2.1. Quantitative Theories of Supposition

Bayesian conditionalization is most commonly understood as a diachronic norm governing the update of probabilistic credence functions. Under that interpretation, when an agent with a prior credence function $c$ learns that some event $E$ has occurred, she should adopt the posterior $c\prime$ matching $c$ conditioned on $E$ so that $c\prime (X)=c(X|E)$ for all $X$ . Conditionalization is defined as follows.

Conditionalization: Given a credence function $c\in \mathcal{C}$ and any $S\in \mathcal{A}$ with $c(s)>0$ , conditioning $c$ by $S$ results in the credence function $c(\cdot \hspace{0.17em}|S)$ defined below.

Given the Bayesian understanding of conditionalization as an account of learning, and the close relationship between rational learning and indicative supposition, it is no surprise that conditionalization has also been understood as a normative quantitative model of indicative supposition. Interestingly, such an interpretation was first suggested by Rev. Thomas Bayes, who wrote, “The probability that two subsequent events will both happen is compounded of the probability of the first and the probability of the second on the supposition the first happens” (1763: 379). There are also more recent examples of this interpretation in the literature. For instance, this interpretation is explicitly endorsed by evidential decision theorists in their account of ex ante evaluations of option-outcomes.

The most popular alternative to evidential decision theory—causal decision theory—replaces the use of indicative suppositions in the calculation with subjunctive suppositions. The debate between evidentialists and causalists in decision theory boils down to a dispute about which type of supposition is relevant for ex ante evaluations of options.⁶ The standard treatments of quantitative subjunctive supposition derive from the imaging rule mentioned in the previous section. Although a number of different versions of imaging have been developed in the literature, we will focus on its best known (and simplest) version, first proposed by Lewis. On an intuitive level, the difference between conditionalization and imaging can be understood in terms of the type of minimal change they encode. We mentioned earlier that conditionalization relies on a global measure of similarity, where imaging uses a local one. This point is elegantly explained by Lewis:

Imaging $P$ on $A$ gives a minimal revision in this sense: unlike all other revisions of $P$ to make $A$ certain, it involves no gratuitous movement of probability from worlds to dissimilar worlds. Conditionalizing $P$ on $A$ gives a minimal revision in this different sense: unlike any other revisions of $P$ to make $A$ certain, it does not distort the profile of probability ratios, equalities, and inequalities among sentences that imply $A$ . (1976: 311)

To introduce the details of imaging, we will need to impose some extra structure on the space of possible worlds. Specifically, we assume that, for any proposition $X$ and possible world $w$ , there is a unique “closest” world at which the sentence $X$ is true. This notion is captured by using a selection function, $\sigma :W\times \mathcal{A}\mapsto W$ . Intuitively, $\sigma (w,X)$ picks out the “closest” or “most similar” possible world to $w$ that satisfies $X$ . Our selection function will be subject to two basic conditions.

Centering: If $w\models X$ , then $\sigma (w,X)=w$ .

This first condition requires that each world is the unique closest world to itself, i.e. if $X$ is true at $w$ , then there is no closer world where $X$ is true.

Uniformity: If $\sigma (w,X)\models Y$ and $\sigma (w,Y)\models X$ , then $\sigma (w,X)=\sigma (w,Y)$ .

This second condition says that whenever the closest $X$ -world satisfies $Y$ and the closest $Y$ -world satisfies $X$ , they are one and the same. In order to illustrate the conceptual motivation for this constraint, we will take a brief but necessary detour into an important philosophical application of selection functions—namely, the semantics of subjunctive conditionals.

Under what conditions are subjunctive conditionals such as ‘If Richard Nixon had pressed the button, there would have been a nuclear war’ true? According to the proposal by Stalnaker (1968), this question is best answered in a semantics that utilises selection functions of the kind described above. The idea, roughly put, is that the subjunctive conditional in the example above is true just in case the closest possible world in which Richard Nixon did push the button is one where there was a nuclear war. The suggestion is that the truth value of the subjunctive conditional ‘if $X$ were true, $Y$ would be true’ at a world $w$ is given by the following definition:

Stalnaker conditional ( $\to$ ): The truth-conditions for the Stalnaker conditional, $X\to Y$ , are given by the semantic clause below.

As should be clear from its definition, the Stalnaker conditional is non-truth functional, since the truth-value of $X\to Y$ at a world $w$ does not supervene on the truth-values of its components at $w$ . Rather, it is true at $w$ just in case the closest world to $w$ at which its antecedent is true is also one at which its consequent is true. For present illustrative purposes, we take subjunctive conditionals such as ‘If Richard Nixon had pressed the button, there would have been a nuclear war’ to be adequately modelled using the Stalnaker conditional.

