Force and Choice

Abstract

Some utterances of imperative clauses have directive force—they impose obligations. Others have permissive force—they extend permissions. The dominant view is that this difference in force is not accompanied by a difference in semantic content. Drawing on data involving free choice items in imperatives, I argue that the dominant view is incorrect.

Appendix A

A model is a tuple $\langle {\mathcal {D}}, {\mathcal {W}}, {\mathcal {G}}, \llbracket \cdot \rrbracket \rangle$. ${\mathcal {D}}$ is a set of individuals, $d, d',\ldots$; ${\mathcal {W}}$ a set of worlds, $w,w',\ldots$; ${\mathcal {G}}$ a set of assignments, $g,g',\ldots$; and $\llbracket \cdot \rrbracket$ an interpretation function.

Definition 1

• 1. $\sigma , \sigma ',\ldots \in \mathcal {P(W)}$ are context states;

• 2. $\Sigma , \Sigma ',\ldots \in \mathcal {P((P(W)))}$ are context spaces.

We define a language, L, along with a propositional fragment, $L^{-}.$ ${\textsf {x}, \textsf {x'},\ldots }$ and ${\textsf {F},\textsf {F'},\ldots }$ are schematic variables over sets of (non-schematic) variables and (unary) predicates respectively. ${\textsf {A}, \textsf {B},\ldots }$ and $\phi ,\psi ,\ldots$ are schematic variables over $L^{-}$ and L, respectively.

Definition 2

• 1.

  • ${\textsf {F(x)}}\in L^{-}$;

  • If ${\textsf {A}, \textsf {B}}\in L^{-}, \text { then } \ {\lnot \textsf {A}},\ {\forall \textsf {x}\ {\textsf {A}}},\text { and } ({\textsf {A}\wedge \textsf {B}})\in L^{-}$;

  • Nothing else is in $L^{-}$.

• 2.

  • If ${\textsf {A}}\in L^{-},$ then ${\textsf {A}}, \ \textsc {Dir}({\textsf {A}}) \text { and } \textsc {Prm}({\textsf {A}})\in L$;

  • If $\phi , \psi \in L$, then $\lnot \phi , \ \forall {\textsf {x} }\ \phi , \text { and } (\phi \wedge \psi )\in L$;

  • Nothing else is in L.

The denotations of expressions in $L^{-}$ and ${L}/{L}^{-}$ differ in type. The former denote propositions: functions from worlds to truth values (type: $\langle s, t\rangle$). The latter denote context change potentials: functions between conversation systems (type: $\langle \langle st,t\rangle , \langle st,t\rangle \rangle$).

$(\cdot )'$ is a generalized complementation operation over expressions of type $\langle \alpha , \beta \rangle$ (where $\beta$ ‘ends in t’).

$\sqcap$ is a generalized intersection operation over two expressions of type $\langle \alpha , \beta \rangle$ (where $\beta$, again, ‘ends in t’). $\sqcap$ is the derivatively defined n-ary operation.

In interpreting our language, we define a single interpretation function recursively over the entirety of L.

Definition 3

• 1. $\llbracket {\textsf {F(x)}} \rrbracket ^{g}=\lambda w. g(\textsf {x})\in w({\textsf {F}}).$

• 2. $\llbracket {\textsc {Dir}}({\textsf {A}}) \rrbracket ^{g}=\lambda \Sigma \lambda \sigma :\sigma \in \Sigma . \ \sigma \subseteq \llbracket {\textsf {A}}\rrbracket ^{g}$

• 3. $\llbracket {\textsc {Prm}}({\textsf {A}}) \rrbracket ^{g}=\lambda \Sigma \lambda \sigma :\sigma \in \Sigma . \ \sigma \cap \llbracket {\textsf {A}}\rrbracket ^{g}\not =\emptyset$

• 4. $\llbracket \lnot \mathsf{\phi } \rrbracket ^{g}= (\llbracket \mathsf{\phi }\rrbracket ^{g})'$

• 5. $\llbracket \phi \wedge \psi \rrbracket ^{g}=\llbracket \phi \rrbracket ^{g}\sqcap \llbracket \psi \rrbracket ^{g}$

• 6. $\llbracket \forall \textsf {x} \phi \rrbracket ^{g}=\sqcap _{d\in {\mathcal {D}} } \ \llbracket \phi \rrbracket ^{g[{\textsf {x}}\rightarrow d]}$

$\llbracket {\textsf {F(x)}}\rrbracket ^{g}$ is the set of worlds w which map F to an extension which includes $g({\textsf {x}})$. $\llbracket {\textsc {Dir}}({\textsf {A}}) \rrbracket ^{g}$ maps $\Sigma$ to the set of its elements which are subsets of $\llbracket {\textsf {A}}\rrbracket ^{g}$. $\llbracket {\textsc {Prm}}({\textsf {A}}) \rrbracket ^{g}$ maps $\Sigma$ to the set of its elements which have a non-empty intersection with $\llbracket {\textsf {A}}\rrbracket ^{g}$. Given an input $\Sigma$, $\llbracket {\textsc {Dir}}({\textsf {A}})\rrbracket ^{g}$ and $\llbracket {\textsc {Prm}}({\textsf {A}})\rrbracket ^{g}$ are defined on a context state $\sigma$ iff $\sigma \in \Sigma$.

