Questions in Predicate Logic

Abstract

In this chapter, we move on from propositional logic to the richer setting of predicate logic. We describe how classical first-order logic can be enriched with questions, leading to a system $\textsf {InqBQ}$ of inquisitive first-order logic (the Q in the acronym stands for quantification).

In this chapter, we move on from propositional logic to the richer setting of predicate logic. We describe how classical first-order logic can be enriched with questions, leading to a system $\textsf {InqBQ}$ of inquisitive first-order logic (the Q in the acronym stands for quantification). With respect to the inquisitive propositional logic of the previous chapters, this extension is interesting not only because first-order logic, as a logic of statements, is a much more expressive system than propositional logic, but also because, through quantification, many important classes of questions can be formalized, in addition to the propositional ‘whether...or’ questions that we were able to formalize in inquisitive propositional logic. For instance, it will become possible to analyze questions that ask for one or more instances of a property, such as (1-a) and (1-b); questions that ask for the unique individual satisfying of a property, such as (1-c); and questions that ask for the extension of a property, such as (1-d).

Thus, inquisitive first-order logic provides a rich environment to regiment many classes of questions and study their logic—although there are also some prominent question types, notably how many questions like (2), which, while semantically analyzable, are not expressible with the resources of $\textsf {InqBQ}$ (see Grilletti and Ciardelli [1]).

As in the case of propositional logic, we will build our inquisitive system in two steps. In the first step, we will show how classical first-order logic can be given a semantics in terms of support at an information state. In the second step, we will exploit the support semantics to introduce questions into first-order logic, equipping the language with new question-forming operators. In addition to inquisitive disjunction , which will work as in InqB, we will now also have an inquisitive existential quantifier $\mathord {\exists \!\!\exists }$, which  asks for a witness of a certain property.

With the move to predicate logic, some of the subtleties of intensional semantics also come into play, such as the different ways in which terms may refer to objects (rigidly or variably), the interpretation of identity, and the distinction between the entities to which information is attached and the objects that actually exist in the world. As we will see, the modeling choices one makes about these issues have repercussions for the logic of questions.

While many of the central features of inquisitive propositional logic  carry over to the first-order case, there are also some crucial differences. Most importantly, it is no longer the case that a question can be recursively associated with a set of statements that capture the different ways to resolve the question. Mathematically, the full system $\textsf {InqBQ}$ turns out to be a rich and complex system. Indeed, in spite of systematic investigation over the past few years, the main meta-theoretical questions about this logic are currently still open: it is not known whether a complete axiomatization exists, nor whether the logic is entailment-compact (in the sense that whatever follows from a set of premises follows from some finite subset), satisfies analogues of the Löwenheim-Skolem theorems, or has a recursively enumerable set of validities.

At the same time, in recent years there have been important developments in the study of $\textsf {InqBQ}$, especially due to work by Grilletti (see [1,2,3]). One exciting recent result that we will cover in detail is the existence of a broad fragment of $\textsf {InqBQ}$, the classical antecedent fragment, which on the one hand contains all the most important classes of questions expressible in $\textsf {InqBQ}$, and on the other hand turns out to be very well-behaved and to admit an elegant completeness result. This fragment can then be regarded in its own right as a rich logic of questions—much richer than its predecessors, such as the Logic of Interrogation of Groenendijk [4]—that shares many of the the key features of standard first-order logic. In addition to this, at the end of the chapter we will also survey some other interesting recent results, as well as some open problems.

5.1 Support for Classical First-Order Logic

Let us start out by describing how classical first-order predicate logic may be given a support semantics. For ease of exposition, we focus first on a language without identity, and then turn to the treatment of identity in Sect. 5.4.

Language. As usual, our language is based on a signature $\mathcal {S}$, consisting of a set $\mathfrak {R}_\mathcal {S}$ of relation symbols (also called predicates) and a set $\mathcal {F}_\mathcal {S}$ of function symbols, where each of these symbols has a certain arity $n\ge 0$. Relation symbols of arity 0 are called propositional atoms, while function symbols of arity 0 are called individual constants. Moreover, we assume that among the function symbols we have a specified set $\mathcal {F}_\mathcal {S}^R$ of rigid function symbols, whose interpretation is required to be fixed across different possible worlds. We will refer to the remaining function symbols, whose interpretation may vary across different possible worlds, as non-rigid function symbols. We will use sans serif fonts to mark rigidity: thus, a meta-variable $\textsf {f}$ will range over rigid function symbols, while f will range over all function symbols, rigid or non-rigid.

As usual, we also have a countably infinite stock of first-order variables, $\textsf {Var}=\{x_0,x_1,x_2,\dots \}$. The set $\textsf {Ter}(\mathcal {S})$ of terms in the language is given as usual by the inductive definition

$$t\;::=\;x\mid f(t,\dots ,t)$$

where $x\in \textsf {Var}$, $f\in \mathcal {F}_{\mathcal {S}}$, and the number of arguments of f in the inductive clause matches the arity of f. The set of rigid terms is defined analogously by

$$\textsf {t}\;::=\;x\mid \textsf {f}(\textsf {t},\dots ,\textsf {t})$$

where $\textsf {f}\in \mathcal {F}_\mathcal {S}^R$ is a rigid function symbol.

The set of classical first-order formulas in the signature $\mathcal {S}$ is also defined as usual, where we take $\bot ,\wedge$, and $\rightarrow$ as our primitive propositional connectives , and $\forall$ as our primitive quantifier.

Definition 5.1.1

(Classical formulas) The set $\mathcal {L}^{\textsf {Q}}_c(\mathcal {S})$ of classical first-order formulas is defined recursively as follows:

$$\begin{aligned} \varphi \;::=\; R(t_1,\dots ,t_n)\;|\;\bot \;|\;\varphi \wedge \varphi \;|\;\varphi \rightarrow \varphi \;|\;\forall x\varphi \end{aligned}$$

where R is an n-ary relation symbol in $\mathcal {S}$, $t_1,\dots ,t_n\in \textsf {Ter}(\mathcal {S})$, and $x\in \textsf {Var}$.

When there is no need to emphasize the signature $\mathcal {S}$, we will drop reference to it and simply refer to the set of classical first-order formulas as $\mathcal {L}^{\textsf {Q}}_c$. We take the remaining operators of classical first-order logic to be defined as follows:

• $\lnot \varphi :=\varphi \rightarrow \bot$;

• $\varphi \vee \psi :=\lnot (\lnot \varphi \wedge \lnot \psi )$;

• $\varphi \leftrightarrow \psi :=(\varphi \rightarrow \psi )\wedge (\psi \rightarrow \varphi )$;

• $\exists x\varphi :=\lnot \forall x\lnot \varphi$.

It will be useful to introduce some abbreviations: we will write $\overline{t}$ for a sequence $\langle t_1,\dots ,t_n\rangle$ of terms and $\overline{x}$ for a sequence $\langle x_1,\dots ,x_n\rangle$ of variables. Moreover, if Q is a quantifier and $\overline{x}=\langle x_1,\dots ,x_n\rangle$ a sequence of variables, we will write $Q\overline{x}\varphi$ for $Qx_1\dots Qx_n\varphi$.

Free and bound occurrences of a variable x in a formula are defined as usual. Given a formula $\varphi$, we write $FV(\varphi )$ for the set of variables which are free in $\varphi$. Moreover, if $x\in \textsf {Var}$ and $t\in \textsf {Ter}(\mathcal {S})$, we write $\varphi [t/x]$ for the formula that results from replacing each free occurrence of x in $\varphi$ by t. As usual, we say that a term t is free for a variable x in a formula $\varphi$ in case no free occurrence of x in $\varphi$ lies within the scope of a quantifier which binds a variable y occurring in t.

We allow ourselves to drop parentheses whenever convenient, including in the case of atomic sentences (writing, e.g., Rxy instead of R(x, y)). We follow standard conventions about the priority of operators: quantifiers and negation have the highest priority, followed by conjunctions and disjunctions (including inquisitive disjunctions in the full language), while implication has the lowest priority. Thus, e.g., the formula $\varphi \rightarrow \lnot \forall x\psi \wedge \chi$ should be parsed as $\varphi \rightarrow ((\lnot \forall x\psi )\wedge \chi )$.

Models. Let us now turn to the structures that are to serve as models for our language. As in the previous chapters, our models will comprise a universe of possible worlds, each representing a certain state of affairs. Moreover, they will comprise a domain D of individuals that our quantifiers range over. These are the individuals that the information represented by the model is about. The state of affairs corresponding to a given world is then characterized by fixing the denotation of the predicate and function symbols.

Note that taking the domain of quantification to be world-independent means that our models will not be able to represent uncertainty about the domain of quantification itself (except insofar as it stems from uncertainty about identities: see Sect. 5.4). This can be seen as a simplifying assumption that one might want to lift in future work, at the cost of introducing some extra complexity.¹

Definition 5.1.2

(Relational information models) A relational information model is a triple $M=\langle W,D,I\rangle$, where:

• W is a set, the elements of which we call possible worlds;

• D is a non-empty set, the elements of which we call individuals²;

• I is a map assigning to each $w\in W$ a function $I_w$ defined on $\mathcal {S}$ such that:

  • * $I_w(R)\subseteq D^n\,$ for an n-ary predicate R; we write $R_w$ for $I_w(R)$.

  • * $I_w(f):D^n\rightarrow D\,$ for an n-ary function symbol f; we write $f_w$ for $I_w(f)$.

    Rigidity constraint: if $\textsf {f}$ is rigid, then for any $w,w'\in W$, $\textsf {f}_w=\textsf {f}_{w'}$.

Note that with each world w of such an information model we can associate a standard relational structure of our signature.

Definition 5.1.3

(Relational structure associated with a world) Let $M=\langle W,D,I\rangle$ be a relational information model and $w\in W$. The relational structure associated with W is $M_w:=\langle D,I_w\rangle$.

Thus, a relational information model can be seen alternatively as a collection $\{M_w\mid w\in W\}$ of relational structures sharing the same underlying domain.

For an illustration, consider a signature containing a unary predicate P and two rigid constants, $\textsf {a}$ and $\textsf {b}$. Consider a simple model M containing two individuals a and b, denoted rigidly by $\textsf {a}$ and $\textsf {b}$ respectively, and four possible worlds, corresponding to the four possible extensions for the predicate P. This model is depicted in Fig. 5.1.

Fig. 5.1

A relational information model with two individuals a, b, denoted rigidly by constants $\textsf {a}$ and $\textsf {b}$, and four possible worlds, corresponding to the four extensions for P. The label ab in the pictures stands for a world w in which $P_w=\{a,b\}$, the label a stands for a world w in which $P_w=\{a\}$, and so on

Support semantics for classical first-order logic. Let us now see how the standard language of first-order logic can be given a support semantics, which interprets formulas relative to information states $s\subseteq W$ drawn from a relational information model.

As is customary, the semantics is given relative to an assignment function g, which fixes the interpretation of variables. Assignments are defined as usual as functions $g:\textsf {Var}\rightarrow D$. If $d\in D$, we write $g[x\mapsto d]$ for the assignment which maps x to d, and otherwise coincides with g.

We can assign to each term in the language a world-dependent referent in a natural way.

Definition 5.1.4

(Referent of a term) The referent of a term t in a world w under an assignment g is the individual $[t]_g^w\in D$ defined inductively as follows:

• $[x]_g^w=g(x)$;

• $[f(t_1,\dots ,t_n)]_g^w=f_w([t_1]_g^w,\dots ,[t_n]_g^w)$.

Note that if t is a closed term (i.e., if t does not contain variables) the referent is independent of g and can be denoted as $[t]^w$, while if $\textsf {t}$ is rigid, the referent is independent of w and can be denoted as $[\textsf {t}]_g$. If $\textsf {t}$ is both closed and rigid we can drop both parameters and write simply $[\textsf {t}]$.

We are now ready to define the relation of support between states and formulas, which specifies what information it takes to settle a first-order formula.

Definition 5.1.5

(Support for classical first-order formulas) If M is a relational information model, s an information state in M, and g an assignment, we let:

• $M,s\,\models _g\, R(t_1,\dots ,t_n)\iff \text {for all }w\in s,\; \langle [t_1]_g^w,\dots ,[t_n]_g^w\rangle \in R_w$

• $M,s\,\models _g\,\bot \iff s=\emptyset$

• $M,s\,\models _g\,\varphi \wedge \psi \iff M,s\,\models _g\,\varphi \text { and }M,s\,\models _g\,\psi$

• $M,s\,\models _g\,\varphi \rightarrow \psi \iff \text {for all }t\subseteq s,\; M,t\,\models _g\,\varphi \text { implies }M,t\,\models _g\,\psi$

• $M,s\,\models _g\,\forall x\varphi \iff \text {for all }d\in D,\; M,s\,\models _{g[x\mapsto d]}\varphi$.

As usual, atomic formulas $R\overline{t}$ are treated as statements: a state settles that $R\overline{t}$ if the information available in s implies that the tuple of individuals denoted by $\overline{t}$ belongs to the extension of R. The clauses for the propositional connectives are familiar from the previous chapters. A universal $\forall x\varphi (x)$ is settled in s in case $\varphi (x)$ is settled for every value of the variable x.

When no confusion arises, we will drop reference to the model M, and simply write $s\,\models _g\,\varphi$. As usual, we refer to the set of information states supporting $\varphi$ in M (now relative to an assignment g) as the support-set of $\varphi$, notation $[\varphi ]_M^g$. The alternatives for $\varphi$ in M relative to g are the $\subseteq$-maximal elements of $[\varphi ]_M^g$, and the set of these alternatives is denoted $\textsc {Alt}_M^g(\varphi )$. It is easy to check that if $FV(\varphi )\subseteq \{x_1,\dots ,x_n\}$, only the value of g on $x_1,\dots ,x_n$ matters for whether $s\,\models _g\,\varphi$; in this case, we may thus write $s\,\models _{[x_1\mapsto d_1,\dots ,x_n\mapsto d_n]}\,\varphi$ to mean that $s\,\models _g\,\varphi$ where g is an arbitrary assignment mapping $x_i$ to $d_i$. In particular, if $\varphi$ is a sentence we may drop reference to the assignment altogether.

Truth at a world w is defined, as usual, as support at the state $\{w\}$:

$$w\,\models _g\,\varphi \overset{def}{\iff } \{w\}\,\models _g\,\varphi .$$

The truth-set of $\varphi$ in M relative to g is the set of worlds where $\varphi$ is true:

$$|\varphi |_M^g:=\{w\in W\,|\,w\,\models _g\,\varphi \}.$$

It is straightforward to check that truth at a world w, as given by our semantics, coincides exactly with truth in the structure $M_w$ as given by the standard Tarskian semantics for first-order predicate logic.

Proposition 5.1.6

(Truth conditions for classical formulas) For any information model M, any world w in M, and any assignment g:

• $w\,\models _g\,R(t_1,\dots ,t_n)\iff \langle [t_1]_g^{w},\dots ,[t_n]_g^{w}\rangle \in R_w$

• $w\,\not \models _g\,\bot$

• $w\,\models _g\, \varphi \wedge \psi \iff w\,\models _g\,\varphi \text { and }w\,\models _g\,\psi$

• $w\,\models _g\,\varphi \rightarrow \psi \iff w\,\not \models _g\,\varphi \text { or }w\,\models _g\,\psi$

• $w\,\models _g\,\forall x\varphi \iff \text {for all }d\in D,\; w\,\models _{g[x\mapsto d]}\,\varphi$.

Thus, our semantics allows us to retrieve standard truth-conditional semantics as a special case. In order to check that our semantics gives a support-based implementation of classical propositional logic, we just have to check that truth and support are related in accordance with the Truth-Support Bridge (Constraint 2.2.5): a formula is supported by a state iff it is true at each world in the state. This is the content of the following proposition.

Proposition 5.1.7

(Classical formulas are truth conditional) For any $\varphi \in \mathcal {L}^{\textsf {Q}}_c$, state s in an information model M, and assignment g:

$$s\,\models _g\,\varphi \iff w\,\models _g\,\varphi \text { for all }w\in s.$$

Proof

By induction on $\varphi$. The novel case with respect to the proof in the propositional case (Proposition 3.1.8) is the one for $\forall$. We have

$$\begin{aligned} s\,\models _g\,\forall x\varphi\iff & {} \text {for all }d\in D: s\,\models _g\,\varphi \\\iff & {} \text {for all }d\in D,\text { for all }w\in s: w\,\models _{g[x\mapsto d]}\,\varphi \\\iff & {} \text {for all }w\in s, \text { for all }d\in D: w\,\models _{g[x\mapsto d]}\,\varphi \\\iff & {} \text {for all }w\in s: w\,\models _g\,\forall x\varphi \end{aligned}$$

where the second step uses the induction hypothesis.    $\square$

Given that truth and support are related in the appropriate way, it follows from the discussion in Sect.  2.2 that our support semantics is a semantics for classical first-order logic. This is illustrated by Fig. 5.2, which shows the alternatives for some first-order sentences in the model of Fig. 5.1. It follows from Proposition 5.1.7 that each statement has a single alternative, consisting of those worlds where it is classically true.

Fig. 5.2

The alternatives for some InqBQ statements in the relational information model of Fig. 5.1

Having thus re-implemented classical predicate logic based on support, we are now ready for the second step of our strategy: bring questions into play by enriching our language with question-forming logical operators.

5.2 Adding Questions to First-Order Logic

In the propositional case, questions are introduced into the system by means of a new connective, the inquisitive disjunction  . In the first-order case, it is natural to also consider a quantifier counterpart of  , denoted $\mathord {\exists \!\!\exists }$, which we will call inquisitive existential quantifier. The full language of our system is obtained by enriching the language of classical first-order logic with these two operators.

Definition 5.2.1

The set $\mathcal {L}^{\textsf {Q}}(\mathcal {S})$ of first-order formulas of $\textsf {InqBQ}$ is defined recursively as follows:

where R is an n-ary redicate in $\mathcal {S}$, $t_1,\dots ,t_n\in \textsf {Ter}(\mathcal {S})$, and $x\in \textsf {Var}$.

As in propositional logic, we also use a derived operator ‘?’, defined by  . The semantics of    is familiar by now: a state s supports an inquisitive disjunction   in case it supports one of the disjuncts. The semantics of $\mathord {\exists \!\!\exists }$ is the quantifier analogue of this clause: a state s supports an inquisitive existential $\mathord {\exists \!\!\exists }x\varphi (x)$ in case it supports $\varphi (x)$ for some specific value of x.

Definition 5.2.2

(Support for $\textsf {InqBQ}$) The relation of support for $\textsf {InqBQ}$ is obtained by augmenting Definition 5.1.5 with the following two clauses:

• or $\,M,s\,\models _g\,\psi$;

• $M,s\,\models _g\,\mathord {\exists \!\!\exists }x\varphi \iff \text { for some }d \in D,\; M,s\,\models _{g[x\mapsto d]}\,\varphi$.

By combining the inquisitive operators    and $\mathord {\exists \!\!\exists }$ with the support-based versions of the classical operators, in $\textsf {InqBQ}$ we can express a wide range of questions. Let us illustrate this with some examples.

First, just like in the propositional setting, we can use the question mark operator ? to turn a statement into the corresponding polar question.

Example 5.2.3

(Polar questions) Consider the formula $?\forall xPx$, which abbreviates  . Using the fact that $\forall xPx$ and $\lnot \forall xPx$ are classical formulas, we have:

$$\begin{aligned} s\,\models _g{?\forall xPx}\iff & {} s\,\models _g\,\forall xPx\text { or }s\,\models _g\,\lnot \forall xPx\\\iff & {} (\forall w\in s: w\models _g\forall xPx)\text { or }(\forall w\in s:w\,\models _g\,\lnot \forall xPx)\\\iff & {} (\forall w\in s: P_w=D)\text { or }(\forall w\in s:P_w\ne D). \end{aligned}$$

In words, $?\forall xPx$ is settled in a state in case the available information determines whether or not all individuals have property P. Thus, $?\forall xPx$ can be seen as a formalization of the polar question whether everyone has property P. The two alternatives for this question in our example model of Fig. 5.1 are shown in Fig. 5.3a.

In addition to  , in $\textsf {InqBQ}$ we can form questions by means of $\mathord {\exists \!\!\exists }$. The following example illustrates how this operator allows us to formalize an important class of questions, namely, mention-some questions (cf. the discussion in Sect. 2.9.2).

Example 5.2.4

(Mention-some questions) Consider the sentence $\mathord {\exists \!\!\exists }xPx$. We have:

$$\begin{aligned} s\models _g\mathord {\exists \!\!\exists }xPx\iff & {} \text {there is a }d\in D\text { such that } s\models _{g[x\mapsto d]} Px\\\iff & {} \text {there is a }d\in D\text { such that for all }w\in s,\; d\in P_w. \end{aligned}$$

In words, $\mathord {\exists \!\!\exists }xPx$ is settled if for some individual d, the available information implies that d has property P. Thus, the formula $\mathord {\exists \!\!\exists }xPx$ can be seen as a formalization of the question what is an instance of P. In the literature, questions that ask for instances of properties are called mention-some questions.