Given this semantics for subjunctive conditionals, the motivation for Uniformity becomes very clear. When $\sigma (w,X)\models Y$ and $\sigma (w,Y)\models X$ , the subjunctives ‘if $X$ were true, $Y$ would be true’ and ‘if $Y$ were true, $X$ would be true’ are both true on the semantics. Now imagine that $\sigma (w,X)\ne \sigma (w,Y)$ . This implies that there is some $Z$ such that the subjunctive ‘If $X$ were true, $Z$ would be true’ is true, but the subjunctive ‘If $Y$ were true, $Z$ would be true’ is false. Thus, the following sentence comes out as true:

Clearly, this would be a deeply strange and counterintuitive result. For this reason, we assume that our selection function satisfies the Uniformity condition.⁷

We are now ready to introduce Lewis’s imaging rule, which will serve as our representative quantitative theory of subjunctive supposition. Stated formally:

Imaging: Given a credence function $c\in \mathcal{C}$ and any $S\in \mathcal{A}$ , imaging $c$ on $S$ results in the credence function $c_S(\cdot )$ defined below.

Intuitively, when $c$ is imaged on $S$ , each world $w$ consistent with $S$ keeps all of its original probability, while the prior probability assigned to each world that is inconsistent with $S$ is transferred to the closest world satisfying $S$ .⁸

As suggested earlier, conditionalization and imaging differ in whether their recommendations are driven by global or local considerations. Conditionalization recommends the closest credence function that accommodates $S$ where the distance between credence functions is interpreted in terms of their global behaviour. In contrast, imaging operates at the local level by shifting credence from each world to the closest world satisfying $S$ .

2.2. Lockean Theories of Supposition

With our quantitative accounts of indicative and subjunctive supposition in hand, we will now outline our approach to comparing them with the qualitative theories we will introduce later. As mentioned earlier, qualitative and quantitative theories articulate the norms of suppositional judgement in terms of different kinds of doxastic attitude. Qualitative theories rely on agents’ belief corpora to offer binary judgements about whether they should regard propositions as acceptable under a supposition. Quantitative theories on the other hand use an agent’s credences to generate numerical judgments corresponding to how acceptable agents ought to find each proposition under any given supposition. To directly compare the two we need a way to bridge the gap between qualitative and quantitative attitudes.

To do so, we apply a suitably adapted version of the Lockean Thesis, so-called by Foley (1993). As it is traditionally understood, the Lockean Thesis provides a normative bridge principle between beliefs and credences, which requires that an agent believes that $X$ just in case she has “sufficiently high” credence in $X$ . This is standardly understood as saying that an agent should believe a proposition $X$ if and only if her credence in $X$ is at least as great as some Lockean threshold, $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,1]$ . Put formally:

Lockean Thesis (LT^t): For some $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,1]$ : $X\in B\iff c(X)\ge t$ .

This principle will be presupposed as a synchronic coherence requirement used to specify the beliefs that are coherent with an agent’s credences. So, when we are talking about Lockean agents, we will presuppose that they have beliefs and credences satisfying LT^t for some $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,1]$ . There is an extensive literature on the Lockean Thesis and its motivations.⁹ Featured prominently in that literature is the Lottery Paradox, first discussed by Kyburg (1961), and the tension it brings to the surface between LT^t and the popular normative requirements that beliefs be logically consistent and deductively closed. Primarily for space considerations, we will only briefly engage with that literature at a few points in the next section. Instead, we will unreflectively adopt LT^t as a technical tool to aid in our comparative project.

But the Lockean Thesis will play another role in our exploration beyond being a standing synchronic coherence requirement. It will also be used together with the quantitative theories of supposition introduced earlier to construct qualitative suppositional judgments that can be directly compared with the representative qualitative theories of supposition. We begin by introducing the Lockean theory of indicative supposition (LIS) defined below.

LIS: Given a corpus $B\in ℬ$ and some $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,1]$ , the set of acceptable propositions under the indicative supposition $S$ is specified in terms of the operation, $\divideontimes :ℬ\times \mathcal{A}\mapsto ℬ$ , defined below.

Where $B$ and $c$ are respectively a corpus of beliefs and credence function satisfying LT^t and $S$ is any proposition, $B_S^{\divideontimes }$ consists of those propositions assigned conditional credence on $S$ a value at least $t$ . The Lockean theory of subjunctive supposition (LSS) is characterised in an analogous fashion.

LSS: Given a corpus $B\in ℬ$ and some $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,1]$ , the set of acceptable propositions under the subjunctive supposition $S$ is specified in terms of the operation, $♦:ℬ\times \mathcal{A}\mapsto ℬ$ defined below.

Strictly speaking, the two Lockean operations, $\divideontimes$ and $♦$ , are not singular operations, but rather each characterise families of operations—one for each $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,1]$ . When it is useful, we will restrict our attention to certain subsets of Lockean thresholds by letting ${\divideontimes }^{[t,t\prime ]}$ and ${♦}^{[t,t\prime ]}$ denote the family of operators bounded by the closed interval $[t,t\prime ]$ . Analogous conventions will be adopted for the open and half-open intervals.

3.1. LIS and the AGM Postulates

The question now arises: how do the suppositional judgments recommended by LIS relate to those given under the qualitative account based on AGM? A partial answer to this question is given by previously established results. We will complete this picture after surveying the extant results from the literature.

