Soft Presuppositions as Scalar Implicatures in Signaling Games

Abstract

In this paper, I will discuss why soft presuppositions behave differently from hard presuppositions: the former are easily defeasible and project nonuniformly in quantificational sentences. I assume that soft triggers should be associated with alternatives, and thus share many similarities with scalar implicatures. As a generalization of Parikh’s games of partial information, I develop game-theoretic models, which provide a unified account for both scalar implicatures and soft presuppositions. I argue that iterated best response (IBR) reasoning allows us to analyze the behaviors of soft presuppositions in accordance with rational inferences. The models yield the following predictions of soft presuppositions: projection happens unless it is common knowledge that the speaker is ignorant about it, in which case the presupposition is defeasible; projection depends on the type of quantifiers, which may lead to nonuniform behaviors.

Appendices

Appendix 1: Proof of Proposition 2

First consider $S_0$-sequence. From Definition 2,

$$S_0=\left\{ \begin{array}{rcl} t_1&{} \mapsto &{}m_1 \\ t_2&{} \mapsto &{}m_1,m_2\\ t_3&{} \mapsto &{}m_1,m_2, m_3\\ \end{array}\right\} . \,$$

Given (1) and (2), $EU_{H_1}(t_1,m_1)=\frac{6}{11}$, $EU_{H_1}(t_2, m_1)=\frac{3}{11}$, $EU_{H_1}(t_3, m_1)=\frac{2}{11}$, $EU_{H_1}(t_2,m_2)=\frac{3}{5}$, $EU_{H_1}(t_3,m_2)=\frac{2}{5}$, $EU_{H_1}(t_3,m_3)=1$. Given (3),

$$H_1=\left\{ \begin{array}{rcl} m_1&{} \mapsto &{}t_1 \\ m_2&{} \mapsto &{}t_2\\ m_3&{} \mapsto &{}t_3\\ \end{array}\right\} . \,$$

Since $S_2$ will act according to her belief in $H_1$,

$$S_2=\left\{ \begin{array}{rcl} t_1&{} \mapsto &{}m_1 \\ t_2&{} \mapsto &{}m_2\\ t_3&{} \mapsto &{}m_3\\ \end{array}\right\} . \,$$

Evidently, the $S_0$-sequence begins repetition after two round of iteration. In other words, $S^*=S_2$ and $H^*=H_1$. The $H_0$-sequence leads to the same fixed point in a similar way.    $\blacksquare$

Appendix 2: Proof of Proposition 3

First, consider the $S_0$-sequence. From Definition 2,

$$S_0=\left\{ \begin{array}{rcl} t_1&{} \mapsto &{}m_1 \\ t_2&{} \mapsto &{}m_1,m_2\\ t_3&{} \mapsto &{}m_1\\ \end{array}\right\} . \,$$

Given (1) and (2), $EU_{H_1}(t_1,m_1)=\frac{2(1-p)}{3+p}$, $EU_{H_1}(t_2, m_1)=\frac{1-p}{3+p}$, $EU_{H_1}(t_3, m_1)=\frac{4p}{3+p}$, $EU_{H_1}(t_2,m_2)=1$. Given (3),

$$H_1=\left\{ \begin{array}{cl} \left\{ \begin{array}{l} m_1 \mapsto t_1,\\ m_2 \mapsto t_2,\\ \end{array}\right. &{} \text {if}\ p<\frac{1}{3} \\ \left\{ \begin{array}{l} m_1 \mapsto t_3,\\ m_2 \mapsto t_2,\\ \end{array}\right. &{} \text {if}\ p>\frac{1}{3}\\ \end{array}\right\} .\,$$

Since $S_2$ will act according to her belief in $H_1$,

$$S^*=\left\{ \begin{array}{rcl} t_1 \mapsto m_1 \\ t_2 \mapsto m_2\\ t_3 \mapsto m_3\\ \end{array}\right\} . \,$$

Evidently, the $S_0$-sequence begins repetition after two round of iteration. Then, $S^*=S_2$ and $H^*=H_1$. And the $H_0$-sequence leads to the same fixed point in a similar way.    $\blacksquare$