$\llbracket \lnot \mathsf{\phi } \rrbracket ^{g}$ is the generalized complement of $\llbracket { \phi } \rrbracket ^{g}$. $\llbracket \phi \wedge \psi \rrbracket ^{g}$ is the generalized intersection of $\llbracket \phi \rrbracket ^{g}$ and $\llbracket \psi \rrbracket ^{g}$. $\llbracket \forall \textsf {x} \phi \rrbracket ^{g}$ is the generalized n-ary intersection of the set $\{ \llbracket \phi \rrbracket ^{g[\mathsf{x\rightarrow d} ]}\ :\ d\in {\mathcal {D}} \}$.

Adopting post-fix notation, we write $\Sigma \llbracket \phi \rrbracket ^{g}$ for $\llbracket \phi \rrbracket ^{g}(\Sigma ).$ Finally, we define relations of support and entailment.

Definition 4

• 1. iff $\Sigma =\Sigma \llbracket \phi \rrbracket ^{g}$

• 2. iff for all g and all $\Sigma$: .

$\Sigma$ supports $\phi$ (relative to g) iff it is a fixed point of update with $\llbracket \phi \rrbracket ^{g}$. $\phi _{i},\ldots ,\phi _{j}$ entail $\psi$ iff for any assignment g and context space, $\Sigma$, sequential update of $\Sigma$ with $\llbracket \phi _{i}\rrbracket ^{g},\ldots ,\llbracket \phi _{j}\rrbracket ^{g}$ supports $\psi$ (relative to g).

We can now make some observations about our language. Fact 1 says that Prm is the dual of Dir.

Fact 1

Proof

By Definition 3.2, we know that $\Sigma \llbracket \textsc {Dir}(\textsf {A})\rrbracket ^{g}=\{\sigma \in \Sigma \ :\ \sigma \subseteq \llbracket {\textsf {A}}\rrbracket ^{g} \}$. By Definition 3.4, we know that $\llbracket \lnot \textsc {Prm}(\lnot {{\textsf {A}}})\rrbracket ^{g}=(\llbracket \textsc {Prm}(\lnot {\textsf {A}})\rrbracket ^{g})'$, the generalized complement of $\llbracket \textsc {Prm}(\lnot {\textsf {A}})\rrbracket ^{g}$. But observe that $\Sigma (\llbracket \textsc {Prm}(\lnot {\textsf {A}})\rrbracket ^{g})'=\{\sigma \in \Sigma \ : \ \sigma \cap \llbracket { \lnot {\textsf {A}}}\rrbracket ^{g}=\emptyset \}$. Since $\sigma \cap \llbracket \lnot {\textsf {A}}\rrbracket ^{g}=\emptyset$ iff $\sigma \subseteq \llbracket {\textsf {A}}\rrbracket ^{g}$, it follows that for any g: $\llbracket \textsc {Dir}(\textsf {A})\rrbracket ^{g}=\llbracket \lnot \textsc {Prm}(\lnot {\textsf {A}})\rrbracket ^{g}$. A fortiori, for any g, and . $\square$

Fact 2.1 says that $\forall \mathsf{x}$ commutes with Dir. Fact 2.2 says that $\forall \mathsf{x}$ does not commute with Prm.

Fact 2

• 1.
• 2.

Proof

Starting with Fact 2.1, first, note that updating a context $\Sigma$ with $\llbracket \forall {\textsf {x}}\ \textsc {Dir}({\textsf {A}})\rrbracket ^{g}$ returns the new context space $\bigcap _{d\in {\mathcal {D}}}\{\sigma \in \Sigma \ : \ \sigma \subseteq \llbracket {\textsf {A}}\rrbracket ^{g[{\textsf {x}}\rightarrow d]} \}$. Similarly, updating $\Sigma$ with $\llbracket \textsc {Dir}(\forall {\textsf {x}} \ {\textsf {A}})\rrbracket ^{g}$ returns the new context space $\{\sigma \in \Sigma \ : \ \sigma \subseteq \llbracket { \forall {\textsf {x}} \textsf {A}}\rrbracket ^{g} \}$. Note that $\llbracket {\forall {\textsf {x}} \textsf {A}}\rrbracket ^{g}=\bigcap _{d\in {\mathcal {D}}} \llbracket {\textsf {A}}\rrbracket ^{g[{\textsf {x}}\rightarrow d]}$. So $\Sigma \llbracket \textsc {Dir}(\forall {\textsf {x}} \ {\textsf {A}})\rrbracket ^{g}$ is the set $\{\sigma \in \Sigma \ : \ \sigma \subseteq \bigcap _{d\in {\mathcal {D}}} \llbracket {\textsf {A}}\rrbracket ^{g[{\textsf {x}}\rightarrow d]}\}$.

