On the presuppositional strength of interrogative clauses

Abstract

A central question in the study of presuppositions is how a presupposition trigger contributes to the meaning of a complex expression containing it. Two competing answers are found in the literature on quantificational expressions. According to the first, a quantificational expression presupposes that every member of its domain satisfies the presuppositions triggered in its scope, and according to the second, a quantificational expression presupposes that at least one member of its domain satisfies the presuppositions triggered in its scope. The former view implies that an interrogative clause, a kind of quantificational expression, presupposes all of its possible answers’ presuppositions, whereas the latter view implies that an interrogative clause presupposes that the presuppositions of at least one of is possible answers are satisfied. This paper contributes to the debate by showing that ‘alternative’ interrogatives, formed with or, project presuppositions in the same, distinctive manner that other disjunctive constructions do: generally, universally. A theory that treats disjunctive words as restricted variables, bindable by various quantificational operators, is extended to account for the presuppositions of ‘alternative’ interrogatives, disjoined declaratives, and disjoined conditional antecedents in a uniform manner. The paper then explores some ways to reconcile the proposal with two special cases where interrogatives have been claimed to have weaker presuppositions: (1) constituent interrogatives in presupposition-weakening contexts, and (2) polar interrogatives containing bias-inducing scalar particles like even.

Appendices

Appendix 1: Some rules and conventions

Generally following Heim and Kratzer (1998), we assume that for any possible world w and variable assignment g, (i)–(v), (10’), (13’), and (31’) hold.

(i)	If γ is a branching node whose daughters are α and β and 〚α〛w,g ∈ Dom(〚β〛w,g), then:
	〚γ〛w,g = 〚β〛w,g(〚α〛w,g)
(ii)	If γ = [∧ β], then: 〚γ〛w,g = [λw’: β ∈ Dom(〚〛w’,g). 〚β〛w’,g]
	If γ = [∨ β], then: 〚γ〛w,g = 〚β〛w,g(w)
(iii)	If γ = [k β] and k is a numerical index, then:
	〚γ〛w,g = [λx: β ∈ Dom(〚〛w,g[k→x]). 〚β〛w,g[k→x]],
		where Dom(g[k→x]) = (Dom(g) ∪ {k}), g[k→x](k) = x, and for all m ∈ Dom(g[k→x]) such that m ≠ k: g[k→x](m) = g(m)
(iv)	If γ is a pronoun or a trace and k is a numerical index, then:
	〚γk〛w,g is defined only if g(k) is defined. If defined, 〚γk〛w,g = g(k).
(v)	a.		〚γ〛g is defined iff for all w’ and w”, 〚γ〛w’,g and 〚γ〛w”,g are defined and [[γ]]w’,g = 〚γ〛w”,g;
			if 〚γ〛g is defined, then for all w’: 〚γ〛g = 〚γ〛w’,g.
	b.		〚γ〛w is defined iff for all g’ and g”, 〚γ〛w,g’ and 〚γ〛w,g” are defined and 〚γ〛w,g’ = 〚γ〛w,g”;
			if 〚γ〛w is defined, then for all g’: 〚γ〛w = 〚γ〛w,g’.
	c.		〚γ〛 is defined iff for all w’, w”, g’ and g”, 〚γ〛w’,g’ and 〚γ〛w”,g” are defined and
			〚γ〛w’,g’ = 〚γ〛w”,g”;
			if 〚γ〛 is defined, then for all w’ and g’: 〚γ〛 = 〚γ〛w’,g’.

