The scope of provability

Abstract

We explore the relationship between evidence and knowledge when knowledge is described by a standard partition over a finite state space, and evidence is represented by a state-dependent collection of finite sets of messages. When the collection is measurable with respect to the partition, there is evidence for an event only if the event is self-evident—i.e., known at every one of its states. Thus, an event is commonly known in and only if there is mutual evidence that implies that the event has occurred, and all knowledge can be proved only when the agent is informed about the state or knows nothing. The existence of partial provability outside these two extremes hinges on the non monotonic nature of provability.

Appendix

Proof of Proposition 1

Fix any agent i and suppose, contrary to hypothesis, that there exists a model $\mathcal {F}_{i}=(S, \Pi _{i},\textbf{M}_{i})$, a state s, and an event $E\in {\textbf{M}_{i}(s)}$ such that $E\not \in {\mathcal {A}_{\Pi _{i}}(s)}$. Then, there exists some $s'\in {E}$ such that $\Pi _{i}(s')\not \subseteq {E}$. Yet consistency entails that $E\in {\textbf{M}_{i}(s')}$, contradicting that $\textbf{M}_{i}$ is measurable with respect to $\Pi _{i}$. $\square$

Proof of Proposition 2

We first show that 1. and 2. are equivalent. Suppose that a model $\mathcal {F}_i=(S, \Pi _i,\textbf{M}_i)$ satisfies both (IC) and (M). The former implies that $\Pi _i(s)\in {\textbf{M}_i(s)}$ for every s. By (M), it follows that $F\in {\textbf{M}_i(s)}$ for every s and every F such that $\Pi _i(s)\subseteq {F}$. Thus, $\mathcal {F}_i$ is a model of total provability. To show that 1. implies 2., notice that if $\mathcal {F}_i$ is a model of total provability, then $\Pi _i(s)\in {\textbf{M}_i(s)}$ for every s. Hence, $\mathcal {F}_i$ is (IC). To show that $\mathcal {F}_i$ satisfies (M) it is sufficient to consider known events. Since knowledge is monotonic and $\mathcal {F}_i$ is a model of total provability, we must have $F\in {\textbf{M}_i(s)}$ for every s and every F such that $\Pi _i(s)\subseteq {F}$. Hence, it follows that $\mathcal {F}_i$ satisfies (M).

To see that 1. and 3. are equivalent, notice that $\Pi _{i}(s)\in {\mathcal {A}_{\Pi _{i}(s)}}$ for every s. If $\mathcal {A}_{\Pi _{i}}$ is closed under supersets, it follows that $F\in {\mathcal {A}_{\Pi _{i}}(s)}$ for every s and every F such that $\Pi (s)\subseteq {F}$. Thus, $\mathcal {F}_{i}$ is a model of total provability. To prove that 1. implies 3., notice that if $\mathcal {F}_{i}$ is a model of total provability, then for every s and every E such that $\Pi _{i}(s)\subseteq {E}$, there exists some evidence structure $\textbf{M}_{i}$ such that $E\in {\textbf{M}_{i}(s)}$. Thus, by Proposition 1, for every s and every E such that $\Pi _{i}(s)\subseteq {E}$ we have $E\in {\mathcal {A}_{\Pi _{i}}(s)}$. Since the set of events that i knows at any s is closed under supersets, $\mathcal {A}_{\Pi _{i}}(s)$ is closed under supersets for every s. $\square$

Proof of Proposition 3

We show this by contradiction. Suppose that the partition is neither the finest nor the coarsest. The latter implies that there are at least two states, s and $s'$, such that $\Pi _i(s')\cap {\Pi _i(s)}=\emptyset$. In turn, the former implies that, without loss of generality, we can take either $\Pi _i(s)$ or $\Pi _i(s')$ to be non-singleton.¹⁰ Assume that $|\Pi _i(s)| > 1$. Then,

$$\begin{aligned}&\Rightarrow {}&\Pi _i(s')\subseteq {}\Pi _i(s')\cup {\{s\}} \hspace{3mm} \\&\Rightarrow {}&\Pi _i(s')\cup {\{s\}}\in {\mathcal {A}_{\Pi _{i}}(s')} \hspace{3mm} \text{ by } \text{ Proposition } 1.\\&\Rightarrow {}&\Pi _i(s')\cup {\{s\}}\in {\mathcal {A}_{\Pi _{i}}(s)} \hspace{4mm} \text{ by } \text{ consistency. } \\&\Rightarrow {}&\Pi _i(s)\subseteq {}\Pi _i(s')\cup {\{s\}} \hspace{6mm} \text{ by } \text{ measurability. }\\&\Rightarrow {}&\Pi _i(s)=\{s\} \hspace{19mm} \text{ by } \Pi _i(s)\cap {\Pi _i(s')=\emptyset }. \end{aligned}$$

