Syntax

Let $\Sigma $ be a predicate logic signature. For simplicity, we first restrict to the case in which $\Sigma $ is a relational signature, i.e., contains no function symbols. The extension to an arbitrary signature, which involves some subtleties familiar from the classical case [Reference Hodges13], is discussed in Section 3.4. The set $\mathcal {L}$ of formulas of InqBQ over $\Sigma $ is defined as follows, where $R\in \Sigma $ is an n-ary relation symbol:

We will take negation to be a defined operator:

$$ \begin{align*} \neg\varphi:=\varphi\to\bot. \end{align*} $$

Formulas without occurrences of

and $\overline{\exists} $ are referred to as classical formulas and can be identified with standard $\textsf {FOL}$ -formulas, that is, first-order logic formulas. That is, the set $\mathcal {L}_c$ of classical formulas is given by:

$$ \begin{align*} \alpha::= R(x_1,\dots,x_n)\mid (x_1=x_2)\mid \bot\mid \alpha\land\alpha\mid\alpha\to\alpha\mid\forall x.\alpha. \end{align*} $$

As usual, classical formulas may be viewed as formalizing statements, such as, for instance, ‘every object has property P’. In the following, the variables $\alpha ,\beta ,\gamma $ will range over classical formulas, while $\varphi ,\psi ,\chi $ will range over arbitrary formulas. If $\alpha $ and $\beta $ are classical formulas, then we can define: Footnote ¹

• • $\alpha \lor \beta :=\neg (\neg \alpha \land \neg \beta )$

• • $\exists x.\alpha :=\neg \forall x.\neg \alpha $

The connective

and the quantifier $\overline{\exists} $ , referred to respectively as inquisitive disjunction and inquisitive existential quantifier, allow us to form questions. For instance, if $\alpha $ is a classical formula then the formula

represents the question whether $\alpha $ . We abbreviate this formula as $?\alpha $ :

• •

The formula $\overline{\exists}\, x.\alpha (x)$ represents the question what is an object satisfying $\alpha (x)$ ; and the formula $\forall x.?\alpha (x)$ represents the question which objects satisfy $\alpha (x)$ .

Semantics

A model for InqBQ consists of: a set W of worlds, representing possible states of affairs; a set D of individuals, the objects that the first-order variables range over; and an interpretation function I, which determines at each world the extension of all relation symbols, including identity.

Definition 2.1 (Models)

A model for the signature $\Sigma $ is a tuple $M=\langle W,D,I\rangle $ where W and D are sets and I is a function mapping each world $w\in W$ and each n-ary relation symbol $R\in \Sigma \cup \{=\}$ to a corresponding n-ary relation $I_w(R)\subseteq D^n$ —the extension of R at w. The interpretation of identity is subject to the following condition:

$$ \begin{align*} &I_w (=) is \ a \text{ congruence }, i.e., \ an \ equivalence \ relation \sim_w \ such \ that, \ if \ R\in\Sigma \\ &and \ d_i\sim_w \ d_i' \ for \ i\le n, \ then \langle d_1,\dots,d_n\rangle\in \ I_w(R)\iff \langle \ d^{\prime}_1,\dots,d^{\prime}_n\rangle\in \ I_w(R). \end{align*} $$

As discussed in the introduction, in inquisitive logic the semantics of the language specifies when a formula is supported at an information state $s\subseteq W$ , rather than when a formula is true at a possible world $w\in W$ . As usual, to handle open formulas and quantification, the support relation is defined relative to an assignment, which is a function from variables to the set D of individuals; if g is an assignment and $d\in D$ , then $g[x\mapsto d]$ is the assignment which maps x to d and behaves like g on all other variables.

Definition 2.2 (Support)

Let $M=\langle W,D,I\rangle $ be a model and let $s\subseteq W$ .

As usual, if $\varphi (x_1,\dots ,x_n)$ is a formula whose free variables are among $x_1,\dots ,x_n$ , then the value of g on variables other than $x_1,\dots ,x_n$ is irrelevant. If $d_1,\dots ,d_n\in D$ , we can therefore write $M,s\models \varphi (d_1,\dots ,d_n)$ to mean that $M,s\models _g\varphi $ holds with respect to an assignment g that maps $x_i$ to $d_i$ . In particular, if $\varphi $ is a sentence we can drop reference to g altogether. Moreover, we write $M\models \varphi $ as a shorthand for $M,W\models \varphi $ and we say that M supports $\varphi $ .

It is easy to verify that the support relation has the following two basic features:

• • Persistency: if $M,s\models _g\varphi $ and $t\subseteq s$ then $M,t\models _g\varphi $ ;

• • Empty state property: if $M,\emptyset \models _g\varphi $ for all $\varphi $ .

Recovering classical logic

In restriction to classical formulas, the above definition of support gives a non-standard semantics for classical first-order logic. To see why, let us associate to each world $w\in M$ a corresponding relational structure $\mathcal {M}_w$ , having as its domain the quotient $D/\!\sim _w$ and with the interpretation of relation symbols induced by $I_w(R)$ . Then we have the following connection.

Proposition 2.3. For all classical formulas $\alpha \in \mathcal {L}_c$ , all models M, assignments g, and information states s:

$$ \begin{align*}M,s\models_g\alpha\iff\forall w\in s: \mathcal{M}_w\models_{\overline{g}} \alpha\text{ holds in first-order logic}\end{align*} $$

where $\overline g$ is the assignment mapping x to the $\sim _w$ -equivalence class of $g(x)$ .

Thus, as far as the standard fragment of the language is concerned, the relation of support is essentially a recursive definition of global truth with respect to a set of structures sharing the same domain. Notice that the standard definition of truth can be recovered as a special case of support by taking s to be a singleton. We will also write $M,w\models _g\alpha $ as an abbreviation for $M,\{w\}\models _g\alpha $ .

Questions

As we just saw, evaluating a classical formula on an information state s amounts to evaluating it at each world in s and determining whether it is satisfied at each world. The same is not true for formulas that contain the operators and $\overline{\exists} $ ; typically, such formulas allow us to express global requirements on a state, which cannot be reduced to requirements on the single worlds in the state. We will illustrate this point by means of some examples. First take a classical sentence $\alpha $ and consider the formula . We have:

$$ \begin{align*}M,s\models {?\alpha}\iff (\forall w\in s: M,w\models\alpha)\text{ or }(\forall w\in s: M,w\models\neg\alpha).\end{align*} $$

Thus, in order for s to support $?\alpha $ , all the worlds in s must agree on the truth-value of $\alpha $ . In other words, $?\alpha $ is supported at s only if the information available in s determines whether or not $\alpha $ is true. Thus, $?\alpha $ can be taken as a formal representation of the question “is $\alpha $ true?”

Next take $\alpha (x)$ to be a classical formula having only the variable x free, and consider the formula $\overline{\exists}\, x.\alpha (x)$ . We have:

$$ \begin{align*} M,s\models\overline{\exists}\, x.\alpha(x)\iff\exists d\in D\text{ such that }\forall w\in s: M,w\models\alpha(d). \end{align*} $$

Thus, in order for s to support $\overline{\exists}\, x.\alpha (x)$ there must be an individual d which satisfies $\alpha (x)$ at all worlds in s. In other words, $\overline{\exists}\, x.\alpha (x)$ is supported at s if the information available in s implies for some specific individual that it satisfies $\alpha (x)$ —i.e., gives us a specific witness for $\alpha (x)$ . Thus, $\overline{\exists}\, x.\alpha (x)$ can be taken as a formal representation of the question “what is an object satisfying $\alpha (x)$ ?”

Finally, let again $\alpha (x)$ be a classical formula having only x free, and let us denote by $I_w(\alpha )$ the set of objects which satisfy $\alpha (x)$ at w, i.e., $I_w(\alpha ):=\{d\in D\mid M,w\models \alpha (d)\}$ . Consider the formula . We have:

$$ \begin{align*} M,s\models\forall x. ?\alpha(x)\iff\forall w,w'\in s: I_w(\alpha)=I_{w'}(\alpha). \end{align*} $$

Thus, in order for s to support $\forall x.?\alpha (x)$ , all the worlds in s must agree on which objects satisfy $\alpha (x)$ . In other words, $\forall x.?\alpha (x)$ is supported at s if the information available in s determines exactly which individuals satisfy $\alpha (x)$ . Thus, $\forall x.?\alpha (x)$ can be taken as a formal representation of the question “which objects satisfy $\alpha (x)?$ ”

Identity and cardinalities

An aspect of InqBQ which is worth commenting on is the interpretation of identity. In InqBQ, the interpretation of identity may differ at different worlds. This allows us to deal with uncertainty about the identity relation: e.g., one may have information about two individuals, a and b (say, one knows $Pa$ and $Qb$ ) and yet be uncertain whether a and b are distinct individuals, or the same. This also allows for uncertainty about how many individuals there are. Indeed, although a model is based on a fixed set D of epistemic individuals—objects to which information can be attributed—the domain of actual individuals at a world w is given by the equivalence classes modulo $\sim _w$ ; the number of actual individuals that exist at w is the number of such equivalence classes, i.e., the cardinality of the quotient $D_w:=D/\!\!\sim _w$ . Similarly, if $\alpha (x)$ is a classical formula having at most x free, then the number of individuals that satisfy $\alpha (x)$ at w is given by the cardinality of the set $\alpha _w:=I_w(\alpha )/\!\!\sim _w$ .