(10’)	For any, n ≥ 2, sequence S, and any P1, P2, …, and Pn such that P1(w)(S) is of type t and P2, …, and Pn are of the same type as P1:
	〚or∃〛w(P1)…(Pn)(S) = 1 iff {Q| Q(w)(S) = 1 ∧ (Q = P1 ∨ … ∨ Q = Pn)} ≠ ∅
				(cf. (10), Sect. 2)
(13’)	For any k ∈ Dom(g), any n ≥ 2, and any P1, P2, …, Pn of the same type:
	〚ork〛g(P1)…(Pn) = 1 iff g(k) = P1 ∨ … ∨ g(k) = Pn
				(cf. (13), Sect. 2)
(31’)	For any, ${\fancyscript{Q}}$ and ${\fancyscript{P}}$, 〚if〛w(${\fancyscript{Q}}$)(${\fancyscript{P}}$) ∈ {1, 0} only if:
		(i)	{S| S is a sequence and ${\fancyscript{Q}}$(w)(S) ∈ {1, 0}} ≠ ∅, and
		(ii)	for all S such that S is a sequence and ${\fancyscript{Q}}$(w)(S) ∈ {1, 0}:
			SIM(w)({w’| ${\fancyscript{Q}}$(w’)(S) = 1}) ⊆ {w’| ${\fancyscript{P}}$(w’)(S) ∈ {1, 0}}.
	If	〚if〛w(${\fancyscript{Q}}$)(${\fancyscript{P}}$) ∈ {1, 0}, 〚if〛w(${\fancyscript{Q}}$)(${\fancyscript{P}}$) = 1 iff:
		for all S such that S is a sequence and ${\fancyscript{Q}}$(w)(S) ∈ {1, 0}:
		SIM(w)({w’| ${\fancyscript{Q}}$(w’)(S) = 1}) ⊆ {w’| ${\fancyscript{P}}$(w’)(S) = 1}.
				(cf. (31), Sect. 2)

We use the following conventions:

(I)	[λx: A. B] is shorthand for “the smallest function f such that f maps every x such that A to B” or “the smallest function f such that f maps every x such that A to 1 if B and to 0 otherwise” (whichever makes sense).

(II)	For any Z whose type ends in t, m ≥ 0, and m-long sequence S:
	if m = 0: Z(S) ≡ Z;
	if m > 0 and S = (X1, …, Xm): Z(S) ≡ Z(X1)…(Xm).

Appendix 2: Presuppositions of disjunctive conditionals

Disjunctive conditionals, including those with ellipsis in the consequent, are subject to the K–P effect. Thus, in utterance contexts with more or less “normal” laws, both readings of (112a) intuitively presuppose that Mary has a son and a daughter and Sue has a son and a daughter, neither reading of (112b) presupposes that either woman has children, and the sloppy reading of (112b) presupposes that if Mary is a parent, Sue is a parent (we set aside the presuppositions of strict pronoun readings, where the consequent implies that Sue stands in some relation to Mary’s children).

(112)	a.	If Mary avoids her son or her daughter, then Sue does.
			Strict VP reading:
			‘Mary avoids Mary’s son or daughter’ → ‘Sue avoids Sue’s son or daughter’
			Sloppy VP reading:
			(‘Mary avoids Mary’s son’ → ‘Sue avoids Sue’s son’) ∧
			(‘Mary avoids Mary’s daughter’ → ‘Sue avoids Sue’s daughter’)
	b.	If Mary is (either) childless or abusive with her children, then Sue is.
			Strict VP reading:
			‘Mary is childless or abusive with Mary’s children’ →
			‘Sue is childless or abusive with Sue’s children’
			Sloppy VP reading:
			(‘Mary is childless’ → ‘Sue is childless’) ∧
			(‘Mary is abusive with Mary’s children’ → ‘Sue is abusive with Sue’s children’)

This is predicted if the LF of the strict VP reading contains presuppositional if¹ (see (31) in Sect. 2), proposition-level presuppositional or_k and presuppositional ∃ (see (32) and (38)), and the LF of the sloppy VP reading contains presuppositional if² in (113) below and property-level presuppositional OR_k in (114) below. The predictions are as in (115)–(116).

(113)	〚if2	〛w(${\fancyscript{Q}}$)(${\fancyscript{P}}$) ∈ {1, 0} only if:
		(i)	{P| ${\fancyscript{Q}}$(w)(P) ∈ {1, 0}} ≠ ∅, and
		(ii)	for all P such that ${\fancyscript{Q}}$(w)(P) ∈ {1, 0}:
			SIM(w)({w’| ${\fancyscript{Q}}$(w’)(P) = 1}) ⊆ {w’| ${\fancyscript{P}}$ (w’)(P) ∈ {1, 0}}.
	If 〚if 2〛w(${\fancyscript{Q}}$)(${\fancyscript{P}}$) ∈ {1, 0}, 〚if 2〛w(${\fancyscript{Q}}$)(${\fancyscript{P}}$) = 1 iff:
		for all P such that ${\fancyscript{Q}}$(w)(P) ∈ {1, 0}:
		SIM(w)({w’| ${\fancyscript{Q}}$(w’)(P) = 1}) ⊆ {w’| ${\fancyscript{P}}$ (w’)(P) = 1}.

