $\vdash\nabla\nabla\varphi\land\nabla\varphi\to\bullet\nabla\varphi$ (see the remark after Prop. 27), it follows that within the neighborhood context $(c)$ , Rumsfeld ignorance amounts to second-order ignorance plus first-order ignorance, and thus Rumsfeld ignorance is definable in terms of first-order ignorance.

5.2.2 ${\bf EN^{\nabla\bullet}}$

Let ${\bf EN^{\nabla\bullet}}={\bf E^{\nabla\bullet}}+\circ\top$ . From $\circ\top$ and Prop. 27 it follows that $\Delta\top$ is derivable in ${\bf EN^{\nabla\bullet}}$ .

Theorem 38.

${\bf EN^{\nabla\bullet}}$ is sound and strongly complete with respect to the class of all $(n)$ -frames.

Proof.

For soundness, by Prop. 29, it remains only to show the validity of $\circ\top$ over the class of $(n)$ -frames. The validity of $\circ\top$ can be found in Prop. 22.

For completeness, define the canonical model $\mathcal{M}^{c}$ w.r.t. ${\bf EN^{\nabla\bullet}}$ as in Def. 30. By Thm. 32, it suffices to show that $\mathcal{M}^{c}$ possesses $(n)$ . By the construction of ${\bf EN^{\nabla\bullet}}$ , for all $s\in S^{c}$ , we have $\circ\top\land\Delta\top\in s$ , and thus $|\top|\in N^{c}(s)$ , that is, $S^{c}\in N^{c}(s)$ , as desired. ∎

5.2.3 Monotone logic

Let ${\bf M^{\nabla\bullet}}$ be the extension of ${\bf E^{\nabla\bullet}}$ plus the following extra axioms:

\begin{array}[]{ll}\text{M1}&\nabla(\varphi\vee\psi)\land\nabla(\neg\varphi\vee\chi)\to\nabla\varphi\\ \text{M2}&\bullet(\varphi\vee\psi)\land\bullet(\neg\varphi\vee\chi)\to\nabla\varphi\\ \text{M3}&\bullet(\varphi\vee\psi)\land\nabla(\neg\varphi\vee\chi)\to\nabla\varphi\\ \text{M4}&\circ\varphi\land\varphi\to\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\psi)\\ \end{array}

Prop. 39 and Prop. 40 tells us how to derive $\Delta\top$ and $\circ\top$ in ${\bf M^{\nabla\bullet}}$ , respectively. They will used in Section 6.

Proposition 39.

$\Delta\varphi\to\Delta\top$ is provable in ${\bf M^{\nabla\bullet}}$ .

Proof.

We have the following proof sequence in ${\bf M^{\nabla\bullet}}$ .

\begin{array}[]{lll}(1)&\nabla(\varphi\vee\top)\land\nabla(\neg\varphi\vee\top)\to\nabla\varphi&\text{M1}\\ (2)&\top\leftrightarrow\varphi\vee\top&\text{TAUT}\\ (3)&\nabla\top\leftrightarrow\nabla(\varphi\vee\top)&(2),\text{RE}\nabla\\ (4)&\top\leftrightarrow\neg\varphi\vee\top&\text{TAUT}\\ (5)&\nabla\top\leftrightarrow\nabla(\neg\varphi\vee\top)&(4),\text{RE}\nabla\\ (6)&\nabla\top\to\nabla\varphi&(1),(3),(5)\\ (7)&\Delta\varphi\to\Delta\top&(6),\text{Def.\leavevmode\nobreak\ }\Delta\\ \end{array}

∎

In the above proof, $\nabla\top\to\nabla\varphi$ says that if one is ignorant about whether $\top$ holds, then one is ignorant about whether everything holds.

Proposition 40.

$\varphi\land\circ\varphi\to\circ\top$ is provable in ${\bf M^{\nabla\bullet}}$ .

Proof.

We have the following proof sequence in ${\bf M^{\nabla\bullet}}$ .

\begin{array}[]{lll}(1)&(\varphi\vee\top)\leftrightarrow\top&\text{TAUT}\\ (2)&\bullet(\varphi\vee\top)\leftrightarrow\bullet\top&(1),\text{RE}\bullet\\ (3)&(\neg\varphi\vee\top)\leftrightarrow\top&\text{TAUT}\\ (4)&\bullet(\neg\varphi\vee\top)\leftrightarrow\bullet\top&(3),\text{RE}\bullet\\ (5)&\bullet(\varphi\vee\top)\land\bullet(\neg\varphi\vee\top)\to\nabla\varphi&\text{M2}\\ (6)&\bullet\top\to\nabla\varphi&(2),(4),(5)\\ (7)&\Delta\varphi\to\circ\top&(6),\text{Def.\leavevmode\nobreak\ }\Delta,\text{Def.\leavevmode\nobreak\ }\circ\\ (8)&\varphi\land\circ\varphi\to\Delta\varphi&\text{Prop.\leavevmode\nobreak\ }\ref{prop.circimpliesdelta}\\ (9)&\varphi\land\circ\varphi\to\circ\top&(7),(8)\\ \end{array}

∎

Proposition 41.

${\bf M^{\nabla\bullet}}$ is sound with respect to the class of all $(s)$ -frames.

Proof.

By soundness of ${\bf E^{\nabla\bullet}}$ (Prop. 29), it suffices to show the validity of the extra axioms. Let $\mathcal{M}=\langle S,N,V\rangle$ be an arbitrary $(s)$ -model and $s\in S$ .

For M1: suppose that $\mathcal{M},s\vDash\nabla(\varphi\vee\psi)\land\nabla(\neg\varphi\vee\chi)$ . Then $(\varphi\vee\psi)^{\mathcal{M}}\notin N(s)$ and $(\neg\varphi\vee\chi)^{\mathcal{M}}\notin N(s)$ , that is, $\varphi^{\mathcal{M}}\cup\psi^{\mathcal{M}}\notin N(s)$ and $(\neg\varphi)^{\mathcal{M}}\cup\chi^{\mathcal{M}}\notin N(s)$ . Since $\varphi^{\mathcal{M}}\subseteq\varphi^{\mathcal{M}}\cup\psi^{\mathcal{M}}$ and $N(s)$ is closed under supersets, we must have $\varphi^{\mathcal{M}}\notin N(s)$ . Similarly, we can show that $(\neg\varphi)^{\mathcal{M}}\notin N(s)$ . Therefore, $\mathcal{M},s\vDash\nabla\varphi$ , as desired. Similarly, we can show the validity of M2 and M3.

For M4: assume that $\mathcal{M},s\vDash\circ\varphi\land\varphi$ . Then $\varphi^{\mathcal{M}}\in N(s)$ . Since $\varphi^{\mathcal{M}}\subseteq\varphi^{\mathcal{M}}\cup\psi^{\mathcal{M}}=(\varphi\vee\psi)^{\mathcal{M}}$ , by the property $(s)$ , we have $(\varphi\vee\psi)^{\mathcal{M}}\in N(s)$ , and therefore $\mathcal{M},s\vDash\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\psi)$ . ∎

Definition 42.

Let $\Lambda$ be an extension of ${\bf M^{\nabla\bullet}}$ . A triple $\mathcal{M}^{\Lambda}=\langle S^{\Lambda},N^{\Lambda},V^{\Lambda}\rangle$ is a canonical neighborhood model for $\Lambda$ if

•

$S^{\Lambda}=\{s\mid s\text{ is a maximal consistent set for }\Lambda\}$ ,
•

$|\varphi|\in N^{\Lambda}(s)$ iff $\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\psi)\in s$ for all $\psi$ ,
•

$V^{\Lambda}(p)=|p|=\{s\in S^{\Lambda}\mid p\in s\}$ .

We need to show that $N^{\Lambda}$ is well defined.

Proposition 43.

Let $s\in S^{\Lambda}$ as defined in Def. 42. If $|\varphi|=|\varphi^{\prime}|$ , then $\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\psi)\in s$ for all $\psi$ iff $\Delta(\varphi^{\prime}\vee\psi)\land\circ(\varphi^{\prime}\vee\psi)\in s$ for all $\psi$ .

Proof.

Suppose that $|\varphi|=|\varphi^{\prime}|$ , then $\vdash\varphi\leftrightarrow\varphi^{\prime}$ , and thus $\vdash\varphi\vee\psi\leftrightarrow\varphi^{\prime}\vee\psi$ . By $\text{RE}\nabla$ , $\text{RE}\bullet$ , $\text{Def.\leavevmode\nobreak\ }\Delta$ and $\text{Def.\leavevmode\nobreak\ }\circ$ , we infer that $\vdash\Delta(\varphi\vee\psi)\leftrightarrow\Delta(\varphi^{\prime}\vee\psi)$ and $\vdash\circ(\varphi\vee\psi)\leftrightarrow\circ(\varphi^{\prime}\vee\psi)$ , and hence $\vdash\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\psi)\leftrightarrow\Delta(\varphi^{\prime}\vee\psi)\land\circ(\varphi^{\prime}\vee\psi)$ . Therefore, $\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\psi)\in s$ for all $\psi$ iff $\Delta(\varphi^{\prime}\vee\psi)\land\circ(\varphi^{\prime}\vee\psi)\in s$ for all $\psi$ . ∎

Lemma 44.

Let $\mathcal{M}^{\Lambda}=\langle S^{\Lambda},N^{\Lambda},V^{\Lambda}\rangle$ be an arbitrary canonical neighborhood model for any system $\Lambda$ extending ${\bf M^{\nabla\bullet}}$ . Then for all $s\in S^{\Lambda}$ , for all $\varphi\in\mathcal{L}(\nabla,\bullet)$ , we have

\mathcal{M}^{\Lambda},s\vDash\varphi\iff\varphi\in s.

That is, $\varphi^{\mathcal{M}^{\Lambda}}=|\varphi|$ .

Proof.

By induction on $\varphi$ . The nontrivial cases are $\nabla\varphi$ and $\bullet\varphi$ .

For case $\nabla\varphi$ :

Suppose that $\mathcal{M}^{\Lambda},s\nvDash\nabla\varphi$ , to show that $\nabla\varphi\notin s$ . By supposition and induction hypothesis, $|\varphi|\in N^{\Lambda}(s)$ or $S\backslash|\varphi|\in N^{\Lambda}(s)$ (that is, $|\neg\varphi|\in N^{\Lambda}(s)$ ). If $|\varphi|\in N^{\Lambda}(s)$ , then $\Delta(\varphi\vee\psi)\in s$ for all $\psi$ . By letting $\psi=\bot$ , then $\Delta\varphi\in s$ , and thus $\nabla\varphi\notin s$ . If $|\neg\varphi|\in N^{\Lambda}(s)$ , with a similar argument we can show that $\Delta\neg\varphi\in s$ , that is, $\Delta\varphi\in s$ , and we also have $\nabla\varphi\notin s$ .