Notice that, as a special case, we could take $\sim _w$ to be the actual relation of identity on D at each world. A model in which identity is treated in this way is called an id-model.

This section is only intended as a summary of the key definitions and features of InqBQ and as a quick illustration of how questions can be captured by formulas in this logic. With these basic notions in place, let us now turn to the first novel contribution of the paper: an Ehrenfeucht–Fraïssé game for InqBQ.

3.1 The game

A position in an EF game for InqBQ is a tuple

$$ \begin{align*} \left\langle M_0,\, s_0,\, \overline a_0;\; M_1,\, s_1,\, \overline a_1 \right\rangle, \end{align*} $$

where:

• • $M_0=\left \langle W_0,D_0,I_0 \right \rangle $ and $M_1=\left \langle W_1,D_1,I_1 \right \rangle $ are models for InqBQ;

• • $s_0$ and $s_1$ are information states in the models $M_0$ and $M_1$ respectively; and

• • $\overline a_0$ and $\overline a_1$ are tuples of equal length of elements from $D_0$ and $D_1$ respectively.

If not otherwise specified, a game between the models $M_0$ and $M_1$ starts from position $\langle M_0,W_0,\varepsilon ;M_1,W_1,\varepsilon ; \rangle $ , where $\varepsilon $ indicates the empty tuple.

Starting a round from a position $\langle M_0,s_0,\overline a_0; M_1,s_1,\overline a_1 \rangle $ , S can choose between the following possible moves:Footnote ²

• • $\overline{\exists} $ -move: S picks an element $b_0\in D_0$ ; D responds with an element $b_1\in D_1$ ; the game continues from the position $\left \langle M_0,s_0,\overline a_0b_0; M_1,s_1,\overline a_1b_1 \right \rangle $ ;

• • $\forall $ -move: S picks an element $b_1\in D_1$ ; D responds with an element $b_0\in D_0$ ; the game continues from the position $\left \langle M_0,s_0,\overline a_0b_0; M_1,s_1,\overline a_1b_1 \right \rangle $ ; and

• • $\to $ -move: S picks a sub-state $t_1\subseteq s_1$ ; D responds with a sub-state ${t_0\subseteq s_0}$ ; S picks $i\in \{1,0\}$ . The game continues from $\left \langle M_i,t_i,\overline a_i; M_{1-i},t_{1-i},\overline {a}_{1-i} \right \rangle $ .

Notice the asymmetry between the roles of the two models: by performing an $\to $ -move, S can pick an information state from $M_1$ , but not a state in $M_0$ .

With respect to termination condition, we consider different versions of the game. In the bounded version of the game, a pair of numbers $\left \langle i,q \right \rangle \in \mathbb {N}^2$ is fixed in advance. This number constrains the development of the game: in total, S can play only i implication moves and only q quantifier moves (i.e., $\overline{\exists} $ -move or a $\forall $ -move). When there are no more moves available, the game ends. If $\left \langle M_0,s_0,\overline a_0; M_1,s_1,\overline a_1 \right \rangle $ is the final position, the game is won by Player D if the following condition is satisfied, and by player S otherwise:

• • Winning condition for D: for all atomic formulas $\alpha (x_1,\dots ,x_n)$ where n is the size of the tuples $\overline a_0$ and $\overline a_1$ , we have:

  (1) $$ \begin{align} M_0,s_0\models \alpha(\overline{a}_0)\; \implies\; M_1,s_1 \models\alpha(\overline{a}_1). \end{align} $$

In the unbounded version of the game, no restriction is placed at the outset on the number of moves to be performed. Instead, player S has the option to declare the game over at the beginning of each round: in this case, the winner is determined as in the bounded version of the game. If the game never stops, then D is the winner.Footnote ³

Example 3.1. Take the signature $\Sigma = \{P\}$ , where P is a unary predicate symbol. Given the models $M_0$ and $M_1$ in Figure 1, in the table we show a run of the bounded game with $\left \langle i,q \right \rangle = \left \langle 1,2 \right \rangle $ between $M_0$ and $M_1$ . At the end of the run, the position is $\left \langle M_1,\{v_1\},\left \langle e_2,e_1 \right \rangle ; M_0,\{w_0\},\left \langle d_1,d_2 \right \rangle \right \rangle $ . The winner is Spoiler, since

Fig. 1

On the left, two inquisitive models in the signature $\Sigma = \{P\}$ . The top row represents the set of worlds of the models (e.g., $\{w_0,w_1\}$ for the model $M_0$ ) and the left column represents the domain (e.g., $\{d_1,d_2\}$ for $M_0$ ); $I(P)$ is encoded by the entries of the table: a $\bullet $ indicates that P holds, while a $\times $ indicates that P does not hold (e.g., $d_1 \in I_{w_0}(P)$ and $d_2 \notin I_{w_0}(P)$ ). On the right, a run of the Ehrenfeucht-Fraïssé game.

As usual, a winning strategy for a player is a strategy which guarantees victory to them, no matter what the opponent plays. If D has a winning strategy in the EF game of length $\left \langle i,q \right \rangle $ starting from position $\left \langle M_0,s_0,\overline {a}_0; M_1,s_1,\overline {a}_1 \right \rangle $ we write:

$$ \begin{align*} ( M_0,s_0,\overline{a}_0 ) \preceq_{i,q} (M_1,s_1,\overline{a}_1 ). \end{align*} $$

We write $\approx _{i,q}$ for the relation $\preceq _{i,q} \cap \succeq _{i,q}$ . Notice that the game with bounds $i,q$ is finite (since the number of turns is bounded by $i+q$ ), zero-sum (as can be seen from the winning condition) and has perfect information. Therefore, if $(M_0,s_0,\overline {a}_0) \preceq _{i,q} (M_1,s_1,\overline {a}_1)$ does not hold, then it follows from the Gale-Stewart Theorem that Spoiler has a winning strategy in the EF game of length $\langle i,q \rangle $ starting from the position $\langle M_0,s_0,\overline {a}_0; M_1,s_1,\overline {a}_1 \rangle $ .

We write $M_0\preceq _{i,q} M_1$ as a shorthand for $(M_0,W_0,\varepsilon ) \preceq _{i,q} (M_1,W_1,\varepsilon )$ .

The following two propositions follow easily from the definition of the game.

Proof. We prove the result by contraposition. Suppose that Spoiler has a winning strategy in the game $ \operatorname {\mathrm {EF}}_{i',q'}(M_0,s_0,\overline {a}_0;M_1,s_1,\overline {a_1})$ for some $i'\leq i$ and $q'\leq q$ . This means that, no matter which choices Duplicator makes during a run, Spoiler can perform $i'$ implication moves and $q'$ quantifier move and force the game to end up in a position

$$ \begin{align*} &\left\langle M_0, t_0, \overline{a}_0\overline{b}_0 ;\; M_1, t_1, \overline{a}_1\overline{b}_1 \right\rangle &\text{(case 1)} \\ \text{or}\qquad &\left\langle M_1, t_1, \overline{a}_0\overline{b}_0 ;\; M_1, t_1, \overline{a}_1\overline{b}_1 \right\rangle &\text{(case 2)} \end{align*} $$

for which there exists an atomic formula $\alpha (\overline {x},\overline {y})$ such that

$$ \begin{align*} &M_0, t_0 \models \alpha(\overline{a}_0,\overline{b}_0) \;\text{and}\; M_1, t_1 \nvDash \alpha(\overline{a}_1,\overline{b}_1) &&\text{in case 1} \\ \text{or}\qquad &M_1, t_1 \models \alpha(\overline{a}_1,\overline{b}_1) \;\text{and}\; M_0, t_0 \nvDash \alpha(\overline{a}_0,\overline{b}_0) &&\text{in case 2.} \end{align*} $$

We will show that, from this point on, Spoiler can still win the game after performing $i-i'$ additional implication moves and $q-q'$ additional quantifier moves: This amounts to a winning strategy in the game $ \operatorname {\mathrm {EF}}_{i,q}(M_0,s_0,\overline {a}_0;M_1,s_1,\overline {a_1})$ . We will focus here on case 1, since the other case is completely analogous.

• • Firstly, Spoiler performs $i-i'$ implication moves, always picking $t_1$ as the information state and choosing $i := 0$ (that is, maintaining the order of the models). Since Duplicator can only choose substates of $t_0$ , the position of the game after these moves will be of the form:

  $$ \begin{align*} \left\langle M_0, t^{\prime}_0, \overline{a}_0\overline{b}_0 ;\; M_1, t_1, \overline{a}_1\overline{b}_1 \right\rangle \end{align*} $$
  with $t^{\prime }_0 \subseteq t_0$ .

