Explaining presupposition projection in (coordinations of) polar questions

Abstract

This article starts off with the observation that in certain cases, presuppositions triggered by an element inside a question nucleus may fail to project. In fact, in what looks like coordinated structures involving polar questions, presupposition projection patterns are exactly parallel to what is observed when the corresponding assertions are coordinated. The article further shows that these facts do not fall out straightforwardly from existing theories of polar questions, (apparent) coordinations of questions, and presupposition projection. It then proposes a trivalent extension of inquisitive semantics such that the observed pattern can be understood in terms of existing theories of presupposition projection. The proposal has the following properties: (a) apparent coordinations of questions are indeed coordinations of questions, and (b) the semantic denotation of polar questions is asymmetric with respect to the “yes” and “no” answers.

Appendix: Formal companion to Sect. 2.4

This appendix contains a formal development that demonstrates the problems encountered when one combines answer set semantics with either the Transparency Theory or trivalent theories, as cursorily explained in Sect. 2.4. We will derive the fact that the observed presupposition filtering pattern is expected in conjunctive questions if they are analysed as the tripartition, but not if they are analysed as the quadripartition. Meanwhile, the observed pattern in disjunctive questions is not derived in any natural way.

Note that throughout the Appendix we work with question denotations, rather than with declarative sentences containing embedded questions, and therefore the results might only directly apply to matrix questions. In principle, at least under the Transparency Theory, we could predict different filtering patterns under certain embeddings (in particular, we might expect more filtering predictions). While I do not think this would lead to different results in practice, no attempt is made to prove it here.

When the proof of a result is omitted, it is because it is immediate.

1.1 1 Equivalence relations on questions

The Transparency Theory requires a notion of equivalence; we define several natural ones below. As in the main text, we identify questions with Hamblin denotations (sets of propositions). We will use Q, $Q'$, etc. to name our abstract variables representing questions. In order to show that the results extend to Karttunen’s approach, we will also deal with Karttunen denotations (functions from worlds to sets of propositions), which we will refer to as Q̂, $\hat{Q}'$, etc. The following relations let us map each kind of denotation into the other:

Definition 1

Relation between Hamblin and Karttunen denotations

For a question Q, we define:

$$ \hat{Q} := \lambda w.\, \{ p \in Q \,|\, p(w)\} $$

As long as Q does not contain a logical contradiction, the following relation lets us reverse the mapping:

$$ Q = \bigcup _{w} \hat{Q}(w) $$

Note that it is not true in general that all Karttunen questions can be derived from a Hamblin question (i.e., that they can be written as Q̂ for some Q). However, all the questions that we consider here are representable as Hamblin questions.

In the main text, we left the notion of equivalence we were using somewhat implicit. Here we define several reasonable notions of equivalence. These notions are all relativised to a context set C (they are notions of contextual equivalence). The purpose of this wealth of definitions is to show that the issues outlined in the main text do not depend on specific implementational choices within the general framework of H/K semantics. In particular, whether we adopt H-equivalence, which is the most natural approach from the perspective of a Hamblin-style theory, or K-equivalence, which is the most natural approach from the perspective of a Karttunen-style theory, does not matter, showing that our discussion applies to Hamblin- and Karttunen-style accounts equally.

For the notions based on a presuppositional answerhood operator, I implement a variety of ways of dealing with the unique-answer presupposition. The first approach treats it essentially as an entailment, a second one only considers worlds where it is true, and a third one only considers contexts where it is true.⁴⁷

Non-contextual versions of the non-presuppositional notions can be obtained by taking C to be the set of all possible worlds.

Definition 2

Pointwise contextual restriction

• (i) For a proposition p and a context set C, $p_{|C}$ is the function from C to truth values that is identical to p at all points (the contextual restriction of p to C).

• (ii) For a question Q and a context set C, the contextual restriction of Q to C is $Q_{|C}$, where:

  $$ Q_{|C} := \{p_{|C} \,|\, p \in Q\} $$

Definition 3

Hamblin equivalence

Q and $Q'$ are H-equivalent relative to context set C iff $Q_{|C} = Q_{|C}'$.