Conversely, assume that $\mathcal{M}^{\Lambda},s\vDash\nabla\varphi$ , to prove that $\nabla\varphi\in s$ . By assumption and induction hypothesis, $|\varphi|\notin N^{\Lambda}(s)$ and $S\backslash|\varphi|\notin N^{\Lambda}(s)$ , that is, $|\neg\varphi|\notin N^{\Lambda}(s)$ . Then $\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\psi)\notin s$ for some $\psi$ , and $\Delta(\neg\varphi\vee\chi)\land\circ(\neg\varphi\vee\chi)\notin s$ for some $\chi$ . We consider the following cases.

-

$\Delta(\varphi\vee\psi)\notin s$ and $\Delta(\neg\varphi\vee\chi)\notin s$ . That is, $\nabla(\varphi\vee\psi)\in s$ and $\nabla(\neg\varphi\vee\chi)\in s$ . Then by axiom M1, we infer that $\nabla\varphi\in s$ .
-

$\Delta(\varphi\vee\psi)\notin s$ and $\circ(\neg\varphi\vee\chi)\notin s$ . That is, $\nabla(\varphi\vee\psi)\in s$ and $\bullet(\neg\varphi\vee\chi)\in s$ . By axiom M3, $\nabla\neg\varphi\in s$ , that is, $\nabla\varphi\in s$ .
-

$\circ(\varphi\vee\psi)\notin s$ and $\Delta(\neg\varphi\vee\chi)\notin s$ . That is, $\bullet(\varphi\vee\psi)\in s$ and $\nabla(\neg\varphi\vee\chi)\in s$ . Then by axiom M3, we derive that $\nabla\varphi\in s$ .
-

$\circ(\varphi\vee\psi)\notin s$ and $\circ(\neg\varphi\vee\chi)\notin s$ . That is, $\bullet(\varphi\vee\psi)\in s$ and $\bullet(\neg\varphi\vee\chi)\in s$ . By axiom M2, we obtain that $\nabla\varphi\in s$ .

Either case implies that $\nabla\varphi\in s$ , as desired.

For case $\bullet\varphi$ .

Suppose that $\bullet\varphi\in s$ , to show that $\mathcal{M}^{\Lambda},s\vDash\bullet\varphi$ . By supposition and axiom E2, we obtain $\varphi\in s$ , which by induction hypothesis means that $\mathcal{M}^{\Lambda},s\vDash\varphi$ . We have also $|\varphi|\notin N^{\Lambda}(s)$ : otherwise, by definition of $N^{\Lambda}$ , we should have $\circ(\varphi\vee\psi)\in s$ for all $\psi$ , which then implies that $\circ\varphi\in s$ (by letting $\psi=\bot$ ), a contradiction. Then by induction hypothesis, $\varphi^{\mathcal{M}^{\Lambda}}\notin N^{\Lambda}(s)$ . Therefore, $\mathcal{M}^{\Lambda},s\vDash\bullet\varphi$ .

Conversely, assume that $\bullet\varphi\notin s$ (that is, $\circ\varphi\in s$ ), to prove that $\mathcal{M}^{\Lambda},s\nvDash\bullet\varphi$ . For this, suppose that $\mathcal{M}^{\Lambda},s\vDash\varphi$ , by induction hypothesis, we have $\varphi\in s$ , and then $\circ\varphi\land\varphi\in s$ . By axiom M4, $\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\psi)\in s$ for all $\psi$ . By definition of $N^{\Lambda}$ , we derive that $|\varphi|\in N^{\Lambda}(s)$ . Then by induction hypothesis again, we conclude that $\varphi^{\mathcal{M}^{\Lambda}}\in N^{\Lambda}(s)$ . Therefore, $\mathcal{M}^{\Lambda},s\nvDash\bullet\varphi$ , as desired. ∎

Given an extension $\Lambda$ of ${\bf M^{\nabla\bullet}}$ , the minimal canonical neighborhood model for $\Lambda$ , denoted $\mathcal{M}^{\Lambda}_{0}=\langle S^{\Lambda},N^{\Lambda}_{0},V^{\Lambda}\rangle$ , is defined such that $N^{\Lambda}_{0}(s)=\{|\varphi|\mid\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\psi)\in s\text{ for all }\psi\}$ . Note that $\mathcal{M}^{\Lambda}_{0}$ is not necessarily supplemented. Therefore, we define a notion of supplementation, which comes from [4].

Definition 45.

Let $\mathcal{M}=\langle S,N,V\rangle$ be a neighborhood model. The supplementation of $\mathcal{M}$ , denoted $\mathcal{M}^{+}$ , is a triple $\langle S,N^{+},V\rangle$ , in which for every $s\in S$ , $N^{+}(s)$ is the superset closure of $N(s)$ ; namely, for each $s\in S$ ,

N^{+}(s)=\{X\subseteq S\mid Y\subseteq X\text{ for some }Y\in N(s)\}.

One may easily show that $\mathcal{M}^{+}$ is supplemented, that is, $\mathcal{M}^{+}$ possesses $(s)$ . Also, $N(s)\subseteq N^{+}(s)$ . Moreover, the properties of being closed under intersections and containing the unit are closed under the supplementation.

Proposition 46.

Let $\mathcal{M}=\langle S,N,V\rangle$ be a neighborhood model and $\mathcal{M}^{+}$ be its supplementation. If $\mathcal{M}$ possesses $(i)$ , then so does $\mathcal{M}^{+}$ ; if $\mathcal{M}$ possesses $(n)$ , then so does $\mathcal{M}^{+}$ .

In what follows, we will use $(\mathcal{M}^{\Lambda}_{0})^{+}$ to denote the supplementation of $\mathcal{M}^{\Lambda}_{0}$ , namely $(\mathcal{M}^{\Lambda}_{0})^{+}=\langle S^{\Lambda},(N^{\Lambda}_{0})^{+},V^{\Lambda}\rangle$ , where $\Lambda$ extends ${\bf M^{\nabla\bullet}}$ . By the definition of supplementation, $(\mathcal{M}^{\Lambda}_{0})^{+}$ is an $(s)$ -model. To show the completeness of ${\bf M^{\nabla\bullet}}$ over the class of $(s)$ -frames, by Lemma 44, it remains only to show that $(\mathcal{M}^{\Lambda}_{0})^{+}$ is a canonical neighborhood model for $\Lambda$ .

Lemma 47.

Let $\Lambda$ extends ${\bf M^{\nabla\bullet}}$ . For every $s\in S^{\Lambda}$ , we have

|\varphi|\in(N^{\Lambda}_{0})^{+}(s)\iff\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\psi)\in s\text{ for all }\psi.

Proof.

Right-to-Left: Immediate by the definition of $N^{\Lambda}_{0}$ and the fact that $N^{\Lambda}_{0}(s)\subseteq(N^{\Lambda}_{0})^{+}(s)$ .

Left-to-Right: Suppose that $|\varphi|\in(N^{\Lambda}_{0})^{+}(s)$ , to prove that $\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\psi)\in s\text{ for all }\psi.$ By supposition, $X\subseteq|\varphi|$ for some $X\in N^{\Lambda}_{0}(s)$ . Then there must be a $\chi$ such that $X=|\chi|$ , and thus $\Delta(\chi\vee\psi)\land\circ(\chi\vee\psi)\in s$ for all $\psi$ , and hence $\Delta(\chi\vee\varphi\vee\psi)\land\circ(\chi\vee\varphi\vee\psi)\in s$ . From $|\chi|\subseteq|\varphi|$ , it follows that $\vdash\chi\to\varphi$ , and then $\vdash\chi\vee\varphi\vee\psi\leftrightarrow\varphi\vee\psi$ , and thus $\vdash\Delta(\chi\vee\varphi\vee\psi)\leftrightarrow\Delta(\varphi\vee\psi)$ and $\vdash\circ(\chi\vee\varphi\vee\psi)\leftrightarrow\circ(\varphi\vee\psi)$ by $\text{RE}\nabla$ , $\text{RE}\bullet$ , $\text{Def.\leavevmode\nobreak\ }\Delta$ and $\text{Def.\leavevmode\nobreak\ }\circ$ . Therefore, $\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\psi)\in s$ for all $\psi$ . ∎

Based on the previous analysis, we have the following.

Theorem 48.

${\bf M^{\nabla\bullet}}$ is sound and strongly complete with respect to the class of all $(s)$ -frames.

We conclude this part with some results which will be used in Section 6. The following result states that if one is ignorant of the fact that either $\varphi$ holds or one is ignorant whether $\varphi$ holds, then one is either ignorant of the fact that $\varphi$ or ignorant whether $\varphi$ holds.

Proposition 49.

$\bullet(\varphi\vee\nabla\varphi)\to(\bullet\varphi\vee\nabla\varphi)$ is provable in ${\bf M^{\nabla\bullet}}$ .¹⁰¹⁰10In fact, we can show a stronger result: $\bullet(\varphi\vee\nabla\psi)\to\bullet\varphi\vee\nabla\psi$ is provable in ${\bf M^{\nabla\bullet}}$ . But we do not need such a strong result below.

Proof.

By Thm. 48, it suffices to show the formula is valid over the class of $(s)$ -frames. Let $\mathcal{M}=\langle S,N,V\rangle$ be an $(s)$ -model and $s\in S$ .

Suppose, for reductio, that $\mathcal{M},s\vDash\bullet(\varphi\vee\nabla\varphi)$ and $\mathcal{M},s\nvDash\bullet\varphi\vee\nabla\varphi$ . From the former, it follows that $\mathcal{M},s\vDash\varphi\vee\nabla\varphi$ and $(\varphi\vee\nabla\varphi)^{\mathcal{M}}\notin N(s)$ ; from the latter, it follows that $\mathcal{M},s\nvDash\bullet\varphi$ and $\mathcal{M},s\nvDash\nabla\varphi$ . This implies that $\mathcal{M},s\vDash\varphi$ , which plus $\mathcal{M},s\nvDash\bullet\varphi$ gives us $\varphi^{\mathcal{M}}\in N(s)$ . Since $\varphi^{\mathcal{M}}\subseteq(\varphi\vee\nabla\varphi)^{\mathcal{M}}$ , by $(s)$ , we conclude that $(\varphi\vee\nabla\varphi)^{\mathcal{M}}\in N(s)$ : a contradiction, as desired. ∎

The following result says that if one is ignorant of the fact that either non-ignorance of $\varphi$ or non-ignorance whether $\varphi$ holds implies that $\varphi$ , then one is ignorant of the fact that $\varphi$ .

Proposition 50.

$\bullet(\circ\varphi\vee\Delta\varphi\to\varphi)\to\bullet\varphi$ is provable in ${\bf M^{\nabla\bullet}}$ .

Proof.

By Thm. 48, it remains only to prove that the formula is valid over the class of $(s)$ -frames. Let $\mathcal{M}=\langle S,N,V\rangle$ be an $(s)$ -model and $s\in S$ .