• • Secondly, Spoiler performs $q-q'$ quantification moves in an arbitrary way; no matter what Duplicator responds, we end up in a position of the form:

  $$ \begin{align*} \left\langle M_0, t^{\prime}_0, \overline{a}_0\overline{b}_0\overline{c}_0 ;\; M_1, t_1, \overline{a}_1\overline{b}_1\overline{c}_1 \right\rangle \end{align*} $$
  for $\overline {c}_0$ and $\overline {c}_1$ of length $q-q'$ .

This is indeed a winning position for Spoiler, since

$$ \begin{align*} \left\{ \begin{array}{l@{\hspace{2em}}l} M_0, t^{\prime}_0 \models \alpha(\overline{a}_0,\overline{b}_0) &(\text{by persistency, since } t^{\prime}_0 \subseteq t_0) \\ M_1, t_1 \nvDash \alpha(\overline{a}_1,\overline{b}_1). \end{array} \right.\\[-37pt] \end{align*} $$

3.2 IQ degree and types

We define the implication degree ( $ \operatorname {\mathrm {Ideg}}$ ) and quantification degree ( $ \operatorname {\mathrm {Qdeg}}$ ) of a formula by the following inductive clauses, where p stands for an atomic formula:

The combined degree of a formula is defined as $ \operatorname {\mathrm {IQdeg}}(\varphi ) = \left \langle \operatorname {\mathrm {Ideg}}(\varphi ), \operatorname {\mathrm {Qdeg}}(\varphi ) \right \rangle $ . We define a partial order $\leq $ on such degrees by setting:

$$ \begin{align*} \left\langle a,b \right\rangle \leq \left\langle a',b' \right\rangle \iff a\leq a' \text{ and } b \leq b'. \end{align*} $$

We denote by $\mathcal {L}_{i,q}^l$ the set of formulas $\varphi $ such that $ \operatorname {\mathrm {IQdeg}}(\varphi )\le \left \langle i,q \right \rangle $ and the set of free variables in $\varphi $ is included in $\{x_1,\dots ,x_l\}$ . We can then define the key notion of $\left \langle i,q \right \rangle $ -type.

Definition 3.4 $\left \langle i,q \right \rangle $ -types

Let M be a model, s an information state, and $\overline {a}$ a tuple of elements in M of length l. The $\left \langle i,q \right \rangle $ -type of $\left \langle M,s,\overline {a} \right \rangle $ is the set

$$ \begin{align*} \operatorname{\mathrm{tp}}_{i,q}(M,s,\overline{a}) := \left\{ \varphi \in \mathcal{L}^{l}_{i,q} \;\middle|\; M,s \models \varphi(\overline{a}) \right\}. \end{align*} $$

We also define the following notation:

$$ \begin{align*} &(M_0, s_0, \overline{a}_0) \sqsubseteq_{i,q} (M_1, s_1, \overline{a}_1) \overset{def}{\iff} \operatorname{\mathrm{tp}}_{i,q}(M_0,s_0,\overline{a}_0) \subseteq \operatorname{\mathrm{tp}}_{i,q}(M_1,s_1,\overline{a}_1) \\ &(M_0, s_0, \overline{a}_0) \equiv_{i,q} (M_1, s_1, \overline{a}_1) \overset{def}{\iff} \operatorname{\mathrm{tp}}_{i,q}(M_0,s_0,\overline{a}_0) = \operatorname{\mathrm{tp}}_{i,q}(M_1,s_1,\overline{a}_1) \end{align*} $$

Notice that, if the signature is finite, there are only a finite number of non-equivalent formulas of combined degree at most $\left \langle i,q \right \rangle $ , and consequently only a finite number of $\left \langle i,q \right \rangle $ -types. This can be shown inductively as follows:

• • The quotient ${\mathcal {L}^{l}_{0,0}/\!\equiv }\,$ is a distributive lattice under the operations $\land $ and . Moreover, since we are working with a finite relational signature, $\mathcal {L}^{l}_{0,0}$ contains only finitely many atomic formulas, and the equivalence classes of these formulas generate the whole lattice. Since every finitely generated distributive lattice is finite, ${\mathcal {L}^{l}_{0,0}/\!\equiv }\,$ is finite, which means that $\mathcal {L}^{l}_{0,0}$ contains only finitely many formulas up to logical equivalence.

• • Formulas in $\mathcal {L}^{l}_{i,q}$ are equivalent to Boolean combinations of formulas in ${A \cup B}$ , for $A = \left \{ \varphi \to \psi \;\middle |\; \varphi , \psi \in \mathcal {L}^{l}_{i-1,q} \right \}$ and $B = \left \{ \overline{\exists}\, x. \varphi , \forall x.\varphi \;\middle |\; \varphi \in \mathcal {L}^{l+1}_{i,q-1} \right \}$ —where we impose by definition $\mathcal {L}^{l}_{i,q} = \emptyset $ if $i<0$ or $q<0$ . By induction hypothesis, A and B contain only finitely many non-equivalent formulas.

3.3 The EF theorem

What follows is the first main result of the paper: the relations $\preceq _{i,q}$ and $\sqsubseteq _{i,q}$ coincide.

Theorem 3.6. Suppose the signature $\Sigma $ is finite. Then

$$ \begin{align*} (M_0,s_0,\overline{a}_0) \preceq_{i,q} (M_1,s_1,\overline{a}_1) \quad\iff\quad (M_0,s_0,\overline{a}_0) \sqsubseteq_{i,q} (M_1,s_1,\overline{a}_1). \end{align*} $$

Proof. We will prove this by well-founded induction on $\left \langle i,q \right \rangle $ . For the basic case, $\left \langle i,q \right \rangle = \left \langle 0,0 \right \rangle $ , we just have to verify that, if Condition (1) holds for all atomic formulas, then it holds for all formulas $\varphi \in \mathcal {L}^{l}_{0,0}$ . This is straightforward. Next, suppose $\left \langle i,q \right \rangle> \left \langle 0,0 \right \rangle $ and suppose the claim holds for all $\left \langle i',q' \right \rangle <\left \langle i,q \right \rangle $ . For the left-to-right direction, proceed by contraposition. Suppose that for some $\varphi \in \mathcal {L}_{i,q}^l$ the following conditions hold:

$$ \begin{align*} \begin{array}{c@{\qquad}c} M_0,s_0 \models \varphi(\overline{a}_0) & M_1,s_1 \not\models \varphi(\overline{a}_1). \end{array} \end{align*} $$

We proceed by induction on the structure of $\varphi $ ; some cases are easy to consider:

• • If $\varphi $ is an atom, it follows $(M_0,s_0,\overline {a}_0) \not \preceq _{0,0} (M_1,s_1,\overline {a}_1)$ ; so, by Proposition 3.2 also $(M_0,s_0,\overline {a}_0) \not \preceq _{i,q} (M_1,s_1,\overline {a}_1)$ . Thus, in this case the conclusion follows.

• • If $\varphi $ is a conjunction $\psi \land \chi $ then we have:

  $$ \begin{align*} \left\{ \begin{array}{l} M_0, s_0 \models \psi(\overline{a}_0) \land \chi(\overline{a}_0) \implies M_0, s_0 \models \psi(\overline{a}_0) \text{ and } M_0, s_0 \models \chi(\overline{a}_0) \\ M_1, s_1 \not\models \psi(\overline{a}_1) \land \chi(\overline{a}_1) \implies M_1, s_1 \not\models \psi(\overline{a}_1) \text{ or } M_1, s_1 \not\models \chi(\overline{a}_1). \\ \end{array} \right. \end{align*} $$
  So, either $\psi $ or $\chi $ is a less complex witness of $(M_0,s_0,\overline {a}_0) \not \sqsubseteq _{i,q} (M_1,s_1,\overline {a}_1)$ .

• • If $\varphi $ is a disjunction , we can reach a conclusion analogous to the one we reached for conjunction.

The remaining cases are those in which $\varphi $ is of the form $\psi \to \chi $ , $\forall x. \psi $ or $\overline{\exists}\, x.\psi $ (cases $\Rightarrow ^{1}$ , $\Rightarrow ^{2}$ , $\Rightarrow ^{3}$ respectively). Let us consider the three cases separately.

Case $\Rightarrow ^{1}$ : $\varphi $ is an implication $\psi \to \chi $ . In this case we have:

$$ \begin{align*} &M_1, s_1 \not\models \psi(\overline{a}_1) \to \chi(\overline{a}_1) \implies (\exists t_1 \subseteq s_1)\left[ M_1, t_1 \models \psi(\overline{a}_1) \text{ and } M_1, t_1 \not\models \chi(\overline{a}_1)\right], \\ &M_0, s_0 \models \psi(\overline{a}_0) \to \chi(\overline{a}_0) \implies (\nexists t_0 \subseteq s_0) \left[ M_0, t_0 \models \psi(\overline{a}_0) \text{ and } M_0, t_0 \not\models \chi(\overline{a}_0) \right]. \end{align*} $$

Thus there exists a state $t_1 \subseteq s_1$ with a different $\langle i-1,q \rangle $ -type than every $t_0\subseteq s_0$ —either because it supports $\psi $ or because it does not support $\chi $ . So by induction hypothesis, if S performs a $\to $ -move and chooses $t_1$ , for every choice $t_0$ of D we have $(M_0,t_0,\overline {a}_0) \not \approx _{i-1,q} (M_1,t_1,\overline {a}_1)$ . It follows by Proposition 3.3 that $(M_0,s_0,\overline {a}_0) \not \preceq _{i,q} (M_1,s_1,\overline {a}_1)$ as wanted.

