Knowledge Is a Diamond

Abstract

In the standard epistemic logic, the knowledge operator is represented as a box operator, a universal quantifier over a set of possible worlds. There is an alternative approach to the semantics of knowledge, according to which an agent a knows $\alpha$ iff a has a reliable (e.g. sensory) evidence that supports $\alpha$. In this interpretation, knowledge is viewed rather as an existential, i.e. a diamond modality. In this paper, we will propose a formal semantics for substructural logics that allows to model knowledge on the basis of this intuition. The framework is strongly motivated by a similar semantics introduced in [3]. However, as we will argue, our framework overcomes some unintuitive features of the semantics from [3]. Most importantly, knowledge does not distribute over disjunction in our logic.

V. Punčochář—The work on this paper was supported by grant no. 16-07954J of the Czech Science Foundation.

Appendix

In the Appendix, we will provide proofs of the results of the paper. For the proof of Theorem 1-(a), see [10]. The proof of 1-(b) is contained in [8].

Proof of Theorem 2: We are proving that $\mathcal {M}^{i}= \langle A, +, \times , \cdot , 0, 1, C^{i}, V^{i} \rangle$ is an information model, so we have to verify that all conditions from the definition of information models are satisfied. Since $+$ is union and $\times$ intersection, it is immediate that $\langle A, +, \times \rangle$ is a distributive lattice. Since 0 is the empty set, we also immediately get $0 + a = 0$ and $0 \cdot a = 0$. The definition of $\cdot$ leads also directly to its distributivity over union. Moreover, $\cdot$ is commutative, since Rxyz implies Ryxz.

We will verify that $1 \cdot a = a$. We have to show that for any upward closed set u it holds that $L \cdot u = u$. First, assume that $z \in L \cdot u$. So, there is $x \in L$ and $y \in u$ such that Rxyz. Then it holds that $y \le z$. Since u is upward closed, it follows that $z \in u$. Second, assume $z \in u$. Since $z \le z$, there is some $x \in L$ such that Rxzz. It follows that $z \in L \cdot u$.

The conditions related to the compatibility relation can be easily verified. Moreover, since the ordering in $\mathcal {M}^{i}$ is inclusion, it holds for any atomic formula p that $V^{i}(p)$ is the principal ideal generated by V(p).

Proof of Theorem 3: We are proving that for every upward closed set of states u it holds that $u \vDash \alpha$ in $\mathcal {M}^{i}$ iff for all $x \in u$, $x \Vdash \alpha$ in $\mathcal {M}$. We can proceed by induction on the complexity of $\alpha$. The case of atomic formulas as well as the constants $\bot , \top , t$ is straightforward. As the induction hypothesis we assume that the statement holds for some $\mathcal {L}$-formulas $\alpha$ and $\beta$.

Negation: First, assume that $u \nvDash \lnot \beta$. Then for some v, $uC^{i}v$ and $v \vDash \beta$, i.e. for all $x \in v$, $x \Vdash \beta$. Then for some v there are $y \in u$ and $z \in v$ such that yCz and $z \Vdash \beta$. So for some $y \in u$, $y \nVdash \lnot \beta$.

Second, assume that for some $y \in u$, $y \nVdash \lnot \beta$, i.e. for some z, yCz and $z \Vdash \beta$. Take $v =\{ w \in W; z \le w \}$. Then $uC^{i}v$ and for all $x \in v$, $x \Vdash \beta$, i.e. $v \vDash \beta$. So $u \nvDash \lnot \beta$.

Implication: First, assume that $u \nvDash \alpha \rightarrow \beta$. This means that there is an upward closed set v such that $v \vDash \alpha$ but $u \cdot v \nvDash \beta$. Then it follows from the induction hypothesis that every state of v supports $\alpha$ in $\mathcal {M}$ and there is a state $z \in u \cdot v$ such that $z \nVdash \beta$. So, there are $x \in u$ and $y \in v$ such that Rxyz. Since $y \Vdash \alpha$ and $z \nVdash \beta$, it follows that $x \nVdash \alpha \rightarrow \beta$.

Second, assume that for some $x \in u$, $x \nVdash \alpha \rightarrow \beta$. That means that there are y, z such that Rxyz, and $y \Vdash \alpha$ and $z \nVdash \beta$. Let v be the set $\{ w \in W; y \le w \}$. Then $z \in u \cdot v$ and it follows that $u \cdot v \nvDash \beta$. Moreover, $v \vDash \alpha$, so $u \nvDash \alpha \rightarrow \beta$.