Assume, for reductio, that $\mathcal{M},s\vDash\bullet(\circ\varphi\vee\Delta\varphi\to\varphi)$ and $\mathcal{M},s\nvDash\bullet\varphi$ . The former implies $\mathcal{M},s\vDash\circ\varphi\vee\Delta\varphi\to\varphi$ and $(\circ\varphi\vee\Delta\varphi\to\varphi)^{\mathcal{M}}\notin N(s)$ ; the latter entails that $\mathcal{M},s\vDash\circ\varphi$ . Then $\mathcal{M},s\vDash\varphi$ , and thus $\varphi^{\mathcal{M}}\in N(s)$ . One may easily verify that $\varphi^{\mathcal{M}}\subseteq(\circ\varphi\vee\Delta\varphi\to\varphi)^{\mathcal{M}}$ . Then by $(s)$ , we conclude that $(\circ\varphi\vee\Delta\varphi\to\varphi)^{\mathcal{M}}\in N(s)$ : a contradiction. ∎

5.2.4 Regular logic

Define ${\bf R^{\nabla\bullet}}:={\bf M^{\nabla\bullet}}+\text{R1}+\text{R2}$ , where

\begin{array}[]{ll}\text{R1}&\Delta\varphi\land\Delta\psi\to\Delta(\varphi\land\psi)\\ \text{R2}&\circ\varphi\land\circ\psi\to\circ(\varphi\land\psi)\\ \end{array}

Proposition 51.

${\bf R^{\nabla\bullet}}$ is sound with respect to the class of quasi-filters.

Proof.

By soundness of ${\bf M^{\nabla\bullet}}$ , it remains to prove the validity of R1 and R2. The validity of R1 has been shown in [8, Prop. 3(iv)], and the validity of R2 has been shown in [11, Thm. 5.2]. ∎

Proposition 52.

Let $\Lambda$ extends ${\bf R^{\nabla\bullet}}$ . Then the minimal canonical model $\mathcal{M}^{\Lambda}_{0}$ has the property $(i)$ . As a corollary, its supplementation is a quasi-filter.

Proof.

Suppose that $X,Y\in N^{\Lambda}_{0}(s)$ , to show that $X\cap Y\in N^{\Lambda}_{0}(s)$ . By supposition, there exist $\varphi$ and $\chi$ such that $X=|\varphi|$ and $Y=|\chi|$ , and then $\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\psi)\in s$ for all $\psi$ , and $\Delta(\chi\vee\psi)\land\circ(\chi\vee\psi)\in s$ for all $\psi$ . By axioms R1 and R2, we can obtain that $\Delta((\varphi\land\chi)\vee\psi)\land\circ((\varphi\land\chi)\vee\psi)\in s$ for all $\psi$ , which implies that $|\varphi\land\chi|\in N^{\Lambda}_{0}(s)$ , that is, $X\cap Y\in N^{\Lambda}_{0}(s)$ . ∎

Theorem 53.

${\bf R^{\nabla\bullet}}$ is sound and strongly complete with respect to the class of quasi-filters.

5.2.5 ${\bf K^{\nabla\bullet}}$

Define ${\bf K^{\nabla\bullet}}:={\bf R^{\nabla\bullet}}+\circ\top$ .

Again, like the case of ${\bf EN^{\nabla\bullet}}$ (Sec. 5.2.2), $\Delta\top$ is derivable from $\circ\top$ and Prop. 27. This hints us that the inference rule R1 in [7, Def. 12] is actually dispensable. (Fact 13 therein is derivable from axiom A1 and axiom A6. Then by R2, we have $\vdash\varphi$ implies $\vdash\circ\varphi\land\varphi$ , and then $\vdash\Delta\varphi$ . Thus we derive R1 there.)

Theorem 54.

${\bf K^{\nabla\bullet}}$ is sound and strongly complete with respect to the class of filters.

Proof.

For soundness, by Prop. 51, it suffices to show the validity of $\circ\top$ over the class of filters. This follows immediately from Prop. 22.

For completeness, define $\mathcal{M}^{\Lambda}_{0}$ as before w.r.t. ${\bf K^{\nabla\bullet}}$ . By Prop. 52 and Prop. 46, it remains only to show that $N^{\Lambda}_{0}(s)$ possesses $(n)$ . By $\circ\top$ and the derivable formula $\Delta\top$ , we have $\vdash\circ\top$ and $\vdash\Delta\top$ . Then by $\vdash(\top\vee\psi)\leftrightarrow\top$ , $\text{RE}\nabla$ , $\text{RE}\bullet$ , $\text{Def.\leavevmode\nobreak\ }\Delta$ and $\text{Def.\leavevmode\nobreak\ }\circ$ , we infer that for all $s\in S^{\Lambda}$ , $\Delta(\top\vee\psi)\land\circ(\top\vee\psi)\in s$ for all $\psi$ , and thus $|\top|\in N^{\Lambda}(s)$ , that is, $S\in N^{\Lambda}_{0}(s)$ , as desired. ∎

Inspired by the definition of $N^{\Lambda}$ , one may define the canonical relation for the extensions of ${\bf K^{\nabla\bullet}}$ as follows:

$sR^{N}t$ iff for all $\varphi$ , if $\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\psi)\in s$ for all $\psi$ , then $\varphi\in t$ .

Recall that the original definition of canonical relation given in [7, Def. 18] is as follows:

$sR^{K}t$ iff there exists $\delta$ such that $(a)$ $\bullet\delta\in s$ , and $(b)$ for all $\varphi$ , if $\Delta\varphi\land\circ(\neg\delta\to\varphi)\in s$ , then $\varphi\in t$ .

One may ask what the relationship between $R^{N}$ and $R^{K}$ is. As we shall see, they are equal to each other. Before this, we need some preparation.

Proposition 55.

$\vdash\bullet\delta\land\Delta\varphi\land\circ(\neg\delta\to\varphi)\to\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\chi)$

Proof.

By Thm. 54, it remains only to show that this formula is valid over the class of filters.

Let $\mathcal{M}=\langle S,N,V\rangle$ be a filter and $s\in S$ . Suppose that $\mathcal{M},s\vDash\bullet\delta\land\Delta\varphi\land\circ(\neg\delta\to\varphi)$ , to show $\mathcal{M},s\vDash\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\chi)$ . By $\mathcal{M},s\vDash\bullet\delta$ , we have $\mathcal{M},s\vDash\delta$ and $\delta^{\mathcal{M}}\notin N(s)$ . By $\mathcal{M},s\vDash\Delta\varphi$ , we infer that $\varphi^{\mathcal{M}}\in N(s)$ or $S\backslash\varphi^{\mathcal{M}}\in N(s)$ . Since $\mathcal{M},s\vDash\delta$ , we derive that $\mathcal{M},s\vdash\neg\delta\to\varphi$ . Then by $\mathcal{M},s\vDash\circ(\neg\delta\to\varphi)$ , we get $(\neg\delta\to\varphi)^{\mathcal{M}}\in N(s)$ , that is, $\delta^{\mathcal{M}}\cup\varphi^{\mathcal{M}}\in N(s)$ . If $S\backslash\varphi^{\mathcal{M}}\in N(s)$ , then as $N(s)$ has the property $(i)$ , $(\delta^{\mathcal{M}}\cup\varphi^{\mathcal{M}})\cap(S\backslash\varphi^{\mathcal{M}})\in N(s)$ , viz. $\delta^{\mathcal{M}}\cap(S\backslash\varphi^{\mathcal{M}})\in N(s)$ . Since $N(s)$ possesses the property $(s)$ and $\delta^{\mathcal{M}}\cap(S\backslash\varphi^{\mathcal{M}})\subseteq\delta^{\mathcal{M}}$ , it follows that $\delta^{\mathcal{M}}\in N(s)$ : a contradiction. This entails that $S\backslash\varphi^{\mathcal{M}}\notin N(s)$ , and thus $\varphi^{\mathcal{M}}\in N(s)$ . Note that $\varphi^{\mathcal{M}}\subseteq\varphi^{\mathcal{M}}\cup\psi^{\mathcal{M}}=(\varphi\vee\psi)^{\mathcal{M}}$ and $\varphi^{\mathcal{M}}\subseteq\varphi^{\mathcal{M}}\cup\chi^{\mathcal{M}}=(\varphi\vee\chi)^{\mathcal{M}}$ . Using $(s)$ again, we conclude that $(\varphi\vee\psi)^{\mathcal{M}}\in N(s)$ and $(\varphi\vee\chi)^{\mathcal{M}}\in N(s)$ , and therefore $\mathcal{M},s\vDash\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\chi)$ , as desired. ∎

Proposition 56.

Let $\Lambda$ be an extension of ${\bf K^{\nabla\bullet}}$ . Then for all $s,t\in S^{\Lambda}$ , $sR^{N}t$ iff $sR^{K}t$ .

Proof.

Suppose that $sR^{N}t$ , to show that $sR^{K}t$ . By supposition, for all $\varphi$ , if $\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\psi)\in s$ for all $\psi$ , then $\varphi\in t$ . Letting $\varphi=\bot$ , we can infer that $\Delta\psi\land\circ\psi\notin s$ for some $\psi$ . If $\Delta\psi\notin s$ , that is, $\nabla\psi\in s$ , then by axiom E3, we derive that $\bullet\psi\in s$ or $\bullet\neg\psi\in s$ . If $\circ\psi\notin s$ , then we have $\bullet\psi\in s$ . Either case implies that $\bullet\delta\in s$ for some $\delta$ . Now suppose for any $\varphi^{\prime}$ that $\Delta\varphi^{\prime}\land\circ(\neg\delta\to\varphi^{\prime})\in s$ . By Prop. 55, we infer that $\Delta(\varphi^{\prime}\vee\chi)\land\circ(\varphi^{\prime}\vee\chi)\in s$ for all $\chi$ . Then by supposition again, we conclude that $\varphi^{\prime}\in t$ . Therefore, $sR^{K}t$ .