Case $\Rightarrow ^{2}$ : $\varphi $ is a universal $\forall x. \psi $ . In this case we have:

$$ \begin{align*} &M_1, s_1 \not\models \forall x. \psi(\overline{a}_1,x) \implies (\exists b_1 \in D_1)\; M_1, s_1 \not\models \psi(\overline{a}_1,b_1), \\ &M_0, s_0 \models \forall x. \psi(\overline{a}_0,x) \implies (\forall b_0 \in D_0)\; M_0, s_0 \models \psi(\overline{a}_0,b_0). \end{align*} $$

Thus if S performs a $\forall $ -move and chooses $b_1$ , for every choice $b_0$ of D, by induction hypothesis we have

$$ \begin{align*} (M_0,s_0,\overline{a}_0b_0) \;\not\preceq_{i,q-1} (M_1,s_1,\overline{a}_1b_1). \end{align*} $$

It follows by Proposition 3.3 that $(M_0,s_0,\overline {a}_0) \not \preceq _{i,q} (M_1,s_1,\overline {a}_1)$ as wanted.

Case $\Rightarrow ^{3}$ : $\varphi $ is an inquisitive existential $\overline{\exists}\, x. \psi $ . This case is similar to the previous one: S can perform an $\overline{\exists} $ -move and pick an element $b_0$ in $ D_0$ with no counterpart in $D_1$ , and by Proposition 3.3 we get the result.

This completes the proof of the left-to-right direction of the inductive step. Now consider the converse direction. Again, we proceed by contraposition. Suppose that S has a winning strategy in the EF game of length $\langle i,q\rangle $ starting from $(M_0,s_0,\overline {a}_0;M_1,s_1,\overline {a}_1)$ . We consider again three cases, depending on the first move of the winning strategy (cases $\Leftarrow ^{1}$ , $\Leftarrow ^{2}$ , $\Leftarrow ^{3}$ respectively).

Case $\Leftarrow ^{1}$ : the first move is a $\to $ -move. Suppose S starts by choosing $t_1\subseteq s_1$ . As this is a winning strategy for S, for every choice $t_0\subseteq s_0$ of D we have

$$ \begin{align*} (M_0,t_0,\overline{a}_0) \not\preceq_{i-1,q} (M_1,t_1,\overline{a}_1) \quad\text{ or }\quad (M_1,t_1,\overline{a}_1) \not\preceq_{i-1,q} (M_0,t_0,\overline{a}_0). \end{align*} $$

By inductive hypothesis, this translates to

$$ \begin{align*} \exists \psi_{t_0} \in \operatorname{\mathrm{tp}}(t_0) \setminus \operatorname{\mathrm{tp}}(t_1) \quad\text{or}\quad \exists \theta_{t_0} \in \operatorname{\mathrm{tp}}(t_1) \setminus \operatorname{\mathrm{tp}}(t_0), \end{align*} $$

where $ \operatorname {\mathrm {tp}}(t_0) := \operatorname {\mathrm {tp}}_{i-1,q}(M_0,t_0,\overline {a}_0)$ and $ \operatorname {\mathrm {tp}}(t_1) := \operatorname {\mathrm {tp}}_{i-1,q}(M_1,t_1,\overline {a}_1)$ .

Given this, there exist two families $\{ \psi _{t_0}\mid t_0\subseteq s_0\}$ and $\{\theta _{t_0}\mid t_0\subseteq s_0\}$ such that:

$$ \begin{align*} \left\{ \begin{array}{ll} \psi_{t_0} \in \operatorname{\mathrm{tp}}(t_0) \setminus \operatorname{\mathrm{tp}}(t) &\text{if } \operatorname{\mathrm{tp}}(t_0)\setminus\operatorname{\mathrm{tp}}(t)\ne\emptyset \\ \psi_{t_0} := \bot &\text{otherwise} \end{array} \right. \\ \left\{ \begin{array}{ll} \theta_{t_0} \in \operatorname{\mathrm{tp}}(t) \setminus \operatorname{\mathrm{tp}}({t_0}) &\text{if } \operatorname{\mathrm{tp}}(t)\setminus\operatorname{\mathrm{tp}}({t_0})\ne\emptyset \\ \theta_{t_0} := \top &\text{otherwise.} \end{array} \right. \end{align*} $$

Moreover, we can suppose the two families to be finite, as there are only a finite number of formulas of degree $\left \langle i-1,q \right \rangle $ up to logical equivalence (see Section 3.2). Define now the formula $\varphi $ as follows:

We have: (i) $ \operatorname {\mathrm {IQdeg}}(\varphi ) \leq \left \langle i,q \right \rangle $ , (ii) $\varphi \notin \operatorname {\mathrm {tp}}_{i,q}(M_0,s_0,\overline {a}_0)$ (since by construction $\varphi $ is falsified at $t_1\subseteq s_1$ ) and (iii) $\varphi \in \operatorname {\mathrm {tp}}_{i,q}(M_1,s_1,\overline {a}_1)$ (since by construction $\varphi $ holds at every state $t_0\subseteq s_0$ ). Thus we have $(M_0,t_0,\overline {a}_0) \not \sqsubseteq _{i-1,q} (M_1,t_1,\overline {a}_1)$ , as we wanted.

Case $\Leftarrow ^{2}$ : the first move is a $\forall $ -move. Suppose S starts by choosing $b_1 \in D_1$ . As this is a winning strategy for S, for every choice $b_0\in D_0$ of D we have

$$ \begin{align*} (M_0,s_0,\overline{a}_0b_0)\; \not\preceq_{i,q-1} (M_1,s_1,\overline{a}_1b_1). \end{align*} $$

By induction hypothesis, the above translates to

$$ \begin{align*} \exists \psi_{b_0}\in \operatorname{\mathrm{tp}}(b_0)\setminus\operatorname{\mathrm{tp}}(b_1), \end{align*} $$

where $ \operatorname {\mathrm {tp}}(b_0):= \operatorname {\mathrm {tp}}_{i,q-1}(M_0,s_0,\overline {a}_0b_0)$ and $ \operatorname {\mathrm {tp}}(b_1):= \operatorname {\mathrm {tp}}_{i,q-1}(M_1,s_1,\overline {a}_1b_1)$ .

Now the formula

has IQ-degree at most $\left \langle i,q \right \rangle $ , and by construction we have $\varphi \in \operatorname {\mathrm {tp}}_{i,q}(M_0,s_0,\overline {a}_0)$ and $\varphi \notin \operatorname {\mathrm {tp}}_{i,q}(M_1,s_1,\overline {a}_1)$ . Thus, we have $(M_0,t_0,\overline {a}_0) \not \sqsubseteq _{i-1,q} (M_1,t_1,\overline {a}_1)$ .

Case $\Leftarrow ^{3}$ : the first move is a $\overline{\exists} $ -move. Reasoning as in the previous case, we find that there exists a $b_0\in D_0$ —the element chosen by S—such that for every $b_1\in D_1$

$$ \begin{align*} \exists \theta_{b_1} \in \operatorname{\mathrm{tp}}(t_0)\setminus\operatorname{\mathrm{tp}}(t_1). \end{align*} $$

In particular, it follows that the formula

$$ \begin{align*} \varphi := \overline{\exists}\, x. \bigwedge_{b_1\in D_1} \psi_{b_1} \end{align*} $$

is a formula of complexity at most $\left \langle i,q \right \rangle $ such that $\varphi \in \operatorname {\mathrm {tp}}_{i,q}(M_0,s_0,\overline {a}_0)$ and $\varphi \notin \operatorname {\mathrm {tp}}_{i,q}(M_1,s_1,\overline {a_1})$ . Again, it follows that $(M_0,t_0,\overline {a}_0) \not \sqsubseteq _{i-1,q} (M_1,t_1,\overline {a}_1)$ .

As a corollary, we also get a game-theoretic characterization of the distinguishing power of formulas in the $\langle i,q\rangle $ -fragment of InqBQ.

Corollary 3.7. For a finite signature $\Sigma $ , we have:

$$ \begin{align*} (M_0,s_0,\overline{a}_0) \approx_{i,q} (M_1,s_1,\overline{a}_1) \quad\iff\quad (M_0,s_0,\overline{a}_0) \equiv_{i,q} (M_1,s_1,\overline{a}_1). \end{align*} $$

3.4 Extending the result to function symbols

The results we just obtained assume that the signature $\Sigma $ is relational. However, it is not hard to extend them to the case in which $\Sigma $ contains function symbols (including nullary function symbols, i.e., constant symbols). In InqBQ, function symbols are interpreted rigidly: if $f\in \Sigma $ is an n-ary function symbol, then the interpretation function I of a model M must assign to all worlds w in the model the same function $I_w(f):D^n\to D$ .Footnote ⁴

As in the case of classical logic [Reference Hodges13] Section 3.3, the presence of function symbols requires some care in formulating the EF game. The reason is that allowing atomic formulas to contain arbitrary occurrences of function symbols allows us to generate with a finite number of choices in the game an infinite sub-structure of the model—which spoils the crucial locality feature of the game. Technically, a simple way to circumvent the problem this is to follow [Reference Hodges13] Section 3.3 and work with formulas which are unnested.