In our model M, which contains just two individuals, the formula $\mathord {\exists \!\!\exists }xPx$ has two distinct alternatives, depicted in Fig. 5.3b. These alternatives correspond to the classical formulas $P\textsf {a}$ and $P\textsf {b}$, which provide just enough information to establish an instance of property P.

It is interesting to pause to point out that the difference between the semantics of the classical existential $\exists xPx$ and its inquisitive counterpart $\mathord {\exists \!\!\exists }xPx$ is one of relative scope of two quantifiers:

$$\begin{aligned} s\models _g\exists xPx\iff & {} \text {for all }w\in s,\text { some }d\in D\text { is such that }d\in I_w(P);\\ s\models _g\mathord {\exists \!\!\exists }xPx\iff & {} \text {some }d\in D\text { is such that for all }w\in s,~ d\in I_w(P). \end{aligned}$$

In words, $\exists xPx$ is supported if the available information implies that some individual has property P, while $\mathord {\exists \!\!\exists }xPx$ is supported if for some individual, the available information implies that it has property P.

The idea illustrated by this example generalizes: if R is a binary relation symbol, then the sentence $\mathord {\exists \!\!\exists }x\mathord {\exists \!\!\exists }yR(x,y)$ is supported in s just in case s establishes of a specific pair $\langle d,d'\rangle$ that it belongs to the extension of R. Thus, $\mathord {\exists \!\!\exists }x\mathord {\exists \!\!\exists }yR(x,y)$ formalizes a question asking for an instance of a pair which stands in the relation R.

Fig. 5.3

The alternatives for some InqBQ questions in the relational information model of Fig. 5.1

In the previous examples, we considered questions that can be formed directly by using the inquisitive operators. But notice that further questions can be expressed by embedding such basic questions under  the classical operators $\wedge ,\rightarrow ,$ and $\forall$, whose semantics is now generalized in such a way that they can operate on questions as well. In Chap. 2, we have already discussed in detail the effect of embedding questions under conjunction and implication. These operations apply in much the same way to the richer repertoire of questions available in the current setting. By means of conjunction we can, e.g., form conjunctive questions like $\mathord {\exists \!\!\exists }xP(x)\wedge \mathord {\exists \!\!\exists }xQ(x)$ which asks at once for an instance of property P and an instance of property Q. By means of implication we can form conditional questions like $\exists xP(x)\rightarrow \mathord {\exists \!\!\exists }xP(x)$, which asks for an instance of property P under the assumption that there is one; we may also form questions such as $\mathord {\exists \!\!\exists }xP(x)\rightarrow \mathord {\exists \!\!\exists }xQ(x)$, which may be seen as asking for a method for turning an instance of property P into an instance of property Q.

The novelty introduced by $\textsf {InqBQ}$ is that it also becomes possible to universally quantify over questions. The following example shows that by doing so we can formalize another important class of questions: mention-all questions.

Example 5.2.5

(Mention-all questions) Consider the sentence $\forall x?P(x)$: this sentence is supported in a state s in case s settles the polar question ?P(x) for all values of x, that is, in case s establishes of every individual $d\in D$ whether or not d has property P. This means that, in order to settle $\forall x?P(x)$, a state s must settle precisely what the extension of P is. More formally, using the fact that P(x) is a classical formula we have:

$$\begin{aligned} s\models _g\!\forall x{?P(x)}\!\!\!\iff & {} \!\!\text {for all }d\in D:\;s\models _{g[x\mapsto d]}?P(x)\\\iff & {} \!\!\text {for all }d\in D:\;s\models _{g[x\mapsto d]}P(x)\text { or }s\models _{g[x\mapsto d]}\lnot P(x)\\\iff & {} \!\!\text {for all }d\in D:\;(\text {for all }w\in s:\,d\in P_w)\text { or }\\{} & {} \!\!\phantom {\text {for all }d\in D:\;\;}(\text {for all }w\in s:\,d\not \in P_w)\\\iff & {} \!\!\text {for all }d\in D,\text { for all }w,w'\in s:\; (d\in P_w\iff d\in P_{w'})\\\iff & {} \!\!\text {for all }w,w'\in s,\text { for all }d\in D:\; (d\in P_w\iff d\in P_{w'})\\\iff & {} \!\!\text {for all }w,w'\in s: P_w=P_{w'}. \end{aligned}$$

Thus, $\forall x{?P(x)}$ can be seen as formalizing the question which individuals have property P, or equivalently, what is the extension of P. Questions that ask for the extension of a property or relation are known as mention-all questions.

In our toy model, this sentence has four alternatives, depicted in Fig. 5.3c: each alternative corresponds to one possibility for the extension of P.

The example generalizes: for instance, if R is a binary relation symbol, the sentence $\forall x\forall y{?R(x,y)}$ is supported by a state s iff s determines the extension of the relation R—i.e., iff the extension is the same at each world in s. Thus, e.g., in a domain consisting only of humans, $\forall x\forall y{?R(x,y)}$ formalizes the mention-all reading of the question who is R-related to whom.

In Sect. 5.7.3 we will see that the questions that can be obtained by universally quantifying over polar questions are precisely those which ask for the extension of a relation defined by a standard formula of classical first-order logic. As we will see, these coincide exactly with the questions expressible in the Logic of Interrogation of Groenendijk [4].

Summing up, then, in $\textsf {InqBQ}$ we can regiment a range of interesting question types. In particular, we can formalize polar questions like (3-a), mention-some wh-questions like (3-b), and mention-all wh-questions like (3-c), in addition to conjunctive and conditional questions derived from such questions.

This discussion does not by any means provide an exhaustive survey of the kinds of questions expressible in $\textsf {InqBQ}$. On the contrary, many more question types can be expressed,³ and even more will become expressible once we introduce identity. But hopefully these examples already illustrate the richness of $\textsf {InqBQ}$ as a logical framework for regimenting questions, and show how naturally many important question types can be expressed by using only a small set of semantically simple logical operators.⁴

5.3 Basic Features of InqBQ

Let us now take a look at some basic features of the system $\textsf {InqBQ}$ which we have defined. We will see that, while many features of the propositional system InqB carry over, there are also some interesting differences.

5.3.1 Support and Alternatives

Let us start by examining the features of the support relation. As we expect, support is persistent, and the empty state trivially supports every formula.

Proposition 5.3.1

For any model M, states s, t, assignment g and formula $\varphi \in \mathcal {L}^{\textsf {Q}}$, we have:

• Persistence property: $s\models _g\varphi$ and $t\subseteq s$ implies $t\models _g\varphi$;

• Empty state property: $\emptyset \models _g\varphi$.

As a consequence of persistence, we also get that the truth-set of a formula always coincides with the union of its support set.

Proposition 5.3.2

For any model M, assignment g, and $\varphi \in \mathcal {L}^{\textsf {Q}}$: $|\varphi |_M^g=\bigcup [\varphi ]_M^g$.

Moreover, as in the propositional case, our semantics is local. That is, support at a state depends exclusively on the features of the worlds in the state.

Proposition 5.3.3

(Locality) Given a relational information model $M=\langle W,D,I\rangle$ and a state s in M, let $M_{|s}$ be the restriction of M to s, i.e., the model $M_{|s}=\langle s,D,I_{|s}\rangle$, where $I_{|s}$ is the restriction of I to worlds in s. For any assignment g and formula $\varphi$ we have:

$$M,s\models _g\varphi \iff M_{|s},s\models _g\varphi .$$

One feature of the propositional system InqB that does not carry over to the first-order setting is normality, i.e., the identity $[\varphi ]_M=\textsc {Alt}_M(\varphi )^\downarrow$.

Proposition 5.3.4

(Failure of normality) There is a model M, a state s, and a sentence $\varphi \in \mathcal {L}^{\textsf {Q}}$ such that $s\models \varphi$ but s is not included in any alternative for $\varphi$ in M.

Proof

Consider a signature consisting only of a predicate symbol P, and a model M given as follows:

$$W=\{w_n\,|\,n\in \mathbb {N}\},\qquad D=\mathbb {N},\qquad P_{w_n}=\{i\in \mathbb {N}\mid i\ge n\}.$$

For any number $k\in \mathbb {N}$, let us define an information state $s_k=\{w_0,\dots ,w_k\}$. Now consider the formula $\varphi :=\mathord {\exists \!\!\exists }xPx$. We have:

$$\begin{aligned} s\models \mathord {\exists \!\!\exists }xPx\iff & {} \text {for some }k\in \mathbb {N}:s\models _{x\mapsto k}Px\\\iff & {} \text {for some }k\in \mathbb {N},\text { for all }w_n\in s:w_n\models _{x\mapsto k}Px\\\iff & {} \text {for some }k\in \mathbb {N},\text { for all }w_n\in s:k\in P_{w_n}\\\iff & {} \text {for some }k\in \mathbb {N},\text { for all }w_n\in s:k\ge n\\\iff & {} \text {for some }k\in \mathbb {N}: s\subseteq s_k. \end{aligned}$$

Thus, we have a sequence $s_0\subset s_1\subset s_2\subset \dots$ of information states, each properly included in the next, such that:

• every state $s_k$ in the sequence supports $\varphi$;

• every state s supporting $\varphi$ is included in some element $s_k$ of the sequence.

From this it follows that in our model, there is no maximal supporting state for $\varphi$. For suppose s supports $\varphi$. Then $s\subseteq s_k$ for some k, and since $s_k\subset s_{k+1}$, s is strictly included in the information state $s_{k+1}$, which also supports $\varphi$. Hence s is not a maximal supporting state.

Thus, there are no alternatives for $\varphi$ in M. Since there are supporting states for $\varphi$ in M (for instance, each state $s_k$) this is a violation of normality.    $\square$

This shows that, unlike in the propositional setting, in the first-order setting the proposition expressed by a sentence in a model is not, in general, fully captured by its set of alternatives.⁵ There are, however, significant syntactic fragments of $\textsf {InqBQ}$ for which normality still holds, as we will see in Sect. 5.7.1.

5.3.2 Truth-Conditional Formulas

Recall that we call a formula $\varphi$ truth-conditional if in any state s of any model, $\varphi$ is supported at s just in case it is true at every world in s. Also recall that we refer to truth-conditional formulas as statements, and to formulas that are not truth-conditional as questions.

We saw above that classical formulas $\varphi \in \mathcal {L}^{\textsf {Q}}_c$ are always truth-conditional. Conversely, as in the propositional case, we can show that any truth-conditional formula in $\textsf {InqBQ}$ is equivalent to a classical formula. In order to show this, we first associate to each formula $\varphi$ a classical formula $\varphi ^{cl}$ having the same truth conditions as $\varphi$.

Definition 5.3.5

(Classical variant of a first-order formula) If $\varphi \in \mathcal {L}^{\textsf {Q}}$, the classical variant of $\varphi$ is the formula $\varphi ^{cl}\in \mathcal {L}^{\textsf {Q}}_c$ obtained by replacing each occurrence of    by $\vee$, and each occurrence of $\mathord {\exists \!\!\exists }$ by $\exists$.

Proposition 5.3.6

For any $\varphi \in \mathcal {L}^{\textsf {Q}}$, model M, world w and assignment g:

$$\;w\models _g\varphi \iff w\models _g\varphi ^{cl}.$$

If $\varphi$ itself is truth-conditional, this implies that $\varphi$ and $\varphi ^{cl}$ are equivalent. Thus, any truth-conditional formula is equivalent to a classical formula. Conversely, if a formula $\varphi$ is equivalent to a classical formula, then since classical formulas are truth-conditional, $\varphi$ must be truth-conditional as well.

Proposition 5.3.7

The following are equivalent for any $\varphi \in \mathcal {L}^{\textsf {Q}}$:

• $\varphi$ is truth-conditional;

• $\varphi \equiv \varphi ^{cl}$;

• $\varphi \equiv \alpha$ for some $\alpha \in \mathcal {L}^{\textsf {Q}}_c$.

This shows that, while our question-forming operators    and $\mathord {\exists \!\!\exists }$ obviously add to the expressive power of the language, enabling us to express questions, they do not allow us to express any new statements.⁶

In Chap. 2, we saw that, like classical formulas, negations are always truth-conditional. Since negation works in exactly the same way in the first-order setting, this is still true in $\textsf {InqBQ}$.

Proposition 5.3.8

For any $\varphi \in \mathcal {L}^{\textsf {Q}}$, $\lnot \varphi$ is truth-conditional.

In particular, for any formula $\varphi$, $\lnot \lnot \varphi$ is truth-conditional. Moreover, since truth conditions work in the standard way, $\lnot \lnot \varphi$ has the same truth-conditions as $\varphi$. Thus, $\lnot \lnot \varphi$ must be equivalent with the classical variant $\varphi ^{cl}$.

Proposition 5.3.9

For any $\varphi \in \mathcal {L}^{\textsf {Q}}$, $\lnot \lnot \varphi \equiv \varphi ^{cl}$.

This shows that negations, just like classical formulas, are representative of all truth-conditional formulas in $\textsf {InqBQ}$.

Proposition 5.3.10

The following are equivalent for any $\varphi \in \mathcal {L}^{\textsf {Q}}$:

• $\varphi$ is truth-conditional;

• $\varphi \equiv \lnot \lnot \varphi$;

• $\varphi \equiv \lnot \psi$ for some $\psi \in \mathcal {L}^{\textsf {Q}}$.

Now consider questions, which by definition are not truth-conditional. If $\mu$ is a question, the classical variant $\mu ^{cl}$ is not equivalent to $\mu$: rather, $\mu ^{cl}$ is a statement which expresses the presupposition of $\mu$ (cf. Sect. 2.6). As in Chap. 3, we will refer to $\mu ^{cl}$ as the presupposition of $\mu$. Let us illustrate this notion by means of two examples.

First, consider the formula $\mathord {\exists \!\!\exists }xPx$, which as we saw captures the question what is an instance of a P. Its presupposition is the formula $\exists xPx$. This is a statement that captures precisely the conditions under which the question admits a resolution: it is in principle possible to provide an instance of a P if and only if there are objects satisfying P.

Next, consider the formula $\forall x{?Px}$, which captures the question what is the extension of P Spelling out the question mark operator, the formula is , and thus its presupposition is the formula $\forall x(Px\vee \lnot Px)$, which is a tautology. This corresponds to the fact that the question $\forall x{?Px}$ can be settled under any circumstances.

5.3.3 Resolutions?

A key property of the propositional logic InqB is that we can associate any formula $\varphi$ with a set $\mathcal {R}(\varphi )$ of classical formulas such that to settle $\varphi$ is to establish that $\alpha$ is true for some $\alpha \in \mathcal {R}(\varphi )$. Is something similar possible the first-order case? That is, can we define for each formula $\varphi \in \mathcal {L}^{\textsf {Q}}$ a set $\mathcal {R}(\varphi )$ of classical formulas with the property that for any M and g, the connection

$$s\models _g\varphi \iff s\models _g\alpha \text { for some }\alpha \in \mathcal {R}(\varphi )$$

holds? The answer is negative. One reason is that we are not assured a priori to have the means to rigidly designate every individual in the given model. Thus, even if a state s settles of some individual d that it has property P—thus supporting the question $\mathord {\exists \!\!\exists }xPx$—we may not be able to trace this to the support of a classical formula, because we may lack a name for the individual d.

We may try to obviate this problem by extending the language with rigid names for all entities in the domain of the given model, and then by giving a set of resolutions $\mathcal {R}_M(\varphi )$ relativized to M, which has the above property for states s in M. But even this does not give us for all formulas $\varphi$ a set $\mathcal {R}_M(\varphi )$ with the required properties. To see this, consider a mention-all question $\forall x{?Px}$, which as we saw asks for the extension of property P. Even if we have names for all individuals in the domain, we do not in general have the syntactic means to describe all possible extensions for P: if the domain D is countably infinite and all possibilities for the extension of P are instantiated in the model, there will be uncountably many such extensions, but only countably many formulas in our language if our initial signature is countable. Now suppose we have no formula in our language stating that the extension of P is a certain set $X\subseteq D$: if s is a non-empty information state containing all and only the worlds in which the extension of P is X, then the question $\forall x{?Px}$ will be supported at s, but this will not be traceable to the support of any classical formula in our language.

5.4 Adding Identity

Definitions. Let us now see how identity can be introduced into the picture. Syntactically, this is straightforward: given a signature $\mathcal {S}$, we consider languages $\mathcal {L}^{\textsf {Q}=}_{c}(\mathcal {S})$ and $\mathcal {L}^{\textsf {Q}=}(\mathcal {S})$ which are defined just like $\mathcal {L}^{\textsf {Q}}_c(\mathcal {S})$ and $\mathcal {L}^{\textsf {Q}}(\mathcal {S})$, but with the addition of atomic formulas of the form $(t=t')$, where $t,t'\in \textsf {Ter}(\mathcal {S})$.

Semantically, to interpret identity we equip a relational information model with a function $\sim$ which assigns to each world $w\in W$ the extension of the identity relation at w, denoted $\sim _w$. This is required to be an equivalence relation on D and a congruence with respect to the interpretation of function and relation symbols. That is, we require that for each world w in the model:

• for any n-ary function symbol f, if $d_1\sim _w d_1', \dots , d_n\sim _w d_n'$, then

  $$f_w(d_1,\dots ,d_n)\sim _w f_w(d_1',\dots ,d_n');$$

• for any n-ary relation symbol R, if $d_1\sim _w d_1', \dots , d_n\sim _w d_n'$, then

  $$\langle d_1,\dots ,d_n\rangle \in R_w\iff \langle d_1',\dots ,d_n'\rangle \in R_w.$$

The semantics of identity atoms is then parallel to that of other atomic sentences:

• $M,s\,\models _g\,(t=t')\iff \text {for all }w\in s,\;[t]_g^w\sim _w[t']_g^w.$

All the facts that we have stated so far about $\textsf {InqBQ}$ carry over immediately to the language extended with identity .

Conceptual motivation. Our formal treatment of identity is built around the idea that there can be uncertainty about the extension identity. At first, this might seem strange. Surely we know a priori what the extension of identity is: every individual is identical to itself, and not identical to any other individual.

Things, however, become more subtle in a setting where the objects to which information is attributed in an information state are not necessarily in a one-to-one correspondence with the objects that actually exist in the world. Let us call the former epistemic individuals, and the latter ontic individuals.

The importance of distinguishing the two is well illustrated by Frege’s puzzle. Consider an ancient astronomer who has just discovered the identity of Hesperus and Phosphorus. Before the discovery, this astronomer had some information about Hesperus (say, that it is visible in the evening), and some information about Phosphorus (say, that it is visible in the morning), yet he did not know whether Hesperus and Phosphorus were in fact two distinct objects, or one and the same object. Thus, our astronomer was in an information state s such that:

$$s\models \text {evening}(\textsf {h}),\qquad s\models \text {morning}(\textsf {p}),\qquad s\not \models {?(\textsf {h}=\textsf {p})}.$$

Note that the astronomer’s uncertainty is not about linguistic facts; in fact, we need not even suppose that the astronomer has named the relevant objects. Instead, it is about astronomical facts: he is uncertain about what astronomical objects there actually are—whether there are two of them or just one.

We can conceptualize the astronomer’s information state s as involving two epistemic individuals h and p, each of which is known to have certain properties, and as leaving open the issue of whether these individuals are in fact the same. Thus, s contains worlds w such that $h\sim _w p$ (that is, worlds where h and p are in fact the same object) as well as worlds v such that $h\not \sim _v p$ (that is, worlds where h and p are in fact distinct). An example of such an information state is shown in Fig. 5.4.

Fig. 5.4

The information state representing the uncertainty of the ancient astronomer prior to learning whether Hesperus is Phosphorus

Epistemic versus ontic individuals. Our perspective in this section involves thinking of the domain D of an information model as a set of epistemic individuals: objects to which information is attached and which may be found out to be identical, or to be distinct, as more information about the world is acquired. The actual individuals existing at a possible world w are not the objects in D themselves, but instead the equivalence classes $\{[d]_{\sim _w}\,|\,d\in D\}$. Thus, even though we think of the domain D as fixed and common to all worlds, it is still possible to model situations in which one is uncertain about how many objects there are, as a result of uncertainty about the identity relation. Notice also that, if two individuals are in fact the same at a world, then they must of course have the same properties at that world, and applying a function to them should give identical results: this is what motivates the requirement that the relation $\sim _w$ be a congruence with respect to predicates and function symbols.