Conversely, assume that $sR^{K}t$ , then there exists $\delta$ such that $(a)$ $\bullet\delta\in s$ , and $(b)$ for all $\varphi$ , if $\Delta\varphi\land\circ(\neg\delta\to\varphi)\in s$ , then $\varphi\in t$ . It remains to prove that $sR^{N}t$ . For this, suppose for any $\varphi$ that $\Delta(\varphi\vee\psi)\land\circ(\varphi\vee\psi)\in s$ for all $\psi$ . By letting $\psi=\bot$ , we obtain that $\Delta\varphi\in s$ ; by letting $\psi=\delta$ , we infer that $\circ(\neg\delta\to\varphi)\in s$ . Thus $\Delta\varphi\land\circ(\neg\delta\to\varphi)\in s$ . Then by $(b)$ , we conclude that $\varphi\in t$ , and therefore $sR^{N}t$ , as desired. ∎

6 Updating neighborhood models

In this section, we extend the previous results to the dynamic case of public announcements. Syntactically, we add the constructor $[\varphi]\varphi$ into the previous languages $\mathcal{L}(\nabla)$ , $\mathcal{L}(\bullet)$ and $\mathcal{L}(\nabla,\bullet)$ , and denote the obtained extensions by $\mathcal{L}(\nabla,[\cdot])$ , $\mathcal{L}(\bullet,[\cdot])$ , $\mathcal{L}(\nabla,\bullet,[\cdot])$ , respectively. $[\psi]\varphi$ is read “after every truthfully public announcement of $\psi$ , $\varphi$ holds”. Also, as usual, $\langle\psi\rangle\varphi$ abbreviates $\neg[\psi]\neg\varphi$ . Semantically, we adopt the intersection semantics in the literature (e.g. [37, 21, 22]). In detail, given a monotone neighborhood model $\mathcal{M}=\langle S,N,V\rangle$ and a state $s\in S$ ,

\begin{array}[]{lll}\mathcal{M},s\vDash[\psi]\varphi&\iff&\mathcal{M},s\vDash\psi\text{ implies }\mathcal{M}^{\cap\psi},s\vDash\varphi\\ \end{array}

where $\mathcal{M}^{\cap\psi}$ is the intersection submodel $\mathcal{M}^{\cap\psi^{\mathcal{M}}}$ , and the notion of intersection submodels is defined as below.

Definition 57.

[21, Def. 3] Let $\mathcal{M}=\langle S,N,V\rangle$ be a monotone model, and $X$ a nonempty subset of $S$ . Define the intersection model $\mathcal{M}^{\cap X}=\langle X,N^{\cap X},V^{X}\rangle$ induced from $X$ in the following.

•

for every $s\in X$ , $N^{\cap X}=\{Y\mid Y=P\cap X\text{ for some }P\in N(s)\}$ ,
•

$V^{X}(p)=V(p)\cap X$ .

Proposition 58.

[21, Prop. 2] The neighborhood property $(s)$ is preserved under taking the intersection submodel. That is, if $\mathcal{M}$ is a monotone neighborhood model with the domain $S$ , then for any nonempty subset $X$ of $S$ , the intersection submodel $\mathcal{M}^{\cap X}$ is also monotone.

The following lists the reduction axioms of $\mathcal{L}(\nabla,\bullet,[\cdot])$ and its sublanguages $\mathcal{L}(\nabla,[\cdot])$ and $\mathcal{L}(\bullet,[\cdot])$ under the intersection semantics.

\begin{array}[]{ll}\text{AP}&[\psi]p\leftrightarrow(\psi\to p)\\ \text{AN}&[\psi]\neg\varphi\leftrightarrow(\psi\to\neg[\psi]\varphi)\\ \text{AC}&[\psi](\varphi\land\chi)\leftrightarrow([\psi]\varphi\land[\psi]\chi)\\ \text{AA}&[\psi][\chi]\varphi\leftrightarrow[\psi\land[\psi]\chi]\varphi\\ \text{A}\nabla&[\psi]\nabla\varphi\leftrightarrow(\psi\to\nabla[\psi]\varphi\land\nabla[\psi]\neg\varphi)\\ \text{A}\bullet&{[\psi]\bullet}\varphi\leftrightarrow(\psi\to\bullet[\psi]\varphi)\\ \end{array}

The following reduction axioms are derivable from the above reduction axioms.

\begin{array}[]{ll}\text{A}\Delta&[\psi]\Delta\varphi\leftrightarrow(\psi\to\Delta[\psi]\varphi\vee\Delta[\psi]\neg\varphi)\\ \text{A}\circ&[\psi]{\circ\varphi}\leftrightarrow(\psi\to\circ[\psi]\varphi)\\ \end{array}

Theorem 59.

Let $\Lambda$ be a system of $\mathcal{L}(\nabla)$ (resp. $\mathcal{L}(\bullet)$ , $\mathcal{L}(\nabla,\bullet)$ ). If $\Lambda$ is sound and strongly complete with respect to the class of monotone neighborhood frames, then so is $\Lambda$ plus AP, AN, AC, AA and $\text{A}\nabla$ (resp. plus AP, AN, AC, AA and $\text{A}\bullet$ , plus AP, AN, AC, AA, $\text{A}\nabla$ and $\text{A}\bullet$ ) under intersection semantics.

Proof.

The validity of axioms AP, AN, AC, AA can be found in [21, Thm. 1], [22, Thm. 2, Thm. 3] and [37, Prop. 3.1]. The validity of $\text{A}\bullet$ has been shown in [11], where the axiom is named $\texttt{A}{\bullet\texttt{Int}}$ . The validity of $\text{A}\nabla$ is shown as follows. Let $\mathcal{M}=\langle S,N,V\rangle$ be an $(s)$ -model and $s\in S$ .

To begin with, suppose that $\mathcal{M},s\vDash[\psi]\nabla\varphi$ and $\mathcal{M},s\vDash\psi$ , to show that $\mathcal{M},s\vDash\nabla[\psi]\varphi\land\nabla[\psi]\neg\varphi$ . By supposition, we have $\mathcal{M}^{\cap\psi},s\vDash\nabla\varphi$ , which implies $\varphi^{\mathcal{M}^{\cap\psi}}\notin N^{\cap\psi}(s)$ and $\psi^{\mathcal{M}}\backslash\varphi^{\mathcal{M}^{\cap\psi}}\notin N^{\cap\psi}(s)$ .

We claim that $\mathcal{M},s\vDash\nabla[\psi]\varphi$ , that is, $([\psi]\varphi)^{\mathcal{M}}\notin N(s)$ and $S\backslash([\psi]\varphi)^{\mathcal{M}}\notin N(s)$ . If $([\psi]\varphi)^{\mathcal{M}}\in N(s)$ , then $([\psi]\varphi)^{\mathcal{M}}\cap\psi^{\mathcal{M}}\in N^{\cap\psi}(s)$ . As $([\psi]\varphi)^{\mathcal{M}}\cap\psi^{\mathcal{M}}\subseteq\varphi^{\mathcal{M}^{\cap\psi}}$ , by $(s)$ , we have $\varphi^{\mathcal{M}^{\cap\psi}}\in N^{\cap\psi}(s)$ : a contradiction. If $S\backslash([\psi]\varphi)^{\mathcal{M}}\in N(s)$ , then $(S\backslash([\psi]\varphi)^{\mathcal{M}})\cap\psi^{\mathcal{M}}\in N^{\cap\psi}(s)$ . Note that $(S\backslash([\psi]\varphi)^{\mathcal{M}})\cap\psi^{\mathcal{M}}\subseteq\psi^{\mathcal{M}}\backslash\varphi^{\mathcal{M}^{\cap\psi}}$ : for any $x\in(S\backslash([\psi]\varphi)^{\mathcal{M}})\cap\psi^{\mathcal{M}}$ , $x\notin([\psi]\varphi)^{\mathcal{M}}$ , thus $x\in\psi^{\mathcal{M}}$ and $x\notin\varphi^{\mathcal{M}^{\cap\psi}}$ , and hence $x\in\psi^{\mathcal{M}}\backslash\varphi^{\mathcal{M}^{\cap\psi}}$ . By $(s)$ again, $\psi^{\mathcal{M}}\backslash\varphi^{\mathcal{M}^{\cap\psi}}\in N^{\cap\psi}(s)$ : a contradiction again.

We also claim that $\mathcal{M},s\vDash\nabla[\psi]\neg\varphi$ , that is, $([\psi]\neg\varphi)^{\mathcal{M}}\notin N(s)$ and $S\backslash([\psi]\neg\varphi)^{\mathcal{M}}\notin N(s)$ . If $([\psi]\neg\varphi)^{\mathcal{M}}\in N(s)$ , then $([\psi]\neg\varphi)^{\mathcal{M}}\cap\psi^{\mathcal{M}}\in N^{\cap\psi}(s)$ . As $([\psi]\neg\varphi)^{\mathcal{M}}\cap\psi^{\mathcal{M}}\subseteq\psi^{\mathcal{M}}\backslash\varphi^{\mathcal{M}^{\cap\psi}}$ , we infer by $(s)$ that $\psi^{\mathcal{M}}\backslash\varphi^{\mathcal{M}^{\cap\psi}}\in N^{\cap\psi}(s)$ : a contradiction. If $S\backslash([\psi]\neg\varphi)^{\mathcal{M}}\in N(s)$ , then $(S\backslash([\psi]\neg\varphi)^{\mathcal{M}})\cap\psi^{\mathcal{M}}\in N^{\cap\psi}(s)$ . Since $(S\backslash([\psi]\neg\varphi)^{\mathcal{M}})\cap\psi^{\mathcal{M}}\subseteq\varphi^{\mathcal{M}^{\cap\psi}}$ , by $(s)$ again, we derive that $\varphi^{\mathcal{M}^{\cap\psi}}\in N^{\cap\psi}(s)$ : a contradiction again.

Conversely, assume that $\mathcal{M},s\vDash\psi\to\nabla[\psi]\varphi\land\nabla[\psi]\neg\varphi$ , to prove that $\mathcal{M},s\vDash[\psi]\nabla\varphi$ . For this, we suppose that $\mathcal{M},s\vDash\psi$ , it remains only to show that $\mathcal{M}^{\cap\psi},s\vDash\nabla\varphi$ , that is, $\varphi^{\mathcal{M}^{\cap\psi}}\notin N^{\cap\psi}(s)$ and $\psi^{\mathcal{M}}\backslash\varphi^{\mathcal{M}^{\cap\psi}}\notin N^{\cap\psi}(s)$ . By assumption and supposition, we obtain $\mathcal{M},s\vDash\nabla[\psi]\varphi\land\nabla[\psi]\neg\varphi$ . This follows that $([\psi]\varphi)^{\mathcal{M}}\notin N(s)$ and $S\backslash([\psi]\varphi)^{\mathcal{M}}\notin N(s)$ , and $([\psi]\neg\varphi)^{\mathcal{M}}\notin N(s)$ and $S\backslash([\psi]\neg\varphi)^{\mathcal{M}}\notin N(s)$ .

We claim that $\varphi^{\mathcal{M}^{\cap\psi}}\notin N^{\cap\psi}(s)$ . Otherwise, that is, $\varphi^{\mathcal{M}^{\cap\psi}}\in N^{\cap\psi}(s)$ , we have $\varphi^{\mathcal{M}^{\cap\psi}}=P\cap\psi^{\mathcal{M}}$ for some $P\in N(s)$ . This implies that $P\subseteq([\psi]\varphi)^{\mathcal{M}}$ : for any $x\in P$ , we would have $x\in([\psi]\varphi)^{\mathcal{M}}$ , since if $x\in\psi^{\mathcal{M}}$ , then $x\in P\cap\psi^{\mathcal{M}}=\varphi^{\mathcal{M}^{\cap\psi}}$ . By $(s)$ , $([\psi]\varphi)^{\mathcal{M}}\in N(s)$ : a contradiction.