Definition 3.8 (Unnested formula)

An unnested atomic formula is a formula of one of the following forms:

$$ \begin{align*} x = y \hspace{5em} c = y \hspace{5em} f(\overline{x}) = \overline{y} \hspace{5em} R(\overline{x}). \end{align*} $$

An unnested formula is a formula that contains only unnested atoms.

Examples of nested formulas—i.e., non-unnested formulas—are $f(x) = g(y)$ , $R(f(x))$ and $f(c) = x$ .

We can now make the following amendments to the definition above: (i) the winning conditions for the game are determined by looking at whether Equation (1) is satisfied for all unnested atomic formulas, and (ii) the $\langle i,q\rangle $ -types are re-defined as sets of unnested formulas of degree at most $\langle i,q\rangle $ . Other than that, the statement of the result and the proof are the same as above.

Using identity we can turn an arbitrary formula into an equivalent unnested one (e.g., replacing $P(f(x))$ with $\forall y.((y=f(x))\to Py)$ ) so the restriction to unnested formula is not a limitation to the generality of the game-theoretic characterization; rather, it can be seen as an indirect way of assigning formulas containing function symbols with the appropriate $\langle i,q\rangle $ -degree—making explicit a quantification which is implicit in the presence of a function symbol.

3.5 A symmetric version of the game

As noticed before, a difference between the Ehrenfeucht–Fraïssé game for classical logic and the game introduced in Section 3 is that the latter is asymmetric—the two models under consideration do not play the same role. This allows us to study the relation $\sqsubseteq _{i,q}$ in addition to the relation $\equiv _{i,q}$ . This contrast with the situation in classical first-order logic, where the relations $\sqsubseteq _q$ and $\equiv _q$ coincide due to the semantics of negation. A natural question is whether we can define a symmetric version of the game which directly characterizes the relation $\equiv _{i,q}$ . In what follows we will consider a naïve modification of the game to obtain a symmetric version and study the induced equivalence relation between models. The symmetric way is just like the game introduced above, except that we replace the $\to $ -move and the winning condition as follows.

• • $\to $ -move. S picks a sub-state $t_i\subseteq s_i$ for $i=0$ or $1$ . D responds with a sub-state $t_{1-i}\subseteq s_{1-i}$ . The game continues from $\langle M_0,t_0,\overline {a}_0;M_1,t_1,\overline {a}_1 \rangle $ .

  Thus, in this version S is free to play a substate move in either model. This obviates the need for swapping roles of the two models.

• • Winning condition for D: for all atomic formulas $\alpha (x_1,\dots ,x_n)$ where n is the size of the tuples $\overline a_0$ and $\overline a_1$ , we have:

  $$ \begin{align*} M_0,s_0\models \alpha(\overline{a}_0)\; \iff\; M_1,s_1 \models\alpha(\overline{a}_1). \end{align*} $$

  Notice that, compared to the original version, the implication has become a biconditional.

We will indicate with $M_0, s_0, \overline {a}_0 \approx ^{s}_{i,q} M_1, s_1 ,\overline {a}_1$ the existence of a winning strategy for Duplicator in the symmetric game from position $\left \langle M_0,s_0,\overline {a}_0; M_1,s_1,\overline {a}_1 \right \rangle $ with bounds $\langle i,q\rangle $ ; we indicate by $M_0, s_0, \overline {a}_0 \approx ^{s} M_1, s_1 ,\overline {a}_1$ the existence of a winning strategy for D in the unbounded version of the symmetric game. We will also use notational conventions analogous to the ones introduced for the original game. Notice that, as anticipated, the roles of the two models in the game are interchangeable:

Comparing this version with the original one, we clearly made Spoiler’s life much easier: now he can perform $\to $ -moves without any restrictions on the model from which he can choose the state; and the winning condition for Duplicator is more restrictive than the original one. So the following result should not come as a surprise:

As an immediate corollary we obtain the following result.

Corollary 3.11. Suppose the signature $\Sigma $ is finite. Then

$$ \begin{align*} (M_0,s_0,\overline{a}_0) \approx^s_{i,q} (M_1,s_1,\overline{a}_1) \quad\implies\quad (M_0,s_0,\overline{a}_0) \equiv_{i,q} (M_1,s_1,\overline{a}_1). \end{align*} $$

However, the converse of Corollary 3.11 does not hold in general.

Proposition 3.12.

Proof. Consider the models M and N in Figure 2a. We have:

• • $M \preceq _{0,1} N$ : the winning strategy for Duplicator is described in Table 2b;

• • $M \succeq _{0,1} N$ : the winning strategy for Duplicator is described in Table 2c;

• • $M \not \approx ^s_{0,1} N$ : if Spoiler picks the element $d_2$ , every move of Duplicator leads to Spoiler’s victory.

Fig. 2

Thus, unlike the original version, the symmetric version of the game does not provide us with an exact characterization of the relation $\equiv _{i,q}$ determined by the logic. Nevertheless, the symmetric game is still useful: if we can show that Duplicator has a winning strategy in the symmetric game with bounds $\langle i,q\rangle $ between two models, this suffices to show that these models are indistinguishable by formulae of degree $\langle i,q \rangle $ . This is convenient, since strategies are often easier to describe in the symmetric game than in the original game, since we do not have to keep track of how the role of the models gets swapped in the course of the game. Indeed, we use this strategy in the proof of Theorem 4.5.

4.1 Cardinality quantifiers in classical first-order logic

In classical logic, a formula $\alpha (x)$ , with at most the variable x free, determines, relative to a model $\mathcal {M}$ , a corresponding set of individuals:

$$ \begin{align*} \alpha_{\mathcal{M}}:=\{d\in D\mid \mathcal{M}\models\alpha(d)\}. \end{align*} $$

Let K be a class of cardinals. This is an operator that can be added to classical first-order logic by stipulating that if $\alpha (x)$ is a classical formula with at most x free, then $Q_Kx.\alpha (x)$ is a formula, with the following semantics ( $\#$ denotes the cardinality of a set):Footnote ⁵

$$ \begin{align*} \mathcal{M}\models Q_Kx.\alpha(x)\iff \#\alpha_{\mathcal{M}}\in K. \end{align*} $$

By a cardinality quantifier we mean a quantifier which is of the form $Q_K$ for some class of cardinals K. Notice that the existential quantifier $\exists $ is a cardinality quantifier, since $\exists = Q_{\textsf {Card} \setminus \{0\}}$ , for $\textsf {Card}$ the class of all cardinals. By contrast, the universal quantifier $\forall $ is not a cardinality quantifier, since the condition $\mathcal {M} \models \forall x.P(x)$ , namely, $P_{\mathcal {M}} = D$ , cannot be formulated solely in terms of $\# P_{\mathcal {M}}$ .Footnote ⁶

Let $\chi _K[P]$ be a FOL-formula (thus, not containing $Q_K$ ). We say that $\chi _K[P]$ defines $Q_K$ if $Q_Kx.Px\equiv \chi _K[P]$ . It is not hard to see that if this is the case, then for every formula $\alpha (x)$ we have $Q_Kx.\alpha (x)\equiv \chi _K[\alpha ]$ . We say that the quantifier $Q_K$ is definable in FOL if there is a FOL-formula which defines it.

The statements in (1) can all be seen as having the form $Qx.Px$ , where Q is a cardinality quantifier. Indeed, we have the following characterizations, where $[3,\dots )$ is the class of cardinals $\ge 3$ ; $\textsf {Even}$ is the set of even natural numbers; Inf is the class of infinite cardinals.

1. (3)

   $$ \begin{align*} \hspace{-4pc} \begin{array}{llll} \\[-2.5pc] \text{a.} &\mathcal {M}\models (1.a) &\iff P_{\mathcal {M}}=\emptyset &\iff \# P_{\mathcal {M}}\in \{0\}\\ \text{b.} &\mathcal {M}\models (1.b) &\iff \#P_{\mathcal {M}}\ge 3 &\iff \#P_{\mathcal {M}}\in [3,\dots )\\ \text{c.} &\mathcal {M}\models (1.c) &\iff \#P_{\mathcal {M}}\text { is even} &\iff \#P_{\mathcal {M}}\in \textsf {Even}\\ \text{d.} &\mathcal {M}\models (1.d) & \iff \#P_{\mathcal {M}}\text { is infinite} &\iff \# P_{\mathcal {M}}\in \textsf {Inf.}\end{array} \end{align*} $$

What cardinality quantifiers are definable in classical first-order logic? That is, for what classes K of cardinals is the quantifier $Q_K$ definable? The answer is given by the following theorem, which is an easy application of EF-games for FOL (and seems, to the best of our knowledge, to be folklore).