Definition 4

Karttunen equivalence

Q and $Q'$ are K-equivalent relative to context set C iff:

$$ \forall w \in C.\, [\hat{Q}(w)]_{|C} = [\hat{Q}'(w)]_{|C} $$

Definition 5

Unique-answer answerhood operator

We define:

Definition 6

Unique-answer equivalence: entailed presupposition

Q and $Q'$ are U-equivalent relative to a context set C iff:

• (i) the unique-answer presuppositions of Q and $Q'$ are equivalent in C,

• (ii) calling $C'$ the subset of C where the presuppositions are met, we have:

  

Definition 7

Unique-answer Strawson equivalence — world-level version

Q and $Q'$ are W-equivalent relative to a context set C iff for any world w in C such that the unique-answer presuppositions of both Q and $Q'$ are true at w, we have:

Definition 8

Unique-answer Strawson equivalence — context-level version

Q and $Q'$ are C-equivalent relative to a context set C iff for any subset $C'$ of C where the unique-answer presuppositions of both Q and $Q'$ are true, we have

Result 1

H-equivalence and K-equivalence are almost the same thing

• (i) If C is a context set and Q and $Q'$ are two questions, then if Q and $Q'$ are H-equivalent in C, they are K-equivalent in C.

• (ii) If C is a context set and Q and $Q'$ are two questions that do not contain any proposition that is false throughout C, then if Q and $Q'$ are K-equivalent in C, they are H-equivalent in C.

Result 2

Order of the answer-based equivalences by strength

If C is a context set and Q and $Q'$ are two questions, then if Q and $Q'$ are U-equivalent in C, they are W-equivalent in C, and if they are W-equivalent in C, they are C-equivalent in C.

Result 3

K-equivalence is stronger than unique-answer equivalences

If C is a context set and Q and $Q'$ are two questions, then if Q and $Q'$ are K-equivalent in C, they are U-equivalent and therefore also W-equivalent and C-equivalent in C.

Result 4

Collapse of the equivalences

If C is a context set and Q and $Q'$ are two questions whose unique-answer presuppositions are satisfied in C, then for any pair of letters α, β within H, K, U, W, and C, Q and $Q'$ are α-equivalent in C iff they are β-equivalent in C.

1.2 2 The Transparency Theory

Here we derive the announced results under Schlenker’s (2008) Transparency Theory. Schlenker (2008) proposes that in a context C, a presuppositional clause is acceptable if and only if the proposition being presupposed is transparent in the clause’s position. Transparency is defined as follows (some details are simplified or left implicit; the reader is referred to Schlenker 2008):

Definition 9

Transparency; Schlenker 2008

Let αβγ be a sentence where β is an embedded clause, and δ be another clause denoting a proposition d. d is transparent in the position of β if and only if for any completion $\gamma '$ that makes $\alpha \beta \gamma '$ well-formed, and for any clause $\beta '$, the two sentences $\alpha \beta ' \gamma '$ and $\alpha (\text{$\delta $ and $\beta '$}) \gamma$ are contextually equivalent in C.

The definition of Transparency depends on a notion of contextual equivalence; it is straightforward to apply it to questions as long as we have defined equivalence over them. Thus we can in principle define H-transparency, K-transparency, and so on in terms of the definitions above.

The relations between our notions of equivalence immediately translate into relations between our notions of transparency:

Result 5

Relations between the transparencies

• (i) H-transparency implies K-transparency.

• (ii) K-transparency implies U-transparency.

• (iii) U-transparency implies W-transparency.

• (iv) W-transparency implies C-transparency.

We can now derive the results that we are interested in. The first one below establishes that under the quadripartitive account, no presupposition filtering should ever be observed when the trigger is in the second conjunct of a conjunctive question.

Result 6

The quadripartition

Let C be a context set, and p a proposition such that there exists a proposition $q_{0}$ that is not related by contextual entailment to p or ¬p.⁴⁸ A proposition d is H-transparent (as well as K/U/W/C-transparent) in the position of q in if and only if C supports d.