We also claim that $\psi^{\mathcal{M}}\backslash\varphi^{\mathcal{M}^{\cap\psi}}\notin N^{\cap\psi}(s)$ . Otherwise, that is, $\psi^{\mathcal{M}}\backslash\varphi^{\mathcal{M}^{\cap\psi}}\in N^{\cap\psi}(s)$ , we infer that $\psi^{\mathcal{M}}\backslash\varphi^{\mathcal{M}^{\cap\psi}}=P\cap\psi^{\mathcal{M}}$ for some $P\in N(s)$ . It then follows that $P\subseteq([\psi]\neg\varphi)^{\mathcal{M}}$ : for any $x\in P$ , we have $x\in([\psi]\neg\varphi)^{\mathcal{M}}$ , since if $x\in\psi^{\mathcal{M}}$ , then $x=P\cap\psi^{\mathcal{M}}=\psi^{\mathcal{M}}\backslash\varphi^{\mathcal{M}^{\cap\psi}}$ , and so $x\in(\neg\varphi)^{\mathcal{M}^{\cap\psi}}$ . By $(s)$ again, $([\psi]\neg\varphi)^{\mathcal{M}}\in N(s)$ : a contradiction, as desired. ∎

For the sake of reference, we use ${\bf M^{\nabla\bullet[\cdot]}}$ to denote the extension of ${\bf M^{\nabla\bullet}}$ with all the above reduction axioms. By dropping $\text{A}\bullet$ from ${\bf M^{\nabla\bullet[\cdot]}}$ , we obtain the system ${\bf M^{\nabla[\cdot]}}$ ; by dropping $\text{A}\nabla$ from ${\bf M^{\nabla\bullet[\cdot]}}$ , we obtain the system ${\bf M^{\bullet[\cdot]}}$ .

In what follows, we will focus on some successful formulas in our languages. A formula is said to be successful, if it still holds after being announced; in symbols, $\vDash[\varphi]\varphi$ . Recall that $\neg{\bullet p}$ is shown to be successful under the relational semantics in [7, Prop. 39] and under the intersection semantics in [11, Prop. 6.5]. We will follow this line of research and say much more. As we shall show, any combination of $p$ , $\neg p$ , ${\neg\bullet}p$ , and $\neg\nabla p$ via conjunction (or, via disjunction) is successful under the intersection semantics.¹¹¹¹11It is shown in [7, Prop. 38] that under Kripke semantics, $\bullet p$ is self-refuting and $\neg{\bullet p}$ is successful.

To begin with, we show that, provably, any combination of $p$ , $\neg p$ , ${\neg\bullet}p$ , and $\neg\nabla p$ via conjunction is successful under the intersection semantics.

Proposition 60.

$p$ is successful under the intersection semantics. That is, $[p]p$ is provable in ${\bf M^{\bullet[\cdot]}}$ .

Proof.

Straightforward by AP. ∎

Proposition 61.

$\neg p$ is successful under the intersection semantics. That is, $[\neg p]\neg p$ is provable in ${\bf M^{\bullet[\cdot]}}$ .

Proof.

Straightforward by AN and AP. ∎

Proposition 62.

${\neg\bullet}p$ is successful under the intersection semantics. That is, $[{\neg\bullet}p]{\neg\bullet}p$ is provable in ${\bf M^{\bullet[\cdot]}}$ .

Proof.

Refer to [11, Prop. 6.5]. ∎

Proposition 63.

$\neg\nabla p$ is successful under the intersection semantics. That is, $[\neg\nabla p]\neg\nabla p$ is provable in ${\bf M^{\nabla[\cdot]}}$ .

Proof.

We have the following proof sequence in ${\bf M^{\nabla[\cdot]}}$ .

\begin{array}[]{lll}&[\neg\nabla p]\neg\nabla p\\ \leftrightarrow&(\neg\nabla p\to\neg[\neg\nabla p]\nabla p)&\text{AN}\\ \leftrightarrow&(\neg\nabla p\to\neg(\neg\nabla p\to\nabla[\neg\nabla p]p\land\nabla[\neg\nabla p]\neg p)&\text{A}\nabla\\ \leftrightarrow&(\neg\nabla p\to\neg(\nabla(\neg\nabla p\to p)\land\nabla(\neg\nabla p\to\neg(\neg\nabla p\to p))))&\text{AP},\text{AN}\\ \leftrightarrow&(\nabla(\neg\nabla p\to p)\land\nabla(\neg\nabla p\to\neg p)\to\nabla p)&\text{TAUT},\text{RE}\nabla\\ \leftrightarrow&(\nabla(p\vee\nabla p)\land\nabla(\neg p\vee\nabla p)\to\nabla p)&\text{TAUT},\text{RE}\nabla\\ \leftrightarrow&\top&\text{M1}\\ \end{array}

Therefore, $[\neg\nabla p]\neg\nabla p$ is provable in ${\bf M^{\nabla[\cdot]}}$ . ∎

Intuitively, $[\neg\nabla p]\neg\nabla p$ means that “after being told that one is not ignorant whether $p$ , one is still not ignorant whether $p$ .” In other words, one’s non-ignorance about a fact cannot be altered by being announced.

Proposition 64.

$p\land\neg p$ is successful under the intersection semantics.

Proof.

Note that $p\land\neg p$ is equivalent to $\bot$ , and $\bot$ is successful. ∎

Proposition 65.

$p\land{\neg\bullet}p$ is successful under the intersection semantics. That is, $[p\land{\neg\bullet}p](p\land{\neg\bullet}p)$ is provable in ${\bf M^{\bullet[\cdot]}}$ .

Proof.

We have the following proof sequence in ${\bf M^{\bullet[\cdot]}}$ .

\begin{array}[]{lll}&[p\land{\neg\bullet}p](p\land{\neg\bullet}p)&\\ \leftrightarrow&([p\land{\neg\bullet}p]p\land[p\land{\neg\bullet}p]{\neg\bullet}p)&\text{AC}\\ \leftrightarrow&(p\land{\neg\bullet}p\to p)\land(p\land{\neg\bullet}p\to\neg[p\land{\neg\bullet}p]{\bullet}p)&\text{AP},\text{AN}\\ \leftrightarrow&(p\land{\neg\bullet}p\to\neg(p\land{\neg\bullet}p\to{\bullet[p\land{\neg\bullet}p]}p)&\text{A}\bullet\\ \leftrightarrow&(p\land{\neg\bullet}p\to{\neg\bullet}[p\land{\neg\bullet}p]p)&\text{TAUT}\\ \leftrightarrow&(p\land{\neg\bullet}p\to{\neg\bullet}(p\land{\neg\bullet}p\to p))&\text{AP}\\ \leftrightarrow&(p\land{\neg\bullet}p\to{\neg\bullet}\top)&\text{TAUT},\text{RE}\bullet\\ \leftrightarrow&(p\land\circ p\to\circ\top)&\text{Def.\leavevmode\nobreak\ }\circ\\ \leftrightarrow&\top&\text{Prop.\leavevmode\nobreak\ }\ref{prop.impliescirctop}\\ \end{array}

Therefore, $[p\land{\neg\bullet}p](p\land{\neg\bullet}p)$ is provable in ${\bf M^{\bullet[\cdot]}}$ . ∎

Proposition 66.

$p\land\neg\nabla p$ is successful under the intersection semantics. That is, $[p\land\neg\nabla p](p\land\neg\nabla p)$ is provable in ${\bf M^{\nabla[\cdot]}}$ .

Proof.

We have the following proof sequence in ${\bf M^{\nabla[\cdot]}}$ .

\begin{array}[]{lll}&[p\land\neg\nabla p](p\land\neg\nabla p)&\\ \leftrightarrow&([p\land\neg\nabla p]p\land[p\land\neg\nabla p]\neg\nabla p)&\text{AC}\\ \leftrightarrow&(p\land\neg\nabla p\to p)\land(p\land\neg\nabla p\to\neg[p\land\neg\nabla p]\nabla p)&\text{AP},\text{AN}\\ \leftrightarrow&(p\land\neg\nabla p\to\neg(p\land\neg\nabla p\to\nabla[p\land\neg\nabla p]p\land\nabla[p\land\neg\nabla p]\neg p))&\text{A}\nabla\\ \leftrightarrow&(p\land\neg\nabla p\to\neg(\nabla[p\land\neg\nabla p]p\land\nabla[p\land\neg\nabla p]\neg p))&\text{TAUT}\\ \leftrightarrow&(p\land\neg\nabla p\to\neg(\nabla\top\land\nabla[p\land\neg\nabla p]\neg p))&\text{AP},\text{RE}\nabla\\ \leftrightarrow&(p\land\neg\nabla p\to\neg\nabla\top\vee\neg\nabla(p\land\neg\nabla p\to\neg p))&\text{AN},\text{AP},\text{RE}\nabla\\ \leftrightarrow&(p\land\Delta p\to\Delta\top\vee\Delta(p\land\Delta p\to\neg p))&\text{Def.\leavevmode\nobreak\ }\Delta\\ \end{array}

By Prop. 39, $\Delta p\to\Delta\top$ is provable in ${\bf M^{\nabla}}$ , so is the last formula in the above proof sequence, and thus $[p\land\neg\nabla p](p\land\neg\nabla p)$ is provable in ${\bf M^{\nabla[\cdot]}}$ . ∎

Proposition 67.

$\neg p\land{\neg\bullet}p$ is successful under the intersection semantics.

Proof.

By E2, $\neg p\land{\neg\bullet}p$ is equivalent to $\neg p$ . And we have already known from Prop. 61 that $\neg p$ is successful. ∎

Proposition 68.

$\neg p\land\neg\nabla p$ is successful under the intersection semantics. That is, $[\neg p\land\neg\nabla p](\neg p\land\neg\nabla p)$ is provable in ${\bf M^{\nabla[\cdot]}}$ .

Proof.

