Questions & Indexicality

Abstract

The truth conditions of sentences with indexicals like ‘I’ and ‘here’ cannot be given directly, but only relative to a context of utterance. Something similar applies to questions: depending on the semantic framework, they are given truth conditions relative to an actual world, or support conditions instead of truth conditions. Two-dimensional semantics can capture the meaning of indexicals and shed light on notions like apriority, necessity and context-sensitivity. However, its scope is limited to statements, while indexicals also occur in questions. Moreover, notions like apriority, necessity and context-sensitivity can also apply to questions. To capture these facts, the frameworks that have been proposed to account for questions need refinement. Two-dimensionality can be incorporated in question semantics in several ways. This paper argues that the correct way is to introduce support conditions at the level of characters, and develops a two-dimensional variant of both proposition-set approaches and relational approaches to question semantics.

Appendix

Proof

By induction on the complexity of $\varphi$. The proof relies on the fact that the truth value of a ?-free sentence depends only on the second index, which follows immediately from the truth conditions above. The only non-trivial cases are those for negation and question operator:

($\lnot$) ($\Rightarrow$):

Suppose $s\models \lnot \varphi$. Then for all $k\in s$, $\{k\}\not \models \varphi$. Take any $j,i\in s$. From $\{i\}\not \models \varphi$ and the induction hypothesis we obtain $i,i\not \models \varphi$. Since $\varphi$ is ?-free, it follows that $j,i\not \models \varphi$ and by the pair semantics for negation that $j,i\models \lnot \varphi$.

($\Leftarrow$):

Suppose $s\not \models \lnot \varphi$. Then there is some $i\in s$ such that $\{i\}\models \varphi$. By the induction hypothesis we obtain $i,i\models \varphi$, so $i,i\not \models \lnot \varphi$. So we found a pair that makes $\lnot \varphi$ false, as required.

(?) ($\Rightarrow$):

Suppose $s\models {\mathord {?}}\varphi$. Then $s\models \varphi$ or $s\models \lnot \varphi$. Take any $j,i\in s$. In the first case, we have $j,i\models \varphi$ and $j,j\models \varphi$ by the induction hypothesis. In the second case, it follows from the semantics of negation that for all $k\in s$, $\{k\}\not \models \varphi$. By the induction hypothesis we obtain $j,j\not \models \varphi$ and $i,i\not \models \varphi$. Since $\varphi$ is ?-free, it follows that $j,i\not \models \varphi$ too. In both cases we can conclude that $j,i\models {\mathord {?}}\varphi$

($\Leftarrow$):

Suppose $s\not \models {\mathord {?}}\varphi$. Then $s\not \models \varphi$ and $s\not \models \lnot \varphi$. First, by the induction hypothesis there exists a pair $k,i\in s$ such that $k,i\not \models \varphi$. Second, by the semantics of negation we have some $j\in s$ such that $\{j\}\models \varphi$. By the induction hypothesis it follows that $j,j\models \varphi$. Since $\varphi$ is ?-free, it follows from $k,i\not \models \varphi$ that $j,i\not \models \varphi$. So we found a pair j, i such that $j,i\not \models {\mathord {?}}\varphi$.

$\square$

Proof

By induction on the complexity of $\varphi$. The semantics of atoms, negation and necessity are already formulated in terms of individual pairs. Conjunction and universal quantification are immediate by unpacking the support conditions. I give only the case for the actuality operator A:

($\Rightarrow$):

Suppose $s\models A\varphi$. Then $\{\langle c,w_c \rangle ~|~ \langle c,i\rangle \in s \}\models \varphi$. Take any $\langle c,i\rangle \in s$. By the induction hypothesis, we obtain $\{\langle c,w_c\rangle \}\models \varphi$. It follows that $\{\langle c,i\rangle \}\models A\varphi$.

($\Leftarrow$):

Suppose $s\not \models A\varphi$. Then $\{\langle c,w_c \rangle ~|~ \langle c,i\rangle \in s \}\not \models \varphi$. By the induction hypothesis there must be some pair $\langle c,i\rangle \in s$ such that $\{\langle c,w_c\rangle \}\not \models \varphi$. It follows that $\{\langle c,i\rangle \}\not \models A\varphi$.

$\square$

Proof

By induction on the complexity of $\varphi$. The atomic case follows from $\varphi$ being indexical-free. Conjunction, universal quantification and question are immediate by unpacking the support conditions. I give the case for the necessity operator $\Box$:

($\Rightarrow$):

Suppose $s\models \varphi$. Then for all $\langle c,i \rangle \in s$ we have that $\{c\} \times W \models \varphi$ and by the induction hypothesis that $C \times W \models \varphi$. So either s is empty or $C \times W\models \varphi$. In the former case, $C \times \textsf {indices}(s)$ is also empty and $\Box \varphi$ is trivially supported. In the latter case, $\varphi$ is supported in all information states, so $C \times \textsf {indices}(s)\models \Box \varphi$.

($\Leftarrow$):

Suppose $s\not \models \Box \varphi$. Since $s\subseteq C\times \textsf {indices}(s)$, it follows by downward closure that $C\times \textsf {indices}(s)\not \models \Box \varphi$.

$\square$