This discussion implies that, in the context of a model $M=\langle W,D,I,\sim \rangle$ for the language including identity, the relational structure associated to a world is not given simply by the structure $M_w=\langle D,I\rangle$, but rather by the quotient of this structure modulo the congruence $\sim _w$. Let us make this precise.

Definition 5.4.1

Let $M=\langle W,D,I,\sim \rangle$ be a model for the language $\mathcal {L}^{\textsf {Q}=}$. The relational structure associated with a world w is $M_w^{\sim }=\langle D_w^\sim ,I_w^{\sim }\rangle$, where:

• $D_w^{\sim }=\{[d]_{\sim _w}\,|\,d\in D\}$ is the set of equivalence classes modulo $\sim _w$;

• $I_w^{\sim }(f)([d_1]_{\sim _w},\dots ,[d_n]_{\sim _w})=[I_w(f)(d_1,\dots ,d_n)]_{\sim _w}$;

• $\langle [d_1]_{\sim _w},\dots ,[d_n]_{\sim _w}\rangle \in I_w^\sim (R)\iff \langle d_1,\dots ,d_n\rangle \in I_w(R)$.

Since $\sim _w$ is a congruence, this model is well-defined, in the sense that the definitions do not depend on the choice of representatives within an equivalence class. The fact that our semantics generalizes the standard Tarskian semantics can then be stated as follows in the setting of the language $\mathcal {L}^{\textsf {Q}=}$.

Proposition 5.4.2

For any world w in a relational  information model M, any assignment g into M, and any classical formula $\alpha \in \mathcal {L}^{\textsf {Q}=}_c$:

$$M,w\models _g\alpha \iff M_w^\sim \models _{g_w} \alpha \text { in standard Tarskian semantics},$$

 where $g_w$ is the assignment given by $g_w(x)=[g(x)]_{\sim _w}$.

Proof

Straightforward by induction on $\varphi$, using the definition of the structure $M_w^\sim$ for the atomic case and Proposition 5.1.6 for complex formulas.   $\square$

The following observation about identity will turn out useful in the following: once it is settled that $t=t'$, replacement of t by $t'$ in a formula preserves support.

Proposition 5.4.3

(Substitution of known identicals) Let $\varphi \in \mathcal {L}^{\textsf {Q}=}$ and  let $t,t'$ be two terms free for x in $\varphi$. For any M, s and g:

$$s\,\models _g\,(t=t')\quad \Longrightarrow \quad (\;s\models _g\varphi [t/x]\iff s\models _g\varphi [t'/x]\;).$$

Proof

Straightforward by induction on $\varphi$, using the fact that $\sim _w$ is required to be a congruence.   $\square$

id -Models. While we allowed for the possibility of uncertainty about the identity relation, it is often the case that one wants to model scenarios where identity is not at stake—situations in which one knows what individuals there are and is merely uncertain about their properties. Such scenarios are captured by models in which the relation $\sim _w$ at each world simply coincides with the identity $id_D=\{\langle d,d\rangle \,|\,d\in D\}$. If our information model M is of this sort, we will say that M is an id-model.

Definition 5.4.4

(id -models) An information model $M\!=\!\langle W,V,I,\sim \rangle$ is an id-model if for all $w\!\in \! W$, $\sim _w=id_D$.

In the setting of id-models, the semantics of identity atoms can be simplified:

• $M,s\,\models _g\,t=t'\iff \forall w\in s:[t]_g^w=[t']_g^w$.

As we shall see in Sect. 5.5.5, restricting to id-models has repercussions on the logic.

Questions involving identity. By means of identity, some interesting classes of questions become expressible, in addition to those discussed in Sect. 5.2. Let us look at some examples.

Example 5.4.5

(Identification questions) Consider the sentence $\mathord {\exists \!\!\exists }x(x=t)$, where t is a term not containing x. We have:

$$\begin{aligned} s\models _g\mathord {\exists \!\!\exists }x(x=t)\iff & {} \text {there is a }d\in D\text { such that } s\models _{g[x\mapsto d]} (x=t)\\\iff & {} \text {there is a }d\in D\text { such that for all }w\in s,\; [t]_g^w\sim _w d. \end{aligned}$$

Thus, our sentence is supported in a state s in case s establishes of some speicific d that it is identical to the referent of the term t. In an id-model, this clause can be simplified as follows:

$$\begin{aligned} s\models _g\mathord {\exists \!\!\exists }x(x=t)\iff & {} \text {there is a }d\in D\text { such that for all }w\in s,\; [t]_g^w=d\\\iff & {} \text {for all }w,w'\in s,\; [t]_g^w=[t]_g^{w'}. \end{aligned}$$

Thus, $\mathord {\exists \!\!\exists }x(x=t)$ is a question which asks to identify the referent of t. For instance, a question such as “who is Bob’s sister?” can be rendered formally by the formula $\mathord {\exists \!\!\exists }x(x=s(b))$.

It will be convenient to introduce a notation for such identification questions. If t is a term, we let

$$\lambda t:= \mathord {\exists \!\!\exists }x(x=t)$$

where x is a variable not occurring in t.⁷

Notice that if $\textsf {t}$ is a rigid term, then in any model, the referent of $\textsf {t}$ is bound to be the same at every world. So, in this case the support conditions for $\mathord {\exists \!\!\exists }x(x=\textsf {t})$ are always satisfied. Thus, for rigid terms, identification questions are trivial. Note that this does not mean that identity questions $?(\textsf {t}=\textsf {t}')$ involving such terms are trivial, as illustrated by our discussion of Frege’s puzzle above.

Example 5.4.6

(Unique-instance questions) For another interesting example of the sort of questions expressible by means of identity, let us introduce the following abbreviation, analogous to the one commonly used for the classical existential quantifier:

$$\mathord {\exists \!\!\exists }!x\varphi (x)\;:=\;\mathord {\exists \!\!\exists }x\forall y(\varphi (y)\leftrightarrow y=x).$$

Now consider the formula $\mathord {\exists \!\!\exists }!xP(x)$, where P is a predicate symbol. We have:

$$\begin{aligned} s\models _g\mathord {\exists \!\!\exists }! xP(x)\iff & {} \text {there is a }d\in D\text { such that for all }w\in s:\\{} & {} d\in P_w\text { and for all }d'\in P_w, d'\sim _w d\\\iff & {} \text {there is a }d\in D\text { s.t.\ for all }w\in s: \; I_w^{\sim }(P)=\{[d]_{\sim _w}\}. \end{aligned}$$

In the setting of id-models, this can be simplified as follows:

$$\begin{aligned} s\models _g\mathord {\exists \!\!\exists }! xP(x)\iff & {} \text {there is a }d\in D\text { such that for all }w\in s: \; P_w=\{d\}\\\iff & {} \text {the extension of }P\text { is the same singleton at each }w\in s. \end{aligned}$$

Thus, the sentence $\mathord {\exists \!\!\exists }!xP(x)$ is supported in a state s in case s establishes of some individual $d\in D$ that d is the individual who has property P.⁸ Thus, $\mathord {\exists \!\!\exists }x!P(x)$ formalizes the question ‘who is the P?’, which presupposes that exactly one individual has property P and asks for the identity of this individual. Note that, as we expect, the presupposition of this question is $\exists !xPx$ (short for $\exists x\forall y(Px\leftrightarrow x=y)$), the statement that exactly one individual satisfies P.

If we regard the model of Fig. 5.3 as an id-model, where the two individuals a and b are distinct at every world, then the formula $\mathord {\exists \!\!\exists }!xP(x)$ has two alternatives, as depicted in Fig. 5.3d. These alternatives coincide with the truth-sets of the classical formulas $\forall x(P(x)\leftrightarrow x=\textsf {a})$ and $\forall x(P(x)\leftrightarrow x=\textsf {b})$, each of which provides just enough information to establish of some individual that it is the unique P.⁹

Example 5.4.7

(Mention-nquestions) By means of the identity predicate, one can also form questions that ask for more than one instance of objects satisfying a given predicate. For instance, consider the formula

$$\mathord {\exists \!\!\exists }x\mathord {\exists \!\!\exists }y(Px\wedge Py\wedge \lnot (x=y)).$$

It is easy to check that this formula is supported at a state s in case there are two individuals $d,d'$ such that s implies that both d and $d'$ have property P ($d,d'\in P_w$ for all $w\in s$) and that d and $d'$ are distinct ($d\not \sim _w d'$ for all $w\in s$). Thus the above sentence is supported by a state s just in case the information in s provides us with at least two instances of property P, and it can be seen as a formalization of the mention-two question ‘What are two individuals that have property P?’. Of course, the idea can be extended straightforwardly to mention-n questions for $n>2$.

Example 5.4.8

(Cardinality questions) As a last example of the sort of questions that we can form by using identity, consider the following sentence:

$$?\exists !xPx\;:=\;{?\exists x\forall y(Py\leftrightarrow y=x)}.$$

This is a polar question that asks whether the statement $\exists !xPx$ is true. The statement is true at a world w just in case there is exactly one (ontic) individual that has property P (i.e., if the actual extension of P at world w, given by the set $P_w/\!\sim _w$, has cardinality 1). Thus, the above question asks whether exactly one object is P. Similarly, for any natural number n it is easy to write polar questions asking whether at most/at least/exactly n objects are P. On the other hand, as we will discuss in Sect. 5.8, it is not possible to write a formula that expresses the related question “How many things are P?/How many P are there?”, which is settled in a state s just in case the information in s determines exactly how many individuals have property P.

5.5 Entailment

Let us now turn to the entailment relation in $\textsf {InqBQ}$, which is defined in the obvious way: $\Phi$ logically entails $\psi$ (notation: $\Phi \,\models \,\psi$) if for any information model M, information state s and assignment g, if $M,s\models _g\Phi$ then $M,s\models _g\psi$.

Logical equivalence and validity are defined in terms of entailment as usual. Two formulas $\varphi$ and $\psi$ are logically equivalent, denoted $\varphi \equiv \psi$, if they entail each other—which amounts to $\varphi$ and $\psi$ having the same support conditions. A formula $\varphi$ is valid in $\textsf {InqBQ}$, denoted $\models \varphi$, if it is entailed by the empty set—which amounts to $\varphi$ being supported by any information state in any model under any assignment.

Contextual entailment is also defined as usual, modulo a relativization to assignments: $\Phi$ entails $\psi$ in the context of an information state s and relative to an assignment g (notation: $\Phi \models _{s,g}\psi$) in case for every information state $t\subseteq s$, if $t\models _g\Phi$ then $t\models _g\psi$. As usual, implication is tightly connected to contextual entailment: $\varphi \rightarrow \psi$ is supported in a state s just in case $\varphi$ entails $\psi$ in the context of s:

$$s\models _g\varphi \rightarrow \psi \;\iff \; \varphi \models _{s,g}\psi .$$

5.5.1 Illustration

In order to appreciate the sort of logical facts that can be captured as entailments in the logic $\textsf {InqBQ}$, we will first look at some examples. Then in the next section we will examine more closely the formal properties of the entailment relation.

Example 5.5.1

Consider two unary predicates, P and Q. Given the information that P is the complement of Q, the extension of P determines the extension of Q. This fact is an instance of logical dependency that can be captured as a case of entailment in $\textsf {InqBQ}$. The assumption that P is the complement of Q can be formalized as usual by the formula $\forall x(Px\leftrightarrow \lnot Qx)$. The mention-all questions what is the extension of P and what is the extension of Q are formalized, as discussed in Sect. 5.2, by the formulas $\forall x?Px$ and $\forall x?Qx$. Thus, the logical dependency that we observed above amounts to the validity of the following entailment:

$$\forall x(Px\leftrightarrow \lnot Qx),\forall x?Px\models \forall x?Qx.$$

We can see that this entailment is valid by reasoning as follows. Take a state s which supports the premises. In order to support the second premise, the extension of P must be the same at every world in s. In order to support the first premise, the extension of Q must be the complement of the extension of P at every world in s. It follows that the extension of Q is the same at every world in s, which means that s supports the conclusion.

Example 5.5.2

Consider a unary predicate P. Given the information which individuals have property P we can in particular determine whether or not all individuals have property P. This is fact is captured by the entailment:

$$\forall x?Px\models {?\forall xPx}.$$

We can see that this is valid as follows. Suppose a state s supports $\forall x?Px$. Then the extension of P is the same at every world in s. If this extension of P is the entire domain D at every world, then s supports $\forall xPx$. Otherwise, the extension of P is different from D at every world, and then s supports $\lnot \forall xPx$. Either way, s supports $?\forall xPx$.

Notice that the entailment is valid because the domain of quantification is fixed: if uncertainty about the domain of quantification were allowed, then even given a specification of the extension of P, one could still be uncertain about whether or not the relevant extension is the entire domain.

Example 5.5.3

Consider a unary predicate P. Under the assumption that the extension of P is non-empty, from the information about what the extension of P is we can obtain an instance of an object that has property P. Here, the assumption that P is non-empty is captured as usual by $\exists xPx$. The question of what is the extension of P is expressed by $\forall x?Px$, and the question of what is an instance of P is expressed by $\mathord {\exists \!\!\exists }xPx$. The above observation then amounts to the entailment:

$$\exists xPx,\forall x?Px\models \mathord {\exists \!\!\exists }xPx.$$

We can see that this entailment is valid by reasoning as follows. Take a state s that supports the premises (we may assume $s\ne \emptyset$, since the empty set supports every formula). Then the extension of P must be the same set X of individuals at every world in s (second premise) and this set must be non-empty (first premise). If we then take an object $d\in X$, d is in the extension of P at every world in s, and so the conclusion is supported.

Example 5.5.4

Consider a rigid binary function symbol $\textsf {f}$ and three non-rigid individual constants a, b, c. The rigidity of $\textsf {f}$ means that we are assuming the denotation of $\textsf {f}$ to be known. Then if we are given the information that $a=\textsf {f}(b,c)$ as well as information identifying b and c, it follows that we can identify a. Recall that we abbreviate the identification question $\mathord {\exists \!\!\exists }x(x=a)$ as $\lambda a$, and similarly for b and c. Then the above fact is captured by the following entailment:

$$a=\textsf {f}(b,c),\;\lambda b,\;\lambda c\,\models \,\lambda a.$$

While the entailment is valid in general, its validity is particularly easy to verify in the setting of an id-model. In this setting, suppose s is a state that supports $\lambda b$ and $\lambda c$. Then the referent of b and c must be the same individuals $d_b$ and $d_c$ in every world in s. Since $\textsf {f}$ is rigid, it denotes the same function F at all worlds in s. Therefore, the term $\textsf {f}(b,c)$ must also denote the same individual $d'=F(d_b,d_c)$ at every world w. If s also supports $a=\textsf {f}(b,c)$, the denotation of a at each world must be $d'$, and thus in particular it must be the same at every world in s. This guarantees that the conclusion $\lambda a$ is supported.

In the case of a model with variable identity, the argument is essentially the same, but the relevant identities have to be computed locally at each world $w\in s$, using the relation $\sim _w$. The details are left to the reader.

In case we are dealing with a non-rigid function symbol f, the above entailment is no longer valid: even if we are given the values of b and c and the information that $a=f(b,c)$, if we do not know what function f denotes we will not in general be able to identify the value of a. However, we can retrieve the entailment if we add the explicit assumption that for every x and y we can identify the value of f(x, y), which is captured by the formula $\forall x\forall x\mathord {\exists \!\!\exists }z(z=f(x,y))$. This results in the following valid entailment:

$$\forall x\forall y\mathord {\exists \!\!\exists }z(z=f(x,y)),\; a=f(b,c),\;\lambda b,\;\lambda c\;\models \;\lambda a.$$

5.5.2 Entailments with Truth-Conditional Conclusions

Many of the features of inquisitive entailment that we discussed in the setting of propositional logic carry over straightforwardly to the first-order case. To start with, entailment towards truth-conditional formulas is truth-conditional.

Proposition 5.5.5

(Entailment to a truth-conditional conclusion) Let $\Phi \cup \{\alpha \}\subseteq \mathcal {L}^{\textsf {Q}=}$, where $\alpha$ is truth-conditional. We have:

$$\Phi \models \alpha \iff \text {for any model }M,\text { world }w,\text { assignment }g: w\models _g\!\Phi \text { implies }w\models _g\!\psi .$$

The proof is the same as in the propositional case (cf. Proposition 3.7.2). In particular, since classical formulas are truth-conditional, and since their conditions are the standard ones (Proposition 5.1.6), entailment among classical formulas coincides with entailment in classical first-order logic. Thus, $\textsf {InqBQ}$ is a conservative extension of classical first-order logic.

Proposition 5.5.6

(Conservativity over classical first-order logic) If $\Gamma \cup \{\alpha \}\subseteq \mathcal {L}^{\textsf {Q}=}_c$, then $\;\Gamma \models \alpha \iff \Gamma$ entails $\alpha$ in classical first-order logic.

As in the propositional case, Proposition 5.5.5 implies that, when the conclusion is truth-conditional, any assumption $\varphi$ may just as well be replaced by its classical variant $\varphi ^{cl}$.

Proposition 5.5.7

If $\alpha$ is truth-conditional, for any $\Phi$ we have

$$\Phi \models \alpha \iff \Phi ^{cl}\models \alpha .$$

It follows, in particular, that every formula entails its classical variant.

Corollary 5.5.8

For every $\varphi \in \mathcal {L}^{\textsf {Q}=}$, $\varphi \models \varphi ^{cl}$.

Moreover, Proposition 5.5.7 implies that a question $\mu$ entails all and only the statements that follow from its presupposition. Thus, for instance, the question $\mathord {\exists \!\!\exists }xPx$ entails its presupposition $\exists xPx$, any statements that follow from it, and no other statements. For another example, consider the mention-all question $\forall x{?Px}$: we saw that the presupposition of this question is a tautology; it follows that tautologies are the only statements entailed by $\forall x{?Px}$.

5.5.3 Entailments with Truth-Conditional Premises

Let us now turn to entailments from truth-conditional premises. The characterization given for propositional logic carries over: a set $\Gamma$ of truth-conditional formulas entails a formula $\varphi$ in case, in any model M and with respect to any assignment g, $\varphi$ is settled in the specific state $|\Gamma |_M^g=\{w\in W_M\mid w\models _g\Gamma \}$ which corresponds to the information that the formulas in $\Gamma$ are true.

Proposition 5.5.9

(Entailment from truth-conditional assumptions) Let $\Gamma \cup \{\varphi \}\subseteq \mathcal {L}^{\textsf {Q}}$, where all formulas in $\Gamma$ are truth-conditional. We have:

$$\Gamma \models \varphi \iff \text {for all models }M\text { and assignments }g,\;\,|\Gamma |_M^g\!\models _g\varphi .$$

In particular, as we saw in Chap. 1, to say that a statement $\alpha$ entails a question $\mu$ is to say that in any model, the information that $\alpha$ is true suffices to settle the question. For instance, suppose $\textsf {t}$ is a rigid term: the statement $P\textsf {t}$ entails the question $\mathord {\exists \!\!\exists }xPx$, since in any model, the information that $P\textsf {t}$ is true suffices to identify an individual which has property P (namely, that individual which is referent of t in the model), and thus suffices to settle the question $\mathord {\exists \!\!\exists }xPx$.

The property that we called specificity (cf. Proposition 3.7.12) also carries over: if $\Gamma$ is a set of truth-conditional formlas, then $\Gamma$ entails $\varphi$ in the context s just in case extending s with the information that all formulas in $\Gamma$ are true leads to a state that supports $\varphi$. The proof is the same as in the propositional case.

Proposition 5.5.10

(Specificity) Let $\Gamma \cup \{\varphi \}\subseteq \mathcal {L}^{\textsf {Q}=}$, where $\Gamma$ is a set of truth-conditional formulas. For any model M, state s, and assignment g:

$$\Gamma \models _{s,g}\varphi \iff s\cap |\Gamma |_M^g\models _g\psi .$$

Using this fact, it is immediate to check that the local split property still holds for  , and an analogous property holds for $\mathord {\exists \!\!\exists }$  as well.

Proposition 5.5.11

(Local split properties) Let $\Gamma$ be a set of truth-conditional formulas, and let $\varphi ,\psi$ be arbitrary formulas. Then for any model M, information state s and assignment g we have:

• ;

• $\Gamma \models _{s,g}\mathord {\exists \!\!\exists }x\varphi \iff \text {for some }d\in D:\Gamma \models _{s,g[x\mapsto d]}\varphi$, provided $x\not \in FV(\Gamma )$.