Consider again the statements in (1), repeated below for convenience with the corresponding classes of cardinals given on the right. It follows immediately from the characterization that the first two statements are expressible in classical first-order logic, while the third and fourth are not.

1. (4)

   $$ \begin{align*} \hspace{-9.4pc} \begin{array}{lll} \\[-2.5pc]\text{a.} &\textrm{There is no } P. &K=\{0\}\\ \text{b.} &\textrm{There are at least three } P. &K=[3,\dots )\\ \text{c.} &\textrm{The number of } P \textrm{ is even}. &K=\textsf {Even}\\ \text{d.} &\textrm{There are infinitely many } P. &K=\textsf {Inf}\end{array} \end{align*} $$

4.2 Cardinality quantifiers in InqBQ

Let us now turn to the inquisitive case. A model M for inquisitive first-order logic represents a variety of states of affairs, one for each possible world w. At each world w, the state of affairs is represented by the first-order structure $\mathcal {M}_w$ , having as its domain the set $D_w:=D/\!\!\sim _w$ . Let $\alpha (x)$ be a classical formula with at most the variable x free. Relative to each world w, $\alpha (x)$ determines an extension $\alpha _w$ , which is a set of individuals from $D_w$ :

$$ \begin{align*} \alpha_{w}:=\{ d\in D_w\mid \mathcal{M}_w\models\alpha(d)\}. \end{align*} $$

Therefore, relative to an information state s, the formula $\alpha (x)$ determines a corresponding set of cardinals, $\{\#\alpha _{w}\mid w\in s\}$ . We refer to this set of cardinals as the cardinality trace of $\alpha (x)$ in s.

Definition 4.2 (Cardinality trace)

Let M be a model, s an information state, and $\alpha (x)$ a classical formula where at most the variable x occurs free. The cardinality trace of $\alpha (x)$ in s is the set of cardinals:

$$ \begin{align*} \mathrm{tr}_{s}(\alpha)=\{\#\alpha_{w}\mid w\in s\}. \end{align*} $$

A cardinal $\kappa $ is in $\mathrm {tr}_s(\alpha )$ if, according to the information available in s, $\kappa $ might be the number of elements satisfying $\alpha (x)$ ; that is, if it might be the case that the extension of $\alpha (x)$ has cardinality $\kappa $ . Thus, $\mathrm {tr}_s(\alpha )$ captures exactly the information available in s about the number of individuals satisfying $\alpha (x)$ .

Now let $\mathbb {K}$ be a class of sets of cardinals. We associate with $\mathbb {K}$ a corresponding quantifier $Q_{\mathbb {K}}$ . We can add this quantifier to InqBQ by stipulating that if $\alpha (x)$ is a classical formula with at most x free, then $Q_{\mathbb {K}} x.\alpha (x)$ is a formula, interpreted by the following clauseFootnote ⁷ :

$$ \begin{align*} M,s\models Q_{\mathbb{K}} x.\alpha(x)\iff \mathrm{tr}_s(\alpha)\in \mathbb{K}. \end{align*} $$

A cardinality quantifier is a quantifier which is of the form $Q_{\mathbb {K}}$ , where $\mathbb {K}$ is a class of sets of cardinals.

Let $\chi _{\mathbb {K}}[P]$ be an InqBQ-formula (thus, without cardinality quantifiers). We say that $\chi _{\mathbb {K}}[P]$ defines the quantifier $Q_{\mathbb {K}}$ if $Q_{\mathbb {K}} x.Px\equiv \chi _{\mathbb {K}}[P]$ . Again, it is not hard to see that if this holds, then for every classical formula $\alpha (x)$ we have $Q_{\mathbb {K}} x.\alpha (x)\equiv \chi _{\mathbb {K}}[\alpha ]$ . We say that $Q_{\mathbb {K}}$ is definable in InqBQ if there is an InqBQ-formula that defines it.

In order to make the notion of a cardinality quantifier more concrete, let us see how the statements in (1) and the questions in (2) can be seen as instantiating the form $Qx.Px$ where Q is a cardinality quantifier in the sense of inquisitive logic.

Consider first the statements in (1). In general, in inquisitive semantics a statement $\alpha $ is supported by a state s iff the information available in s implies that $\alpha $ is true. This means that $\alpha $ is true at all worlds $w\in s$ . Keeping this in mind, we can see that the statements in (1) have the following semantics:

1. (5)

   $$ \begin{align*} \hspace{-0.2pc} \begin{array}{llll} \\[-2.5pc] \text{a.} &M,s\models (1.a) &\iff \forall w\in s: P_w=\emptyset &\iff \mathrm {tr}_s(P)\subseteq \{0\}\\ \text{b.} &M,s\models (1.b) &\iff \forall w\in s: \#P_w\ge 3 &\iff \mathrm {tr}_s(P)\subseteq [3,\dots )\\ \text{c.} &M,s\models (1.c) &\iff \forall w\in s: \#P_w\text { is even} &\iff \mathrm {tr}_s(P)\subseteq \textsf {Even}\\ \text{d.} &M,s\models (1.d) &\iff \forall w\in s: \#P_w\text { is infinite} &\iff \mathrm {tr}_s(P)\subseteq \textsf {Inf}\end{array} \end{align*} $$

Let us now check that all these statements correspond to statements of the form $Qx.Px$ for Q a cardinality quantifier. For this, we introduce a useful notation.

Thus, if K is a set, then $K^{\downarrow }=\wp (K)$ . However, if K is a proper class, then $K^{\downarrow }$ will not be a set either; moreover, $K^{\downarrow }$ will not contain K, since K is not a set.

Now consider the cardinality quantifiers $Q_{1}$ – $Q_4$ determined by the following classes:

$$ \begin{align*} \mathbb{K}_{1}=\{0\}^{\downarrow}\qquad\mathbb{K}_{2}=[3,\dots)^{\downarrow}\qquad\mathbb{K}_{3}=\textsf{Even}^{\downarrow}\qquad\mathbb{K}_{4} =\textsf{Inf}^{\downarrow}. \end{align*} $$

We have:

$$ \begin{align*} \begin{array}{l @{\;\iff\;} l @{\;\iff\;} l @{\;\iff\;} l} M,s\models Q_{1}x.Px &\mathrm{tr}_s(P)\in\mathbb{K}_{1} &\mathrm{tr}_s(P)\subseteq\{0\} &M,s\models (1.a)\\ M,s\models Q_{2}x.Px &\mathrm{tr}_s(P)\in\mathbb{K}_{2} &\mathrm{tr}_s(P)\subseteq[3,\dots) &M,s\models (1.b)\\ M,s\models Q_{3}x.Px &\mathrm{tr}_s(P)\!\in \mathbb{K}_{3} &\mathrm{tr}_s(P)\subseteq\textsf{Even} &M,s\models (1.c) \\ M,s\models Q_{4}x.Px &\mathrm{tr}_s(P)\in \mathbb{K}_{4} &\mathrm{tr}_s(P)\subseteq\textsf{Inf} &M,s\models (1.d) \end{array} \end{align*} $$

Next, consider the questions in (2). Start with (2.a), the question whether there are any P. This question is settled in an information state s in case the information in s implies that there are no P, or it implies that there are some P. The former is the case if the extension of P is empty in all worlds $w\in s$ . The latter is the case if the extension of P is non-empty in all worlds $w\in s$ . Thus, the semantics of (2.a) is as follows.

$$ \begin{align*} M,s\models (2.a) &\iff (\forall w\in W: P_w=\emptyset)\text{ or }(\forall w\in W: P_w\neq\emptyset) \\ &\iff (\forall w\in W: \#P_w=0)\text{ or }(\forall w\in W: \#P_w\ge 1) \\ &\iff \mathrm{tr}_s(P)=\{0\}\text{ or }\mathrm{tr}_s(P)\subseteq [1,\dots). \end{align*} $$

Second, consider the question (2.b), how many individuals are P. This question is settled in an information state s if the information available in s determines exactly how many individuals are P. This is the case if there is a cardinal $\kappa $ such that at every world $w\in s$ , the extension $P_w$ contains $\kappa $ elements.Footnote ⁸

$$ \begin{align*} M,s\models (2.b) &\iff \exists\kappa\forall w\in W: \#P_w=\kappa \\ &\iff \mathrm{tr}_s(P) \;\text{contains at most one element} \\ &\iff \mathrm{tr}_s(P)\subseteq\{\kappa\} \;\text{for some cardinal}\; \kappa. \end{align*} $$

Next, consider (2.c), the question whether the number of P is even or odd. This is settled in an information state s in case the information available in s implies that the number of P is even, or that the number of P is odd.Footnote ⁹ The former holds if the extension of P is even at every world in s. The latter holds if the extension of P is odd al every world in s.

$$ \begin{align*} M,s\models (2.c) &\iff (\forall w\in W: \#P_w\text{ is even})\text{ or }(\forall w\in W: \#P_w\text{ is odd}) \\ &\iff \mathrm{tr}_s(P)\subseteq\textsf{Even}\text{ or }\mathrm{tr}_s(P)\subseteq\textsf{Odd.} \end{align*} $$