Conjunctions: The step for the conjunction $\wedge$ is straightforward. We will show the step for the intentional conjunction&. First, assume that $u \vDash \alpha \&\beta$. So, there are $v_1$, $v_2$ such that $v_1 \vDash \alpha$, $v_2 \vDash \beta$, and $u \subseteq v_1 \cdot v_2$. Take an arbitrary state $x \in u$. Then $x \in v_1 \cdot v_2$, which means that there are $y \in v_1$ (so $y \Vdash \alpha$) and $z \in v_2$ (so $z \Vdash \beta$) such that Ryzx. Therefore, $x \Vdash \alpha \&\beta$.

Second, assume that for every $x \in u$, $x \Vdash \alpha \&\beta$. So, for every $x \in u$, there are $y, z \in W$ such that $y \Vdash \alpha$, $z \Vdash \beta$, and Ryzx. For every $x \in u$, let us select some y, z with this property and denote them $y_x$ and $z_x$. Now we define

$$\begin{aligned} v_1 = \mathop {\bigcup }\nolimits _{x \in u} \{w \in W; y_x \le w \},\\ v_2 = \mathop {\bigcup }\nolimits _{x \in u} \{w \in W; z_x \le w \}. \end{aligned}$$

Then $v_1 \vDash \alpha$ and $v_2 \vDash \beta$. Moreover, $u \subseteq v_1 \cdot v_2$. It follows that $u \vDash \alpha \&\beta$.

Disjunction: First, assume that $u \vDash \alpha \vee \beta$. So, there are $v_1$, $v_2$ such that $v_1 \vDash \alpha$, $v_2 \vDash \beta$, and $u \subseteq v_1 \cup v_2$. Take an arbitrary state $x \in u$. Then $x \in v_1$ or $x \in v_2$, so $x \Vdash \alpha$ or $x \Vdash \beta$. Therefore, $x \Vdash \alpha \vee \beta$.

Second, assume that for every $x \in u$, $x \Vdash \alpha \vee \beta$. So, for every $x \in u$, $x \Vdash \alpha$ or $x \Vdash \beta$. Take $v_1 = || \alpha ||_{\mathcal {M}}$ and $v_2 = || \beta ||_{\mathcal {M}}$. Then $v_1 \vDash \alpha$ and $v_2 \vDash \beta$. Moreover, $u \subseteq v_1 \cup v_2$. It follows that $u \vDash \alpha \vee \beta$.

Proof of Theorem 4: To show that $\mathcal {N}_{\lambda }^K$ is an epistemic information model we have to verify the conditions (a)–(c) from Definition 8 which is straightforward. We will show the inductive step for K in the proof of the claim that $\varDelta \vDash \alpha$ iff $\alpha \in \varDelta$. Assume that the claim holds for some $\mathcal {L}_K$-formula $\beta$. We have to prove that $K\beta \in \varDelta$ iff there is a $\lambda$-theory $\varGamma$ such that $\varGamma S_{\lambda } \varDelta$ and $\beta \in \varGamma$.

First, assume that there is a $\lambda$-theory $\varGamma$ such that $\varGamma S_{\lambda } \varDelta$ and $\beta \in \varGamma$. Then we have directly from the definition of $S_{\lambda }$ that $K\beta \in \varDelta$.

Second, assume $K\beta \in \varDelta$. We define $\varGamma = \{ \gamma ; \beta \rightarrow \gamma \in \lambda \}$. $\varGamma$ is a $\lambda$-theory, due to the application of the rules: $\beta \rightarrow \gamma , \gamma \rightarrow \delta /\beta \rightarrow \delta$ and $\beta \rightarrow \delta , \beta \rightarrow \epsilon /\beta \rightarrow (\delta \wedge \epsilon )$. Moreover, due to the axiom $\beta \rightarrow \beta$, $\beta \in \varGamma$. Now, it will suffice to prove that $\varGamma S_{\lambda } \varDelta$. Assume that $\gamma \in \varGamma$, i.e. $\beta \rightarrow \gamma \in \lambda$. Then $K \beta \rightarrow K \gamma \in \lambda$ due to R1. Since $K \beta \in \varDelta$, then also $K \gamma \in \varDelta$, since $\varDelta$ is a $\lambda$-theory.

Proof of Theorem 5: Proving soundness is a matter of a mechanical verification of the respective axioms in the respective class of models. We will discuss only completeness. (a) is a consequence of Theorem 4. To prove (b), we have to show that the canonical model of $\lambda =\lambda ^{K}_{1}$ is intended, i.e. it holds: if $\varDelta S_{\lambda } \varGamma$, then $\varDelta \subseteq \varGamma$. Assume that $\varDelta S_{\lambda } \varGamma$ and $\alpha \in \varDelta$. Then $K \alpha \in \varGamma$ and since we assume that $K \alpha \rightarrow \alpha \in \lambda$, we have $\alpha \in \varGamma$.