Proof

The direction “If C supports d, then d is transparent” is immediate.

For the other direction, it suffices to show that if d is C-transparent, then C supports d, as C-transparency is the weakest form of transparency. Note that the definedness condition of is always met if the answers form a logical partition as they do here, and therefore we do not need to consider subcontexts at all.

Assume then that d is C-transparent. We will write ; the fact that d is C-transparent means that is equivalent in C to for any q and any w∈C (taking $\beta '$ to be q, and $\gamma '$ to be the empty string).

It follows from the existence of $q_{0}$ that p is not trivially true or false in C. Let then w, $w'$ be two worlds such that p(w)=1 and $p(w') = 0$. We have and .

• If $d(w) = d(w') = 0$, then , from which it follows that p∧¬d is equivalent to p, i.e. that p entails ¬d. We also have , from which it follows similarly that ¬p entails ¬d. The only way these two entailments can hold is if d is a trivial falsehood. But then, we have , which is equivalent to p, so p is equivalent to , which is either $p \land q_{0}$ or $p \land \neg q_{0}$. This contradicts the assumption that p and $q_{0}$/$\neg q_{0}$ are not related by entailment.

• If d(w)=1 and $d(w') = 0$, then and . It follows that p entails d and that ¬p entails ¬d. Hence p is equivalent to d. Then, we have , which is equivalent to ¬p, so ¬p is equivalent to , which is either $\neg p \land q_{0}$ or $\neg p \land \neg q_{0}$. It follows that $\neg q_{0}$ or $q_{0}$ entails ¬p, against assumption.

• If d(w)=0 and $d(w') = 1$, the same reasoning as in the previous case applies, replacing any occurrence of p with ¬p and vice versa, and replacing $w'$ with w.

The only remaining possibility is that $d(w) = d(w') = 1$. Since w can be any p-world and $w'$ can be any ¬p-world, d has to be a trivial truth in C. □

The next result shows that if conjunctive questions denote the tripartition, then we predict the observed pattern of presupposition filtering in the second conjunct.

Result 7

The tripartition

Let C be a context set, and p a proposition such that there exists a proposition $q_{0}$ that is not related by contextual entailment to p or ¬p. A proposition d is H-transparent (as well as K/U/W/C-transparent) in the position of q in if and only if C supports the material conditional p→d.

Proof

We define: .

Assume that C supports p→d, and let q be an arbitrary proposition. p∧d∧q is equivalent in C to p∧q, and this relation can equivalently be written as $[p \land d \land q]_{|C} = [p \land q]_{|C}$. Moreover, we have p∧¬(d∧q)=(p∧¬d)∨(p∧¬q), which is equivalent to p∧¬q (as p∧¬d is a contextual contradiction). It follows that Q(q) and Q(d∧q) are H-equivalent in C. Then, for any completion γ, Q(q)γ and Q(d∧q)γ are also H-equivalent in C.⁴⁹ Therefore, d is H-transparent in the position of q in Q(q), and it is also K/U/W/C-transparent.

As for the other direction, assume that d is C-transparent in the position of q in Q(q) in C. As before, it follows from the existence of $q_{0}$ that p is not trivially true or false in C. Let us then take w∈C such that p(w)=1. If d(w)=0, since and , p and p∧¬d are contextually equivalent, which is equivalent to saying that C supports p→¬d. Assume without loss of generality (as we could replace $q_{0}$ by $\neg q_{0}$) that $q_{0}(w) = 1$. Then, we have . Since p entails ¬d, it also entails $\neg d \lor q_{0}$, and therefore is equivalent to p. By C-transparency, p is thus equivalent to , i.e. p is equivalent to $p \land q_{0}$, or equivalently, p contextually entails $q_{0}$, against assumptions. Therefore this case is impossible, and d(w)=1. Since this holds for any w such that p(w)=1, C supports p→d. □

Moving on to disjunction, the next result shows that no filtering is predicted to be possible with H/K/U-transparency. The subsequent result is an extension to W-transparency (that an additional condition is needed is essentially a bug in the definition). A final result shows that C-transparency derives a degenerate pattern where everything is transparent.