Since $\Pi _i(s)$ is non-singleton by hypothesis, we reach a contradiction. $\square$

Proof of Proposition 4

The proof of this statement hinges on a couple of auxiliary results: $\square$

Claim 1

For every s such that $\Pi _i(s)$ is not a singleton, every $s'$, and every E,

$$\begin{aligned} E\in {\textbf{M}_i(s')}\Rightarrow {\Pi _i(s)\subseteq {E}}. \end{aligned}$$

(1)

Claim 2

For every s such that $\Pi _i(s)$ is not a singleton, and every $s'$

$$\begin{aligned} \textbf{M}_i(s')\subseteq {\textbf{M}_i(s)}. \end{aligned}$$

(2)

Proof of Claim 1

Take any s such that $\Pi _i(s)$ is not a singleton and assume, contrary to hypothesis, that there is some state $s'$ and event E such that $E\in {\textbf{M}_i(s')}$ but $\Pi _i(s)\not \subseteq {E}$. The latter implies that there is some $\bar{s}\in {\Pi _i(s)}$ such that $\bar{s}\not \in {E}$. In fact, we must have $\Pi _i(s)\cap {E}=\emptyset$. To see this, suppose that there is some $\hat{s}\in {\Pi _i(s)}$ such that $\hat{s}\in {E}$. Since $E\in {\textbf{M}_i(s')}$, consistency entails that $E\in {\textbf{M}_i(\hat{s})}$. Thus, $\Pi _i(\hat{s})\subseteq {E}$. But since $\hat{s}\in {\Pi _i(s)}$, it follows that $\Pi _i(s)\subseteq {E}$. Contradiction.

Since $E\in {\textbf{M}_i(s')}$, monotonicity implies that $E\cup {\{\bar{s}\}}\in {\textbf{M}_i(s')}$ for every $\hat{s}\in {\Pi (s)}$. By consistency, we must then have that $E\cup {\{\bar{s}\}}\in {\textbf{M}_i(\hat{s})}$ for every $\hat{s}\in {\Pi _i(s)}$. Thus, it follows that $\Pi _i(s)\subseteq {E\cup {\{s\}}}$. But since $\Pi _i(s)\cap {E}=\emptyset$, it must follow that $\Pi _i(s)=\{s\}$. Contradiction. $\square$

Proof of Claim 2

To see this, take any $E\in {\textbf{M}_i(s')}$. By Claim 1, we must have that $\Pi _i(s)\subseteq {E}$. Hence, $s\in {E}$. Thus, consistency implies that $E\in {\textbf{M}_i(s)}$.

Armed with these two claims, it is easy to see that the only-if part holds since both (M) and (SM) are trivially satisfied in cheap-talk models. To see the if part, notice that, by hypothesis, there is a state s at which $\Pi _i(s)$ is not a singleton. Then, (SM) implies that for every s such that $\Pi _i(s)$ is not a singleton we must actually have that $\textbf{M}_i(s')=\textbf{M}_i(s)$ for every $s'$. Suppose that this is not the case. Then, for some states $s'$ and s such that $|\Pi _i(s)|> 1$, $\textbf{M}_i(s') \subset {\textbf{M}_i(s)}$. It follows that $\vert {\textbf{M}_i(s)}\vert {}>\vert {\textbf{M}_i(s')}\vert {}$. Then, by (SM) we have that $\vert {\Pi _i(s')}\vert {}\ge {}\vert {\Pi _i(s)}\vert {}$. Since $\Pi _i(s)$ is not a singleton, $\Pi _i(s')$ is also not a singleton. But then, Claim 2 implies that $\textbf{M}_i(s)\subseteq {\textbf{M}_i(s')}$, a contradiction. Thus, whenever $\Pi _i$ is not the finest partition it follows that the model is a cheap-talk model. $\square$