Beginning with their synchronic requirements, there is an immediate tension between LT^t and Cogency that has been extensively discussed in the literatures on the Preface and Lottery Paradoxes—these same issues straightforwardly apply to the synchronic requirements imposed by $\ast$ 1 and $\ast$ 5. The remaining basic Gärdenfors postulates have been considered from a Lockean perspective by Shear and Fitelson (2019).¹³ LIS satisfies both of the remaining AGM synchronic coherence requirements, $\ast$ 2 and $\ast$ 6. Neither result is surprising: LIS satisfies $\ast$ 2 in virtue of the fact that $c(S|S)=1$ , while the satisfaction of $\ast$ 6 is secured by the extensional character of conditionalization.

The situation is more interesting for the diachronic requirements given by $\ast$ 3 and $\ast$ 4. Interestingly, $\ast$ 3 is satisfied by LIS in full generality. The reason why is relatively easy to see. It is a theorem of the probability calculus that $c(S\supset X)\ge c(X|S)$ . Thus, whenever $X\in B_S^{\divideontimes }$ it follows that $S\supset X\in B$ , and so $B_S^{\divideontimes }\subseteq \hspace{0.17em}\hspace{0.17em}Cn(B\cup \{S\})$ . Turning to the final basic postulate, $\ast$ 4, we see that in general LIS can violate this requirement. The basic reason why is relatively clear, though there are some subtleties that we will discuss. As the characteristic postulate of AGM, $\ast$ 4 says that an agent’s beliefs should remain acceptable under any supposition that is logically consistent with her corpus. However, when an agent is not fully certain of one of her beliefs (say $X$ ), it is possible for that some supposition ( $S$ ) might be logically consistent with her corpus but still count as counter-evidence to $X$ in the sense that . This allows for the possibility that $c(X)\ge t$ even though and, thus, that $B⊬\neg S$ but $B⊈B_S^{\divideontimes }$ . Still, there are some further constraints that can be imposed under which LIS can be made to satisfy $\ast$ 4.

The explanation immediately above is suggestive of the first situation in which LIS will be guaranteed to satisfy $\ast$ 4. Indeed, Gärdenfors (1988) established a result, which implies that when belief is taken to imply certainty (i.e. when t = 1), LIS will satisfy $\ast$ 4. Moreover, Gärdenfors’s result actually implies that LIS will satisfy all of the AGM postulates. One might wonder then: is the resulting satisfaction of $\ast$ 4 a consequence of the fact that $\ast$ 1 and $\ast$ 5 are satisfied when t = 1?

Shear and Fitelson show that the answer to this question is no, LIS can violate $\ast$ 4 even under the further assumption of Cogency. However, they establish the more surprising result that, assuming Cogency, LIS can only violate $\ast$ 4 when the Lockean threshold is relatively high. In particular, such violations are only possible when the Lockean threshold is at least the inverse of the Golden ratio (i.e. when $t\in ({\varphi }^{-1},1)$ , where ${\varphi }^{-1}\approx 0.618$ ). As an immediate corollary, assuming both Cogency and that $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,{\varphi }^{-1}]$ , LIS satisfies all of the basic Gärdenfors postulates, $\ast$ 1 – $\ast$ 6.

But this only tells part of the story about the import of the “Golden threshold” at ${\varphi }^{-1}$ .¹⁴ This is because LIS exhibits interesting behaviour relative to the two weakened variants of Preservation provided below.

The first of these postulates, Very Weak Preservation ( $\ast$ 4^v), requires that taking something that you already believe as a supposition for the sake of argument should not lead you to reject any of your other beliefs under that supposition. The second, Weak Preservation ( $\ast$ 4^w), says that under the same conditions, you should accept anything that you believe.

Although imposing the assumption of Cogency on LIS was not sufficient to guarantee the satisfaction of full Preservation ( $\ast$ 4), it turns out that it is sufficient to ensure that LIS will satisfy both of the weaker requirements, $\ast$ 4^v and $\ast$ 4^w. However, there is another way to guarantee that LIS will satisfy Very Weak Preservation: if the Lockean threshold is at least ${\varphi }^{-1}$ , then LIS will satisfy $\ast$ 4^v (even without the help of Cogency). These results are summarised in table 2 below.

The import of these results will depend on how you regard $\ast$ 4^v, $\ast$ 4^w, $\ast$ 4, and Cogency. We regard $\ast$ 4^v as eminently reasonable: it would seem very strange to believe both $P$ and $Q$ , but reject $Q$ under the supposition that $P$ . After all, that would mean that $P$ ’s certain truth would provide sufficient evidence to accept $\neg Q$ —that would seem to be ruled out by your concurrent beliefs that $P$ and that $Q$ . For the die-hard Lockeans who reject Cogency, this gives reason to maintain that the Lockean threshold must be a sufficiently high ( $t>{\varphi }^{-1}$ ) so as to rule out this possibility. The import of the remaining results is up for debate. A Lockean who finds $\ast$ 4^w plausible will be forced into adopting Cogency. However, this would be harder to motivate for a Lockean since once we accept that rational belief need not require certainty, there is no obvious argument in favour of $\ast$ 4^w. Still, proponents of AGM who find LIS attractive may take solace in the realisation that their preferred account can be reconciled with LIS through the acceptance of a sufficiently low threshold.¹⁵

Thus far, we have presented a number of results concerning LIS and the basic Gärdenfors postulates, $\ast$ 1 – $\ast$ 6, but have not addressed two remaining supplementary postulates, $\ast$ 7 and $\ast$ 8. Shear and Fitelson only mention these postulates in passing, since their primary concern was with the diachronic requirements governing single-step belief change rather than the dispositional requirements that provide bridges between different potential revisions. However, in the context of supposition, dispositional requirements are more obviously relevant. Accordingly, we will now complete the picture by reporting some new results establishing that the relationship between LIS and $\ast$ 3 and $\ast$ 4 carries over to their generalisations given by $\ast$ 7 and $\ast$ 8.