(114)	For any x of type e, and P1 and P2 of type (s, et), 〚ORk〛w,g(P1)(P2)(x) ∈ {1, 0} iff:
	a.	g(k)(w)(x) ∈ {1, 0},
	b.	g(k) = P1 ∨ g(k) = P2, and
	c.	(P1(w)(x) ∈ {1, 0} ∨ SIM(w)({w’| P2(w’)(x) = 0}) ⊆ {w’| P1(w’)(x) ∈ {1, 0}}) ∧
		(P2(w)(x) ∈ {1, 0} ∨ SIM(w)({w’| P1(w’)(x) = 0}) ⊆ {w’| P2(w’)(x) ∈ {1, 0}})
	If 〚ORk〛w,g(P1)(P2)(x) ∈ {1, 0}, 〚ORk〛w,g(P1)(P2)(x) = 1 iff g(k)(w)(x) = 1.

This analysis comes at a cost. While if¹ and if² can be merged into one item (see their generalized counterpart (31’) in Appendix 1), proposition-level presuppositional or_k and property-level presuppositional OR_k cannot be merged into one item: both or_k and OR_k have a (generalizable) conditional presupposition, but or_k asserts ‘g(k) = p₁ ∨ g(k) = p₂’, whereas OR_k presupposes ‘g(k) = P₁ ∨ g(k) = P₂’ and asserts ‘g(k)(w)(x) = 1’. It is worth noting that the generalized presuppositional disjunctive word in (117), with the necessary adjustment of existential closure in (118), covers both property-level and proposition-level disjunction and makes the same predictions as or_k and OR_k (see Sharvit 2020, for a similar proposal).

(117)	For any n ≥ 2, P1, P2, …, and Pn, sequence S, and numerical index k:
	〚or♠k〛w,g(P1)…(Pn)(S) ∈ {1, 0} iff:
	a.	g(k)(w)(S) ∈ {1, 0},
	b.	g(k) = P1 ∨ … ∨ g(k) = Pn, and
	c.	for all P ∈ {P1, …, Pn}: P(w)(S) ∈ {1, 0} ∨
			{D| D ⊆ {P1, …, Pn} ∧ {w’| D ⊆ {Q| Q(w’)(S) ∈ {1, 0}} ⊈ {w’| P(w’)(S) ∈ {1, 0}} ∧
			SIM(w)({w’| D ⊆ {Q| Q(w’)(S) = 0}}) ⊆ {w’| P(w’)(S) ∈ {1, 0}}} ≠ ∅.
	If 〚or♠ k〛w,g(P1)…(Pn)(S) ∈ {1, 0}, 〚or♠k〛w,g(P1)…(Pn)(S) = 1 iff g(k)(w)(S) = 1.

(118)	〚∃♠〛(X) ∈ {1, 0} iff Dom(X) ≠ ∅.
	If 〚∃♠〛(X) ∈ {1, 0}, 〚∃♠〛(X) = 1 iff {Z ∈ Dom(X)| X(Z) = 1} ≠ ∅.

Adopting (117) requires adopting the view that the extension of an interrogative is a set of true possible answers as in Karttunen (1977), and not a set of merely possible answers as in Hamblin (1973). Presupposition projection in polar and ‘alternative’ interrogatives are still accounted for under (117), as questions are still required to have satisfied presuppositions in order to be issued. However, the semantic presuppositions of wh-interrogatives and their pragmatics (see Sect. 3.3) would need to be reworked if Karttunen (1977) is adopted.

Appendix 3: Functional restrictors

It has been proposed (see, for example, von Fintel 1994) that quantificational restrictors can be “functional”. For example, in (119), where the subject of the prejacent of even varies across possible answers, the restrictor of even is the complex f₂-pro₄. Who binds its trace, t₄, and the co-indexed pro₄, which is the argument of the free pronoun f₂, whose value is determined by the context.