Result 8

No filtering in alternative questions (H/K/U)

Let C be a context set. A proposition d is H/K/U-transparent in the position of q in ?p∨?q in C if and only if C supports d.

Proof

The direction “If C supports d, then d is transparent” is immediate.

We define Q(q): = ?p∨?q = {p,q}. Assume that d is U-transparent in C. The unique-answer presupposition of Q(⊤) is equivalent to ¬p, from which it follows that the unique-answer presupposition of Q(d) is equivalent to ¬p, i.e. that p⊻d is equivalent to ¬p (⊻ represents an exclusive disjunction). This can be verified to be equivalent to the fact that d is trivially true in C. □

Result 9

No filtering in alternative questions (W)

Let C be a context set, and let d be a proposition that does not contextually entail p.⁵⁰d is W-transparent in the position of q in ?p∨?q if and only if C supports d.

Proof

Once again, define Q(q): = ?p∨?q = {p,q}.

The direction “If C supports d, then d is transparent” is immediate.

Assume that d is W-transparent. Let w be a world such that p(w)=0 and d(w)=1 (such a world exists by assumption). and , so d is contextually equivalent to ⊤. □

Result 10

C-equivalence derives a degenerate pattern in alternative questions

Let C be a context set, and let d be a proposition. d is C-transparent in the position of q in ?p∨?q .

Proof

Once again, define Q(q): = ?p∨?q = {p,q}.

Take q to be an arbitrary proposition, and call $C'$ the set of worlds in C such that p∨(d∧q) and ¬(p∧q) are true. $C'$ is the biggest subset of C such that the unique-answer presupposition of both Q(q) and Q(d∧q) is met. What we need to prove is that and define the same two-place predicate over $C'$, or equivalently that for any $w \in C'$, both operators return equivalent propositions when applied to w, with equivalence being taken in $C'$.⁵¹ Take then w in $C'$.

• If p(w)=1, then .

• If p(w)=0, from the definition of $C'$ we have d(w)=q(w)=1. It follows that and . Due to the way $C'$ is defined, both q and d∧q are equivalent to ¬p in $C'$ and therefore to one another, as desired.

Thus, d is C-transparent. □

These results are essentially preserved if we look at {p,q,¬(p∨q)} rather than {p,q}; the proofs are omitted to save space.

1.3 3 Trivalent theories: a compositional approach

We now turn to trivalent theories. The simplest way of adapting trivalent theories to our problem is to consider that conjunction and disjunction of propositions are trivalent, and that question conjunction and disjunction just do what they usually do in H/K semantics, except that questions are sets of trivalent propositions. Below we define such a system, called the Simple Trivalent Answer set Theory (STAT). It is straightforward to verify that under STAT, essentially the same results as with the Transparency Theory obtain: the conjunctive case can be dealt with through the tripartition, while the disjunctive case remains puzzling.⁵²

Concretely, assume the following:

Definition 11

STAT

• (i) Questions are sets of trivalent propositions.

• (ii) Question conjunction is pointwise Middle Kleene conjunction.

• (iii) Question disjunction is set union.

• (iv) Questions are felicitous only if all their answers return 0 or 1 at all worlds in the context.

Result 11

Predictions of STAT

Under the assumptions given in Definition 11:

• (i) If a conjunctive question (schema: ?p∧?q) denotes the quadripartition {p∧q,p∧¬q,¬p∧q,¬p∧¬q}, it should presuppose π(p) and π(q).

• (ii) If a conjunctive question (schema: ?p∧?q) denotes the tripartition {p∧q,p∧¬q,¬p}, it should presuppose π(p) and p→π(q).

• (iii) Whether a disjunctive question (schema: ?p∨?q) denotes {p,q} or {p,q,¬p∧¬q}, it should presuppose π(p) and π(q).