Finally, consider (2.d), the question whether there are infinitely many P. This is settled in an information state s in case the information available in s implies that there are infinitely many P, or it implies that the aren’t infinitely many P. The former is the case if the extension of P is infinite at every world $w\in s$ , while the latter is the case if the extension of P is finite at every world $w\in s$ .

$$ \begin{align*} M,s\models (2.d) &\iff (\forall w\in s: \#P_w\text{ is finite})\text{ or }(\forall w\in s: \#P_w\text{ is infinite}) \\ &\iff \mathrm{tr}_s(P)\subseteq\textsf{Fin}\text{ or }\mathrm{tr}_s(P)\subseteq\textsf{Inf.} \end{align*} $$

Now consider four cardinality quantifiers, $Q_{5}$ – $Q_{8}$ , determined by the following classes:

1. (6)

   1. a. $\mathbb {K}_{5}=\{0\}^{\downarrow }\cup [1,\dots )^{\downarrow }$

   2. b. $\mathbb {K}_{6}=\bigcup \{\{\kappa \}^{\downarrow }\mid \kappa \text { a cardinal}\}$

   3. c. $\mathbb {K}_{7}=\textsf {Even}^{\downarrow }\cup \textsf {Odd}^{\downarrow }$

   4. d. $\mathbb {K}_{8}=\textsf {Fin}^{\downarrow }\cup \textsf {Inf}^{\downarrow }$

Then we have:

$$ \begin{align*} \begin{array}{lll} M,s\models Q_{5}x.Px &\;\iff\; \mathrm{tr}_s(P)\in\mathbb{K}_{5} &\;\iff\; \mathrm{tr}_s(P)\subseteq\{0\}\text{ or }\mathrm{tr}_s(P)\subseteq[1,\dots) \\ &&\;\iff\; M,s\models (2.a) \\[.5em] M,s\models Q_{6}x.Px &\;\iff\; \mathrm{tr}_s(P)\in\mathbb{K}_{6} &\;\iff\; \mathrm{tr}_s(P)\subseteq\{\kappa\}\text{ for some }\kappa \\ &&\;\iff\; M,s\models (2.b) \\[.5em] M,s\models Q_{7}x.Px &\;\iff\; \mathrm{tr}_s(P)\in \mathbb{K}_{7} &\;\iff\; \mathrm{tr}_s(P)\subseteq\textsf{Even}\text{ or }\mathrm{tr}_s(P)\subseteq\textsf{Odd} \\ &&\;\iff\; M,s\models (2.c) \\[.5em] M,s\models Q_{8}x.Px &\;\iff\; \mathrm{tr}_s(P)\in \mathbb{K}_{8} &\;\iff\; \mathrm{tr}_s(P)\subseteq\textsf{Fin}\text{ or }\mathrm{tr}_s(P)\subseteq\textsf{Inf} \\ &&\;\iff\; M,s\models (2.d) \end{array} \end{align*} $$

So, in the inquisitive setting, a new range of “inquisitive” cardinality quantifiers come into play, which combine with a property to yield questions like those exemplified in (2). In addition to standard cardinality quantifiers like ‘no’, ‘at least three’, ‘infinitely many’, we also have new, question-forming cardinality quantifiers like ‘how many’ and ‘whether finitely or infinitely many’.

4.3 Characterization

What cardinality quantifiers can be expressed in InqBQ? Given that, in the inquisitive setting, cardinality quantifiers are in one-to-one correspondence with classes of sets of cardinals, this question can be made precise as follows.

The next theorem provides an answer to this question. In essence, what the theorem says is that the cardinality quantifiers definable in InqBQ are all and only the inquisitive disjunctions of cardinality quantifiers definable in classical first-order logic.Footnote ¹⁰ Before stating the Theorem, let us fix some useful notations. For any natural number n, we let:

• • $[0,n] := \{m\in \textsf {Card}\mid m \leq n\}$

• • $[n,\dots ) := \{ \kappa \in \textsf {Card} \,|\, \kappa \geq n \}$

Moreover, we introduce an equivalence relation $=_n$ that disregards differences between cardinals larger than n. More precisely, if $\kappa $ and $\kappa '$ are two cardinals:

$$ \begin{align*} \kappa=_n\kappa'\iff \kappa=\kappa' \;\text{or}\; \kappa,\kappa'>n. \end{align*} $$

If A and B are sets of cardinals, we write $A=_n B$ if A and B are the same set, modulo identifying all cardinals larger than n:

$$ \begin{align*} A=_nB\iff &\forall \kappa\in A\exists\kappa'\in B \;\text{such that}\; \kappa=_n\kappa' \;\text{and}\\ &\forall \kappa'\in B\exists\kappa\in A \;\text{such that}\; \kappa=_n\kappa'. \end{align*} $$

Moreover, we say that a class of sets of cardinals $\mathbb {K}$ is:

• • $=_n$ -invariant, if whenever $B\in \mathbb {K}$ and $A=_nB$ we have $A\in \mathbb {K}$ and

• • downward-closed, if whenever $B\in \mathbb {K}$ and $A\subseteq B$ we have $A\in \mathbb {K}$ .

We can now state our second main result.

Proof. We show that $2\Rightarrow 1\Rightarrow 3\Rightarrow 2$ .

$\pmb {[2 \Rightarrow 1]}$ Suppose $\mathbb {K}= K_1^{\downarrow }\cup \dots \cup K_n^{\downarrow }$ where for each $K_i$ there exists a natural number m such that $K_i$ contains either all or none of the cardinals $\kappa \ge m$ . By Theorem 4.1, for each $K_i$ we have a classical formula $\chi _{i}$ such that, in classical first-order logic:

$$ \begin{align*} \mathcal{M}\models \chi_i\iff \# P_{\mathcal{M}}\in K_i. \end{align*} $$

These formulas are also formulas of InqBQ, and it follows from Proposition 2.3 that we have:

$$ \begin{align*} M,s\models\chi_i&\iff \forall w\in s: \mathcal{M}_w\models\chi_i\\ &\iff\forall w\in s: \#{P_{\mathcal{M}_w}} \in K_i\\ &\iff\mathrm{tr}_s(P)\subseteq K_i\\ &\iff\mathrm{tr}_s(P)\in K_i^{\downarrow}. \end{align*} $$

Now consider the inquisitive disjunction

. We have:

This shows that the InqBQ formula

defines the quantifier $Q_{\mathbb {K}}$ .

$\pmb {[1 \Rightarrow 3]}$ Next, consider the implication from 1 to 3. Suppose $Q_{\mathbb {K}}$ is definable in InqBQ by a formula $\varphi _{\mathbb {K}}$ . We need to show that $\mathbb {K}$ is downward closed and $=_m$ invariant for some natural number m.

We firstly show that $\mathbb {K}$ is downward closed. Suppose $A\subseteq B\in \mathbb {K}$ . This means that there exists a model M and an information state s such that $M,s\vDash \varphi _{\mathbb {K}}$ and $\mathrm {tr}_s(P) = B$ . Consider now the state $t := \{ w\in s \;|\; \#P_w \in A \} \subseteq s$ . By definition we have $\mathrm {tr}_t(P) = A$ ; and by persistency $M, t \vDash \varphi _{\mathbb {K}}$ . Thus $A \in \mathbb {K}$ , as wanted.

Next, we show that $\mathbb {K}$ is closed under $=_m$ for some m. We want to show that the condition above holds for $m = q$ , where q is the quantifier degree of the defining formula $\varphi _{\mathbb {K}}$ . So, suppose $A\in \mathbb {K}$ and $A=_{q} B$ . If we find two information models $M,N$ such that $\mathrm {tr}_{W^M}(P) = A$ , $\mathrm {tr}_{W^N}(P) = B$ and $M \approx _{i,q} N$ then we are done, since in this case:

$$ \begin{align*} A \in \mathbb{K} \quad\iff\quad M \models \varphi_{\mathbb{K}} \quad\iff\quad N \models \varphi_{\mathbb{K}} \quad\iff\quad B \in \mathbb{K}. \end{align*} $$

Consider enumerations of the sets A and B:Footnote ¹¹ $A := \left \{ \kappa _{\alpha } \;\middle |\; \alpha < \lambda \right \}$ and $B := \left \{ \kappa ^{\prime }_{\alpha } \;\middle |\; \alpha < \lambda \right \}$ which both start with the same initial sequence $\langle \kappa _1,\dots ,\kappa _l\rangle =\langle \kappa _1',\dots ,\kappa _l'\rangle $ enumerating $A \cap [0,q] = B \cap [0,q]$ . Let $M,N$ be the models defined by the following clauses:

$$ \begin{align*} \begin{array}{l@{\hspace{5em}}l} W^M := \left\{ w_{\alpha} \;\middle|\; \alpha < \lambda \right\} &W^N := \left\{ w_{\alpha} \;\middle|\; \alpha < \lambda \right\} = W^M \\[.8em] D^M := \left\{ d^\alpha_\beta \;\middle|\; \alpha<\lambda \;\&\; \beta<\kappa_{\alpha} \right\} &D^N := \left\{ e^\alpha_\beta \;\middle|\; \alpha<\lambda \;\&\; \beta<\kappa^{\prime}_{\alpha} \right\} \\[.8em] I^M_{w_\gamma}(P)\left(d^\alpha_\beta\right) \;\iff\; \alpha = \gamma &I^N_{w_\gamma}(P)\left(e^\alpha_\beta\right) \;\iff\; \alpha = \gamma \\[.8em] M \text{ is an } \textsf{id}\text{-model} &N \text{ is an } \textsf{id}\text{-model.} \end{array} \end{align*} $$

An example of these models is given in Figure 3. Notice that $\# I^M_{w_{\alpha }}(P) = \kappa _{\alpha }$ and $\# I^N_{w_{\alpha }}(P) = \kappa ^{\prime }_{\alpha }$ . In particular, it follows that $\mathrm {tr}_{W^M}(P) = A$ and $\mathrm {tr}_{W^N}(P) = B$ . So if we show that $M \approx _{i,q} N$ then we are done. In order to show this, we present here a winning strategy for Duplicator in the symmetric version of the EF-game between M and N (cf. Section 3.5):

• • If Spoiler plays an implication move and chooses an information state s from either of the models, then Duplicator responds by choosing the same state s from the other model (this is possible since $W^M = W^N$ ).