(119)	a.	Who even ate the CAKE?
	b.	〚who-C1〛g(λxλp. p = 〚∧[even-f2-pro4 [∧[t4 ate the cakeF] ~ f2-pro4]]〛g[4→x])
		Which x ∈ C1 is such that ‘x ate the cake’ is least likely in f2(x)?
	c.	C1 ⊇ {j, b, k}
		f2(j)	⊆	{〚∧[John ate the cake]〛, 〚∧[John ate the candy]〛, 〚∧[John ate the chips]〛, …}
		f2(b)	⊆	{〚∧[Bill ate the cake]〛, 〚∧[Bill ate the candy]〛, 〚∧[Bill ate the chips]〛, …}
		f2(k)	⊆	{〚∧[Kat ate the cake]〛, 〚∧[Kat ate the candy]〛, 〚∧[Kat ate the chips]〛, …}

This raises the possibility that the ‘alternative’ interrogative in (120a) below (= (85b) in Sect. 4) could, in principle, have a satisfiable presupposition: f₃ in (120b) has a different silent pronominal argument in each of the disjuncts: the first is anaphoric to the cake and the second to the candy.

We claim that (120b) is still excluded by DU, since the functional restrictors f₃-pro₁ and f₃-pro₂ are not identical.

Yet DU alone does not suffice to rule out (120a) once we acknowledge the availability of functional restrictors, because functional restrictors are not generally banned from constructions that involve ellipsis. For example, thanks to the assumption that Q(uantifier) R(aising) is available, the elided quantifier in (121) is licit.

Nevertheless, the QR option is not available for (120a), for independent reasons. In other words, an independent constraint bans (122), where the focus-associates of even are traces bound from above even by the cake and the candy. That constraint does not ban either (119b) or (121b), which do not contain illicit traces.

(122)	*[5 [	∧[the cake 1 [even-f3-pro1 [∧[John ate [t1]F] ~ f3-pro1]]] or5
		∧[the candy 1 [even-f3-pro1 [∧[John ate [t1]F] ~ f3-pro1]]]]]

That such a constraint is indeed at play is corroborated by the fact that a QR-ed phrase cannot bind a trace that is the focus-associate of the focus-sensitive superlative operator est, despite the fact that a wh-phrase can bind such a trace. While (123a,b) are both acceptable (Szabolcsi 1986; Heim 1999), the reading of (124b) where every student binds a trace that is the focus-associate of est is not available.

(123)	a.	Who gave the most books to Joe? (cf. ANN gave the most books to Joe)
		〚who-C1〛g(λzλp. p =  〚∧[est-f3-pro1 [∧[2 [t1]F gave d2-many books to Joe] ~ f3-pro1]]〛g[1→z])
		Which z ∈ C1 is such that ∀x ≠ z[x ∈ f3(z)]: z gave Joe more books than x gave Joe?
	b.	Who did Ann give the most books to? (cf. Ann gave the most books to JOE)
		〚who-C1〛g(λzλp. p =  〚∧[est-f3-pro1 [∧[2 Ann gave d2-many books to [t1]F] ~ f3-pro1]]〛g[1→z])
		Which z ∈ C1 is such that ∀x ≠ z[x ∈ f3(z)]: Ann gave z more books than Ann gave x?

(124)	a.	JANE gave the most books to every student.
		every student [1 est-f3-pro1 [∧[2 JaneF gave d2-many books to t1] ~ f3-pro1]]
		For every student x and every z in f3(x) such that z ≠ Jane:
			Jane gave x more books than z gave x.
	b.	Jane gave the most books to EVERY STUDENT/every student.
		*every student [1 est-f3-pro1 [∧[2 Jane gave d2-many books to [t1]F] ~ f3-pro1]]
		For every student x and every z in f3(x) such that z ≠ x:
			Jane gave x more books than Jane gave z.

(125)	Where C is a set of degree-properties, 〚est〛w(C)(P) ∈ {1, 0} only if P ∈ C and {d| P(w)(d) = 1} ≠ ∅.
	If 〚est〛w(C)(P) ∈ {1, 0}, 〚est〛w(C)(P) = 1 iff {d| {Q ∈ C| Q(w)(d) = 1} = {P}} ≠ ∅.
	(Howard 2014; cf. Heim 1999)

Finally, it is worth noting that (100a) in Sect. 4—If Mary is even dancing or swimming, then Sue is—has a narrow-scope-even strict VP reading. Its DU-compliant LF in (126) below contains functional restrictors but does not contain illicit focused traces.