Proof

The proof of the first item is identical to the one given in the propositional case. For the second item, we have

$$\begin{aligned} \Gamma \models _{s,g}\mathord {\exists \!\!\exists }x\varphi\iff & {} s\cap |\Gamma |_M^g\models _g\mathord {\exists \!\!\exists }x\varphi \\\iff & {} \text {for some }d\in D: s\cap |\Gamma |_M^g\models _{g[x\mapsto d]}\varphi \\\iff & {} \text {for some }d\in D: s\cap |\Gamma |_M^{g[x\mapsto d]}\models _{g[x\mapsto d]}\varphi \\\iff & {} \text {for some }d\in D: \Gamma \models _{s,g[x\mapsto d]}\varphi \end{aligned}$$

where the first and last step use Specificity, while the third step uses the fact that $|\Gamma |_M^g=|\Gamma |_M^{g[x\mapsto d]}$ because x does not occur free in $\Gamma$.   $\square$

As in the propositional case, these properties amount to the validity of certain logical equivalences, which allow us to distribute a truth-conditional antecedent over an inquisitive consequent.

Proposition 5.5.12

(Split equivalences) Let $\alpha$ be a truth-conditional formula, and let $\varphi ,\psi$ be arbitrary formulas. We have:

•   split: ;

• $\mathord {\exists \!\!\exists }$  split: if $x\not \in \text {FV}(\alpha )$, .

Proof

Again, we spell out only the case for $\mathord {\exists \!\!\exists }$ , since the one for    is the same as in propositional logic. Take any model M, state s, and assignment g. Using the previous proposition as well as the connection between implication and contextual entailment, we have:

We also have logical counterparts of the split properties. For the case of disjunction, the relevant property is the same as in propositional logic. The proof strategy is also the same: we take two countermodels and combine them into a single one by a disjoint union construction. However, in the current setting the relevant construction must be a bit more subtle, due to the requirement to get a single common domain for the two models being combined.

Theorem 5.5.13

(Logical split property for  $|\vee$ (Grilletti [2]))

Let $\Gamma \cup \{\varphi ,\psi \}\subseteq \mathcal {L}^{\textsf {Q}=}$, where $\Gamma$ is a set of truth-conditional formulas. Then:

Proof

Given two relational information models, $M^A=\langle W^A,D^A,I^A,\sim ^A\rangle$ and $M^B=\langle W^B,D^B,I^B,\sim ^B\rangle$, we define a new model $M^B\oplus M^B=\langle W,D,I,\sim \rangle$, where:

• W is the disjoint union of $W^A$ and $W^B$: $W=W^A\uplus W^B$.

• D is the Cartesian product of $D^A$ and $D^B$: $D=D^A\times D^B$.

• For every relation symbol R and world w:

  • if $w\in W^A$, $I_w(R)(\langle a_1,b_1\rangle ,\dots ,\langle a_n,b_n\rangle )\iff I_w^A(R)(a_1,\dots ,a_n)$;

  • if $w\in W^B$, $I_w(R)(\langle a_1,b_1\rangle ,\dots ,\langle a_n,b_n\rangle )\iff I_w^B(R)(b_1,\dots ,b_n)$.

• Similarly, for any world w:

  • if $w\in W^A$, $\langle a_1,b_1\rangle \sim _w\langle a_2,b_2\rangle \iff a_1\sim _w^A a_2$;

  • if $w\in W^B$, $\langle a_1,b_1\rangle \sim _w\langle a_2,b_2\rangle \iff b_1\sim _w^B b_2$.

• For a non-rigid function symbol f:

  $$I_w(f)(\langle a_1,b_1\rangle ,\dots ,\langle a_n,b_n\rangle )=\left\{ \begin{array}{l l} \langle I_w^A(f)(a_1,\dots ,a_n),b_0\rangle &{}\text {if }w\in W^A\\ \langle a_0,I_w^B(f)(b_1,\dots ,b_n)\rangle &{}\text {if }w\in W^B \end{array}\right. $$

  where $a_0$ is an arbitrary element of $D^A$ and $b_0$ an arbitrary element of $D^B$.

• For a rigid function  symbol $\textsf {f}$:

  $$I_w(\textsf {f})(\langle a_1,b_1\rangle ,\dots ,\langle a_n,b_n\rangle )=\langle F^A(a_1,\dots ,a_n),F^B(b_1,\dots ,b_n)\rangle $$

  where $F^A=I_{w'}^A(\textsf {f})$ for an arbitrary world $w'\in W^A$, and $F^B=I_{w''}^B(\textsf {f})$ for an arbitrary world $w''\in W^B$ (since $\textsf {f}$ is rigid, the choice of $w'$ and $w''$ does not matter). Note that the interpretation of f given in this way is indeed rigid, i.e., it yields the same individual for any world w in the model.

It is straightforward to check that $\sim _w$ is indeed a congruence at every world, which guarantees that $M^A\oplus M^B$ is a relational information model.

The crucial feature of the sum model $M^A\oplus M^B$ is that it behaves like the model $M^A$ on states $s\subseteq W^A$, and like the model $M^B$ on states $s\subseteq W^B$. To make this precise, let $\pi _1: D^A\times D^B\rightarrow D^A$ and $\pi _2:D^A\times D^B\rightarrow D^B$ be the natural projection functions. Then for any assignment g into $M^A\oplus M^B$, any information state s in this model, and any formula $\chi$ we have:

• if $s\subseteq W^A$ then: $\;M^A\oplus M^B,s\models _g\chi \iff M^A, s\models _{\pi _1\circ g}\chi$;

• if $s\subseteq W^B$ then: $\;M^A\oplus M^B,s\models _g\chi \iff M^B, s\models _{\pi _2\circ g}\chi$.

The inductive verification of the claim is left as an exercise.

With this model-theoretic construction at hand, we are ready to prove our theorem. By contraposition, suppose $\Gamma \not \models \varphi$ and $\Gamma \not \models \psi$. Then we can find models $M^A=\langle W^A,D^A,I^A,\sim ^A\rangle$ and $M^B=\langle W^B,D^B,I^B,\sim ^B\rangle$, and corresponding information states $s^A,s^B$ and assignments $g^A$ and $g^B$ such that:

• $M^A,s^A\models _{g^A}\Gamma ,\quad M^A,s^A\not \models _{g^A}\varphi$;

• $M^B,s^B\models _{g^B}\Gamma ,\quad M^B,s^B\not \models _{g^B}\psi$.

Now consider the model $M^A\oplus M^B$ and the information state $s=s^A\uplus s^B$ obtained as the disjoint union of $s^A$ and $s^B$. Also, define a valuation function g by setting $g(x)=\langle g^A(x),g^B(x)\rangle$ and notice that $g^A=\pi _1\circ g$ and $g^B=\pi _2\circ g$. By the above property of the sum model, we have:

• $M^A\oplus M^B,s^A\models _{g}\Gamma ,\quad M^A\oplus M^B,s^A\not \models _{g}\varphi$;

• $M^A\oplus M^B,s^B\models _{g}\Gamma ,\quad M^A\oplus M^B,s^B\not \models _{g}\psi$.

By persistency, since $s^A$ and $s^B$ are substates of $s^A\uplus s^B$ we have $M^A\oplus M^B,s^A\uplus s^B\not \models _{g}\varphi$ and $M^A\oplus M^B,s^A\uplus s^B\not \models _{g}\psi$, and therefore also:

At the same time, consider an arbitrary formula $\alpha \in \Gamma$ and a world $w\in s^A\uplus s^B$. Suppose $w\in s^A$: since $M^A\oplus M^B,s^A\models _{g}\alpha$, by persistency we have $M^A\oplus M^B,w\models _{g}\alpha$. The same conclusion can be reached in case $w\in s^B$, using the fact that $M^A\oplus M^B,s^B\models _{g}\alpha$. Thus, for all $w\in s^A\uplus s^B$ we have $M^A\oplus M^B,w\models _{g}\alpha$. Since $\alpha$ is truth-conditional, it follows that $M^A\oplus M^B,s^A\uplus s^B\models _{g}\alpha$. And since this is the case for all $\alpha \in \Gamma$, we have: $M^A\oplus M^B,s^A\uplus s^B\models _{g}\Gamma$.

Thus, we can conclude that

.    $\square$

We can also prove a logical split property for the inquisitive existential quantifier: whenever a set $\Gamma$ of truth-conditional formulas entails an inquisitive existential formula $\mathord {\exists \!\!\exists }x\varphi (x)$, this is traceable to the fact that it entails $\varphi (\textsf {t})$ for some rigid term $\textsf {t}$. However, the proof of this fact is complex, involving non-trivial model-theoretic constructions. The interested reader is referred to Grilletti [2] for the details.

Theorem 5.5.14

(Logical split property for $\mathord {\exists \!\!\exists }$  (Grilletti [2]))

Let $\Gamma \cup \{\varphi \}\subseteq \mathcal {L}^{\textsf {Q}=}$, where all formulas in $\Gamma$ are truth-conditional. Then:

$$\Gamma \models \mathord {\exists \!\!\exists }x\varphi \iff \Gamma \models \varphi [\textsf {t}/x]\text { for some rigid term }\textsf {t}\text { free for }x\text { in }\varphi .$$

Note that by taking $\Gamma =\emptyset$ in the above theorems we obtain for the inquisitive operators the disjunction and existence property familiar from intuitionistic logic and arithmetic.

Corollary 5.5.15

(Disjunction and existence property) For all $\varphi ,\psi \in \mathcal {L}^{\textsf {Q}=}$:

• if   is valid then $\varphi$ is valid or $\psi$ is valid;

• if $\mathord {\exists \!\!\exists }x\varphi$ is valid then $\varphi [\textsf {t}/x]$ is valid for some rigid term t.

Note also that if a certain signature $\Sigma$ does not contain any rigid function symbols, the only rigid terms in the language are variables. Thus, in this case if $\Gamma \,\models \,\mathord {\exists \!\!\exists }x\varphi$ we can conclude $\Gamma \models \varphi [y/x]$ for some variable y free for x in $\varphi$. If additionally $\Gamma$ is a set of sentences, we have $\Gamma \models \varphi [y/x]$ only in case $\Gamma \models \forall x\varphi$. So, we also have the following corollary.

Corollary 5.5.16

Let $\Gamma \cup \{\psi \}\subseteq \mathcal {L}^{\textsf {Q}=}(\Sigma )$, where $\Sigma$ contains no rigid function symbols and $\Gamma$ is a set of truth-conditional sentences. Then:

$$\Gamma \models \mathord {\exists \!\!\exists }x\varphi \iff \Gamma \models \forall x\varphi .$$

5.5.4 The Role of Rigidity

The logical properties of a term depend crucially on whether or not the term is rigid. Rigid terms yield witnesses for the inquisitive existential quantifier, and allow instantiation from the universal quantifier.

Proposition 5.5.17

For any $\varphi (x)\in \mathcal {L}^{\textsf {Q}=}$ and any rigid term $\textsf {t}$ which is free for x in $\varphi$ we have:

• $\varphi (\textsf {t})\models \mathord {\exists \!\!\exists }x\varphi (x)$;

• $\forall x\varphi (x)\models \varphi (\textsf {t})$.

Proof

Consider an arbitrary information state s and assignment g. Since $\textsf {t}$ is rigid, there is an object $d_{\textsf {t}}\in D$ such that in all worlds $w\in s$, $[\textsf {t}]_w^g=d_{\textsf {t}}$. This means that in every world $w\in s$, $[\textsf {t}]_w^g=[x]_w^{g[x\mapsto d_{\textsf {t}}]}$. Using this fact, it is straightforward to check by induction that for any formula $\varphi (x)$ we have:

$$s\models _g\varphi (\textsf {t})\iff s\models _{g[x\mapsto {d_{\textsf {t}}}]}\varphi (x).$$

Now suppose $s\models _g\varphi (\textsf {t})$. Then $s\models _{g[x\mapsto {d_{\textsf {t}}}]}\varphi (x)$, and so . This proves the first entailment.

Next, suppose $s\models _g\forall x\varphi (x)$. Then in particular we have $s\models _{g[x\mapsto d_{\textsf {t}}]}\varphi (x)$, and thus also $s\models _g\varphi (\textsf {t})$. This shows the second entailment.   $\square$

To see that the above proposition does not hold in general if t is non-rigid, consider a unary predicate symbol P and a non-rigid constant c. We have A counterexample to the entailment is given by the model in Fig. 5.5a: the state $\{w,v\}$ in the model supports P(c) but not $\mathord {\exists \!\!\exists }xP(x)$. Intuitively, the entailment fails because establishing that c has property P is not sufficient to identify an object with property P, if we do not know what object c refers to. For instance, suppose c stands for ‘the thief’ and P for the property ‘having stolen the jewels’. Having the information that the thief stole the jewels (P(c)) does not allow us to settle who stole the jewels ($\mathord {\exists \!\!\exists }xP(x)$) unless we know who the thief is (i.e., unless our state supports $\mathord {\exists \!\!\exists }x(x=c)$).

Fig. 5.5

Two countermodels showing the invalidity of the schemata $\varphi (t)\models \mathord {\exists \!\!\exists }x\varphi (x)$ and $\forall x\varphi (x)\models \varphi (t)$ when t is non-rigid. In both models, the domain is $D=\{a,b\}$, c is a non-rigid constant and P a unary predicate

Similarly, if c is non-rigid we also have:

$$\forall x{?P(x)}\not \models {?P(c)}.$$

A counterexample is given by the model of Fig. 5.5b: the information state $\{w,v\}$ in the model supports $\forall x?P(x)$ but not ?P(c). Again, this failure is intuitively motivated: knowing the extension of P is not enough to know whether the object denoted by c is in this extension, unless we know what this object is. For instance, suppose again that c stands for ‘the thief’ and P for ‘being in this room’. We may know that only Alice is in this room, and so know exactly what the extension of P is. However, we may not know whether the thief is in this room, since we may not know whether Alice is the thief.

On the other hand, since we saw that $\textsf {InqBQ}$ is a conservative extension of classical first-order logic, it follows that all terms, rigid or not, bear the standard relation to the classical quantifiers, in the context of classical formulas. Conceptually, this is because the semantics of classical formulas can be assessed in a point-wise way, world by world, and at the level of a single world, there is no difference between rigid and non-rigid terms.

Proposition 5.5.18

For any classical formula $\alpha (x)\in \mathcal {L}^{\textsf {Q}=}_c$ and any term t which is free for x in $\varphi$, whether rigid or not, we have:

• $\alpha (t)\models \exists x\alpha (x)$; (notice the classical quantifier!)

• $\forall x\alpha (x)\models \alpha (t)$.

In fact, for the case of $\exists$ the restriction  to classical formulas is inessential, as the following proposition shows.

Proposition 5.5.19

For any formula $\varphi (x)\in \mathcal {L}^{\textsf {Q}=}$ and any term t free for x in $\varphi$ we have $\varphi (t)\models \exists x\varphi (x)$.

Proof

Recall that $\exists x\varphi (x)$ is an abbreviation for $\lnot \forall x\lnot \varphi (x)$. By Proposition 5.3.8, negations are always truth-conditional, and by Proposition 5.3.7, truth-conditional formulas are logically equivalent to their classical variant. Thus, $\exists x\varphi (x)\equiv (\exists x\varphi (x))^{cl}$ $=\exists x\varphi ^{cl}(x).$ By Corollary 5.5.8, $\varphi (t)\models \varphi ^{cl}(t)$. Since $\varphi ^{cl}$ is a classical formula, by the previous proposition we have $\varphi ^{cl}(t)\models \exists x\varphi (x)$. Putting everything together, we have $\varphi (t)\models \varphi ^{cl}(t)\models \exists x\varphi ^{cl}(x)\equiv \exists x\varphi (x).$    $\square$

5.5.5 id-Entailment

If we are only interested in cases where the extension of identity is not at stake, we will want to work with a stronger notion of entailment, one that only takes id-models into account. We will refer to this stronger notion as id-entailment.

Definition 5.5.20

(id -entailment) $\Phi \models _\textsf {id}\psi \iff$ for all id-models M, all states s in M, and all assignments g, $\;M,s\models _g\Phi \;$ implies $\;M,s\models _g\psi$.

We refer to the corresponding notions of equivalence and validity as id-equivalence (denoted $\equiv _\textsf {id}$) and id-validity.

In an id-model, matters concerning identities of individuals are assumed to be settled a priori. This is captured by the fact that the question $\forall x\forall y?(x=y)$, which asks to specify the extension of the identity relation, is an id-validity.

Proposition 5.5.21

$\forall x\forall y?(x=y)$ is id-valid.

Proof

Spelling out the support conditions of the question, we have:

$$\begin{aligned} s\models \forall x\forall y?(x=y)\iff & {} \text {for all }w,w'\in s: \;\; \sim _w=\,\sim _{w'}. \end{aligned}$$

This condition is always satisfied in an id-model, since in such a model $\sim _w$ coincides with the identity relation at every $w\in W$.    $\square$

Since universals can always be validly instantiated to a rigid term (Proposition 5.5.17), this also gives the following corollary.

Corollary 5.5.22

For any rigid terms $\textsf {t, t{\,}'}$, $?(\textsf {t}=\textsf {t}{\,}')$ is id-valid.

This corollary can be generalized into the following proposition, which says that the truth value of a statement concerning only identities between rigid terms is settled a priori in an id-model.

Proposition 5.5.23

Let $\alpha$ be a classical formula built up from identity atoms of the form $(\textsf {t}=\textsf {t}{\,}')$ where $\textsf {t,t}{\,}'$ are rigid. Then $?\alpha$ is id-valid.

Proof

Consider any id-model M. If $\textsf {t,t{\!}'}$ are rigid then all worlds in M assign the same truth-value to the atom $(\textsf {t}=\textsf {t}')$ relative to every assignment. Since truth conditions for classical formulas can be computed recursively in the standard way, all worlds in M assign the same truth value to any classical formula $\alpha$ built up from such atoms. This means that every state s in M supports the polar question $?\alpha$ (recall that $?\alpha$ is supported precisely if all worlds in the state agree on the truth value of $\alpha$).   $\square$

A consequence of this proposition is that, in the setting of an id-model, characterizing a predicate in terms of identities with rigid terms counts as settling its extension.

Proposition 5.5.24

Let $\alpha$ and $\beta$ be classical formulas, where $\alpha$ is as in the previous proposition, and let $\overline{x}$ be a sequence of variables. Then:

$$\forall \overline{x}(\alpha \leftrightarrow \beta )\models _\textsf {id}\forall \overline{x}{?\beta }.$$

Proof

It is easy to verify that the following entailment is generally valid, and thus also id-valid:

$$\forall \overline{x}(\alpha \leftrightarrow \beta ),\forall \overline{x}{?\alpha }\models \forall \overline{x}{?\beta }.$$

By the previous proposition, the formula ${?\alpha }$ is id-valid, and thus so is its universal closure $\forall \overline{x}{?\alpha }$. Obviously, adding id-valid premises does not make a difference to the validity of an id-entailment, so our claim follows.   $\square$

As notable special cases we have the following, where $\textsf {t}_1,\dots ,\textsf {t}_n$ are rigid:

• $\forall x(P(x)\;\leftrightarrow \; (x=\textsf {t}_1\vee \dots \vee x=\textsf {t}_n))\models _\textsf {id}\forall x{?P(x)}$;

• $\forall x(P(x)\;\leftrightarrow \; (x\ne \textsf {t}_1\wedge \dots \wedge x\ne \textsf {t}_n))\models _\textsf {id}\forall x{?P(x)}$.

To get a sense of the differences between general entailment and id-entailment, it is helpful to consider a couple of further examples.

Example 5.5.25

Take any natural number $n\ge 1$. Consider the standard first-order sentence that says that there are exactly n individuals in the domain:

$$\chi _n:=\exists x_1\dots \exists x_n\left( \bigwedge _{i< j}(x_i\ne x_j)\wedge \forall y(y=x_1\vee \dots \vee y=x_n)\right) .$$

In an id-model, this is either true at every world (if the cardinality of D is n) or false at every world (if the cardinality of D is not n). In the former case, $\chi _n$ is supported at any state in the model; in the latter case, $\lnot \chi _n$ is supported at any state in the model. Either way, the polar question $?\chi _n$ is supported at any state in the model. This shows that this question is id-valid: $\models _\textsf {id}{?\chi _n}$.

By contrast, the polar question $?\chi _n$ is not generally valid in $\textsf {InqBQ}$. To see this, consider a model M with set of possible worlds $W=\{w_1,w_2,w_3,\dots \}$, domain $D=\mathbb {N}$, and such that $x\sim _{w_n} y$ holds iff x and y are equivalent modulo n. In this model, the sentence $\chi _n$ is true at world $w_n$ but false at world $w_m$ for $m\ne n$. Thus, the polar question $?\chi _n$ is not supported by the state W.

Example 5.5.26

Let $\textsf {a}_1,\textsf {a}_2,\textsf {b}_1,\textsf {b}_2$ be rigid constants, and let $\Gamma$ be the set of the following classical sentences:

• $\forall x(Px\rightarrow (x\!=\!\textsf {a}_1\vee x\!=\! \textsf {a}_2))$,

• $P\textsf {a}_1\rightarrow Q\textsf {b}_1$,

• $P\textsf {a}_2\rightarrow Q\textsf {b}_2$.