Proposition 1. LIS must satisfy  $\ast$ 7. That is, the following is satisfied for any  $B$ , $S,S\prime$ , and $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,1]$ :

Proof. Let $X\in B_{S\wedge S\prime }^{\divideontimes }$ , i.e.  $c(X|S\wedge S\prime )\ge t$ . Then, letting $c^S(\cdot ):=c(\cdot |S)$ , we get:

Thus, $c(S\prime \supset X|S)\ge t$ and so $S\prime \supset X\in B_S^{\divideontimes }$ . From this we conclude $X\in Cn(B_S^{\divideontimes }\cup \{S\prime \})$ .              $\square$

Proposition 2. In the absence of Cogency, LIS can violate  $\ast$ 8 for any $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,1)$ . That is, if $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,1)$ , it is possible that:

Proof. Let $c$ be any credence function satisfying the conditions below, where $\epsilon >0$ is arbitrarily small:

It is simple to see that case provides the basis for a counterexample to $\ast$ 8 for any threshold $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,1)$ in the absence of Cogency.              $\square$

Proposition 3. The twin requirements of Cogency and $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,{\varphi }^{-1}]$ are necessary and sufficient to guarantee that LIS satisfies  $\ast$ 8.

Proof. Supposing Cogency, we let $S\prime$ be consistent with $B_S^{\divideontimes }$ and $X\in Cn(B_S^{\divideontimes }\cup \{S\prime \})$ , and define $c$ as a vector on the assignments below.

We start by showing $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,{\varphi }^{-1}]$ only if $X\in B_{S\wedge S\prime }^{\divideontimes }$ , and hence that $\ast$ 8 is satisfied. For contradiction, suppose that $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,{\varphi }^{-1}]$ , but . Since $X\in Cn(B_S^{\divideontimes }\cup \{S\prime \})$ implies $S\prime \supset X\in B_S^{\divideontimes }$ , our assumptions imply

First, note that since $S\prime \supset X\in B_S^{\divideontimes }$ , by Cogency  $S\prime \supset \neg X\in B_S^{\divideontimes }$ would imply that $(S\prime \supset X)\wedge (S\prime \supset \neg X)\in B_S^{\divideontimes }$ . This is equivalent to $\neg S\prime \in B_S^{\divideontimes }$ thus contradicting our assumption that $S\prime$ is consistent with $B_S^{\divideontimes }$ . So, $S\prime \supset \neg X\notin B_S^{\divideontimes }$ (i.e.  ) which gives us

Next, observe that $S\prime \in B_S^{\divideontimes }$ would imply by Cogency that $S\prime \wedge X\in B_S^{\divideontimes }$ , since $S\prime \supset X\in B_S^{\divideontimes }$ . But then $c(X|S\prime \wedge S)\ge c(X\wedge S\prime |S)\ge t$ , which contradicts 1. So , which implies that

Taken together, 3 and 4 give us 5, which combined with 3 lets us infer 6.

Now, since $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,{\varphi }^{-1}]$ , we can use the special fact about the Golden Ratio that $t\le {\varphi }^{-1}$ iff $t^2\le 1-t$ to infer $\frac{\alpha }{\alpha +\beta }\ge t$ , which contradicts our assumption 1. Thus, our initial assumptions were inconsistent and we infer that assuming $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,{\varphi }^{-1}]$ together with Cogency suffices to guarantee that LIS satisfies $\ast$ 8.

To see that LIS can violate $\ast$ 8 for any $t\in ({\varphi }^{-1},1)$ —even under the assumption of Cogency—consider any credence function $c$ satisfying the following constraints, where $\epsilon >0$ is arbitrarily small:

By construction, we have that $c(S\prime \supset X)>t$ and , which shows that $X\in Cn(B_S^{\divideontimes }\cup \{S\prime \})$ but $X\notin B_{S\wedge S\prime }^{\divideontimes }$ , as desired. Note also that since $c(S)=1$ , $B_S^{\divideontimes }=B$ . Furthermore, it can be verified that $B=Cn(S\wedge X)$ holds for every (and only) $t>{\varphi }^{-1}$ , which establishes Cogency and confirms that $S\prime$ is consistent with $B_S^{\divideontimes }$ .              $\square$

This completes our assessment of the relationship between the theories of suppositions provided by LIS and AGM. A full summary of the results from this section is given in table 3 below. In the next section, we turn our attention to the relationship between the subjunctive theories.

Table 3

Table 3

LIS and the AGM Postulates.

