The two faces of compatibility with justified beliefs

Abstract

When discussing knowledge, two relations are of interest: justified doxastic accessibility $S$ (for all the agent is justified in believing in $x$, she is in $y$) and justification equivalence $E$ (the agent would have in $y$ exactly the same justified beliefs that she has in $x$). Speaking of compatibility with the agent’s justified beliefs is potentially ambiguous: either of the two relations $S$ or $E$ can be meant. I discuss the possibility of identifying the relation of epistemic accessibility $R$ (for all the agent knows in $x$, she is in $y$) with the union of $S$ and $E$. Neither Gettier’s examples nor the ‘fake barn’ cases contradict this identification. However, the proposal leads to justification equivalent scenarios being symmetric with respect to knowledge: we cannot know a true proposition in a scenario if it is false in a justification equivalent scenario. This analysis may appear to render non-trivial knowledge impossible. This conclusion follows if the extra premise is granted that for all relevant true propositions there is a justification equivalent scenario in which the proposition is false. I provide a meaning-theoretic argument against this premise. I conclude by pointing out problems that would ensue from giving up the proposed connection between $S$, $E$ and $R$ and allowing asymmetry of justification equivalent scenarios relative to knowledge.

Appendix

In this Appendix all formulas have ‘$p$’ as their sole propositional variable. Augmented models are triples $\langle W, S, p \rangle$, where $\langle W, S \rangle$ is a model and $p \subseteq W$. It is understood that ‘$p$’ stands for $p$. A formula $\varphi$ is valid in $\langle W, S \rangle$ (denoted $\langle W, S \rangle \models \varphi$) if for all augmented models $M=\langle W, S, p \rangle$ and all $w \in W$, the proposition expressed by $\varphi$ in $M$ is true at $w$. (‘Valid in a model’ means what is usually termed ‘valid in a frame’. The terminology is deviant because I follow Williamson (2013a) in not incorporating a valuation in a model.) I write ${ Kp }$ for $\{v:$ for all $x$, if $vR^*x$, then $x \in p\}$. A model $\langle W, S \rangle$ is justificatorily homogeneous if all sets $S(w)$ with $w \in W$ are justificatorily homogeneous. We recall from Sect. 5 that $S(w)$ is justificatorily homogeneous if $S(v)=S(w)$ for all $v \in S(w)$.

Fact 1

A model $\langle W, S \rangle$ is justificatorily homogeneous $\Leftrightarrow$ the relation $S$ is transitive (whenever $xSy$ and $ySz$, then $xSz$) and Euclidean (whenever $xSy$ and $xSz$, then $ySz$) $\Leftrightarrow$ $\langle W, S \rangle \models { Jp } \rightarrow K{ Jp }$ and $\langle W, S \rangle \models \lnot { Jp } \rightarrow K\lnot { Jp }$.

Proof

We prove two more specific claims:

• (a) $S$ is transitive $\Leftrightarrow$ $S(y) \subseteq S(x)$ for all $x \in W$ and $y \in S(x)$ $\Leftrightarrow$ $\langle W, S \rangle \models { Jp } \rightarrow KJp$.

• (b) $S$ is Euclidean $\Leftrightarrow$ $S(x) \subseteq S(y)$ for all $x \in W$ and $y \in S(x)$ $\Leftrightarrow$ $\langle W, S \rangle \models \lnot { Jp } \rightarrow K\lnot { Jp }$.

Now, $S$ is transitive $\Leftrightarrow$ for all $x$, $y$, $z$, we have: if $y \in S(x)$ and $z \in S(y)$, then $z \in S(x)$ $\Leftrightarrow$ for all $x$ and $y$: if $y \in S(x)$, then $S(y) \subseteq S(x)$. Further, $S$ is Euclidean $\Leftrightarrow$ for all $x$, $y$, $z$: if $y \in S(x)$ and $z \in S(x)$, then $z \in S(y)$ $\Leftrightarrow$ for all $x$ and $y$: if $y \in S(x)$, then $S(x) \subseteq S(y)$. Let us, then, consider the second equivalence of claim $(a)$. Let $M$ be any augmented model in which $S$ is transitive and let $x$ be any scenario. Suppose $S(x) \subseteq p$. Let $y$ be any scenario with $xR^*y$, whence $xSy$ or $xEy$. Using the first equivalence of $(a)$ and the definition of $E$, we conclude that $S(y) \subseteq S(x)$. It follows that $S(y) \subseteq p$ and indeed $x \in { KJp }$. Conversely, assume $\langle W, S \rangle \models { Jp } \rightarrow { KJp }$. Suppose for contradiction there are $x,y,z$ such that $xSy$ and $ySz$ but not: $xSz$. Consider the augmented model $\langle W,S,p \rangle$ where $p=S(x)$. Then $x \in { Jp }$. On the other hand, $S(y) \not \subseteq p$ since $z \notin S(x)$. As $xR^*y$, it follows that $x \notin { KJp }$. Finally, we consider the second equivalence of claim $(b)$. Let $M$ be any augmented model in which $S$ is Euclidean and let $x$ be any scenario. Suppose $S(x) \not \subseteq p$. Let $y$ be any scenario with $xR^*y$, whence $xSy$ or $xEy$. Using the first equivalence of $(b)$ and the definition of $E$, we may infer that $S(x) \subseteq S(y)$. It follows that $S(y) \not \subseteq p$. Conversely, assume $\langle W, S \rangle \models \lnot { Jp } \rightarrow K\lnot { Jp }$. Suppose for contradiction there are $x,y,z$ such that $xSy$ and $xSz$ but not: $ySz$. Consider the augmented model $\langle W,S,p \rangle$ where $p=W \setminus \{z\}$. Then $x \in \lnot { Jp }$. Yet $x \notin K\lnot { Jp }$, since $xR^*y$ but $S(y) \subseteq p$. The main claim follows immediately from the auxiliary claims $(a)$ and $(b)$. $\square$

Fact 2

There are models in which $S$ is neither transitive nor Euclidean and which refute both introspection principles ${ Jp } \rightarrow K{ Jp }$ and $\lnot { Jp } \rightarrow K\lnot { Jp }$.