Given $\Gamma$, it might at first seem that from a witness for P we can derive a witness for Q, and so, that we should have $\Gamma ,\mathord {\exists \!\!\exists }xPx\models \mathord {\exists \!\!\exists }xQx$. However, this is not the case. For suppose we are given the information that d is a witness of P: we know from $\Gamma$ that either $d=[\textsf {a}_1]$, in which case $[\textsf {b}_1]$ is a witness for Q, or $d=[\textsf {a}_2]$, in which case $[\textsf {b}_2]$ is a witness for Q. But if we cannot tell which of these two cases applies, we are in effect unable to produce a witness for Q. And indeed, it is not hard to show that:

$$\Gamma ,\;\mathord {\exists \!\!\exists }xPx\;\not \models \;\mathord {\exists \!\!\exists }xQx.$$

On the other hand, in the context of an id-model, the relevant dependency holds, since it is guaranteed that given the information that d has property P, we will be able to tell whether $d=[\textsf {a}_1]$ or $d=[\textsf {a}_2]$, and thus to produce a corresponding witness for Q. And indeed, it is easy to check that:

$$\Gamma ,\;\mathord {\exists \!\!\exists }xPx\;\models _\textsf {id}\;\mathord {\exists \!\!\exists }xQx.$$

Failure of the logical split properties. It is interesting to note that the disjunction property for and the existence property for $\mathord {\exists \!\!\exists }$, given by Corollary 5.5.15 in the context of general entailment, both fail for id-entailment.

For the disjunction property, note that the formula $?(x=y)$ is id-valid and amounts to the disjunction , but obviously neither $x=y$ nor $\lnot (x=y)$ are id-valid. Alternatively, consider again the cardinality sentence $\chi _n$ from Example 5.5.25, which says that there are exactly n distinct individuals: we saw above that $?\chi _n$ is id-valid, but again neither $\chi _n$ nor $\lnot \chi _n$ is id-valid.

We can also use the same idea to build a counterexample to the existence property for $\mathord {\exists \!\!\exists }$ . For instance, consider the formula

$$\mathord {\exists \!\!\exists }x(((x=y)\wedge \chi _2)\vee ((x=z)\wedge \lnot \chi _2)).$$

To see that this is valid on id-models, take an arbitrary id-model M, an information state s, and an assignment g. There are two possibilities: either the cardinality of the domain D is 2, or it is different from 2. In the first case, since M is an id-model we have $s\,\models _g\,\chi _2$. Then by choosing $d=g(y)$ we have that $s\models _{g[x\mapsto d]}(x=y)\wedge \chi _2$, from which it follows that $s\models _g\mathord {\exists \!\!\exists }x(((x=y)\wedge \chi _2)\vee ((x=z)\wedge \lnot \chi _2))$. If the cardinality of D is different from 2, then again since M is an id-model we have $s\models _g\lnot \chi _2$. Then by choosing $d=g(z)$ we have $s\models _{g[x\mapsto d]}(x=z)\wedge \lnot \chi _2$, and again it follows that $s\models _g\mathord {\exists \!\!\exists }x(((x=y)\wedge \chi _2)\vee ((x=z)\wedge \lnot \chi _2))$. In either case, our formula is supported, which shows that this formula is id-valid.

However, we claim that there is no rigid term $\textsf {t}$ such that the instantiation

$$((\textsf {t}=y)\wedge \chi _2)\vee ((\textsf {t}=z)\wedge \lnot \chi _2)$$

is id-valid. Indeed, let $\textsf {t}$ be any rigid term. If $\textsf {t}\ne y,z$, it is obvious that the resulting formula is not going to be valid, since then we can easily give an id-model and assignment which satisfy $\lnot (\textsf {t}=y)\wedge \lnot (\textsf {t}=z)$, which ensures that our disjunction is not supported. So, we only have to consider the cases $\textsf {t}=y$ and $\textsf {t}=z$. For $\textsf {t}=y$ we get the formula

$$((y=y)\wedge \chi _2)\vee ((y=z)\wedge \lnot \chi _2).$$

This is not a validity: to refute it, take a model M where D contains more than two elements, and an assignment g such that $g(y)\ne g(z)$. Then at any world w in M, $\chi _2$ is false and $(y=z)$ is false, so the above disjunction is false, which implies that this disjunction is not valid.

Similarly, for $\textsf {t}=z$ we get $((z=y)\wedge \chi _2)\vee ((z=z)\wedge \lnot \chi _2)$, which is not a validity either: to refute it, just take a model where D contains exactly two elements and an assignment g such that $g(y)\ne g(z)$.

Thus, the above inquisitive existential is id-valid while no instantiation of it with a rigid term is id-valid—in contrast with the existence property that characterizes our general notion of validity.

General entailment and id -entailment. What is the exact relation between general and id-entailment? To answer this question, let us first introduce the auxiliary notion of a model with decidable identity. This is a model where there is no uncertainty about the extension of the identity relation.

Definition 5.5.27

(Decidable identity) A model $M=\langle W,D,I,\sim \rangle$ has decidable identity if $\forall w,w'\in W: {\sim _w}={\sim _{w'}}$.

Note that the formula $\forall x\forall y?(x=y)$ characterizes a model as having decidable identity, in the following sense.

Remark 5.5.28

$M\text { has decidable identity}\iff M,W\models \forall x\forall y?(x=y)$.

Clearly, id-models have decidable identity. On the other hand, a model M with decidable identity is not necessarily an id-model, since $\sim _w$ might be the same relation at every world while being different from the identity relation on D. However, in this case, one can always simplify M turn it into an id-model by taking its quotient modulo $\sim _w$.

Definition 5.5.29

(Turning a model with decidable identity into an id -model) Let $M=\langle W,D,I,\sim \rangle$ have decidable identity. Let us write $\sim$ for $\sim _w$ where w is an arbitrary world, and let us write [d] for the equivalence class of d modulo $\sim$. The id -contract of M is the id-model $M^\textsf {id}=\langle W,D/{\sim },I^\sim ,\approx \rangle$, where:

• $D/{\sim }=\{[d]\,|,d\in D\}$;

• $I_w^\sim (f)([d_1],\dots ,[d_n])=[I_w(f)(d_1,\dots ,d_n)]$;

• $\langle [d_1],\dots ,[d_n]\rangle \in I_w^\sim (R)\iff \langle d_1,\dots ,d_n\rangle \in I_w(R)$;

• ${\approx _w}$ is the identity relation on $D/{\sim }$ for any $w\in W$.

The fact that $\sim$ is a congruence at each world guarantees that this is a good definition, i.e., that the definition of $I_w^\sim (f)$ and $I_w^\sim (R)$ does not depend on the choice of representatives for each equivalence class.

The following proposition ensures that this transformation does not affect the satisfaction of formulas. The straightforward proof is omitted.

Proposition 5.5.30

Let M have decidable identity and let s be a state in M and g a valuation into M. Let $g^\textsf {id}:\textsf {Var}\rightarrow W/\!\!\sim$ be the valuation $x\mapsto [g(x)]$. For any formula $\varphi \in \mathcal {L}^{\textsf {Q}=}$:

$$M,s\models _g\varphi \iff M^\textsf {id},s\models _{g^\textsf {id}}\varphi .$$

We can now prove that the relation between general entailment and id-entailment is simple: id-entailment can be simulated within general entailment by adding the decidability of identity as an extra premise.

Proposition 5.5.31

(Simulating id-entailment) For any $\Phi \cup \{\psi \}\subseteq \mathcal {L}^{\textsf {Q}=}$,

$$\Phi \,\models _\textsf {id}\,\psi \iff \Phi ,\,\forall x\forall y?(x\!=\!y)\,\models \,\psi .$$

Proof

The right-to-left direction of the theorem follows immediately from the fact that $\forall x\forall y?(x\!=\!y)$ is id-valid. For the converse, we reason by contraposition. Suppose $\Phi ,\forall x\forall y?(x\!=\!y)\not \models \psi$. Then, there must be a model $M=\langle W,D,I,\sim \rangle$, a state s, and an assignment g such that $M,s\models _g\Phi$ and $M,s\models _g\forall x\forall y?(x\!=\!y)$, but $M,s\not \models _g\psi$. Now consider the restriction of M to s, $M_{|s}$. By Locality (Proposition 5.3.3) we have $M,s\models _g\chi \iff M_{|s},s\models _g\chi$ for any formula $\chi$. Since the universe of $M_{|s}$ is s and we have $M_{|s},s\models _g\forall x\forall y?(x\!=\!y)$, by Remark 5.5.28 the model $M_{|s}$ has decidable identity. Thus, as we just saw, by a quotient construction it can be turned into an id-model $(M_{|s})^{\textsf {id}}$ based on the same universe that satisfies the same formulas at every state. In particular, we have $(M_{|s})^{\textsf {id}},s\models _{g^\textsf {id}}\Phi$ but $(M_{|s})^{\textsf {id}},s\not \models _{g^\textsf {id}}\psi$, which shows that $\Phi \not \models _\textsf {id}\psi$.   $\square$

Conversely, it is also possible to simulate general entailment within id-entailment. The trick is to treat non-rigid identity as a new predicate, adding axioms ensuring that it is interpreted as a congruence.

Formally, given a signature $\Sigma$ we can consider a signature $\Sigma ^{\asymp }$ which extends $\Sigma$ with a fresh binary predicate $\asymp$ (we write $t\asymp t'$ instead of ${\asymp }(t,t')$). Now for any formula $\varphi \in \mathcal {L}^{\textsf {Q}=}(\Sigma )$, consider the formula $\varphi ^\asymp \in \mathcal {L}^{\textsf {Q}}(\Sigma ^{\asymp })$ obtained by replacing each identity atom $t=t'$ in $\varphi$ by a corresponding atom $t\asymp t'$.

Now to an arbitrary model M for the signature $\Sigma$ we can associate an id-model $M^{\asymp }$ for the extended signature $\Sigma ^{\asymp }$ where identity is interpreted rigidly as $\{\langle d,d\rangle \mid d\in D\}$, but where the role of non-rigid identity is taken over by $\asymp$, i.e., for all worlds w we have $I_w(\asymp )={\sim _w}$ where $\sim _w$ is the interpretation of identity at w in the model M. It is obvious from the construction that for any information state s in M and any assignment g we have:

$$M,s\models _g\varphi \iff M^\asymp ,s\models _g\varphi ^\asymp .$$

It is easy to check that the map $(\cdot )^{\asymp }$ gives a 1-1 correspondence between the class of models for $\Sigma$ and the class of id-models for $\Sigma ^\asymp$ where $\asymp$ is interpreted as a congruence at each world.

The fact that $\asymp$ is interpreted as a congruence at each world can be captured by a set of formulas $\text {Cong}^\asymp$ in the language $\mathcal {L}^{\textsf {Q}}(\Sigma ^\asymp )$, containing the axioms of an equivalence relation, namely,

• $\forall x(x\asymp x)$,

• $\forall x\forall y(x\asymp y\,\rightarrow \, y\asymp x)$,

• $\forall x\forall y\forall z(x\asymp y\,\wedge \, y\asymp z\,\rightarrow \, x\asymp z)$,

in addition to the following formulas for all predicates R and function symbol f in the signature $\Sigma$:

• $\forall x_1\dots x_n\forall y_1\dots y_n(\bigwedge _{1\le i\le n}(x_i\asymp y_i)\rightarrow (R(x_1,\dots ,x_n)\leftrightarrow R(y_1,\dots ,y_n))$,

• $\forall x_1\dots x_n\forall y_1\dots y_n(\bigwedge _{1\le i\le n}(x_i\asymp y_i)\rightarrow (f(x_1,\dots ,x_n)\asymp f(y_1,\dots ,y_n))$.

We can now state exactly how general entailment can be simulated within id-entailment. The proof follows straightforwardly from the preceding discussion.

Proposition 5.5.32

(Simulating general entailment) For any $\Phi \cup \{\psi \}\subseteq \mathcal {L}^{\textsf {Q}=}$,

$$\Phi \,\models \,\psi \iff \Phi ^\asymp \cup \text {Cong}^\asymp \,\models _{\textsf {id}}\,\psi ^\asymp .$$

5.5.6 Open Problems

Some of the key questions about the meta-theoretical properties of $\textsf {InqBQ}$ are currently open, in spite of much recent progress. In this section we briefly survey some of these questions. As we will see later on, these questions have recently been answered in restriction to several interesting fragments of the logic.

A first major question concerns the compactness of entailment in $\textsf {InqBQ}$.

Open Problem 5.5.33

(Entailment compactness) Let $\Phi \cup \{\psi \}\subseteq \mathcal {L}^{\textsf {Q}=}$. Is it always the case that if $\Phi \models \psi$ there is a finite subset $\Phi _0\subseteq \Phi$ such that $\Phi _0\models \psi$?

It should be noted that compactness is oftentimes formulated not in terms of entailment, but in terms of satisfiability. While the two formulations are equivalent in the context of classical predicate logic, they come apart in the inquisitive setting. In the inquisitive setting, it is natural to call a set $\Phi$ of formulas satisfiable if there are a model M, a non-empty information state s, and an assignment g such that $M,s\models _g\Phi$. It is then easy to show that $\textsf {InqBQ}$ is compact in the sense of satisfiability, as made precise by the following proposition.

Proposition 5.5.34

(Satisfiability compactness) Let $\Phi \subseteq \mathcal {L}^{\textsf {Q}=}$. If every finite subset of $\Phi$ is satisfiable then $\Phi$ is satisfiable.

Proof

First note that by persistency, if $M,s\models _g\Phi$ then for all worlds $w\in s$ we have $M,\{w\}\models _g\Phi$, that is, $M,w\models _g\Phi$. This means that a set $\Phi$ is satisfiable just in case it is true at some world relative to some assignment. In particular, for a set $\Gamma$ of classical formulas, satisfiability is just satisfiability in the sense of classical logic. By Proposition 5.3.6 we know that an arbitrary set of formulas $\Phi$ has the same truth conditions as the set $\Phi ^{cl}$ obtained by replacing each formula $\varphi \in \Phi$ by its classical variant $\varphi ^{cl}$. So, we have that $\Phi$ is satisfiable iff $\Phi ^{cl}$ is satisfiable in classical first-order logic. Using this fact and the compactness of classical first-order logic, for an arbitrary set $\Phi \subseteq \mathcal {L}^{\textsf {Q}=}$ we have:

$$\begin{aligned} \Phi \text { is satisfiable}\iff & {} \Phi ^{cl}\text { is satisfiable }\\\iff & {} \text {for all finite }\Gamma \subseteq \Phi ^{cl}:\Gamma \text { is satisfiable}\\\iff & {} \text {for all finite }\Psi \subseteq \Phi :\Psi ^{cl}\text { is satisfiable}\\\iff & {} \text {for all finite }\Psi \subseteq \Phi :\Psi \text { is satisfiable.}{\qquad \qquad \qquad \qquad \qquad \,}\Box \end{aligned}$$

A second major open problem concerns whether the set of $\textsf {InqBQ}$-validities is recursively enumerable, which is a prerequisite for the existence of a complete proof system (under the desideratum that proofs be finite and verifiable).

Open Problem 5.5.35

(Recursive enumerability) Let $\Sigma$ be a countable signature. Is the set of $\textsf {InqBQ}$-validities from the language $\mathcal {L}^{\textsf {Q}=}(\Sigma )$ recursively enumerable?

One way to address the previous question is to ask if there is a translation from $\textsf {InqBQ}$ to classical first-order logic, in the sense made precise below.

Open Problem 5.5.36

(Existence of a translation to first-order logic) Given a signature $\Sigma$, is there a computable map $(\cdot )^*:\mathcal {L}^{\textsf {Q}=}(\Sigma )\rightarrow \mathcal {L}^{\textsf {Q}=}_c(\Sigma ')$ from the language of $\textsf {InqBQ}$ to the language of classical first-order  logic based on some signature$\Sigma '$ such that for all $\Phi \cup \{\psi \}\subseteq \mathcal {L}^{\textsf {Q}=}(\Sigma )$ we have

$$\Phi \models \psi \iff \Phi ^*\models \psi ^*$$

where $\Phi ^*=\{\varphi ^*\mid \varphi \in \Phi \}$? (Notice that the entailment on the right-hand side of the biconditional is an entailment in classical first-order logic, since all formulas involved are classical.)

If an entailment-preserving translation to classical first-order logic exists, then since the validities of classical first-order logic are recursively enumerable, so are $\textsf {InqBQ}$ validities.

We then have a number of interesting questions inspired by the Löwenheim-Skolem theorem, about the cardinalities of models and countermodels of a given $\textsf {InqBQ}$ formula. I will just mention a simple example of such questions.

Open Problem 5.5.37

(Logic of countable models) Can any formula $\varphi$ which is not valid in $\textsf {InqBQ}$ be refuted in a countable model, i.e., in a model $M=\langle W,D,I,\sim \rangle$ with $\# W \le \aleph _0$ and $\# D \le \aleph _0$?

The difficulties that one faces in answering these questions have a clear source: they stem from the fact that in inquisitive logic, the implication connective $\rightarrow$ introduces a quantification over subsets of the evaluation state. This means that, if we regard an information model as a two-sorted relational structure with domains W and D, the semantic condition on a model defined by a formula $\varphi$ involving implication is not, at least prima facie, a first-order condition. (Whether every $\varphi$ in fact corresponds to a first-order condition is an open question at this stage.) Due to the presence of implication, the features of $\textsf {InqBQ}$ depend in part of the structure of powerset algebras. Mathematical questions concerning powersets are notoriously difficult—when they admit of definite answers at all. This is well-known from the case of set theory: it is not possible to decide, based on the standard ZFC foundation, what is the cardinality of the powerset of a countably infinite set. But there are also examples of this phenomenon closer to home, in logic. Medvedev’s logic [11, 12] arises from a natural formalization of Kolmogorov interpretation of formulas as problems. Model-theoretically, Medvedev’s logic can be characterized as the logic of models based on finite powerset structures deprived of the empty set (see Chagrov and Zakharyaschev [13]). As in inquisitive logic, in Medvedev’s logic implication is understood as a quantifier over subsets of the evaluation state. In fact, there are tight connections between Medvedev’s logic and inquisitive logic (see Miglioli et al. [14], Ciardelli [7], Ciardelli and Roelofsen [15]). In spite of a significant amount of research over the last fifty years, which yielded partial results (see in particular Maksimova et al. [16]), it is still an open question whether Medvedev’s logic admits a recursive axiomatization, and (equivalently) whether its set of validities is recursively enumerable.

5.6 Coherence

In this section, based on Ciardelli and Grilletti [17], we look at an interesting semantic property that $\textsf {InqBQ}$-formulas may or may not have: finite coherence. The idea is simple: a formula $\varphi$ is finitely coherent if there is a natural number n such that, in order to decide whether $\varphi$ is supported at a state, it is sufficient to look at substates of size at most n. We first define a more general notion of $\kappa$-coherence where $\kappa$ is an arbitrary cardinal.¹⁰

Definition 5.6.1

($\kappa$-coherence) For $\kappa$ a cardinal, we say that a formula $\varphi \in \mathcal {L}^{\textsf {Q}=}$ is $\kappa$ -coherent if for any model M, state s, and assignment g we have

$$s\models _g\varphi \iff (t\models _g\varphi \text { for all }t\subseteq s\text { with }\# t\le \kappa )$$

where $\# t$ is the cardinality of t. We call $\varphi$ coherent if it is $\kappa$-coherent for some cardinal $\kappa$, and finitely coherent if it is n-coherent for some natural number n.

In the above definition, the left-to-right direction always holds by persistency, so $\kappa$-coherence amounts to the requirement that the converse hold as well. Note that truth-conditionality is a special case of n-coherence for $n=1$. Furthermore, note that if $\varphi$ is $\kappa$-coherent then it is also $\lambda$-coherent for all $\lambda \ge \kappa$. This justifies the following definition.

Definition 5.6.2

(Coherence degree) The coherence degree of a coherent formula $\varphi$, denoted $d_\varphi$, is the least $\kappa$ such that $\varphi$ is $\kappa$-coherent (if no such $\kappa$ exists, $\varphi$ has no coherence degree).

In many cases, a bound for the coherence degree of a compound can be obtained from the coherence degrees of its components, as the following proposition shows.

Proposition 5.6.3

Suppose $\varphi$ is $\kappa$-coherent and $\psi$ is $\kappa '$-coherent. Then:

• $\varphi \wedge \psi$ is $\lambda$-coherent for $\lambda =\max (\kappa ,\kappa ')$ (thus, $d_{\varphi \wedge \psi }\le \max (d_\varphi ,d_\psi )$);

• is $\lambda$-coherent for $\lambda =\kappa +\kappa '$ (thus, );

• $\chi \rightarrow \varphi$ is $\kappa$-coherent for any formula $\chi$ (thus, $d_{\chi \rightarrow \varphi }\le d_\varphi$);

• $\forall x\varphi$ is $\kappa$-coherent (thus, $d_{\forall x\chi }\le d_\chi$).