We have the following proof sequence in ${\bf M^{\nabla[\cdot]}}$ .

\begin{array}[]{lll}&[\neg p\land\neg\nabla p](\neg p\land\neg\nabla p)&\\ \leftrightarrow&[\neg p\land\neg\nabla p]\neg p\land[\neg p\land\neg\nabla p]\neg\nabla p&\text{AC}\\ \leftrightarrow&(\neg p\land\neg\nabla p\to\neg p)\land(\neg p\land\neg\nabla p\to\neg[\neg p\land\neg\nabla p]\nabla p)&\text{AN},\text{AP}\\ \leftrightarrow&(\neg p\land\neg\nabla p\to\neg[\neg p\land\neg\nabla p]\nabla p)&\text{TAUT}\\ \leftrightarrow&(\neg p\land\neg\nabla p\to\neg(\nabla[\neg p\land\neg\nabla p]p\land\nabla[\neg p\land\neg\nabla p]\neg p)&\text{A}\nabla\\ \leftrightarrow&(\neg p\land\Delta p\to\Delta(\neg p\land\Delta p\to p)\vee\Delta\top)&\text{AP},\text{AN},\text{RE}\nabla\\ \leftrightarrow&\top&\text{Prop.\leavevmode\nobreak\ }\ref{prop.impliesdeltatop}\\ \end{array}

Therefore, $[\neg p\land\neg\nabla p](\neg p\land\neg\nabla p)$ is provable in ${\bf M^{\nabla[\cdot]}}$ . ∎

We have seen that both ${\neg\bullet}p$ and $\neg\nabla p$ are successful. One natural question would be whether their conjunction, viz. ${\neg\bullet}p\land\neg\nabla p$ , is successful. Note that this does not obviously hold, since for instance, both $p$ and $\neg Kp$ are successful, whereas $p\land\neg Kp$ is not, see e.g. [36, Example 4.34].

Proposition 69.

${\neg\bullet}p\land\neg\nabla p$ is successful under the intersection semantics. That is, $[{\neg\bullet}p\land\neg\nabla p]({\neg\bullet}p\land\neg\nabla p)$ is provable in ${\bf M^{\nabla\bullet[\cdot]}}$ .

Proof.

By AC, $[{\neg\bullet}p\land\neg\nabla p]({\neg\bullet}p\land\neg\nabla p)\leftrightarrow([{\neg\bullet}p\land\neg\nabla p]{\neg\bullet}p\land[{\neg\bullet}p\land\neg\nabla p]\neg\nabla p)$ . We show that both $[{\neg\bullet}p\land\neg\nabla p]{\neg\bullet}p$ and $[{\neg\bullet}p\land\neg\nabla p]\neg\nabla p$ are provable in ${\bf M^{\nabla\bullet[\cdot]}}$ .

We have the following proof sequence in ${\bf M^{\nabla\bullet[\cdot]}}$ .

\begin{array}[]{lll}&[{\neg\bullet}p\land\neg\nabla p]{\neg\bullet}p&\\ \leftrightarrow&({\neg\bullet}p\land\neg\nabla p\to\neg[{\neg\bullet}p\land\neg\nabla p]{\bullet p})&\text{AN}\\ \leftrightarrow&({\neg\bullet}p\land\neg\nabla p\to\neg({\neg\bullet}p\land\neg\nabla p\to\bullet[{\neg\bullet}p\land\neg\nabla p]p)&\text{A}\bullet\\ \leftrightarrow&({\neg\bullet}p\land\neg\nabla p\to{\neg\bullet}[{\neg\bullet}p\land\neg\nabla p]p)&\text{TAUT}\\ \leftrightarrow&({\neg\bullet}p\land\neg\nabla p\to{\neg\bullet}({\neg\bullet}p\land\neg\nabla p\to p))&\text{AP},\text{RE}\bullet\\ \leftrightarrow&(\bullet({\neg\bullet}p\land\neg\nabla p\to p)\to(\bullet p\vee\nabla p))&\text{TAUT}\\ \leftrightarrow&(\bullet({\bullet}p\lor\nabla p\vee p)\to(\bullet p\vee\nabla p))&\text{TAUT},\text{RE}\bullet\\ \leftrightarrow&(\bullet(p\vee\nabla p)\to(\bullet p\vee\nabla p))&\text{E2},\text{RE}\bullet\\ \leftrightarrow&\top&\text{Prop.\leavevmode\nobreak\ }\ref{prop.useful}\\ \end{array}

\begin{array}[]{lll}&[{\neg\bullet}p\land\neg\nabla p]\neg\nabla p&\\ \leftrightarrow&({\neg\bullet}p\land\neg\nabla p\to\neg[{\neg\bullet}p\land\neg\nabla p]{\nabla p})&\text{AN}\\ \leftrightarrow&({\neg\bullet}p\land\neg\nabla p\to\neg({\neg\bullet}p\land\neg\nabla p\to\nabla[{\neg\bullet}p\land\neg\nabla p]p\land&\\ &\nabla[{\neg\bullet}p\land\neg\nabla p]\neg p))&\text{A}\nabla\\ \leftrightarrow&({\neg\bullet}p\land\neg\nabla p\to\neg(\nabla[{\neg\bullet}p\land\neg\nabla p]p\land\nabla[{\neg\bullet}p\land\neg\nabla p]\neg p))&\text{TAUT}\\ \leftrightarrow&({\neg\bullet}p\land\neg\nabla p\to\neg(\nabla({\neg\bullet}p\land\neg\nabla p\to p)\land\nabla({\neg\bullet}p\land\neg\nabla p\to\neg p))&\text{AP},\text{AN}\\ \leftrightarrow&(\nabla(p\vee\bullet p\vee\nabla p)\land\nabla(\neg p\vee\bullet p\lor\nabla p)\to(\nabla p\vee\bullet p))&\text{TAUT},\text{RE}\bullet\\ \leftrightarrow&\top&\text{M1}\\ \end{array}

Thus both $[{\neg\bullet}p\land\neg\nabla p]{\neg\bullet}p$ and $[{\neg\bullet}p\land\neg\nabla p]\neg\nabla p$ are provable in ${\bf M^{\nabla\bullet[\cdot]}}$ . Therefore, $[{\neg\bullet}p\land\neg\nabla p]({\neg\bullet}p\land\neg\nabla p)$ is provable in ${\bf M^{\nabla\bullet[\cdot]}}$ . ∎

Intuitively, $[{\neg\bullet}p\land\neg\nabla p]({\neg\bullet}p\land\neg\nabla p)$ says that after being told that one is neither ignorant whether nor ignorant of $p$ , one is still neither ignorant whether nor ignorant of $p$ . In short, one’s non-ignorance whether and non-ignorance of a fact cannot be altered by being announced.

The following two propositions can be shown as in Prop. 64.

Proposition 70.

$p\land\neg p\land{\neg\bullet}p$ is successful under the intersection semantics.

Proposition 71.

$p\land\neg p\land\neg\nabla p$ is successful under the intersection semantics.

Proposition 72.

$p\land{\neg\bullet}p\land\neg\nabla p$ is successful under the intersection semantics. That is, $[p\land{\neg\bullet}p\land\neg\nabla p](p\land{\neg\bullet}p\land\neg\nabla p)$ is provable in ${\bf M^{\nabla\bullet[\cdot]}}$ .

Proof.

By AC, $[p\land{\neg\bullet}p\land\neg\nabla p](p\land{\neg\bullet}p\land\neg\nabla p)\leftrightarrow([p\land{\neg\bullet}p\land\neg\nabla p]p\land[p\land{\neg\bullet}p\land\neg\nabla p]{\neg\bullet}p\land[p\land{\neg\bullet}p\land\neg\nabla p]\neg\nabla p)$ . One may easily verify that $[p\land{\neg\bullet}p\land\neg\nabla p]p$ is provable in ${\bf M^{\nabla\bullet[\cdot]}}$ . It remains only to show that both $[p\land{\neg\bullet}p\land\neg\nabla p]{\neg\bullet}p$ and $[p\land{\neg\bullet}p\land\neg\nabla p]\neg\nabla p$ are provable in the system in question.

We have the following proof sequence in ${\bf M^{\nabla\bullet[\cdot]}}$ .

\begin{array}[]{lll}&[p\land{\neg\bullet}p\land\neg\nabla p]{\neg\bullet}p&\\ \leftrightarrow&(p\land{\neg\bullet}p\land\neg\nabla p\to\neg[p\land{\neg\bullet}p\land\neg\nabla p]{\bullet p})&\text{AN}\\ \leftrightarrow&(p\land{\neg\bullet}p\land\neg\nabla p\to{\neg\bullet}[p\land{\neg\bullet}p\land\neg\nabla p]p)&\text{A}\bullet\\ \leftrightarrow&(p\land{\neg\bullet}p\land\neg\nabla p\to{\neg\bullet}\top)&\text{AP},\text{RE}\bullet\\ \leftrightarrow&(p\land\circ p\land\Delta p\to\circ\top)&\text{Def.\leavevmode\nobreak\ }\circ,\text{Def.\leavevmode\nobreak\ }\Delta\\ \end{array}

By Prop. 40, $p\land\circ p\to\circ\top$ is provable in ${\bf M^{\bullet}}$ , so is the last formula in the above proof sequence, and thus $[p\land{\neg\bullet}p\land\neg\nabla p]{\neg\bullet}p$ is provable in ${\bf M^{\nabla{\bullet[\cdot]}}}$ .

Also, we have the following proof sequence in ${\bf M^{\nabla{\bullet[\cdot]}}}$ .

\begin{array}[]{lll}&[p\land{\neg\bullet}p\land\neg\nabla p]\neg\nabla p&\\ \leftrightarrow&(p\land{\neg\bullet}p\land\neg\nabla p\to\neg[p\land{\neg\bullet}p\land\neg\nabla p]\nabla p)&\text{AN}\\ \leftrightarrow&(p\land{\neg\bullet}p\land\neg\nabla p\to\neg(\nabla[p\land{\neg\bullet}p\land\neg\nabla p]p\land\nabla[p\land{\neg\bullet}p\land\neg\nabla p]\neg p))&\text{A}\nabla\\ \leftrightarrow&(p\land{\circ}p\land\Delta p\to\Delta[p\land{\neg\bullet}p\land\neg\nabla p]p\lor\Delta[p\land{\neg\bullet}p\land\neg\nabla p]\neg p)&\text{Def.\leavevmode\nobreak\ }\circ,\text{Def.\leavevmode\nobreak\ }\Delta\\ \leftrightarrow&(p\land{\circ}p\land\Delta p\to\Delta\top\lor\Delta[p\land{\neg\bullet}p\land\neg\nabla p]\neg p)&\text{AP},\text{RE}\nabla\\ \end{array}

By Prop. 39, $\Delta p\to\Delta\top$ is provable in ${\bf M^{\nabla}}$ , thus the last formula in the above proof sequence is provable in ${\bf M^{\nabla\bullet}}$ . Therefore, $[p\land{\neg\bullet}p\land\neg\nabla p]\neg\nabla p$ is provable in ${\bf M^{\nabla\bullet[\cdot]}}$ .

According to the previous analysis, $[p\land{\neg\bullet}p\land\neg\nabla p](p\land{\neg\bullet}p\land\neg\nabla p)$ is provable in ${\bf M^{\nabla\bullet[\cdot]}}$ . ∎

Proposition 73.

$\neg p\land{\neg\bullet}p\land\neg\nabla p$ is successful under the intersection semantics. That is, $[\neg p\land{\neg\bullet}p\land\neg\nabla p](\neg p\land{\neg\bullet}p\land\neg\nabla p)$ is provable in ${\bf M^{\nabla\bullet[\cdot]}}$ .