4.1. LSS and the KM Postuates

We now proceed to consider how LSS relates to the KM postulates from above. Beginning with the general case where no further constraints are imposed on LSS, we establish which of the KM postulates are satisfied by LSS. As recorded in the proposition below, LSS is guaranteed to satisfy five of the KM postulates: Success ( $◇$ 1), Consistency Preservation ( $◇$ 3), Extensionality ( $◇$ 4), Chernoff ( $◇$ 5), and Primeness ( $◇$ 7).

Proposition 4. LSS must satisfy  $◇$ 1, $◇$ 3, $◇$ 4, $◇$ 5 and  $◇$ 7. That is, each of the following is satisfied for any  $B$ , $S$  and  $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,1]$ :

1. $S\in B_S^{◆}$

2. If  $B⊬\perp$  and  $⊬\neg S$ , then  $B_S^{◆}⊬\perp$

3. If  $⊢S\equiv S\prime$ , then  $B_S^{◆}=B_{\mathrm{S\prime }}^{◆}$

4. $B_{S\wedge S\prime }^{◆}\subseteq Cn(B_S^{◆}\cup \{S\prime \})$

5. If  $B$  is complete, then  $B_{S\vee S\prime }^{◆}\subseteq Cn(B_S^{◆}\cup B_{S\prime }^{◆})$

Proof. Proceeding sequentially:

1. Simply recall that $c_S(S)=1$ to infer $S\in B_S^{◆}$ and, thus, conclude that LIS must satisfy $◇$ 1.              $\square$

2. First, suppose that $B⊬\perp$ and $⊬\neg S$ . Next, note that whenever $B$ is consistent, if $w\in 〚B〛$ , then $c(w)>1-t$ . We prove the contrapositive by first supposing that $B_S^{◆}$ is inconsistent, i.e. $B_S^{◆}⊢\perp$ . That implies that for any $w\in W$ , $\neg w\in B_S^{◆}$ and hence $S\to \neg w\in B$ . But, since $S$ is consistent, there is no $w^{\ast }$ such that $w^{\ast }\models S\to \neg w$ for every $w\in W$ , and therefore $B$ is also inconsistent.               $\square$

3. Suppose that $⊢S\equiv S\prime$ . This implies that $\sigma (w,S)\models S\prime$ just in case $\sigma (w,S\prime )\models S$ . By Uniformity we get $\sigma (w,S)=\sigma (w,S\prime )$ and conclude $B_S^{◆}=B_{S\prime }^{◆}$ . So, LSS must satisfy $◇$ 4.              $\square$

4. To show that LSS must satisfy $◇$ 5, first suppose $X\in B_{Y\wedge Z}^{◆}$ so that $c_{Y\wedge Z}(X)\ge t$ . Now, we show that if $\sigma (w,Y\wedge Z)\models X$ , then $\sigma (w,Y)\models Z\supset X$ . To do so, we assume that $\sigma (w,Y\wedge Z)\models X$ . Then, either $\sigma (w,Y)\models Z$ or $\sigma (w,Y)\models \neg Z$ . In the first case, we may infer $\sigma (w,Y)=\sigma (w,Y\wedge Z)$ , and by our assumption that $\sigma (w,Y)\models X$ , we conclude $\sigma (w,Y)\models Z\supset X$ . In the second case, $\sigma (w,Y)\models \neg Z$ and so $\sigma (w,Y)\models Z\supset X$ . So either way $\sigma (w,Y)\models Z\supset X$ as desired. Applying the definition of imaging gives us

   $c_{Y\wedge Z}(X)={\displaystyle \sum _{\begin{array}{c}w\in W:\\ \sigma (w,Y\wedge Z)\hspace{0.17em}\hspace{0.17em}\models X\end{array}}c}(w)\hspace{1em}\hspace{1em}and\hspace{1em}\hspace{1em}{c}_Y(Z\supset X)={\displaystyle \sum _{\begin{array}{c}w\in W:\\ \sigma (w,Y)\hspace{0.17em}\hspace{0.17em}\models Z\supset X\end{array}}c}(w),$

   which imply $c_Y(Z\supset X)\ge c_{Y\wedge Z}(X)\ge t$ . From this we may then infer $Z\supset X\in B_Y^{◆}$ and thus conclude that $X\in Cn(B_Y^{◆}\cup \{Z\})$ .              $\square$

5. We begin by supposing that $B$ is complete, which means that there is a unique world satisfying all propositions in $B$ —call this $w_B$ . This implies that $c(w_B)\ge t>\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ . Now, let $X\in B_{S\vee S\prime }^{◆}$ and infer $c_{(S\vee S\prime )}(X)\ge t$ . Since $c(w_B)\ge t$ and $c_{S\vee S\prime }(X)\ge t$ , it must be that $\sigma (w_B,S\vee S\prime )\models X$ . Clearly either $\sigma (w_B,S\vee S\prime )\models S$ must satisfy either $S$ or $S\prime$ . Assuming the former, we infer $\sigma (w_B,S)=\sigma (w_B,S\vee S\prime )\models X$ and thus $c_S(X)\ge t$ and so $X\in B_S^{◆}$ . The same reasoning suffices for the latter. Thus, we infer $X\in Cn(B_S^{◆}\cup B_{S\prime }^{◆})$ to conclude that LSS must satisfy $◇$ 7.              $\square$