Proof

We prove the claim for   and leave the other cases as exercises to the reader (Exercise 5.9.6). We need to show that is $\lambda$-coherent for $\lambda =\kappa +\kappa '$. Take an arbitrary model M, information state s, and assignment g. We need to show that if for every subset $t\subseteq s$ of size at most $\lambda$ we have , then . Contrapositively, suppose . Then $s\not \models _g\varphi$ and $s\not \models _g\psi$. Since $\varphi$ is $\kappa$-coherent and $\psi$ is $\kappa '$-coherent, there are substates $t,t'\subseteq s$ with $\# t \le \kappa$, $\# t' \le \kappa '$ such that $t\not \models _g\varphi$ and $t'\not \models _g\psi$. Now consider the state $t\cup t'$: this is a subset of s of cardinality at most $\kappa +\kappa '=\lambda$, and by persistency we have .

Note that by definition of coherence degree, $\varphi$ is $d_\varphi$-coherent and $\psi$ is $d_\psi$-coherent. So by what we have just proved, is $d_\varphi +d_\psi$-coherent. Since is defined as the least cardinal for which is coherent, .    $\square$

Using this result, we can show that many formulas of $\textsf {InqBQ}$ are finitely coherent: in particular, all $\mathord {\exists \!\!\exists }$-free formulas are (this can be strengthened slightly, as we will see in Sect. 5.7.1.)

Proposition 5.6.4

Every $\mathord {\exists \!\!\exists }$-free formula of $\textsf {InqBQ}$ is finitely coherent.

Proof

By induction on the structure of the formula. Atomic formulas and $\bot$ are truth-conditional and so 1-coherent. The previous proposition implies immediately that all operators except for $\mathord {\exists \!\!\exists }$ preserve finite coherence.   $\square$

Formulas involving $\mathord {\exists \!\!\exists }$ , on the other hand, are not in general finitely coherent. In fact, even the simplest examples of such formulas are not $\kappa$-coherent for any $\kappa$, finite or infinite, as the following proposition shows.

Proposition 5.6.5

The formula $\mathord {\exists \!\!\exists }x P x$ is not $\kappa$-coherent for any $\kappa$. That is, for every cardinal $\kappa$, there exists a model M and a state s such that $s \not \models \mathord {\exists \!\!\exists }x P x$ and for all $t\subseteq s$ with $\# t \le \kappa$ we have $t \models \mathord {\exists \!\!\exists }x P x$.

Proof

Consider an arbitrary cardinal $\kappa$, and indicate with $\kappa ^+$ the cardinal successor of $\kappa$. Consider the model $M=\langle W,D,I\rangle$ given by:

• $W = \{ w_i \,|\, i < \kappa ^+ \}$;

• $D = \{ d_j \,|\, j < \kappa ^+ \}$;

• $d_j \in P_{w_i}\iff i \ne j$.

We have $M,W\not \models \mathord {\exists \!\!\exists }x P x$: indeed, for every element $d_j \in D$ we have have $M,W\not \models P(d_j)$, since $d_j\not \in P_{w_j}$. However, given any proper subset $t\subset W$ we have $M,t\models \mathord {\exists \!\!\exists }xPx$: to see this, let $w_j$ be a world such that $w_j\not \in t$; then for any $w_i\in t$ we have $i\ne j$ and so $d_j\in w_i$, which implies $M,t\models P(d_j)$.

Since the cardinality of W is $\kappa ^+>\kappa$, any subset $t \subseteq W$ with $\# t \le \kappa$ will be a proper subset of W and thus will support $\mathord {\exists \!\!\exists }xPx$. Thus, we have found a state where $\mathord {\exists \!\!\exists }xPx$ is not supported, while being supported at all subsets of cardinality up to $\kappa$.   $\square$

The previous result shows that there are formulas that lack a coherence degree. On the other hand, with some work one may produce, for every $n\in \mathbb {N}$, a formula of $\textsf {InqBQ}$ whose coherence degree is exactly n (see Ciardelli and Grilletti [17]). We conjecture that these are the two only possibilities for formulas in $\textsf {InqBQ}$.

Conjecture 5.6.6

(Dichotomy) The coherence degree of a formula $\varphi$ is either finite or undefined.

Finitely coherent formulas have a number of important properties. To start with, they enjoy the following finite universe property.¹¹

Proposition 5.6.7

Let $n\in \mathbb {N}$. If $\Phi \not \models \psi$ and $\psi$ is n-coherent, the entailment can be falsified in a model M based on a finite universe W with $\# W \le n$.

Proof

Suppose $\Phi \not \models \psi$. Then there are M, s and g such that $M,s\models _g\Phi$ but $M,s\not \models _g\psi$. If $\psi$ is n-coherent, there exists a state $t\subseteq s$ of size at most n such that $M,t\not \models _g\psi$. By persistency, $M,s\models _g\Phi$ implies $M,t\models _g\Phi$. Then $M_{|t}$, the restriction of M to t, is a model whose universe is t, and thus contains at most n worlds. By locality we have $M_{|t},t\models _g\Phi$ but $M_{|t},t\not \models _g\psi$.   $\square$

Second, whereas formulas of $\textsf {InqBQ}$ are not in general normal, finitely coherent formulas always are. Thus, it is no coincidence that the same formula $\mathord {\exists \!\!\exists }xPx$ that we used in the proof of Proposition 5.3.4 as a counterexample to normality is also a counterexample to finite coherence.

Proposition 5.6.8

(Finite coherence implies normality) If $\varphi$ is finitely coherent, then it is normal; that is, for every model M, state s and assignment g, if $M,s\models _g\varphi$ then $s\subseteq a$ for some alternative $a\in \textsc {Alt}_M^g(\varphi )$.

Proof

Take an arbitrary model M, information state s, and assignment g such that $M,s\models _g\varphi$. Consider the set S of states containing s and supporting $\varphi$,

$$S=\{t\subseteq W\mid s\subseteq t\text { and }M,t\models _g\varphi \}.$$

We want to show that S contains a maximal element.

For this, we first claim that for every non-empty chain $C\subseteq S$ we have $\bigcup C\in S$. Towards a contradiction, suppose this is not the case. Then we have a non-empty chain $C\subseteq S$ such that $\bigcup C\not \in S$. Since $\bigcup C$ does include s, we must have $M,\bigcup C\not \models _g\varphi$. Since $\varphi$ is finitely coherent, there must be a subset $t\subseteq \bigcup C$ of cardinality at most $d_\varphi$ such that $M,t\not \models _g\varphi$. Since $t\subseteq \bigcup C$, every $w\in t$ is included in some element of the chain, and since t is finite, there must be an element $s'\in C$ of the chain such that $t\subseteq s'$. By persistency, since $M,t\not \models _g\varphi$ we also have $M,s'\not \models _g\varphi$. But this contradicts the hypothesis that $C\subseteq S$.

We have shown that every non-empty chain from S has an upper bound in S. By Zorn’s lemma, S contains a maximal element a. This means that a is a maximal extension of s such that $M,a\models _g\varphi$, i.e., $s\subseteq a$ and $a\in \textsc {Alt}_M^g(\varphi )$.   $\square$

Furthermore, $\textsf {InqBQ}$ entailments with finitely coherent conclusions are compact, in the sense specified by the following proposition.

Theorem 5.6.9

(Compactness for finitely coherent conclusions) If $\Phi \models _{\textsf {InqBQ}}\psi$ and $\psi$ is finitely coherent, there exists a finite subset $\Phi _0\subseteq \Phi$ such that $\Phi _0\models _{\textsf {InqBQ}}\psi$.

In order to prove this result, we develop a family of maps from the language of $\textsf {InqBQ}$ over the given signature $\Sigma$ to the language of classical first-order logic over a modified signature $\Sigma ^*$. These maps allow us to emulate the semantics of $\textsf {InqBQ}$ within standard first-order logic, provided a finite upper bound to the size of information states is fixed in advance. This can then be used in combination with Proposition 5.6.7, which guarantees that given an entailment with a finitely coherent conclusion, such a finite bound on the size of the states can indeed be fixed without affecting the validity of the entailment. The remainder of this section spells out the details of the strategy.

As a first step, we associate to a signature $\Sigma$ a corresponding signature $\Sigma ^*$ over two sorts, $\texttt {w}$ for worlds and $\texttt {e}$ for individuals. $\Sigma ^*$ is given as follows:

• For every n-ary predicate symbol $R\in \Sigma$, $\Sigma ^*$ contains a predicate symbol $R^*$ of arity $n+1$ where the first argument is of sort w and the remaining arguments of sort e.

• For every non-rigid n-ary function symbol $f\in \Sigma$, $\Sigma ^*$ contains a function symbol $f^*$ of arity $n+1$ where the first argument is of sort w and the remaining arguments as well as the output are of sort e.

• For every rigid n-ary function symbol $\textsf {f}\in \Sigma$, $\Sigma ^*$ contains a function symbol $\textsf {f}^*$ of arity n where the arguments and the output are of sort e.

Denote by $\mathcal {L}^{\textsf {FOL}}_{\texttt {w},\texttt {e}}(\Sigma ^*)$ the language of two-sorted first-order logic over $\Sigma ^*$. We use $\textsf {w},\textsf {w}_0,\textsf {w}_1,\dots$ for variables of type $\texttt {w}$ in the latter language, and $x,y,\dots$ for variables of type $\texttt {e}$, which we assume to be the same as the variables of $\mathcal {L}^{\textsf {Q}=}(\Sigma )$.

Next, we associate to a relational information model $M=\langle W,D,I\rangle$ for the signature $\Sigma$ a two-sorted relational structure $M^*=\langle W,D,I^*\rangle$ for $\Sigma ^*$, where:

• For a predicate symbol R: $I^*(R^*)(w,d_1,\dots ,d_n)\iff I_w(R)(d_1,\dots ,d_n)$;

• For a non-rigid function symbol f: $I^*(f^*)(w,d_1,\dots ,d_n)= I_w(f)(d_1,\dots ,d_n)$;

• For a rigid function symbol $\textsf {f}$: $I^*(\textsf {f}^*)(d_1,\dots ,d_n)= I_w(\textsf {f})(d_1,\dots ,d_n)$ for an arbitrary $w\in W$.

It is easy to verify that the map $M\mapsto M^*$ is a bijection between relational information models for $\Sigma$ and two-sorted relational structures for $\Sigma ^*$.

The next step is to translate terms. Given a term t of $\mathcal {L}^{\textsf {Q}=}(\Sigma )$ and a world variable $\textsf {w}$, we define a corresponding term $t_\textsf {w}$ of type e of the language $\mathcal {L}^{\textsf {Q}=}(\Sigma )$ inductively as follows:

• if t is a variable x then $t_\textsf {w}=x$;

• if $t=f(t^1,\dots ,t^n)$ where f is non-rigid then $t_\textsf {w}=f^*(\textsf {w},t^1_\textsf {w},\dots ,t^n_\textsf {w})$;

• if $t=\textsf {f}(t^1,\dots ,t^n)$ where $\textsf {f}$ is rigid then $t_\textsf {w}=\textsf {f}^*(t^1_\textsf {w},\dots ,t^n_\textsf {w})$.

It is straightforward to check that for any relational information model M, assignment g, and term t of $\mathcal {L}^{\textsf {Q}=}(\Sigma )$ we have

$$[t]^M_{w,g}=[t_\textsf {w}]^{M^*}_{g[\textsf {w}\mapsto w]}$$

where $g[\textsf {w}\mapsto w]$ is an arbitrary assignment that coincides with g on the variables of type $\texttt {e}$ and maps the variable $\textsf {w}$ to w.

Finally, let $\textsf {s}=\{\textsf {w}_1,\dots ,\textsf {w}_n\}$ be a finite nonempty set of world variables. We define for each formula $\varphi \in \mathcal {L}^{\textsf {Q}=}(\Sigma )$ a translation $\textsf {tr}_{\textsf {s}}(\varphi )\in \mathcal {L}^{\textsf {FOL}}_{\texttt {w},\texttt {e}}(\Sigma )$ as follows:

We spell out one example by way of illustration. We have

The key property of the map $\textsf {tr}_\textsf {s}$ is given by the following proposition. We omit the proof, which is a matter of straightforward case-by-case verification.

Proposition 5.6.10

Let M be a relational information model, g an assignment, and $s=\{w_1,\dots ,w_n\}$ a finite nonempty state. Let $\textsf {s}=\{\textsf {w}_1,\dots ,\textsf {w}_n\}$ be a set of n world variables and let $g[\textsf {s}\mapsto s]$ be any two-sorted assignment that coincides with g on variables of type $\texttt {e}$ and which maps the world variable $\textsf {w}_i$ to $w_i$ for $i=1,\dots ,n$. For any formula $\varphi \in \mathcal {L}^{\textsf {Q}=}(\Sigma )$ we have:

$$\begin{aligned} M,s\models _g\varphi \;\iff \; M^*\models _{g[\textsf {s}\mapsto s]}\textsf {tr}_\textsf {s}(\varphi ). \end{aligned}$$

The following proposition shows that, although the maps $\textsf {tr}_\textsf {s}$ are not in general translations from $\textsf {InqBQ}$ to standard first-order logic, they preserve the validity of entailments whose conclusion is n-coherent for $n=\#\textsf {s}$.

Proposition 5.6.11

Let $\Phi \cup \{\psi \}\subseteq \mathcal {L}^{\textsf {Q}=}(\Sigma )$ where $\psi$ is n-coherent for $n\in \mathbb {N}$. We have:

$$\begin{aligned} \Phi \models _{\textsf {InqBQ}}\psi \;\iff \; \textsf {tr}_\textsf {s}(\Phi )\models _{\textsf {FOL}}\textsf {tr}_\textsf {s}(\psi ), \end{aligned}$$

where $\models _{\textsf {FOL}}$ denotes entailment in first-order logic, $\textsf {s}=\{\textsf {w}_1,\dots ,\textsf {w}_n\}$ is an arbitrary set of n world variables, and $\textsf {tr}_\textsf {s}(\Phi )=\{\textsf {tr}_\textsf {s}(\varphi )\mid \varphi \in \Phi \}$.

Proof

Suppose $\Phi \not \models _{\textsf {InqBQ}}\psi$ and suppose $\psi$ is n-coherent. By Proposition 5.6.7, we can find a model M, an assignment g, and a state s of cardinality at most n such that $M,s\models _g\Phi$ but $M,s\not \models _g\psi$. In fact, we may assume that the cardinality of s is exactly n (if needed, we may always duplicate some worlds in s). By Proposition 5.6.10 we have $M^*\models _{g[\textsf {s}\mapsto s]}\textsf {tr}_\textsf {s}(\Phi )$ but $M^*\not \models _{g[\textsf {s}\mapsto s]}\textsf {tr}_\textsf {s}(\psi )$, which shows that $\textsf {tr}_\textsf {s}(\Phi )\not \models _{\textsf {FOL}}\textsf {tr}_\textsf {s}(\psi )$.

For the converse direction, suppose $\textsf {tr}_\textsf {s}(\Phi )\not \models _{\textsf {FOL}}\textsf {tr}_\textsf {s}(\psi )$. This means that there is a two-sorted relational structure $M'$ and an assignment $g'$ such that $M'\models _{g'}\textsf {tr}_\textsf {s}(\Phi )$ but $M'\not \models _{g'}\textsf {tr}_\textsf {s}(\psi )$. Now let M be the relational information model such that $M^*=M'$ (which exists since the map $M\mapsto M^*$ is a bijection between relational information models for $\Sigma$ and two-sorted structures for $\Sigma ^*$). Let g be the assignment defined by $g(x)=g'(x)$ for every individual variable x, and let $s=\{w_1,\dots ,w_n\}$ where $w_i=g'(\textsf {w}_i)$. By the previous proposition, for any $\chi \in \mathcal {L}^{\textsf {Q}=}(\Sigma )$ we have

$$\begin{aligned} M,s\models _g\chi \;\iff \; M^*\models _{g[\textsf {s}\mapsto s]}\textsf {tr}_{\textsf {s}}(\chi ) \;\iff \; M'\models _{g'}\textsf {tr}_{\textsf {s}}(\chi ), \end{aligned}$$

where the last biconditional holds because $g'$ and $g[\textsf {s}\mapsto s]$ coincide on all variables which occur free in $\textsf {tr}_{\textsf {s}}(\chi )$. This then implies that $M,s\models _{g}\Phi$ but $M,s\not \models _{g}\psi$, which shows that $\Phi \not \models _{\textsf {InqBQ}}\psi$.   $\square$

Finally, with this result in place we are able to show the compactness of $\textsf {InqBQ}$ towards finitely coherent conclusions.

Proof of Theorem 5.6.9. Suppose $\Phi \models _\textsf {InqBQ}\psi$ and $\psi$ is finitely coherent. By Proposition 5.6.11, choosing $\textsf {s}=\{\textsf {w}_1,\dots ,\textsf {w}_{d_\psi }\}$ we have $\textsf {tr}_\textsf {s}(\Phi )\models _{\textsf {FOL}}\textsf {tr}_\textsf {s}(\psi )$. By the compactness of classical first-order logic, there is a finite subset $\Phi _0\subseteq \Phi$ such that $\textsf {tr}_\textsf {s}(\Phi _0)\models _{\textsf {FOL}}\textsf {tr}_\textsf {s}(\psi )$. Again by Proposition 5.6.11, it follows that $\Phi _0\models _\textsf {InqBQ}\psi$.    $\square$

5.7 Fragments

In this section we discuss three interesting syntactic fragments of $\textsf {InqBQ}$: the restricted existential fragment $\mathcal {L}_{\text {Rex}}$ [17], the classical antecedent fragment $\mathcal {L}_{\text {Clant}}$ [19], and the mention-all fragment $\mathcal {L}_{\text {MA}}$. The first two are interesting as they are the largest syntactic fragments of $\mathcal {L}$ for which the questions posed in Sect. 5.5.6 have been answered in the positive: entailment in these fragments is compact, validities are recursively enumerable, and every non-entailment can be refuted in a countable model. Moreover, in each case a complete proof system has been established. The mention-all fragment, on the other hand, is interesting in that it has exactly the same expressive power as the logic of interrogation of Groenendijk [4] (cf. Sect. 2.9.2)—a predecessor of $\textsf {InqBQ}$ which also extends first-order logic with questions and which was axiomatized by ten Cate and Shan [20]. The inclusions between the three fragments discussed in this section are shown in Fig. 5.6.

Fig. 5.6

Relations between the three fragments discussed in this section: the mention-all fragment $\mathcal {L}_{\text {MA}}$ is included both in the classical antecedent fragment $\mathcal {L}_{\text {Clant}}$ and in the restricted existential fragment $\mathcal {L}_{\text {Rex}}$, while the latter two fragments are incomparable in terms of inclusion

5.7.1 The Restricted Existential (Rex) Fragment

The first fragment of $\textsf {InqBQ}$ that we consider is the restricted existential (in short, rex) fragment of $\textsf {InqBQ}$, obtained by restricting the occurrence of $\mathord {\exists \!\!\exists }$ to antecedents of an implication.

Definition 5.7.1

(Rex fragment of $\textsf {InqBQ}$ [17]) The set $\mathcal {L}_{\text {Rex}}(\Sigma )$ of rex formulas is given by the following syntax:

where p is an atomic sentence from the signature $\Sigma$ and $\psi$ an arbitrary sentence from $\mathcal {L}^{\textsf {Q}=}$, possibly containing occurrences of $\mathord {\exists \!\!\exists }$.

The rex fragment is a broad fragment of $\textsf {InqBQ}$: it contains all classical formulas and it is closed under conjunction, implication, inquisitive disjunction, and the universal quantifier. The key feature of the rex fragment is that every formula is finitely coherent, and an upper bound for its coherence degree is computable from its syntax.

Proposition 5.7.2

(Finite coherence property) For every formula $\varphi \in \mathcal {L}_{\text {Rex}}$ there is a natural number $n_\varphi$, computable from the syntax of $\varphi$, such that $\varphi$ is $n_\varphi$-coherent.

Proof

We define $n_\varphi$ as follows for $\varphi \in \mathcal {L}_{\text {Rex}}$:

• $n_p = 1$ if p is atomic;

• $n_\bot = 1$;

• $n_{\varphi \wedge \psi } = \max (n_\varphi ,n_\psi )$;

• ;

• $n_{\chi \rightarrow \varphi } = n_\varphi$;

• $n_{\forall x\varphi } = n_\varphi$.