Proof.

By the proof of Prop. 67, $\neg p\land{\neg\bullet}p\land\neg\nabla p$ is equivalent to $\neg p\land\neg\nabla p$ . And we have already shown in Prop. 68 that $\neg p\land\neg\nabla p$ is successful under the intersection semantics. ∎

Proposition 74.

$p\land\neg p\land{\neg\bullet}p\land\neg\nabla p$ is successful under the intersection semantics.

Proof.

The proof is similar to that of Prop. 64. ∎

Now we demonstrate that any combination of $p$ , $\neg p$ , ${\neg\bullet}p$ , and $\neg\nabla p$ via disjunction is successful under the intersection semantics. First, one may show that $[\psi](\varphi\vee\chi)\leftrightarrow([\psi]\varphi\vee[\psi]\chi)$ is provable from the above reduction axioms. For the sake of reference, we denote it AD.

Proposition 75.

$p\vee\neg p$ is successful under the intersection semantics.

Proof.

Note that $p\vee\neg p$ is equivalent to $\top$ , and $\top$ is successful. ∎

Proposition 76.

$p\vee{\neg\bullet}p$ is successful under the intersection semantics. That is, $[p\vee{\neg\bullet}p](p\vee{\neg\bullet}p)$ is provable in ${\bf M^{\bullet[\cdot]}}$ .

Proof.

Just note that $p\vee{\neg\bullet}p$ is equivalent to $\bullet p\to p$ , which by E2 is equivalent to $\top$ . And $\top$ is successful. ∎

Proposition 77.

$p\vee\neg\nabla p$ is successful under the intersection semantics. That is, $[p\vee\neg\nabla p](p\vee\neg\nabla p)$ is provable in ${\bf M^{\nabla[\cdot]}}$ .

Proof.

We have the following proof sequence in ${\bf M^{\nabla[\cdot]}}$ .

\begin{array}[]{lll}&[p\vee\neg\nabla p](p\vee\neg\nabla p)&\\ \leftrightarrow&([p\vee\neg\nabla p]p\vee[p\vee\neg\nabla p]\neg\nabla p)&\text{AD}\\ \leftrightarrow&(p\vee\neg\nabla p\to p)\vee(p\vee\neg\nabla p\to\neg[p\vee\neg\nabla p]\nabla p)&\text{AP},\text{AN}\\ \leftrightarrow&(p\vee\neg\nabla p\to p)\vee(p\vee\neg\nabla p\to\neg(\nabla[p\vee\neg\nabla p]p\land\nabla[p\vee\neg\nabla p]\neg p))&\text{A}\nabla\\ \leftrightarrow&(p\vee\neg\nabla p\to p)\vee(p\vee\neg\nabla p\to\neg(\nabla(p\vee\neg\nabla p\to p)\land\nabla(p\vee\neg\nabla p\to\neg p)))&\text{AP},\text{AN}\\ \leftrightarrow&(p\vee\neg\nabla p\to p\vee\neg(\nabla(p\vee\neg\nabla p\to p)\land\nabla(p\vee\neg\nabla p\to\neg p)))&\text{TAUT}\\ \leftrightarrow&(\neg p\land\nabla(p\vee\neg(p\vee\neg\nabla p))\land\nabla(\neg p\vee\neg(p\vee\neg\nabla p))\to\neg p\land\nabla p)&\text{TAUT},\text{RE}\nabla\\ \leftrightarrow&\top&\text{M1}\\ \end{array}

Therefore, $[p\vee\neg\nabla p](p\vee\neg\nabla p)$ is provable in ${\bf M^{\nabla[\cdot]}}$ . ∎

Proposition 78.

$\neg p\vee{\neg\bullet}p$ is successful under the intersection semantics.

Proof.

By E2, $\neg p\vee{\neg\bullet}p$ is equivalent to ${\neg\bullet}p$ , and Prop. 62 has shown that ${\neg\bullet}p$ is successful under the intersection semantics. ∎

Proposition 79.

$\neg p\vee\neg\nabla p$ is successful under the intersection semantics. That is, $[\neg p\vee\neg\nabla p](\neg p\vee\neg\nabla p)$ is provable in ${\bf M^{\nabla[\cdot]}}$ .

Proof.

We have the following proof sequence in ${\bf M^{\nabla[\cdot]}}$ .

\begin{array}[]{lll}&[\neg p\vee\neg\nabla p](\neg p\vee\neg\nabla p)&\\ \leftrightarrow&[\neg p\vee\neg\nabla p]\neg p\vee[\neg p\vee\neg\nabla p]\neg\nabla p&\text{AD}\\ \leftrightarrow&(\neg p\vee\neg\nabla p\to\neg p)\vee(\neg p\vee\neg\nabla p\to\neg[\neg p\vee\neg\nabla p]\nabla p)&\text{AN},\text{AP}\\ \leftrightarrow&(\neg p\vee\neg\nabla p\to\neg p)\vee(\neg p\vee\neg\nabla p\to\neg(\nabla[\neg p\vee\neg\nabla p]p\land\nabla[\neg p\vee\neg\nabla p]\neg p))&\text{A}\nabla\\ \leftrightarrow&(\neg p\vee\neg\nabla p\to\neg p)\vee(\neg p\vee\neg\nabla p\to\neg(\nabla(\neg p\vee\neg\nabla p\to p)\land\nabla(\neg p\vee\neg\nabla p\to\neg p)))&\text{AP},\text{AN}\\ \leftrightarrow&(\neg p\vee\neg\nabla p\to\neg p\vee\neg(\nabla(\neg p\vee\neg\nabla p\to p)\land\nabla(\neg p\vee\neg\nabla p\to\neg p)))&\text{TAUT}\\ \leftrightarrow&p\land\nabla(p\vee\neg(\neg p\vee\neg\nabla p))\land\nabla(\neg p\vee\neg(\neg p\vee\neg\nabla p))\to p\land\nabla p&\text{TAUT}\\ \leftrightarrow&\top&\text{M1}\\ \end{array}

Therefore, $[\neg p\vee\neg\nabla p](\neg p\vee\neg\nabla p)$ is provable in ${\bf M^{\nabla[\cdot]}}$ . ∎

Proposition 80.

${\neg\bullet}p\vee\neg\nabla p$ is successful under the intersection semantics. That is, $[{\neg\bullet}p\vee\neg\nabla p]({\neg\bullet}p\vee\neg\nabla p)$ is provable in ${\bf M^{\nabla\bullet[\cdot]}}$ .

Proof.

We have the following proof sequence in ${\bf M^{\nabla\bullet[\cdot]}}$ .

\begin{array}[]{lll}&[{\neg\bullet}p\vee\neg\nabla p]({\neg\bullet}p\vee\neg\nabla p)&\\ \leftrightarrow&([{\neg\bullet}p\vee\neg\nabla p]{\neg\bullet}p\vee[{\neg\bullet}p\vee\neg\nabla p]{\neg\nabla}p)&\text{AD}\\ \leftrightarrow&({\neg\bullet}p\vee\neg\nabla p\to\neg[{\neg\bullet}p\vee\neg\nabla p]{\bullet}p)\vee&\\ &({\neg\bullet}p\vee\neg\nabla p\to\neg[{\neg\bullet}p\vee\neg\nabla p]{\nabla}p)&\text{AN}\\ \leftrightarrow&({\neg\bullet}p\vee\neg\nabla p\to{\neg\bullet}[{\neg\bullet}p\vee\neg\nabla p]p)\vee&\\ &({\neg\bullet}p\vee\neg\nabla p\to\neg({\nabla}[{\neg\bullet}p\vee\neg\nabla p]p\land&\\ &{\nabla}[{\neg\bullet}p\vee\neg\nabla p]\neg p))&\text{A}\bullet,\text{A}\nabla\\ \leftrightarrow&({\neg\bullet}p\vee\neg\nabla p\to{\neg\bullet}({\neg\bullet}p\vee\neg\nabla p\to p))\vee\\ &({\neg\bullet}p\vee\neg\nabla p\to\neg({\nabla}({\neg\bullet}p\vee\neg\nabla p\to p)\land&\\ &{\nabla}({\neg\bullet}p\vee\neg\nabla p\to\neg p)))&\text{AP},\text{AN}\\ \leftrightarrow&({\neg\bullet}p\vee\neg\nabla p\to{\neg\bullet}({\neg\bullet}p\vee\neg\nabla p\to p)\vee&\\ &\neg({\nabla}({\neg\bullet}p\vee\neg\nabla p\to p)\land{\nabla}({\neg\bullet}p\vee\neg\nabla p\to\neg p)))&\text{TAUT}\\ \leftrightarrow&{\bullet}({\neg\bullet}p\vee\neg\nabla p\to p)\land{\nabla}({\neg\bullet}p\vee\neg\nabla p\to p)\land&\\ &{\nabla}({\neg\bullet}p\vee\neg\nabla p\to\neg p)\to\bullet p\land\nabla p&\text{TAUT}\\ \leftrightarrow&\bullet(\circ p\vee\Delta p\to p)\land\nabla(\circ p\vee\Delta p\to p)\land&\\ &\nabla(\circ p\vee\Delta p\to\neg p)\to\bullet p\land\nabla p&\text{Def.\leavevmode\nobreak\ }\circ,\text{Def.\leavevmode\nobreak\ }\Delta\\ \end{array}

By Prop. 50, $\bullet(\circ p\vee\Delta p\to p)\to\bullet p$ is provable in ${\bf M^{\nabla\bullet}}$ ; by axiom M1 and $\text{RE}\nabla$ , we can show the provability of $\nabla(\circ p\vee\Delta p\to p)\land\nabla(\circ p\vee\Delta p\to\neg p)\to\nabla p$ in ${\bf M^{\nabla\bullet}}$ . Therefore, $[{\neg\bullet}p\vee\neg\nabla p]({\neg\bullet}p\vee\neg\nabla p)$ is provable in ${\bf M^{\nabla\bullet[\cdot]}}$ . ∎

Intuitively, $[{\neg\bullet}p\vee\neg\nabla p]({\neg\bullet}p\vee\neg\nabla p)$ says that one’s either non-ignorance of or non-ignorance whether a fact cannot be altered by being announced: after being told that one is either not ignorant of or not ignorant whether $p$ , one is still either not ignorant of or not ignorant whether $p$ .

Next two propositions are shown as in Prop. 75.

Proposition 81.

$p\vee\neg p\vee{\neg\bullet}p$ is successful under the intersection semantics.

Proposition 82.

$p\vee\neg p\vee{\neg\nabla}p$ is successful under the intersection semantics.

Proposition 83.