Most of these results will not be unexpected. Success ( $◇$ 1) should be validated by any plausible account of supposition, while Extensionality ( $◇$ 4) will hold in any non-hyperintensional account like LSS. The generalisation of ( $\ast$ 3) given by Chernoff ( $◇$ 5) holds in virtue of the fact that the probability of a material conditional cannot be less than the probability of its consequent. The satisfaction of Primeness ( $◇$ 7) is intuitive, since if $B$ is complete it should already decide either $S$ or $S\prime$ and updating by their disjunction should not result in more propositions being accepted than by either disjunct. The only result that is remotely surprising is that LSS satisfies Consistency Preservation ( $◇$ 3). Lockean accounts typically struggle to satisfy consistency requirements. So, it is interesting to note that LSS will not lead you to an inconsistent set of suppositional judgments when your beliefs are consistent.

We turn now to the remaining KM postulates: Closure ( $◇$ 0), Stability ( $◇$ 2), Reciprocity ( $◇$ 6) and Compositionality ( $◇$ 8). When no additional restrictions are imposed, LSS can violate each as shown below.

Proposition 5. LSS can violate  $◇$ 0, $◇$ 2, $◇$ 6, and  $◇$ 8. That is, each of the following is possible:

1. $B_S^{◆}\ne Cn(B_S^{◆})$

2. $S\in B$ , but  $B_S^{◆}\ne B$

3. $S\in B_{S\prime }^{◆}$  and  $S\prime \in B_S^{◆}$ , $B_S^{◆}\ne B_{S\prime }^{◆}$

4. $〚B〛=〚B\prime 〛\cup 〚B″〛$ , but  $〚B_S^{◆}〛\ne 〚B{{\prime }_S}^{◆}〛\cup 〚B{{″}_S}^{◆}〛$

Proof. Proceeding sequentially:

1. To see that LSS can violate $◇$ 0, simply recall that Lockean accounts generally permit violations of deductive closure, as demonstrated in the Lottery Paradox.              $\square$

2. A counterexample showing that LSS can violate $◇$ 2 for any $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,1)$ is generated by the assignments provided on the table below, where $\epsilon >0$ is arbitrarily small.

   $\begin{array}{llll}W\hfill & \phi \hfill & c(\phi )\hfill & c_S(\phi )\hfill \\ w_1\hfill & S\wedge X\hfill & t-\epsilon \hfill & 1-\epsilon \hfill \\ w_2\hfill & S\wedge \neg X\hfill & \epsilon \hfill & \epsilon \hfill \\ w_3\hfill & \neg S\wedge X\hfill & 0\hfill & 0\hfill \\ w_4\hfill & \neg S\wedge \neg X\hfill & 1-t\hfill & 0\hfill \end{array}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\begin{array}{c}\\ \sigma (w_1,S)=w_1\\ \sigma (w_2,S)=w_2\\ \sigma (w_3,S)=w_2\\ \sigma (w_4,S)=w_1\end{array}$

   It is easy to see that $X\notin B$ , but $X\in B_S^{◆}$ .              $\square$

3. Our counterexample showing that LSS can violate $◇$ 6 proceeds by assuming that $W$ contains the following six possible worlds.

   $\begin{array}{l}{w}_1\models S\wedge S\prime \wedge X\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{w}_2\models S\wedge S\prime \wedge \neg X\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{w}_3\models S\wedge \neg S\prime \wedge X\\ w_4\models S\wedge \neg S\prime \wedge \neg X\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{w}_5\models \neg S\wedge S\prime \wedge X\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{w}_6\models \neg S\wedge S\prime \wedge \neg X\end{array}$

   Now, let $c$ be such that $c(w_3)=c(w_4)=c(w_5)=c(w_6)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ and select any $\sigma$ such that

   $\sigma (w_3,S\prime )=\sigma (w_4,S\prime )=w_1\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\sigma (w_5,S)=\sigma (w_6,S)=w_2.$

   This gives us $c_S(S\prime )=c_{S\prime }(S)=1,\hspace{0.17em}\hspace{0.17em}{c}_S(X)=0$ and $c_{S\prime }(X)=1$ , which implies that $S\in B_{S\prime }^{◆}$ and $S\prime \in B_S^{◆}$ , but $X\notin B_S^{◆}$ and $X\in B_{S\prime }^{◆}$ . Note that the choice of $t$ played no role here and this suffices as a counterexample to the postulate for any $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,1]$ .              $\square$

4. To build a counterexample showing that LSS can violate $◇$ 8, fix some threshold $t$ , let $\epsilon >0$ be arbitrarily small, and let $n$ be such that $\frac{t-\epsilon }{n-3}\le 1-t$ . Then where $W=\{w_1,\mathrm{\dots },w_n\}$ , let $\sigma (w_i,w_{n-1}\vee w_n)=w_{n-1}$ for $i\le n-3$ and $\sigma (w_{n-2},w_{n-1}\vee w_n)=w_n$ . The credence functions $c,c\prime$ , and $c″$ are defined piecewise below.