To see that $\varphi$ is $n_\varphi$-coherent it suffices to note that atoms and $\bot$ are 1-coherent (i.e., truth-conditional) and to apply inductively Proposition 5.6.3.    $\square$

Note that the number $n_\chi$ is not necessarily equal to the coherence degree $d_\chi$: for instance, if we have $n_{\chi }=n_{Px}+n_{Px}=2$, but since we have $d_{\chi }=d_{Px}=1$. However, since the coherence degree $d_\chi$ is defined as the least number n for which $\chi$ is n-coherent, we have $n_\chi \ge d_\chi$.

An interesting open problem is whether the rex fragment is expressively complete with respect to finitely coherent properties expressible in $\textsf {InqBQ}$, in the following sense.

Open Problem 5.7.3

(Completeness for finitely coherent properties)

Is every finitely coherent $\varphi \in \mathcal {L}^{\textsf {Q}=}$ logically equivalent to some $\varphi ^*\in \mathcal {L}_{\text {Rex}}$?

Since rex formulas are finitely coherent, all the results from the previous section apply to them. In particular, rex formulas are always normal, and entailments with rex conclusions are compact, that is, if $\Phi \models \psi$ and $\psi$ is a rex formula then $\Phi _0\models \psi$ for some finite $\Phi _0\subseteq \Phi$. Moreover, using the results in the previous section we can show that the set of rex validities is recursively enumerable.

Theorem 5.7.4

(Rex validities are recursively enumerable)

The set $\text {Val}_{\textsf {rex}}=\{\chi \in \mathcal {L}_{\text {Rex}}\mid \chi \text { is valid in }\textsf {InqBQ}\}$ is recursively enumerable.

Proof

We need to show that there is a method to semi-decide whether a given formula $\chi$ belongs to the set $\text {Val}_{\textsf {rex}}$. This amounts to semi-deciding whether the conjunction $(\chi \in \mathcal {L}_{\text {Rex}}\text { and }\models _{\textsf {InqBQ}}\chi )$ holds. For this, we proceed as follows. First, we check whether $\chi$ is a rex formula . This is a decidable matter: we just need to check if all occurrences of an inquisitive existential quantifier are within the antecedent of a conditional. If $\chi$ is not a rex formula , we do not return any output. Otherwise, we need to semi-decide whether $\chi$ is valid in $\textsf {InqBQ}$. For this, we first compute the number $n_\chi$ and then compute the finitary first-order translation $\textsf {tr}_\textsf {s}(\chi )$ for $\textsf {s}$ a set of $n_\chi$ world variables. Since $\chi$ is $n_\chi$-coherent, by Proposition 5.6.11 we have $\models _{\textsf {InqBQ}}\chi \iff \; \models _{\textsf {FOL}}\textsf {tr}_\textsf {s}(\chi )$. Thus, our task reduces to semi-deciding whether $\textsf {tr}_\textsf {s}(\chi )$ is valid in classical first-order logic. This is possible, since validity in first-order logic is semi-decidable.   $\square$

The theorem implies that the set of $\textsf {InqBQ}$-entailments with finitely many premises and a rex conclusion is also recursively enumerable. This is because we have:

$$\varphi _1,\dots ,\varphi _n\models \chi \;\iff \; \models \varphi _1\wedge \dots \wedge \varphi _n\rightarrow \chi .$$

Thus, semi-deciding whether $\varphi _1,\dots ,\varphi _n\models _{\textsf {InqBQ}}\chi$ reduces to semi-deciding the validity of the formula $\varphi _1\wedge \dots \wedge \varphi _n\rightarrow \chi$, which is a rex formula if $\chi$ is.

The fact that entailments with rex conclusions are compact and recursively enumerable suggests that it may be possible to give a proof system for $\textsf {InqBQ}$ which is complete with respect to such entailments, i.e., such that if $\Phi \models \psi$ and $\psi$ is a rex formula then $\psi$ is provable from $\Phi$ in this system. This is indeed the case: one such proof system is obtained by extending the system for $\textsf {InqBQ}$ presented in the next chapter with a coherence rule which allows one to freely use certain cardinality formulas in order to infer a finitely coherent formula; the interested reader is referred to Ciardelli and Grilletti [17] for the details.

5.7.2 The Classical Antecedent (Clant) Fragment

The second fragment of $\textsf {InqBQ}$ that we consider is the classical antecedent (in short, clant) fragment, obtained by allowing only classical formulas as antecedents of implications.

Definition 5.7.5

(Classical antecedent fragment of $\textsf {InqBQ}$ [19]) The set $\mathcal {L}_{\text {Clant}}(\Sigma )$ of clant formulas in a signature $\Sigma$ is given by:

$$\phi ::= p\mid\bot\mid\phi\land\phi\mid\alpha\to\phi\mid\phi|\vee\phi\mid\forall x\phi\mid\exists x\phi,$$

where p ranges over atomic formulas in the signature $\Sigma$ and $\alpha$ over classical formulas.

The clant fragment is a broad fragment of $\textsf {InqBQ}$: it includes all classical formulas, polar questions of the form $?\alpha$ for $\alpha$ classical, and more generally disjunctive questions of the form where the $\alpha _i$ are all classical; mention-some questions of the form $\mathord {\exists \!\!\exists }x_1\dots \mathord {\exists \!\!\exists }x_n\alpha$ with $\alpha$ classical, including as special cases single-instance questions like $\mathord {\exists \!\!\exists }! xP(x)$, and identification questions like $\mathord {\exists \!\!\exists }x(x=t)$; mention-all questions of the form $\forall x_1\dots \forall x_n{?\alpha }$ with $\alpha$ classical; and all questions that can be formed by conjoining questions of the above kinds, or conditionalizing such questions to a classical antecedent. In fact, all examples of questions discussed in this chapter, as well as all examples of $\textsf {InqBQ}$-entailments, involved only clant formulas.

What is not included in the fragment are implications of the form $\mu \rightarrow \mu '$ where both $\mu$ and $\mu '$ contain inquisitive operators, for instance the formulas $\forall x?Px\rightarrow \forall x?Qx$ or $\mathord {\exists \!\!\exists }xPx\rightarrow \mathord {\exists \!\!\exists }xQx$, as well as compounds including such implications.¹² As we discussed in Sect. 2.5, implications of this sort capture the fact that in the evaluation state, a certain dependency holds. Thus, what we cannot generally capture in the clant fragment is what follows from premises stating that certain dependencies hold.

The clant fragment neither contains nor is contained in the rex fragment discussed in the previous section. For instance, $\mathord {\exists \!\!\exists }xPx$ is a clant formula but not a rex formula , while $\forall x?Px\rightarrow \forall x?Qx$ is a rex formula but not a clant formula.

Semantically, the key feature of the classical antecedent fragment is that the second-order quantification associated with implication is neutralized. This is because if $\alpha$ is truth-conditional (and thus in particular if $\alpha$ is classical) the clause for implication can be simplified as follows (cf. Propositions 5.5.10 and 2.5.2):

$$M,s\models _g\alpha \rightarrow \varphi \;\iff \; M,s\cap |\alpha |_M^g\models _g\varphi .$$

Thus, to check if an implication $\alpha \rightarrow \varphi$ holds in a state s it is not necessary to check all subsets of s; it suffices to check one subset of s, namely, $s\cap |\alpha |_M^g$.

One consequence of this fact is that in a clant formula, implication can always be pushed down to the level of classical formulas. Let us make this precise.

Definition 5.7.6

The set of restricted implication (for short, rimp) formulas, is the set of formulas where implication occurs only within classical sub-formulas. More formally, the set of rimp formulas is given by the following grammar:

where $\alpha$ ranges over classical formulas.

Note that every rimp formula is a clant formula. The converse is not the case: for instance, $\exists xPx\rightarrow \forall x?Px$ is a clant formula but not a rimp formula. However, every clant formula can be turned into an equivalent rimp formula.

Proposition 5.7.7

Every clant formula is equivalent to a rimp formula.

Proof

The key to the result lies in the following claim: if $\xi$ is a rimp formula and $\alpha$ is classical, $\alpha \rightarrow \xi$ is equivalent to a rimp formula. This is proved by induction on $\xi$, making crucial use of the split equivalences given by Proposition 5.5.12. We leave this inductive proof as an exercise (Exercise 5.9.7).

Using this claim, we prove our proposition by induction on $\varphi \in \mathcal {L}_{\text {Clant}}$. The interesting case is the one for $\varphi =(\alpha \rightarrow \psi )$, where $\alpha$ is a classical formula. By induction hypothesis, $\psi \equiv \psi ^*$ for some rimp formula $\psi ^*$, so $\varphi \equiv (\alpha \rightarrow \psi ^*)$, and by the above claim, the latter formula is equivalent to a rimp formula.   $\square$

Using this fact, it is possible to define an entailment-preserving translation $(\cdot )^*$ from the clant fragment of $\textsf {InqBQ}$ to classical first-order logic with two sorts for worlds and individuals.

Proposition 5.7.8

Let $\Sigma$ be a signature and let $\mathcal {L}^{\textsf {FOL}}_{\texttt {w},\texttt {e}}(\Sigma ^*)$ be the corresponding two-sorted language as defined in Sect. 5.6. There is a computable translation $\textsf {tr}:\mathcal {L}_{\text {Clant}}(\Sigma )\rightarrow \mathcal {L}^{\textsf {FOL}}_{\texttt {w},\texttt {e}}(\Sigma ^*)$ such that for all $\Phi \cup \{\psi \}\subseteq \mathcal {L}_{\text {Clant}}(\Sigma )$:

$$\Phi \models \psi \iff \textsf {tr}(\Phi )\models _{\textsf {FOL}}\textsf {tr}(\psi ),$$

where $\models _{\textsf {FOL}}$ is entailment in classical first-order logic.

Proof Sketch. We only give the proof strategy, leaving the details as an exercise to the reader (see Exercise 5.9.8). We refer to Sect. 5.6 for the definition of the two-sorted language $\mathcal {L}^{\textsf {FOL}}_{\texttt {w},\texttt {e}}(\Sigma ^*)$ and of the bijection $M\mapsto M^*$ between relational information models for $\Sigma$ and relational structures for $\Sigma ^*$.

• Step 1. Define for each classical formula $\alpha \in \mathcal {L}^{\textsf {Q}}_{c}(\Sigma )$ a two-sorted formula $\alpha ^\star (\textsf {w})$ containing a single free variable $\textsf {w}$ of sort w, in such a way that for every relational information model M, world w, and assignment g we have

  $$M,w\models _g\alpha \iff M^*\models _{\overline{g}[\textsf {w}\mapsto w]}\alpha ^\star ,$$

  where $\overline{g}$ is any assignment for the two-sorted language which coincides with g on variables of sort e.

  Example: $(\forall xPx)^\star =\forall xP^*(\textsf {w},x)$.

• Step 2. Define for each restricted implication formula $\varphi$ a formula $\textsf {tr}(\varphi )\in \mathcal {L}^{\textsf {FOL}}_{\texttt {w},\texttt {e}}(\Sigma ^*)$, without free variables of sort w, such that

  $$M,W\models _g\varphi \iff M^*\models _{\overline{g}}\varphi .$$

  Example: $\textsf {tr}(\mathord {\exists \!\!\exists }xPx)=\exists x\forall \textsf {w}P^*(\textsf {w},x)$.

• Step 3. Extend $\textsf {tr}$ to all clant formulas using Proposition 5.7.7.

• Step 4. Using the fact that $M\mapsto M^*$ is a bijection between relational information models for $\Sigma$ and two-sorted structures for $\Sigma ^*$, show that for all $\Phi \cup \{\psi \}\subseteq \mathcal {L}_{\text {Clant}}(\Sigma )$ we have $\Phi \models \psi \iff \textsf {tr}(\Phi )\models _{\textsf {FOL}}\textsf {tr}(\psi )$.   $\square$

As consequences of the translation and the properties of classical first-order logic, we obtain the following theorems.

Theorem 5.7.9

(Compactness for the clant fragment) Suppose $\Phi \cup \{\psi \}\subseteq \mathcal {L}_{\text {Clant}}$. If $\Phi \models \psi$ then there is a finite $\Phi _0\subseteq \Phi$ such that $\Phi _0\models \psi$.

Proof

Suppose $\Phi \cup \{\psi \}\subseteq \mathcal {L}_{\text {Clant}}$ and $\Phi \models _\textsf {InqBQ}\psi$. By the previous proposition we have $\textsf {tr}(\Phi )\models _{\textsf {FOL}}\textsf {tr}(\psi )$. By the compactness of classical first-order logic, there is a finite subset $\Phi _0\subseteq \Phi$ such that $\textsf {tr}(\Phi _0)\models _{\textsf {FOL}}\textsf {tr}(\psi )$. Again by the previous proposition, $\Phi _0\models _\textsf {InqBQ}\psi$.    $\square$

Theorem 5.7.10

(Recursive enumerability of clant entailments)

The set $\{\langle \varphi _1,\dots ,\varphi _n,\psi \rangle \mid n\ge 0,\,\varphi _1,\dots ,\varphi _n,\psi \in \mathcal {L}_{\text {Clant}}, \;\varphi _1,\dots ,\varphi _n\models \psi \}$

is recursively enumerable.

Proof

We need to show that, given a sequence $\langle \varphi _1,\dots ,\varphi _n,\psi \rangle$ of clant formulas , we can semi-decide whether $\varphi _1,\dots ,\varphi _n\models \psi$ holds. By Proposition 5.7.8, this boils down to semi-deciding whether $\textsf {tr}(\varphi _1),\dots ,\textsf {tr}(\varphi _n)\models _{\textsf {FOL}}\textsf {tr}(\psi )$ holds, i.e., whether $\textsf {tr}(\varphi _1)\wedge \dots \wedge \textsf {tr}(\varphi _n)\rightarrow \textsf {tr}(\psi )$ is a valid formula in first-order logic. Since validity in first-order logic is semi-decidable, the conclusion follows.   $\square$

Note that, while this last theorem implies that the set of clant validities is r.e., it is a strictly stronger claim. This is because an entailment $\varphi _1,\dots ,\varphi _n\models \psi$ among clant formulas does not always correspond to the validity of a single clant formula: although we have $\varphi _1,\dots ,\varphi _n\models \psi$ iff the formula $\varphi _1\wedge \dots \wedge \varphi _n\rightarrow \psi$ is valid, the latter is not in general a clant formula, even when $\varphi _1,\dots ,\varphi _n,\psi$ are.

These positive results suggest that it may be possible to find a complete deduction system for the clant fragment of $\textsf {InqBQ}$. This is indeed the case: following Grilletti [19], in Sect. 6.3 we will describe a natural deduction system for the clant fragment and prove its completeness.

5.7.3 The Mention-All Fragment

The mention-all fragment of $\textsf {InqBQ}$ is given by the following definition.

Definition 5.7.11

(Mention-all fragment) The set $\mathcal {L}_{\text {MA}}$ of mention-all formulas is defined as the union $\mathcal {L}^{\textsf {Q}=}_c\cup \mathcal {L}_{\text {MA?}}$, where $\mathcal {L}^{\textsf {Q}=}_c$ is the set of classical formulas and $\mathcal {L}_{\text {MA?}}=\{\forall x_1\dots \forall x_n?\alpha \mid n\ge 0,\alpha \in \mathcal {L}^{\textsf {Q}=}_c\}$. In words, the formulas in $\mathcal {L}_{\text {MA}}$ are either classical or of the form $\forall \overline{x}?\alpha$, where $\alpha$ is classical and $\overline{x}$ is a possibly empty sequence of variables.

The mention-all fragment is a rather small fragment of $\textsf {InqBQ}$; it is included both in the rex fragment (since mention-all formulas are $\mathord {\exists \!\!\exists }$-free) and in the clant fragment (since implications are only allowed within classical formulas).

One basic feature of the fragment is that every formula in it is 2-coherent.

Proposition 5.7.12

Every $\varphi \in \mathcal {L}_{\text {MA}}$ is 2-coherent.

Proof

If $\varphi$ is classical, then it is 1-coherent and thus also 2-coherent. If $\varphi =\forall \overline{x}?\alpha$ where $\alpha$ is classical, by Proposition 5.6.3 and the truth-conditionality of classical formulas we have $d_{\forall \overline{x}{?\alpha }}\le d_{?\alpha }\le d_\alpha +d_{\lnot \alpha }=1+1=2$.    $\square$

The fact that formulas in this fragment are 2-coherent implies that their semantics is completely encoded at the level of states of size at most 2. This opens the way to the possibility of giving an equivalent semantics for the fragment that evaluates formulas with respect to pairs $\langle w,w'\rangle$ of worlds, rather than with respect to states. We will come back to this point when relating the fragment to the Logic of Interrogation. Before turning to that, let us introduce some useful notation.

Definition 5.7.13

Let M be a model and g an assignment. If $\alpha$ is a classical formula and $\overline{x}=\langle x_1,\dots ,x_n\rangle$ is a tuple of variables, let $\alpha ^{\overline{x}}_g$ be the intensional relation determined by $\alpha$ with respect to $\overline{x}$, i.e., the function which maps any world $w\in W$ to the set of tuples $\overline{d}\in D^n$ that satisfy $\alpha$ in w relative to g.

$$\alpha ^{\overline{x}}_g(w)=\{\overline{d}\in D^n\,|\, w\models _{g[\overline{x}\mapsto \overline{d}]}\alpha \}.$$

We extend this to the case in which the tuple $\overline{x}$ is empty by letting $\alpha ^{\emptyset }_g(w)$ be the truth value of $\alpha$ at w relative to g:

$$\alpha ^{\emptyset }_g(w)=\left\{ \begin{array}{ll} 1 &{} \text {if }w\models _g\alpha \\ 0 &{} \text {if }w\not \models _g\alpha .\\ \end{array} \right. $$

Clearly, if $\alpha$ contains no variables besides those in $\overline{x}$ then the assignment g plays no role, so we can drop reference to it.

The following proposition states that, given a classical formula $\alpha$ and a tuple $\overline{x}$ of variables, the question $\forall \overline{x}{?\alpha }$ asks for the extension of the relation $\alpha ^{\overline{x}}_g$. That is, the question is settled in a state s if any two worlds $w,w'\in s$ agree on the extension of $\alpha ^{\overline{x}}_g$.

Proposition 5.7.14

(Semantics of mention-all questions) Let $\alpha \in \mathcal {L}^{\textsf {Q}=}_c$. For any information model M, state s and assignment g:

$$s\models _{g}\forall \overline{x}{?\alpha }\;\iff \;\text {for all }w,w'\in s: \alpha ^{\overline{x}}_g(w)=\alpha ^{\overline{x}}_g(w').$$

Proof

We have the following sequence of equivalences:

$$\begin{aligned} s\models _g\!\forall \overline{x}{?\alpha }\!\!\!\iff & {} \!\!\text {for all }\overline{d}\in D^n:\;s\models _{g[\overline{x}\mapsto \overline{d}]}\alpha \text { or }s\models _{g[\overline{x}\mapsto \overline{d}]}\lnot \alpha \\\iff & {} \!\!\text {for all }\overline{d}\in D^n:\;(\text {for all }w\in s,\,\overline{d}\in \alpha ^{\overline{x}}_g(w))\text { or }\\{} & {} \!\!\phantom {\text {for all }d\in D^n:\;\;}(\text {for all }w\in s,\,\overline{d}\not \in \alpha ^{\overline{x}}_g(w))\\\iff & {} \!\!\text {for all }\overline{d}\in D^n,\text { for all }w,w'\in s:\; \overline{d}\in \alpha ^{\overline{x}}_g(w)\iff \overline{d}\in \alpha ^{\overline{x}}_g(w')\\\iff & {} \!\!\text {for all }w,w'\in s,\text { for all }\overline{d}\in D^n:\; \overline{d}\in \alpha ^{\overline{x}}_g(w)\iff \overline{d}\in \alpha ^{\overline{x}}_g(w')\\\iff & {} \!\!\text {for all }w,w'\in s: \alpha ^{\overline{x}}_g(w)=\alpha ^{\overline{x}}_g(w'). \end{aligned}$$

It is easy to check that this holds also for the special case in which $\overline{x}$ is empty and the formula under consideration is a polar question ${?\alpha }$.    $\square$

This result allows us to show that questions of the form $\forall \overline{x}?\alpha$ can be seen as inducing partitions of the logical space. Let us make this precise.