• • If Spoiler plays a quantifier move and chooses an element $d^\alpha _\beta $ from the model M, we consider two separate cases:

  1. – If $d^\alpha _\beta = a_i$ for some i, that is, it has already been picked during the run—by either Spoiler or Duplicator—then Duplicator responds by choosing $b_i$ .

  2. – If $d^\alpha _\beta $ has not been previously picked, then Duplicator chooses an element $e^\alpha _\gamma $ (notice that the elements have the same superscript and possibly different subscripts) which has not been previously picked during the run. The fact that duplicator can find such an element is guaranteed by $A=_q B$ : this means that either $\kappa _{\alpha }=\kappa _{\alpha }'$ , or else $\kappa _{\alpha },\kappa _{\alpha }'> q$ . In the former case the number of elements $d^\alpha _\beta $ and $e^\alpha _\gamma $ is exactly the same; in the latter case the number of elements $e^\alpha _\gamma $ is larger than the number of quantifier moves in the game.

• • If Spoiler plays a quantifier move and chooses an element $e^\alpha _\beta $ from the model N, then Duplicator applies the same strategy as in the previous case, swapping the roles of the models M and N.

Fig. 3

Suppose $q=2$ , and consider the sets $A=\{2,3,5\}$ and $B=\{2,4\}$ . Notice that $A=_2B$ . We enumerate these sets as $\langle 2,3,5\rangle $ and $\langle 2,4,4\rangle $ . The figure shows the models M and N derived from this enumeration. These models are indistinguishable in the EF-game with only 2 quantifier moves, regardless of the number of implication moves.

Notice that with this strategy Duplicator ensures that at the end of the run the final position:

1. 1. has the same state s for both models;

2. 2. $a_i = a_j$ if and only if $b_i = b_j$ ;

3. 3. corresponding elements $a_i,b_i$ in the two models have the same superscripts, that is, $a_i$ and $b_i$ are of the form $d^\alpha _\beta $ and $e^\alpha _\beta $ respectively.

This is indeed a winning strategy, since:

$$ \begin{align*} M, s \models P(d^{\alpha}_{\beta}) \;\iff\; s \subseteq \{\alpha\} \;\iff\; N, s \models P(e^{\alpha}_{\gamma}) \\ M, s \models a_i = a_j \;\iff\; N, s \models b_i = b_j. \qquad \quad \end{align*} $$

$\pmb {[3 \Rightarrow 2]}$ Suppose $\mathbb {K}$ is downward closed and $=_m$ -invariant for some number m. Let $A_1,\dots ,A_n$ be the subsets of $[0,m+1]$ which are contained in $\mathbb {K}$ . Now define:

$$ \begin{align*} K_i = \left\{ \begin{array}{ll} A_i &\text{if}\; m+1\notin A_i \\ A_i \cup [m+1,\dots) &\text{if}\; m+1 \in A_i. \end{array} \right. \end{align*} $$

We claim that $\mathbb {K}=K_1^{\downarrow }\cup \dots \cup K_n^{\downarrow }$ . Start with the right-to-left inclusion. Let $B\in K_1^{\downarrow }\cup \dots \cup K_n^{\downarrow }$ . This means that $B\subseteq K_i$ for some i. Now we distinguish two cases.

• • Case 1: $K_i=A_i$ . Then $A_i\in \mathbb {K}$ by definition, and since $\mathbb {K}$ is downward closed, also $B\in \mathbb {K}$ .

• • Case 2: $K_i=A_i\cup [m+1,\dots )$ . We claim that in this case, $A_i=_m A_i\cup B$ : if so, since $A_i\in \mathbb {K}$ and $\mathbb {K}$ is $=_m$ -invariant, we have $A_i\cup B\in \mathbb {K}$ , which in turn by downward closure yields $B\in \mathbb {K}$ . To see that $A_i=_nA_i\cup B$ , the only non-trivial step is to show that for all $\kappa \in B$ there exists some $\kappa '\in A_i$ such that $\kappa '=_m\kappa $ . So, take $\kappa \in B$ : if $\kappa \le m$ then $\kappa \in A_i$ (since $B\subseteq K_i= A_i\cup [m+1,\dots )$ ), so we can take $\kappa '=\kappa $ ; if on the other hand $\kappa> m$ , then $\kappa =_m m+1\in A_i$ .

Either way, we conclude $B\in \mathbb {K}$ , which gives the right-to-left inclusion.

For the converse inclusion, suppose $B\in \mathbb {K}$ . Again, we distinguish two cases.

• • Case 1: $B\subseteq [0,m]$ . In this case, $B=A_i$ for some $i\le n$ , and thus $B\in K_i^{\downarrow }$ .

• • Case 2: $B\ni \kappa $ for some $\kappa>m$ . In this case, $B=_m B\cup \{m+1\}$ , since $\kappa =_m m+1$ . Since $B\in \mathbb {K}$ and $\mathbb {K}$ is $=_m$ -invariant, also $B\cup \{m+1\}\in \mathbb {K}$ . Now take $(B\cup \{m+1\})\cap [0,m+1]$ : by downward closure, this set is in $\mathbb {K}$ , and since it is a subset of $[0,m+1]$ , it coincides with $A_i$ for some $i\le n$ . Notice that $m+1\in A_i$ , and thus, $K_i=A_i\cup [m+1,\dots )$ . Therefore, $B\subseteq K_i$ , which implies $B\in K_i^{\downarrow }$ .

In either case, we conclude that $B\in K_i$ for some $i\le n$ , which gives the left-to-right inclusion.□

Theorem 4.5 allows us to tell immediately which among the questions in (2) are expressible in InqBQ: (2.a), the question whether there is any P, is expressible, since it has the form $Q_{\mathbb {K}} x.Px$ for the class $\mathbb {K}_5=\{0\}^{\downarrow }\cup \,[1,\dots )^{\downarrow }$ , where both $\{0\}$ and $[1,\dots )$ are definable in classical first-order logic. Indeed, the defining formula is simply $?\exists x.Px$ , which abbreviates .

The remaining questions, (2.b), (2.c), and (2.d) are not expressible, since they have the form $Q_{\mathbb {K}} x.Px$ for the following classes $\mathbb {K}$ :

$$ \begin{align*} \mathbb{K}_{6}=\bigcup\{\{\kappa\}^{\downarrow}\mid \kappa\text{ a cardinal}\}\qquad\mathbb{K}_{7}=\textsf{Even}^{\downarrow}\cup\textsf{Odd}^{\downarrow}\qquad\mathbb{K}_{8} =\textsf{Fin}^{\downarrow}\cup\textsf{Inf}^{\downarrow}. \end{align*} $$

Clearly, these classes are not the form $K_1^{\downarrow }\cup \dots \cup K_n^{\downarrow }$ for $K_1,\dots ,K_n$ definable in classical first-order logic. In a similar way, we can see that none of the following questions about the cardinality of P is expressible in InqBQ.

1. (7)

   1. a. How many P are there, modulo k? (for $k\ge 2$ )

   2. b. Is the number of P even, odd, or infinite?

   3. c. Is the number of P a prime number, or a composite one?

   4. d. Are there uncountably many P?

It is worth pausing to remark that, while InqBQ can express the question “what objects are P?” (by means of the formula $\forall x.?Px$ , see Section 2), it cannot express the corresponding cardinality question “how many object are P?” From the perspective of logical modeling of questions, this means that analyzing how many questions—an important class of questions—requires a proper extension of the logic InqBQ. Developing and investigating such an extension is an interesting prospect for future work.

Since the proof of Theorem 4.5 is quite flexible, the characterization result can be seen to hold also when we restrict to certain salient classes of models. For instance, since the proof uses only $\mathsf {id}$ -models, we obtain the following Corollary.

Moreover, it is an easy exercise to adapt the proof to show the following result, concerning the class of finite models and the class of finite $\mathsf {id}$ -models.