   $c(w_i):=\{\begin{array}{ll}\frac{t-\epsilon }{n-3}\hfill & i\le n-3\hfill \\ 1-t+\epsilon \hfill & i=n-2\hfill \\ 0\hfill & \text{otherwise}\hfill \end{array}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}c\prime (w_i):=\{\begin{array}{ll}1\hfill & i=n-2\hfill \\ 0\hfill & \text{otherwise}\hfill \end{array}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}c″(w_i):=\{\begin{array}{ll}1-t\hfill & i=1\hfill \\ t\hfill & i=n-2\hfill \\ 0\hfill & \text{otherwise}\hfill \end{array}$

   Let $B$ , $B\prime$ , $B″$ be the Lockean belief sets corresponding to $c,\hspace{0.17em}\hspace{0.17em}c\prime ,\hspace{0.17em}\hspace{0.17em}c″$ , respectively. It is easy to see that $〚B〛=〚B\prime 〛=〚B″〛=〚B\prime 〛\cup 〚B″〛=\left\{w_{n-2}\right\}$ . Imaging each of these credence functions on $w_{n-1}\vee w_n$ results in the following assignments.

   $\begin{array}{l}{c}_{w_{n-1}\vee w_n}(w_{n-1})=t-\epsilon \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}c{\prime }_{w_{n-1}\vee w_n}(w_{n-1})=0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}c{″}_{w_{n-1}\vee w_n}(w_{n-1})=1-t\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{c}_{w_{n-1}\vee w_n}(w_n)=1-t+\epsilon \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}c{\prime }_{w_{n-1}\vee w_n}(w_n)=1\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}c{″}_{w_{n-1}\vee w_n}(w_n)=t\end{array}$

   Thus, we see that $〚B_S^{◆}〛=\{w_{n-1},w_n\}\ne 〚B{{\prime }_S}^{◆}〛\cup 〚B{{″}_S}^{◆}〛=\left\{w_n\right\}$ .              $\square$

The first three of these results are expected. As Lockean accounts generally fail to require Cogency, we find that LSS similarly may violate $◇$ 0. We also see that LSS can violate $◇$ 2. This postulate is equivalent to the conjunction of $\ast$ 3 and $\ast$ 4. Recall that LIS violated the latter and we find similar behaviour with LSS. Next, the fact that LSS can violate $◇$ 6 is somewhat obvious. The violation of $◇$ 8 is somewhat more surprising. As we briefly discussed earlier, $◇$ 8 is deeply connected to the idea that update proffers a form of ‘local belief change’; and, as we have mentioned, Lewis presents imaging as a method for updating credences by a local dynamics. But, as we will see in the next section, all is not lost.

4.2. Closure under the Stalnaker Conditional Yields Convergence of LSS and KM

When we considered the relationship between the indicative theories given by LIS and AGM, we also saw divergences in the general case—most notably, LIS could violate AGM’s characteristic postulate $\ast$ 4. However, we also saw that the two could be made to converge so long as we assume Cogency and a sufficiently low Lockean threshold. We might then wonder whether there is a similar path towards convergence between LSS and KM.

As we will soon see, there is such a path. However, the requirements involved in establishing convergence between LSS and KM are different. In this case, neither restrictions on the Lockean threshold nor standard Cogency will suffice. Instead, we will augment Cogency with the additional requirement that $B$ is closed under the Stalnaker conditional ( $\to$ ). But this will take some work since our language does not officially include $\to$ . To deal with this, we will augment our finite propositional language $ℒ$ to the “flat fragment” of $ℒ$ extended with the Stalnaker conditional. That is, we introduce $\to$ into the language’s signature to generate ${ℒ}^{+}$ , which only adds conditional sentences of the form $X\to Y$ , where $X,Y\in ℒ$ . The statement of Cogency remains unchanged from earlier. However, the type of logical consequence used in the expression of its requirements (Cn) is richer. We let ‘ $Cogenc{y}^{\to }$ ’ refer to the stronger requirement that results from imposing Cogency with the richer language ${ℒ}^{+}$ . At this stage, there are two important observations to make. Firstly, it is well known that the probability of the Stalnaker conditional $X\to Y$ is given by the probability of $Y$ after imaging on $X$ , $c_X(Y)$ . Thus, the conditions under which Stalnaker conditionals are believed are clear: $X\to Y\in B$ iff $c_X(Y)\ge t$ . Second, observe that the Stalnaker conditional satisfies modus ponens, i.e.  $X\to Y,X⊢Y$ . This means that $Cogenc{y}^{\to }$ requires that $X\to Y,X⊢Y$ and $X\in B$ imply $Y\in B$ .

Surprisingly, we find that in this richer environment where we have $Cogenc{y}^{\to }$ , LSS satisfies all of the KM postulates. We have already shown that LSS will always satisfy $◇$ 1, $◇$ 4, $◇$ 5, and $◇$ 7; it is straightforward to see that Propositions 4 and 5 will carry over to this richer environment. So, it remains only to show that, given $Cogenc{y}^{\to }$ , the remaining postulates are all satisfied.