In order to deal with disjunction, one might think that case (iv) in Definition 11 should be relaxed to an existential presupposition: at least one answer should be defined at a given world in the context. We can in fact unify the constraint with the independent assumption, generally made in H/K theories, that questions presuppose that one of their answers is true:

Definition 12

Existential variant of STAT

• (iv’) Questions presuppose that at least one answer is defined and true at each world in the context.

The observed case of filtering is immediately predicted, but we derive no order effects.

1.4 4 Trivalent deployment of the connectives

Another way of extending trivalent theories to questions is to apply the methodology of “Peters-Kleene deployment”, as described by George (2014), to derive trivalent meanings for the question connectives from their bivalent meanings. Below we provide an implementation of George’s idea, and derive the same results as with the Transparency Theory: presupposition filtering is predicted for the tripartition but not for the quadripartition as far as conjunction is concerned, and not at all as far as disjunction is concerned (only the case of {p,q} is discussed, as using {p,q,¬p∧¬q} instead does not make a difference).

The idea behind Peters-Kleene deployment is as follows: assume that you want to derive how potential undefinedness in p will project in the environment F(p). What you know is the specification of F in the bivalent world (i.e., you know what F(p) is for a total proposition p). Now define the repair set of a trivalent proposition p: it is the set of all propositions that agree with p wherever p is defined.

Definition 13

Repair set of a trivalent proposition

If p is a trivalent proposition, the repair set of p is written as $p^{R}$ and given by:

$$ p^{R} := \{p' \in \{0,1\}^{\Omega}\,|\, \forall w.\, p(w) \in \{0,1\} \rightarrow p'(w) = p(w)\} $$

(Here Ω is the set of possible worlds, and therefore $\{0,1\}^{\Omega}$ is the set of total/bivalent propositions.)

From the repair set of p, we derive the deployment of F, $F^{D}$. Unlike F(p), $F^{D}(p)$ is potentially defined for some trivalent inputs p. Those inputs are those where, no matter how p is repaired to form a bivalent proposition $p'$, $F(p')$ is the same.

Definition 14

Deployment of a functor

If F is a function whose input is a bivalent proposition, the deployment of F is written as $F^{D}$ and given by:

• (i) If there is an output X such that for all $p' \in p^{R}$, $F(p') = X$, then $F^{D}(p) = X$.

• (ii) Otherwise, $F^{D}(p) = \#$.

In our case, F will be the function from q to the denotation of the question schematized as ?p∧?q or ?p∨?q. The desired result is that even if q is potentially undefined, F(q) might be defined in some cases.

Applying this methodology is most naturally done within a Karttunen-style account. Thus, the output of K will be a Karttunen question (type s(st)t). The definition of deployment above presupposes a notion of equality on the outputs of F; the natural choice in our case is the usual definition of set equality.⁵³

We can now prove the results promised above:

Result 12

Deployment of conjunction (quadripartition)

For total propositions p and q, define $F_{p}(q)$ to represent the Karttunen denotation of a quadripartitive conjunctive question:

$$ F_{p}(q) := \lambda w.\, \lambda r_{st}.\, r \in \{p \land q, p \land \neg q, \neg p \land q, \neg p \land \neg q\} \land r(w) $$

Let p be a total proposition that is true in at least two worlds and false in at least two worlds.⁵⁴$F^{D}_{p}(q)$ is defined if and only if q is total.

Proof

It is clear that if q is total, $F^{D}_{p}(q)$ is defined (it is just $F_{p}(q)$).

Assume that q is not total, i.e. that there is w such that q(w)=#. Take $q' \in q^{R}$ such that $q'(w) = 1$, and $q'' \in q^{R}$ that is exactly the same as $q'$, except that $q''(w) = 0$.

• If p(w)=1, then $F_{p}(q')(w) = \{p \land q'\}$ and $F_{p}(g'')(w) = \{p \land \neg q''\}$. Let $w'$ be another world where p is true: we have $[p \land q'](w') = q'(w')$ and $[p \land q''](w') = q''(w')$. By construction, $q'(w') \neq q''(w')$, so $F_{p}(q')(w) \neq F_{p}(g'')(w)$.