But observe that, by elementary set theory, $\bigcap _{d\in {\mathcal {D}}}\{\sigma \in \Sigma \ : \ \sigma \subseteq \llbracket {\textsf {A}}\rrbracket ^{g[{\textsf {x}}\rightarrow d]} \} =\{\sigma \in \Sigma \ : \ \sigma \subseteq \bigcap _{d\in {\mathcal {D}}} \llbracket {\textsf {A}}\rrbracket ^{g[{\textsf {x}}\rightarrow d]}\}.$ It follows that for any g, $\llbracket \forall {\textsf {x}}\ \textsc {Dir}({\textsf {A}})\rrbracket ^{g}=\llbracket \textsc {Dir}(\forall {\textsf {x}} \ {\textsf {A}})\rrbracket ^{g}$. Accordingly, for any g and $\Sigma$, and . $\square$

Turning to Fact 2.2, first, note that updating a context space $\Sigma$ with $\llbracket \forall {\textsf {x}}\ \textsc {Prm}({\textsf {A}})\rrbracket ^{g}$ returns the new context space $\bigcap _{d\in {\mathcal {D}}} \{\sigma \in \Sigma \ : \ \sigma \cap \llbracket {\textsf {A}}\rrbracket ^{g[{\textsf {x}}\rightarrow d]} \not =\emptyset \}$. Similarly, updating $\Sigma$ with $\llbracket \textsc {Prm}(\forall {\textsf {x}} \ {\textsf {A}})\rrbracket ^{g}$ returns the new context space $\{\sigma \in \Sigma \ : \ \sigma \cap \llbracket \forall {\textsf {x}} \ {\textsf {A}}\rrbracket ^{g} \not =\emptyset \}$. Note that $\llbracket {\forall {\textsf {x}}} \textsf {A}\rrbracket ^{g}=\bigcap _{d\in {\mathcal {D}}} \llbracket {\textsf {A}}\rrbracket ^{g[{\textsf {x}}\rightarrow d]}$. So $\Sigma \llbracket \textsc {Prm}(\forall {\textsf {x}} \ {\textsf {A}})\rrbracket ^{g}$ is the set $\{\sigma \in \Sigma \ : \ \sigma \cap (\bigcap _{d\in {\mathcal {D}}} \llbracket {\textsf {A}}\rrbracket ^{g[{\textsf {x}}\rightarrow d]})\not =\emptyset \}$.

We know that . So

However, crucially, $\bigcap _{d\in {\mathcal {D}}} \{\sigma : \ \sigma \cap \llbracket {\textsf {A}}\rrbracket ^{g[{\textsf {x}}\rightarrow d]} \not =\emptyset \}\not \subseteq \{\sigma : \ \sigma \cap (\bigcap _{d\in {\mathcal {D}}} \llbracket \mathsf{A}\rrbracket ^{g[{\textsf {x}}\rightarrow d]})\not =\emptyset \}$. Thus, .

As a countermodel, suppose that ${\mathcal {W}}=\{w,v\}$ and ${\mathcal {D}}=\{d,d'\}$, letting $w({\textsf {F}})=\{d\}$ and $v({\textsf {F}})=\{d'\}$. Consider the context space $\Sigma =\{ \{w,v \}\}$. $\Sigma$ is a fixed point of $\llbracket \forall \mathsf{x}\ \textsc {Prm}(\mathsf{A}) \rrbracket ^{g}$, since $w\in \llbracket {F(\textsf {x})}\rrbracket ^{g[\textsf {x}\rightarrow d]}$ and $v\in \llbracket {\textsf { F(x)}}\rrbracket ^{g[{\textsf {x}}\rightarrow d']}$. However, updating $\Sigma$ with $\llbracket \ \textsc {Prm}(\forall \mathsf{x}\ {\textsf {A}}) \rrbracket ^{g}$ returns $\{\{\}\}$, the absurd space, since $\bigcap _{d\in {\mathcal {D}}} \llbracket {\textsf {F(x)}}\rrbracket ^{g[{\textsf {x}}\rightarrow d]}=\emptyset$.