Proposition 6. Assuming $Cogenc{y}^{\to }$ , LSS must satisfy  $◇$ 0, $◇$ 2, $◇$ 6, and  $◇$ 8. That is, for any  $c$  and $t\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,1]$ , if  $B$  satisfies $Cogenc{y}^{\to }$ , then:

1. $B_S^{◆}=Cn(B_S^{◆})$

2. If  $S\in B$  then  $B=B_S^{◆}$

3. If  $S\in B_{S\prime }^{◆}$ , and  $S\prime \in B_S^{◆}$ , then  $B_S^{◆}=B_{S\prime }^{◆}$

4. If  $〚B〛=〚B\prime 〛\cup 〚B″〛$ , then  $〚B_S^{◆}〛=〚B{{\prime }_S}^{◆}〛\cup 〚B{{″}_S}^{◆}〛$

Proof. As before, we proceed sequentially, where $Cogenc{y}^{\to }$ is taken as a standing assumption:

1. It is an immediate consequence of $Cogenc{y}^{\to }$ that LSS satisfies $◇$ 0.              $\square$

2. Suppose that $S\in B$ to show that $B\subseteq B_S^{◆}$ . Let $X\in B$ and by $Cogenc{y}^{\to }$ infer $S\wedge X\in B$ . This implies $c(S\wedge X)\ge t$ . Since imaging on $S$ won’t lower the probability of any $S\wedge X$ worlds, it follows that $c_S(X)\ge t$ and thus $X\in B_S^{◆}$ . For the other direction, let $X\in B_S^{◆}$ so that $c_S(X)\ge t$ and hence $S\to X\in B$ . By $Cogenc{y}^{\to }$ , we get $X\in B$ as desired and thus conclude that LSS now satisfies $◇$ 2.              $\square$

3. Suppose that $S\in B_{S\prime }^{◆}$ and $S\prime \in B_S^{◆}$ . This gives us $c_S(S\prime )\ge t$ and $c_{S\prime }(S)\ge t$ , from which we infer that $S\to S\prime \in B$ and $S\prime \to S\in B$ and hence $S\leftrightarrow S\prime \in B$ . Now, letting $X\in B_S^{◆}$ , we infer $S\to X\in B$ . By Uniformity and $Cogenc{y}^{\to }$ , $S\to X\in B$ and $S\leftrightarrow S\prime \in B$ jointly entail $S\prime \to X\in B$ . Thus we infer $X\in B_{S\prime }^{◆}$ and hence $B_S^{◆}\subseteq B_{S\prime }^{◆}$ . The same argument shows the converse. Thus, given $Cogenc{y}^{\to }$ , LSS will satisfy $◇$ 6.

4. To show that LSS will now satisfy $◇$ 8, let $〚B〛=〚B\prime 〛\cup 〚B″〛$ , and suppose that $B$ , $B\prime$ and $B″$ are all $cogen{t}^{\to }$ , and satisfy LT^t with respect to the credence functions $c$ , $c\prime$ and $c″$ . Let $w\in 〚B_S^{◆}〛$ , $w\notin 〚B{{\prime }_S}^{◆}〛\cup 〚B{{″}_S}^{◆}〛$ . This implies that $S\to \neg w\notin B$ , $S\to \neg w\in B\prime ,B″$ , and hence that $\forall w\prime \in 〚B\prime 〛\cup 〚B″〛$ , $w\prime ⊢S\to \neg w$ , i.e. $\forall w\prime \in 〚B〛$ , $w\prime ⊢S\to \neg w$ . Since $B$ is cogent, we have

   $t\le c(\wedge B)=c(\vee 〚B〛)\le c(S\to \neg w).$

   This implies that $S\to \neg w\in B$ , which is a contradiction. So $w\in 〚B_S^{◆}〛$ implies $w\in 〚B{{\prime }_S}^{◆}〛\cup 〚B{{″}_S}^{◆}〛$ , i.e.  $〚B_S^{◆}〛\subseteq 〚B{{\prime }_S}^{◆}〛\cup 〚B{{″}_S}^{◆}〛$ . Conversely, let $w\notin 〚B_S^{◆}〛$ , $w\in 〚B{{\prime }_S}^{◆}〛\cup 〚B{{″}_S}^{◆}〛$ . For argument’s sake, let $w\in 〚B{{\prime }_S}^{◆}〛$ . This implies that $S\to \neg w\in B$ , $S\to \neg w\notin B\prime$ . and hence that $\forall w\prime \in 〚B〛$ , $w\prime ⊢S\to \neg w$ , i.e. $\forall w\prime \in 〚B\prime 〛$ , $w\prime ⊢S\to \neg w$ . Since $B\prime$ is cogent, we have

   $t\le c\prime (\wedge B\prime )=c\prime (\vee 〚B\prime 〛)\le c\prime (S\to \neg w)$

   This implies that $S\to \neg w\notin B\prime$ , which is a contradiction. So $w\in 〚B{{\prime }_S}^{◆}〛\cup 〚B{{″}_S}^{◆}〛$ implies $w\in 〚B_S^{◆}〛$ , i.e.  $〚B{{\prime }_S}^{◆}〛\cup 〚B{{″}_S}^{◆}〛\subseteq 〚B_S^{◆}〛$ .              $\square$