Definition 5.7.15

(Partition formulas) We say that $\varphi \in \mathcal {L}^{\textsf {Q}=}$ is a partition formula if given any information model M and assignment g there is a partition $\Pi ^\varphi$ of W such that for every state $s\subseteq W$:

$$s\models _g\varphi \iff s\subseteq a\text { for some }a\in \Pi ^\varphi .$$

Equivalently, $\varphi$ is a partition formula if for any model M and assignment g there is an equivalence relation $\approx ^\varphi$ on W such that for every state $s\subseteq W$:

$$s\models _g\varphi \iff \forall w,w'\in s: w\approx ^\varphi w'.$$

Proposition 5.7.16

If $\alpha$ is a classical formula, $\forall \overline{x}?\alpha$ is a partition formula. Indeed, given any model M and assignment g, the set $\textsc {Alt}_M^g(\forall \overline{x}?\alpha )$ forms a partition of the logical space W, and for any information state $s\subseteq W$ we have

$$s\models _g\forall \overline{x}?\alpha \iff s\subseteq a\text { for some }a\in \textsc {Alt}_M^g(\forall \overline{x}?\alpha ).$$

The proof of this proposition is left as an exercise to the reader (Exercise 5.9.9).

A natural question is whether $\mathcal {L}_{\text {MA?}}$ is expressively complete with respect to partition formulas. We will leave this as an open problem.

Open Problem 5.7.17

Is every partition formula in $\textsf {InqBQ}$ equivalent to one of the form $\forall \overline{x}?\alpha$ with $\alpha$ classical?

As we discussed in Sect. 2.9.2, equivalence relations on the logical space—and the partitions they induce—are precisely the objects taken to capture question meanings in the theory of questions of Groenendijk and Stokhof [21]. A logical system based on this theory, called the Logic of Interrogation (LoI) has been developed by Groenendijk [4], and axiomatized by ten Cate and Shan [20]. This system is the most important predecessor of $\textsf {InqBQ}$. We will now introduce LoI more precisely and show that it is essentially equivalent to the mention-all fragment of $\textsf {InqBQ}$.

The language $\mathcal {L}_\textsf {LoI}$ of the Logic of Interrogation consists of two kinds of formulas: declaratives, which are simply classical formulas $\alpha \in \mathcal {L}^{\textsf {Q}=}_c$, and interrogatives, which are of the form $Q\overline{x}\alpha$, where Q is a special question-forming quantifier, $\overline{x}$ is a possibly empty sequence of variables, and $\alpha$ is a classical formula.¹³ Intuitively $Q\overline{x}\alpha$ stands for the mention-all question asking which tuples $\overline{x}$ satisfy $\alpha$. As a special case, when $\overline{x}$ is empty, $Q\alpha$ stands for the polar question asking whether $\alpha$ is true or false.

The semantics of LoI is given relative to models that are essentially the same as our relational information models. The original setup for LoI is slightly less general in two ways: first, only rigid constant symbols are allowed in the language; second, the semantics is restricted to id-models. Since it is straightforward to see how both restrictions can be lifted, I present LoI without these restrictions, so as to bring out the connections to $\textsf {InqBQ}$ in greater generality.

In its original formulation [4], LoI is presented as a dynamic semantics (in the tradition of Groenendijk et al. [22], Jäger [23], Hulstijn [24], among others). However, as pointed out by ten Cate and Shan, the dynamic coating is inessential: one can give a simple static semantics for LoI in which formulas are interpreted with respect to ordered pairs of worlds. A classical formula $\alpha$ is satisfied at a pair $\langle w,w'\rangle$ in case w and $w'$ agree that $\alpha$ is true, while a question $Q\overline{x}\alpha$ is satisfied in case the worlds w and $w'$ agree on the extension of the relation defined by $\alpha$ with respect to $\overline{x}$—i.e., agree on the answer to the question $Q\overline{x}\alpha$.

Definition 5.7.18

If M is a relational information model and $w,w'$ are worlds in M (not necessarily distinct) the semantics of LoI is given by:

• $\langle w,w'\rangle \models _g^{\textsf {LoI}}\alpha \iff w\models _g\alpha$ and $w'\models _g\alpha$;

• $\langle w,w'\rangle \models _g^{\textsf {LoI}} Q\overline{x}\alpha \iff \alpha _g^{\overline{x}}(w)=\alpha _g^{\overline{x}}(w')$.

Entailment in LoI can be defined in the expected way in terms of this semantics.

Definition 5.7.19

(Entailment in the Logic of Interrogation) Let $\Phi \cup \{\psi \}\subseteq \mathcal {L}_\textsf {LoI}$. $\Phi \models _\textsf {LoI}\psi$ in case for all models M, all pairs $\langle w,w'\rangle$ of worlds, and all assignments g, if $\langle w,w'\rangle \models _g\varphi$ for all $\varphi \in \Phi$ then $\langle w,w'\rangle \models _g\psi$.

LoI and the mention-all fragment of $\textsf {InqBQ}$ are equivalent in a natural sense. This can be made precise by defining the following translations between $\mathcal {L}_\textsf {LoI}$ and $\mathcal {L}_{\text {MA}}$.

Definition 5.7.20

(Translations) We define two translations $(\cdot )^\sharp :\mathcal {L}_{\text {MA}}\rightarrow \mathcal {L}_\textsf {LoI}$ and $(\cdot )^\flat :\mathcal {L}_\textsf {LoI}\rightarrow \mathcal {L}_{\text {MA}}$, as follows:

• $\alpha ^\sharp =\alpha$

• $(\forall \overline{x}{?\alpha })^\sharp = Q\overline{x}\alpha$

• $\alpha ^\flat =\alpha$

• $(Q\overline{x}\alpha )^\flat =\forall \overline{x}{?\alpha }$

Clearly, the two translations are inverse to each other, i.e., we have $\varphi ^{\sharp \flat }=\varphi$ and $\varphi ^{\flat \sharp }=\varphi$. The semantic connections between a sentence and its translation are captured by the following proposition. The proof is immediate by inspecting the translations and by Proposition 5.7.14.

Proposition 5.7.21

Let M be an id-model, g an assignment. Then:

• for any $\varphi \in \mathcal {L}_{\textsf {LoI}}$ and worlds $w,w'$: $\langle w,w'\rangle \models _g^{\textsf {LoI}}\varphi \iff \{w,w'\}\models _g\varphi ^{\flat }$;

• for any $\varphi \in \mathcal {L}_{\text {MA}}$ and worlds $w,w'$: $\{w,w'\}\models _g\varphi \iff \langle w,w'\rangle \models _g^{\textsf {LoI}}\varphi ^\sharp$;

• for any $\varphi \in \mathcal {L}_{\text {MA}}$ and state s: $s\models _g\varphi \iff \forall w,w'\in s:\langle w,w'\rangle \models _g^{\textsf {LoI}}\varphi ^\sharp$.

This connection ensures that both translations are entailment-preserving.

Proposition 5.7.22

(Translations are entailment-preserving)

• For all $\Phi \cup \{\psi \}\subseteq \mathcal {L}_{\text {MA}}$, $\Phi \models \psi \iff \Phi ^\sharp \models _\textsf {LoI}\psi ^\sharp$.

• For all $\Phi \cup \{\psi \}\subseteq \mathcal {L}_{\textsf {LoI}}$, $\Phi \models _\textsf {LoI}\psi \iff \Phi ^\flat \models \psi ^\flat$.

Proof

Consider the first item. If $\Phi \not \models \psi$, then we can find a relational  information model M, an assignment g, and an information state s such that $s\models _g\Phi$ and $s\not \models _g\psi$. Since formulas in $\mathcal {L}_{\text {MA}}$ are 2-coherent, we can find worlds $w,w'\in s$ such that such that $\{w,w'\}\not \models _g\psi$. Since $\{w,w'\}\subseteq s$, by persistency we have $\{w,w'\}\models _g\Phi$. By the previous proposition, in the semantics of LoI we have $\langle w,w'\rangle \models _g^{\textsf {LoI}}\Phi ^\sharp$ and $\langle w,w'\rangle \not \models _g^{\textsf {LoI}}\psi ^\sharp$, which shows that $\Phi ^\sharp \not \models _\textsf {LoI}\psi ^\sharp$. Reasoning similarly we can show that if $\Phi ^\sharp \not \models _{\textsf {LoI}}\psi ^\sharp$ then $\Phi \not \models \psi$.

The second item follows from the first and the fact that the translations are inverse to each other, as we have $\Phi ^\flat \models \psi ^\flat \iff \Phi ^{\flat \sharp }\models _\textsf {LoI}\psi ^{\flat \sharp }\iff \Phi \models _\textsf {LoI}\psi$.   $\square$

Thus, the Logic of Interrogation can be identified with a fragment of our inquisitive first-order logic $\textsf {InqBQ}$. One insight that emerges from this connection is that the primitive question quantifier Q of LoI, whose logical features seem quite complex and unfamiliar (cf. the axiomatization in ten Cate and Shan [20]), can in fact be analyzed in terms of a combination of operators which are logically simple and familiar. Indeed, in $\textsf {InqBQ}$ the formula $Q\overline{x}\alpha$ corresponds to the compound

where $\lnot$ is standard negation on statements,  is a constructive disjunction, and $\forall$ is a generalization to questions of the standard universal quantifier.¹⁴

Another repercussion of the translation is that it is possible to transform ten Cate and Shan’s [20] completeness result for LoI into a completeness result for the mention-all fragment of $\textsf {InqBQ}$ in which proofs use only formulas from the fragment. The interested reader is referred to Sect. 4.8 of Ciardelli [26].¹⁵

Finally, it is worth noting that $\textsf {InqBQ}$, and even its well-behaved classical antecedent fragment, is much more expressive than LoI. As we saw, in the clant fragment of $\textsf {InqBQ}$ we can express, among others, mention-some questions such as $\mathord {\exists \!\!\exists }xP(x)$, unique-instance questions such as $\mathord {\exists \!\!\exists }!xP(x)$, disjunctive questions such as , and conditional questions such as $\exists xP(x)\rightarrow \forall x?P(x)$. None of these questions has a counterpart in LoI.

5.8 ‘How Many’ Questions and Generalized Quantifiers

An important class of questions that we have not discussed so far is given by ‘how many’ questions, asking about the number of objects satisfying a certain property. The semantics of such questions can be analyzed in a natural way in our setting. To better see the idea, let us focus on a special case. Suppose P is a unary predicate. Consider first the case of an id-model. In this case, the objects that have property P at a world w are those in the extension $P_w$. The number of objects that have property P at w is thus captured by the cardinality $\# P_w$ of this set. An information state s settles how many objects have property P if s implies for some cardinal $\kappa$ that there are $\kappa$ objects with property P, i.e., if there is $\kappa$ such that $\# P_w=\kappa$ for all $w\in s$. This, in turn, happens if and only if the cardinality of the extension $P_w$ is the same in all $w\in s$.

In the context of id-models, we can thus say that a sentence $\varphi _{hm}$ expresses the question how many objects have property P if it has the following semantics:

$$\begin{aligned} s\models \varphi _{hm}\iff & {} \text {there is a cardinal }\kappa \text { such that }\forall w\in s:\# P_w=\kappa \\\iff & {} \forall w,w'\in s:\# P_w=\# P_{w'}. \end{aligned}$$

In the more general setting of a model M that has variable identity, things are slightly more subtle. In this case, the set of actual individuals having property P at a world w is not given directly by $P_w$, but by the quotient modulo the local identity $\sim _w$ (cf. the discussion in Sect. 5.4). Thus, the number of individuals having property P at w is given by the cardinal:

$$\text {num}_{w}(P):=\#(P_w/\!\!\sim _w).$$

In the general setting, we can thus say that a sentence $\varphi _{hm}$ expresses the question how many objects have property P if it has the following semantics:

$$\begin{aligned} s\models \varphi _{hm}\iff & {} \text {there is a cardinal }\kappa \text { such that }\forall w\in s:\text {num}_w(P)=\kappa \\\iff & {} \forall w,w'\in s:\text {num}_w(P)=\text {num}_{w'}(P). \end{aligned}$$

It is then natural to ask: is there in fact a sentence of $\textsf {InqBQ}$ with the required semantics? If not, we could make our demands more modest and ask: is there a sentence of $\textsf {InqBQ}$ that has the required semantics at least in restriction to models where D is finite? Or, even more modestly, in restriction to id-models where D is finite?

In order to study questions like this one, concerning the expressive power of $\textsf {InqBQ}$, [1] have recently developed an inquisitive counterpart of the classical Ehrenfeucht-Fraïssé game for first-order logic. The game is shown to characterize the expressive power of the logic, in the following sense: if two information states are distinguished by a formula $\varphi$ of $\textsf {InqBQ}$, then in the game there is a strategy to bring out the difference between the states within a finite number of moves determined by the number of quantifier nestings and implication nestings in $\varphi$; conversely, if there is a strategy to bring out the difference between two information states within a finite number of moves in the game, then there is a formula $\varphi$ of $\textsf {InqBQ}$ that distinguishes those states.

As in the classical case, the game is a powerful tool to show that certain properties are not expressible in $\textsf {InqBQ}$. In particular, it can be used to give a negative answer to the question we posed above, even in its less demanding version. We refer to Grilletti and Ciardelli [1] for the proof.

Theorem 5.8.1

(Inexpressibility of ‘how many’ questions in $\textsf {InqBQ}$) There is no sentence $\varphi _{hm}\in \mathcal {L}^{\textsf {Q}=}$ such that for all models M and states s:

$$s\models \varphi _{hm}\iff \forall w,w'\in s:\text {num}_w(P)=\text {num}_{w'}(P).$$

What is more, there is no sentence that satisfies this condition even in restriction to id-models where D is finite.

This result shows that $\textsf {InqBQ}$ does not have the resources to express how many questions. Given the importance of this class of questions, this result points to an interesting project for future research: add a how many inquisitive quantifier to $\textsf {InqBQ}$ and study the resulting logic.

Open Problem 5.8.2

(Extending $\textsf {InqBQ}$ with a ‘how many’ quantifier) Consider a logic InqBQH whose language is like that of $\textsf {InqBQ}$ but with the additional clause that if $\alpha$ is a classical formula and $\overline{x}$ a sequence of variables then $\textsf {H}\overline{x}\alpha$ is a formula. Intuitively, $\textsf {H}$ is the inquisitive quantifier ‘how many’, and $\textsf {H}\overline{x}\alpha$ stands for the question how many values of the sequence $\overline{x}$ satisfy $\alpha$. The number of such values at a world w is given by:

$$\text {num}_{w,g}^{\overline{x}}(\alpha )\;:=\;\#({\alpha }_g^{\overline{x}}(w)/\sim _w)$$

where $\alpha _g^{\overline{x}}(w)$ is the extension of $\alpha$ as given by Definition 5.7.13. We can then let the new formulas be interpreted by the following clause:

$$s\models _g\textsf {H}\overline{x}\alpha \iff \forall w,w'\in s: \text {num}_{w,g}^{\overline{x}}(\alpha )=\text {num}_{w',g}^{\overline{x}}(\alpha ).$$

What are the properties of the resulting logic InqBQH?

This open problem is an instance of a more general enterprise that awaits to be pursued: investigate generalized quantifiers in the setting of inquisitive predicate logic. As Grilletti and Ciardelli [1] discuss, the inquisitive setting gives rise to a new and more general notion of a quantifier, which encompasses not only standard quantifiers like ‘some x’, ‘at least three x’ or ‘infinitely many x’, but also properly inquisitive quantifiers like ‘which x’, ‘whether finitely or infinitely many x’, or ‘how many x’. Grilletti and Ciardelli [1] make a first step in the study of such quantifiers, giving a precise characterization of those unary cardinality quantifiers that are expressible in $\textsf {InqBQ}$. The undefinability of how many questions, given by Theorem 5.8.1, is a corollary of this characterization. An interesting aspect of the characterization is that, at least with respect to matters of cardinality, $\textsf {InqBQ}$ turns out to pattern with classical first-order logic, and not with second-order logic, in that each formula is only able to draw distinctions up to a fixed finite cardinal n.

5.9 Exercises

Exercise 5.9.1

(Formalizing questions in $\textsf {InqBQ}$ ) Consider a signature $\Sigma$ containing a binary predicate D and a rigid constant $\textsf {a}$. Suppose the domain of quantification consists of guests at a party, D(x, y) stands for “x danced with y” and $\textsf {a}$ denotes Alice. Give a formalization in InqBQ of the following natural language questions:

Exercise 5.9.2

(Formalizing questions in $\textsf {InqBQ}$ ) Consider a variation of Exercise 2.10.1: somone picked a secret code consisting of two natural numbers. Consider a signature $\Sigma$ consisting of a rigid constant symbol $\textsf {n}$ for each number $n\in \mathbb {N}$, a rigid binary function symbol +, and a binary relation symbol C, where Cxy is read intuitively as “the code is $\langle x,y\rangle$”. Suppose we formalize the scenario as a relational id-model M for this signature, where:

• $W=\{w_{ij}\mid i,j\in \mathbb {N}\}$ (so, worlds can be arranged as in the picture below);

• $D=\mathbb {N}$ (the set of natural numbers);

• $I_w(\textsf {n})=n$ for all $w\in W$;

• $I_w(+)$ is the sum operation, for all $w\in W$;

• $I_{w_{ij}}(C)=\{\langle i,j\rangle \}$ (that is, $w_{ij}$ is a world where the code is $\langle i,j\rangle$).

Write formulas of $\mathcal {L}^{\textsf {Q}=}$ that, in the context of this model, express the following statements and questions.

Exercise 5.9.3

(Formalizing questions in $\textsf {InqBQ}$ ) Given a unary predicate P and a number $n\in \mathbb {N}$, show how to write a formula $\varphi _n$ that expresses the question whether the number of objects satisfying P is less than n, exactly n, or more than n. Thus, $\varphi _n$ should be a formula with the following semantics (for the definition of $\text {num}_w(P)$, see p. xx):

$$\begin{aligned} s\models \varphi _n\iff & {} \forall w\in s: \text {num}_w(P)< n,\text { or }\\{} & {} \forall w\in a:\text {num}_w(P)= n,\text { or }\\{} & {} \forall w\in s:\text {num}_w(P)> n. \end{aligned}$$

Exercise 5.9.4

(Entailment in InqBQ) Let P, Q be unary predicates. Are the following entailments valid? Give a proof or a countermodel.

• 1. $\forall x?Px\wedge \forall x?Qx\models \forall x?(Px\wedge Qx)$

• 2. $\forall x?(Px\wedge Qx)\models \forall x?Px\wedge \forall x?Qx$

• 3.
• 4.

• 5. $\forall x(?Px\rightarrow {?Qx})\models \forall x?Px\rightarrow \forall x?Qx$

• 6. $\forall x?Px\rightarrow \forall x?Qx\models \forall x(?Px\rightarrow {?Qx})$

Exercise 5.9.5

(Entailment in InqBQ) Show that a necessary condition for an entailment to be valid in $\textsf {InqBQ}$ is that the classical counterpart of the entailment be valid in classical first-order logic. That is, show that for any $\Phi \cup \{\psi \}\subseteq \mathcal {L}^{\textsf {Q}=}$ we have:

$$\Phi \models \psi \;\Longrightarrow \;\Phi ^{cl}\models \psi ^{cl}$$

where $\varphi ^{cl}$ is the classical variant of a formula, as given by Definition 5.3.5, and if $\Phi ^{cl}=\{\varphi ^{cl}\mid \varphi \in \Phi \}$. Give an example in which the converse implication fails.

Exercise 5.9.6

(Coherence) Complete the proof of Proposition 5.6.3 by showing the following claims:

• if $\varphi$ is $\kappa$-coherent and $\psi$ is $\kappa '$ coherent then $\varphi \wedge \psi$ is $\max (\kappa ,\kappa ')$-coherent;

• if $\varphi$ is $\kappa$-coherent  then so is $\chi \rightarrow \varphi$ for any formula $\chi$;

• if $\varphi$ is $\kappa$-coherent then so is $\forall x\varphi$.

Exercise 5.9.7

(Turning clant formulas into rimp formulas) Complete the proof of Proposition 5.7.7 showing the following claim: if $\xi$ is a rimp formula (cf. Definition 5.7.6) and $\alpha$ is a classical formula, $\alpha \rightarrow \xi$ is equivalent to a rimp formula.

Hint: use the split equivalences given by Proposition 5.5.12.

Exercise 5.9.8

(First-order translation of clant formulas) Execute the strategy outlined in the proof sketch for Theorem 5.7.8 to show that there is a translation from the clant fragment of $\textsf {InqBQ}$ to classical two-sorted first-order logic.

Exercise 5.9.9

(Mention-all questions) Prove Proposition 5.7.16.

Hint: given $\varphi =(\forall \overline{x}?\alpha )$, take $\approx ^{\varphi }$ to be defined by $w\approx ^{\varphi } w'\iff \alpha _g^{\overline{x}}(w)=\alpha _g^{\overline{x}}(w')$. Clearly, $\approx ^{\varphi }$ is an equivalence relation on the set of worlds. Show that $\textsc {Alt}_M^g(\varphi )$ is precisely the set of equivalence classes of worlds under $\approx ^{\varphi }$ and that a state supports $\varphi$ just in case it is included in one of these equivalence classes.