$p\vee{\neg\bullet}p\vee\neg\nabla p$ is successful under the intersection semantics. That is, $[p\vee{\neg\bullet}p\vee\neg\nabla p](p\vee{\neg\bullet}p\vee\neg\nabla p)$ is provable in ${\bf M^{\nabla\bullet[\cdot]}}$ .

Proof.

By the proof of Prop. 76, $p\vee{\neg\bullet}p$ is equivalent to $\top$ , so is $p\vee{\neg\bullet}p\vee\neg\nabla p$ . And $\top$ is successful. ∎

Proposition 84.

$\neg p\vee{\neg\bullet}p\vee\neg\nabla p$ is successful under the intersection semantics.

Proof.

By E2, $\neg p\vee{\neg\bullet}p\vee\neg\nabla p$ is equivalent to ${\neg\bullet}p\vee\neg\nabla p$ , and we have shown in Prop. 80 that ${\neg\bullet}p\vee\neg\nabla p$ is successful under the intersection semantics. ∎

Proposition 85.

$p\vee\neg p\vee{\neg\bullet}p\vee\neg\nabla p$ is successful under the intersection semantics.

Proof.

The proof is similar to that of Prop. 75. ∎

7 Conclusion and Future work

In this paper, we investigated the bimodal logic of Fitchean ignorance and first-order ignorance $\mathcal{L}(\nabla,\bullet)$ under the neighborhood semantics. We compared the relative expressivity between $\mathcal{L}(\nabla,\bullet)$ and the logic of (first-order) ignorance $\mathcal{L}(\nabla)$ and the logic of Fitchean ignorance $\mathcal{L}(\bullet)$ , and between $\mathcal{L}(\nabla,\bullet)$ and standard epistemic logic $\mathcal{L}(\Diamond)$ . It turns out that over the class of models possessing $(c)$ or $(t)$ , all of these logics are equally expressive, whereas over the class of models possessing either of other eight neighborhood properties, $\mathcal{L}(\nabla,\bullet)$ is more expressive than both $\mathcal{L}(\nabla)$ and $\mathcal{L}(\bullet)$ , and over the class of models possessing either of eight neighborhood properties except for $(d)$ , $\mathcal{L}(\nabla,\bullet)$ is less expressive than $\mathcal{L}(\Diamond)$ . We explored the frame definability of the bimodal logic, which turns out that all ten frame properties except for $(n)$ are undefinable in $\mathcal{L}(\nabla,\bullet)$ . We axiomatized the bimodal logic over various classes of neighborhood frames, which among other results includes the classical logic, the monotone logic, and the regular logic. Last but not least, by updating the neighborhood models via the intersection semantics, we found suitable reduction axioms and thus reduced the public announcement operators to the bimodal logic. This gives us good applications to successful formulas, since as we have shown, any combination of $p$ , $\neg p$ , ${\neg\bullet}p$ and $\neg\nabla p$ via conjunction (or, via disjunction) is successful under the intersection semantics. We also partly answers open questions raised in [9, 11].

For future work, we hope to know whether $\mathcal{L}(\nabla,\bullet)$ is less expressive than $\mathcal{L}(\Diamond)$ over the class of $(d)$ -models. We conjecture the answer is positive, but the model constructions seems hard, where the desired models both needs at least three points. Moreover, as we have seen, the proofs of the expressivity and frame definability results involve nontrivial (if not highly nontrivial) constructions of neighborhood models and frames, we thus also hope to find the bisimulation notion for $\mathcal{L}(\nabla,\bullet)$ under the neighborhood semantics.

References

[1] R. Bett. Socratic ignorance. In Donald R. Morrison, editor, The Cambridge Companion to Socrates, pages 215–236. Cambridge University Press, 2011.
[2] J. C. Bjerring, J. U. Hansen, and N. J. L. L. Pedersen. On the rationality of pluralistic ignorance. Synthese, 191(11):2445–2470, 2014.
[3] S. Bonzio, V. Fano, P. Graziani, and M. Pra Baldi. A logical modeling of severe ignorance. Journal of Philosophical Logic, 52:1053–1080, 2023.
[4] B. F. Chellas. Modal Logic: An Introduction. Cambridge University Press, 1980.
[5] J. Fan. Removing your ignorance by announcing group ignorance: A group announcement logic for ignorance. Studies in Logic, 9(4):4–33, 2016.
[6] J. Fan. Neighborhood contingency logic: A new perspective. Studies in Logic, 11(4):37–55, 2018.
[7] J. Fan. Bimodal logics with contingency and accident. Journal of Philosophical Logic, 48:425–445, 2019.
[8] J. Fan. A family of neighborhood contingency logics. Notre Dame Journal of Formal Logic, 60(4):683–699, 2019.
[9] J. Fan. Bimodal logic with contingency and accident: Bisimulation and axiomatizations. Logica Universalis, 2021. https://doi.org/10.1007/s11787-021-00270-9.
[10] J. Fan. A logic for disjunctive ignorance. Journal of Philosophical Logic, 2021. https://doi.org/10.1007/s10992-021-09599-4.
[11] J. Fan. Unknown truths and false beliefs: Completeness and expressivity results for the neighborhood semantics. Studia Logica, 110:1–45, 2022. https://doi.org/10.1007/s11225-021-09950-5.
[12] J. Fan and H. van Ditmarsch. Neighborhood contingency logic. In M. Banerjee and S. Krishna, editors, Logic and Its Application, volume 8923 of Lecture Notes in Computer Science, pages 88–99. Springer, 2015.
[13] J. Fan, Y. Wang, and H. van Ditmarsch. Almost necessary. In Advances in Modal Logic, volume 10, pages 178–196, 2014.
[14] J. Fan, Y. Wang, and H. van Ditmarsch. Contingency and knowing whether. The Review of Symbolic Logic, 8(1):75–107, 2015.
[15] K. Fine. Ignorance of ignorance. Synthese, 195:4031–4045, 2018. https://doi.org/10.1007/s11229-017-1406-z.
[16] V. Goranko. On relative ignorance. Filosofiska Notiser, 8(1):119–140, 2021.
[17] J. Hintikka. Knowledge and Belief. Cornell University Press, Ithaca, NY, 1962.
[18] K. Konolige. Circumscriptive ignorance. In AAAI, pages 202–204, 1982.
[19] E. Kubyshkina and M. Petrolo. A logic for factive ignorance. Synthese, pages 1–12, 2019.
[20] P. Le Morvan and R. Peels. The nature of ignorance: Two views. In R. Peels and M. Blaauw, editors, The epistemic dimensions of ignorance. Cambridge: Cambridge University Press, 2016.
[21] M. Ma and K. Sano. How to update neighborhood models. In International Workshop on Logic, Rationality and Interaction, pages 204–217. Springer, 2013.
[22] M. Ma and K. Sano. How to update neighbourhood models. Journal of Logic and Computation, 28(8):1781–1804, December 2018. https://doi.org/10.1093/logcom/exv026.
[23] J. Marcos. Logics of essence and accident. Bulletin of the Section of Logic, 34(1):43–56, 2005.
[24] R. Montague. Universal grammar. Theoria, 36:373–398, 1970.
[25] P. Le Morvan. On ignorance: A reply to Peels. Philosophia, 39(2):335–344, 2011.
[26] Hubert J O’Gorman and Stephen L Garry. Pluralistic ignorance — a replication and extension. Public Opinion Quarterly, 40(4):449–458, 1976.
[27] E. J. Olsson and C. Proietti. Explicating ignorance and doubt: A possible worlds approach. The Epistemic Dimensions of Ignorance, 2015.
[28] E. Pacuit. Neighborhood semantics for modal logic. Springer, 2017.
[29] R. Peels. Ignorance is Lack of True Belief: A Rejoinder to Le Morvan. Philosophia, 39(2):344–355, 2011.
[30] C. Proietti and E. J. Olsson. A ddl approach to pluralistic ignorance and collective belief. Journal of Philosophical Logic, 43(2):499–515, 2014.
[31] D. Scott. Advice on modal logic. Philosophical Problems in Logic: Some Recent Developments, pages 143–173, 1970.
[32] Y. Shoham. Chronological ignorance. In Proceedings of the Fifth National Conference on Artificial Intelligence, pages 389–393, 1986.
[33] C. Steinsvold. A note on logics of ignorance and borders. Notre Dame Journal of Formal Logic, 49(4):385–392, 2008.
[34] W. van der Hoek and A. Lomuscio. Ignore at your peril - towards a logic for ignorance. In Proc. of 2nd AAMAS, pages 1148–1149. ACM, 2003.
[35] W. van der Hoek and A. Lomuscio. A logic for ignorance. Electronic Notes in Theoretical Computer Science, 85(2):117–133, 2004.
[36] H. van Ditmarsch, W. van der Hoek, and B. Kooi. Dynamic Epistemic Logic, volume 337 of Synthese Library. Springer, 2007.
[37] J. Zvesper. Playing with information. PhD thesis, Institute for Logic, Language and Computation, University of Amsterdam, 2010.

Fitchean Ignorance and First-order Ignorance: A Neighborhood Look

Abstract

1 Introduction

2 Syntax and Neighborhood Semantics

Definition 1 (Languages).

Definition 2 (Neighborhood structures).

Definition 3 (Neighborhood properties).

Definition 4 (Semantics).

3 Expressivity

Definition 5.

Proposition 6.

Proof.

Proposition 7.

Proof.

Proposition 8.

Proof.

Remark 9.

Proposition 10.

Proof.

Proposition 11.

Proof.

Proposition 12.

Proof.

Proposition 13.

Proof.

Proposition 14.

Proof.

Proposition 15.

Proof.

Proposition 16.

Proof.

Corollary 17.

Proposition 18.

Proof.

Proposition 19.

Proof.

Proposition 20.

Proof.

4 Frame Definability

Definition 21.

Proposition 22.

Proof.

Proposition 23.

Proof.

Proposition 24.

Proof.

Proposition 25.

Proof.

5 Axiomatizations

5.1 Classical logic

5.1.1 Proof system and Soundness

Definition 26.

Proposition 27.

Proof.

Proposition 28.

Proof.

Proposition 29.

Proof.

5.1.2 Completeness

Definition 30.

Lemma 31.

Proof.

Theorem 32.

5.2 Extensions

5.2.1 𝐄𝐜∇⁣∙{\bf E_{c}^{\nabla\bullet}}

Definition 33.

Lemma 34.

Proof.

Proposition 35.

Proof.

Theorem 36.

Proposition 37.

5.2.2 𝐄𝐍∇⁣∙{\bf EN^{\nabla\bullet}}

Theorem 38.

Proof.

5.2.3 Monotone logic

Proposition 39.

Proof.

Proposition 40.

Proof.

Fitchean Ignorance and First-order Ignorance:
A Neighborhood Look

5.2.1 ${\bf E_{c}^{\nabla\bullet}}$

5.2.2 ${\bf EN^{\nabla\bullet}}$

5.2.5 ${\bf K^{\nabla\bullet}}$