To prove (c), we have to show that the canonical model of $\lambda =\lambda ^{K}_{12}$ is strong. So we have to verify that in that case if $\varDelta S_{\lambda } \varGamma$ and $\varGamma \ne \mathcal {L}_K$, then $\varDelta C \varGamma$. Assume $\varDelta S_{\lambda } \varGamma$ and $\varGamma \ne \mathcal {L}_K$. For the contradiction, assume that it does not hold that $\varDelta C \varGamma$. Then it also does not hold that $\varGamma C \varDelta$, so there is an $\mathcal {L}_K$-formula $\alpha$ such that $\lnot \alpha \in \varGamma$ and $\alpha \in \varDelta$. Then $K \alpha \in \varGamma$, and since we assume A2, it follows $\bot \in \varGamma$. Then $\varGamma = \mathcal {L}_K$, which is a contradiction. The proof of (d) is similar.

Proof of Theorem 6: Let a be a state of a model on a source-distributive epistemic information frame and let $\alpha$ be any $\mathcal {L}_K$-formula. Assume that $a \vDash K(\alpha \vee \beta )$. Then there is some $b \in S(a)$ such that $b \vDash \alpha \vee \beta$. So, there are c, d such that $c \vDash \alpha$, $d \vDash \beta$, and $b \le c + d$. It follows that $(c+d)Sa$. Since the model is on a source-distributive frame, there are $c'$, $d'$ such that $cSc'$, $dSd'$, and $a \le c' + d'$. Then $c' \vDash K \alpha$ and $d' \vDash K \beta$, and, as a consequence, $a \vDash K \alpha \vee K \beta$.

We have proved that A3 is valid in every model on any source-distributive epistemic information frame. Now consider an arbitrary epistemic information model on a frame that is not source-distributive. So, there are some states a, b, c such that $(a+b)Sc$ and there are no d, e such that aSd, bSe, and $c \le d+e$. Define a valuation V on the model in such a way that $V(p)=\{d; d \le a\}$ and $V(q)=\{e ; e \le b \}$. It holds that $a+b \vDash p \vee q$, so $c \vDash K(p \vee q)$. We want to show that $c \nvDash Kp \vee Kq$. For the sake of contradiction, assume that $c \vDash Kp \vee Kq$. Then there are d, e such that $d \vDash Kp$, $e \vDash Kq$, and $c \le d + e$. But then there are $d' \in S(d)$ and $e' \in S(e)$ such that $d' \vDash p$ and $e' \vDash q$. It follows that aSd and bSe, which is in contradiction with our assumption.

Proof of Theorem 7: This result is an extension of Theorem 2. We have to verify that $\mathcal {D}^{i}$ satisfies the conditions (a)–(e) from Definition 8 and the condition from Definition 11. The conditions (a)–(c) are quite straightforward. To prove (e), we have to show that if $uS^{i}v$, then $v \subseteq u$. So, assume that $uS^{i}v$ and $x \in v$. Then there is $y \in u$ such that ySx. But then $y \le x$, and so $x \in u$, which is what we wanted to show. Now we will show that $\mathcal {D}^{i}$ is source distributive. Assume $(u \cup v)S^{i}w$. Let us define

$$\begin{aligned} u' = \{ y; \text { there is }x \in u\text { such that }xSy\}, \\ v' = \{ y; \text { there is }x \in v\text { such that }xSy\}. \end{aligned}$$

Both $u'$ and $v'$ are upward closed. Moreover, $uS^{i}u'$, $vS^{i}v'$, and $w \subseteq u' \cup v'$.

Proof of Theorem 8: This result is an extension of Theorem 3. So, we are proving that for every upward closed set of states u it holds that $u \vDash \alpha$ in $\mathcal {D}^{i}$ iff for all $x \in u$, $x \Vdash \alpha$ in $\mathcal {D}$, and we have to show the inductive step for K. Assume that the statement holds for some $\mathcal {L}_K$-formula $\beta$. First, assume that $u \vDash K\beta$ in $\mathcal {D}^{i}$. So, there is $v \in S^{i}(u)$ such that $v \vDash \beta$. Then for every $x \in u$ there is some $y \in v$ such that ySx, and for every $y \in v$, $y \Vdash \beta$. Therefore, for every $x \in u$ there is y such that ySx, and $y \Vdash \beta$, that is, for every $x \in u$, $x \Vdash K \beta$.

Second, suppose for every $x \in u$, $x \Vdash K \beta$, that is, for every $x \in u$ there is $x'$ such that $x'Sx$, and $x' \Vdash \beta$. Define $v = \{z ;$ there is $x \in u$ such that $x' \le z$}. Obviously, v is upward closed, and for all $z \in v$, $z \Vdash \beta$. Moreover, for every $x \in u$ there is $z \in v$ (namely $x'$) such that zSx. It follows that there is v such that $vS^{i}u$ and $v \vDash \beta$. Therefore, $u \vDash K \beta$ in $\mathcal {D}^{i}$.