• If p(w)=0, then $F_{p}(q')(w) = \{\neg p \land q'\}$ and $F_{p}(q'')(w) =\{\neg p \land \neg q''\}$. Let $w'$ be another world where p is false: we have $[\neg p \land q'](w') = q'(w')$ and $[\neg p \land q''](w') = q''(w')$. By construction, $q'(w') \neq q''(w')$, so $F_{p}(q')(w) \neq F_{p}(g'')(w)$.

Either way, $F_{p}^{D}(q)$ is not defined. By contraposition, if $F_{p}^{D}(q)$ is defined, q is total. □

Result 13

Deployment of conjunction (tripartition)

For total propositions p and q, define $F_{p}(q)$ to represent the Karttunen denotation of a tripartitive conjunctive question:

$$ F_{p}(q) := \lambda w.\, \lambda r_{st}.\, r \in \{p \land q, p \land \neg q, \neg p\} \land r(w) $$

Let p be a total proposition that is true in at least two worlds. $F^{D}_{p}(q)$ is defined if and only if all #-worlds for q are 0-worlds for p.⁵⁵

Proof

Assume that q is only undefined at worlds where p is false. Take $q', q''$ in $q^{R}$, and let w be a world.

• If p(w)=0, then $[p \land q'](w) = [p \land q''](w) = 0$, and $[p \land \neg q'](w) = [p \land \neg q''](w) = 0$

• If p(w)=1, then q is defined at w and $q'(w) = q''(w) = q(w)$. If q(w)=1, we have $[p \land q'](w) = [p \land q''](w) = 1$ and $[p \land \neg q'](w) = [p \land \neg q''](w) = 0$. If q(w)=0, we have $[p \land q'](w) = [p \land q''](w) = 0$ and $[p \land \neg q'](w) = [p \land \neg q''](w) = 1$.

Then, $p \land q' = p \land q''$ and $p \land \neg q' = p \land \neg q''$, from which the fact that $F_{p}(q') = F_{p}(q'')$ immediately follows. Therefore $F_{p}^{D}(q)$ is defined.

As for the other direction, the same reasoning as in the previous result shows that if there is w such that q(w)=# and p(w)=1, then $F_{p}^{D}(q)$ is not defined. By contraposition, if $F_{p}^{D}(q)$ is defined, there is no such w. □

Result 14

Deployment of disjunction

For total propositions p and q, define $F_{p}(q)$ to represent the Karttunen denotation of a disjunctive question:

$$ F_{p}(q) := \lambda w.\, \lambda r_{st}.\, r \in \{p, q\} \land r(w) $$

Let p be a total proposition that is true in at least two worlds. $F^{D}_{p}(q)$ is defined if and only if q is total.

Proof

It is clear that if q is total, $F^{D}_{p}(q)$ is defined.

Assume that there is w such that q(w)=#. Take $q' \in q^{R}$ such that q(w)=1, and $q'' \in q^{R}$ that is the same as $q'$, except that $q''(w) = 0$.

• If p(w)=0, we have $F_{p}(q')(w) = \{q'\}$ and $F_{p}(q'')(w) = \varnothing$.

• If p(w)=1, we have $F_{p}(q')(w) = \{p, q'\}$ and $F_{p}(q'')(w) = \{p\}$. If $q' \neq p$, $F_{p}(q')(w) \neq F_{p}(q')(w)$. If $q' = p$, take $w'$ such that $p(w') = 1$. We have $F_{p}^{D}(q')(w') = \{p, q'\} = \{p\}$ and $F_{p}(q'')(w') = \{p, q''\}$ (recall that $q''$ agrees with $q'$, and therefore with p, at $w'$). Since $q'' \neq q'$ by construction, $F_{p}(q')(w') \neq F_{p}(q'')(w')$.

Either way, $F_{p}^{D}(q)$ is not defined. By contraposition, if $F_{p}^{D}(q)$ is defined, q is total. □