Proof

We saw in Sect. 5 that there are models and scenarios $x$, $y$ such that $S(x) \not \subseteq S(y)$ and $S(y) \not \subseteq S(x)$. The claims follow by the proof of Fact 1. $\square$

Fact 3

$R^*$ is transitive $\Leftrightarrow$ all $w \in W$ and $v \in S(w)$ satisfy:

$$\begin{aligned} S(v) \cup E(v) \subseteq S(w)\, or\, S(v) = S(w). \end{aligned}$$

Proof

Assume first that $S(v) \cup E(v) \subseteq S(w)$ or $S(v) = S(w)$ for all $w \in W$ and $v \in S(w)$. Suppose $xR^*y$ and $yR^*z$. That is, [$xEy$ and $yEz$] or [$xSy$ and $ySz$] or [$xEy$ and $ySz$] or [$xSy$ and $yEz$]. In the first case we have trivially $xEz$ and in the third case trivially $xSz$, so in both cases $xR^*z$. In the second case we have $xSz$: namely, given our assumption, $y \in S(x)$ entails in particular that $S(y) \subseteq S(x)$, whence $z \in S(y) \subseteq S(x)$. In the fourth case, consider the two subcases: either $S(y) \cup E(y) \subseteq S(x)$ or $S(y) = S(x)$ (at least one of them holds, since $xSy$). In the former subcase, we have $xSz$ because $z \in E(y) \subseteq S(x)$. In the latter subcase, we have $xEz$ because $S(x)=S(y)$ and $S(y)=S(z)$. Thus, we have $xR^*z$ also in the second and the fourth case.

Conversely, assume $R^*$ is transitive. Suppose for contradiction that there are $w \in W$ and $v \in S(w)$ such that either $S(w) \ne S(v) \not \subseteq S(w)$ or $E(v) \not \subseteq S(w) \ne S(v)$. In the former case, there is $u$ such that $vSu$ but not: $wSu$. Since $wSv$, $S$ is not transitive. This is impossible as $S \subseteq R^*$ and $R^*$ is transitive. Let us proceed to consider the latter case. Here we may assume that $S(v) \subseteq S(w)$. Because $E(v) \not \subseteq S(w)$, there is $u$ such that $vEu$ and $u \notin S(w)$. Since $S(v) \subseteq S(w)$, in particular $u \notin S(v)$. Now, $wSv$ and $vEu$, whence $wR^*v$ and $vR^*u$, from which it follows by the transitivity of $R^*$ that $wR^*u$. So either $wSu$ or $wEu$. The latter option is blocked by the fact that $S(w) \ne S(v) = S(u)$. The former option is likewise blocked: $u \notin S(w)$. This is a contradiction. $\square$

Fact 4

There are models in which $S$ is transitive but $R^*$ is not.

Proof

Define a model $\langle W,S \rangle$ by setting $W=\{0,1,-1,2\}$ and $S=\{(0,1),(1,2), (0,2),$ $(-1,2)\}$. We note that $S$ is transitive and that $E=\{(x,x) : x \in W\} \cup \{(1,-1), (-1,1)\}$. Now $1 \in S(0)$, but $\{-1,1\}=E(1) \not \subseteq S(0)=\{1,2\}\ne \{2\}=S(1)$. It follows by Fact 3 that $R^*$ is not transitive. $\square$

Fact 5

Transitivity of $R^*$ does not trivialize the analysis (4).

Proof

Consider the model $\langle \mathbb {N},S \rangle$ with $S=\{(x,y) \in \mathbb {N}^2 : x < y\}$, whence $E=\{(x,x) : x \in \mathbb {N}\}$. Consequently $R^*=\{(x,y) \in \mathbb {N}^2 : x \le y\}$ and in particular $R^*$ is transitive. Now, $(0,0) \in E \setminus S$. Further, $(0,1) \in S \setminus E$. It follows that $E \ne R^*\ne S$. $\square$

Fact 6

The propositions ${ Jp }$ and $\lnot K \lnot Kp$ are logically independent of each other.

Proof

Consider a model in which $S(w)$ is justificatorily homogeneous, but there is $x \notin S(w)$ with $wEx$. Letting $p = S(w)$, we have $w \in { Jp }$. Suppose, then, that $wR^*v$. We have $wEv$, because $S(w)$ is justificatorily homogeneous. Consequently $vEx$ and therefore $x \in R^*(v)$. Yet $x \notin p$, whence $w \notin \lnot K \lnot Kp$. That is, ${ Jp }$ does not entail $\lnot K \lnot Kp$. Recall, then, the example of Alice and the tomato (Sect. 5). Let $w$ (respectively $v$) be a scenario in which the tomato weighs 3.0 oz (respectively 3.5 oz). Let $p = R^*(v)$. If $u$ is a scenario in which the tomato weighs 2.25 oz, neither $uEv$ nor $u \in S(v)$. Consequently, $u \in S(w)$ but $u \notin R^*(v)$. Therefore $S(w) \not \subseteq R^*(v)=p$, whence $w \notin { Jp }$. On the other hand, we have $w \in \lnot K \lnot Kp$, because $wR^*v$ and $v \in { Kp }$. That is, $\lnot K \lnot Kp$ does not entail ${ Jp }$. $\square$