Subset Spaces for Conditional Norms

Abstract

We introduce two notions of conditionals, forward conditional for deductive implication and backward conditional for abductive implication. The former is in regard to Lewis [16]’s conditional, while the latter is treated as a binary window modality. We introduce logics of forward and backward conditionals, interpreted over a point-set semantics (with explicit likelihood) from the logic of subset spaces. These conditionals and their logics have applications in the studies of conditional norms.

A Some Proofs of Proposition 3

Clause 1. We show this by giving a counter model. Consider $M=(\{w\},\mathscr {O},V)$ such that $\mathscr {O}= \{\emptyset ,\{w\}\}$, $V(p)=\emptyset$ and $V(q)=\{w\}$. We show . It suffices to show that if

$$\begin{aligned} \hbox {for all} \, O'\in \mathscr {O}_w, \, \text {if} \, M,w, O'\models q \, \ \text {then} \ \, {\llbracket p\rrbracket }_{O'}\in \text {sub}(\{w\}), \end{aligned}$$

(1)

then

$$\begin{aligned} \hbox {for all} \, O'\in \mathscr {O}_w, \, \text {if} \, {\llbracket q\rrbracket }_{O'}\in \text {sub}(\{w\}) \, \ \text {then} \, \, M,w, O'\models p. \end{aligned}$$

(2)

Since there is only one open neighborhood of w. Clause (1) is equivalent to:

$$\begin{aligned} \text {if} \, \, M,w,\{w\} \models q, \text {then} \, \, {\llbracket p\rrbracket }_{\{w\}}\in \text {sub}(\{w\}) \end{aligned}$$

(3)

By definition, $M,w,\{w\}\models q \iff w\in V(q)$ which is true, and ${\llbracket p\rrbracket }_{\{w\}}=\emptyset \in \text {sub}(\{w\})$ is also true. So clause (3) holds.

Similarly, clause (2) is equivalent to the following:

$$\begin{aligned} \text {if} \, \ {\llbracket q\rrbracket }_{\{w\}}\in \text {sub}(\{w\}), \, \text {then} \, \ M,w,\{w\}\models p. \end{aligned}$$

(4)

While ${\llbracket q\rrbracket }_{\{w\}}=\{w\}\in \text {sub}(\{w\})$ holds, $M,w,\{w\}\models p$ does not. So clause (4) does not hold. Which shows that the given model M is such that .

Clause 2. Consider the formula $\bot \multimap \bot$. Given a model $M=(W,\mathscr {O},V)$, $w\in W$ and $O\in \mathscr {O}_w$, by definition,

$$\begin{array}{lll} M,w,O\models \bot \multimap \bot &{}\iff &{} \text {for all} \, O'\in \mathscr {O}_w, \, \text {if} \, \ {\llbracket \bot \rrbracket }_{O'}\in \text {sub}(O) \, \ \text {then} \, \ M,w,O'\models \bot \\ &{}\iff &{}\text {for all} \, O'\in \mathscr {O}_w, \text {if} \, \ \emptyset \in \mathscr {O}\ \ \text {then} \, \ M,w,O'\models \bot \\ &{}\iff &{}\text {for all} \, O'\in \mathscr {O}_w, \emptyset \notin \mathscr {O}\\ \end{array}$$

To find a counter model, all we need is to make the empty set an open in a model and make sure that there is an open neighborhood of a world w. A candidate of such a model is $(\{w\},\{\emptyset , \{w\}\}, V)$, where the valuation V is arbitrary.

This shows that $\varphi \multimap \varphi$ is not valid in general, and further that $\alpha \multimap \alpha$ is not valid even when $\alpha$ is propositional.

Clause 5. Consider the fomrula . Given a model $M=(W,\mathscr {O},V)$, $w\in W$ and $O\in \mathscr {O}_w$, by definition,

This is clearly not valid. We can easily find a counterexample, as long as $W\notin \mathscr {O}$.