The Logic of Observation and Belief Revision in Scientific Communities

Abstract

Scientists collect evidence in order to confirm or falsify scientific theories. Unfortunately, scientific evidence may sometimes be false or deceiving and as a consequence lead individuals to believe in a false theory. By interaction between scientists, such false beliefs may spread through the entire community. There is currently a debate about the effect of various network configurations on the epistemic reliability of scientific communities (e.g. Zollman 2010 and Rosenstock et al. 2017). To contribute to this debate from a logical perspective, this paper introduces an epistemic logical framework of observation, interaction and belief revision in scientific communities. The presented sound and complete system provides the formal tools for qualitative analysis of the social dynamics of scientific inquiry. Furthermore, this paper includes detailed suggestions for future applications of the framework.

Appendices

Appendices

1. Multisets

Here, some definitions on multisets are presented. This may be helpful to understand the nature of observation sets $O_i$.

Definition 1

(Multiset) A multisetA is a pair $\langle {\mathscr {A}},f\rangle$ where ${\mathscr {A}}$ is a regular set and $f:{\mathscr {A}}\rightarrow {\mathbb {N}}$ a function, giving the multiplicity of each $a\in {\mathscr {A}}$. That is, f will tell how often a is included in the multiset. Unlike for regular sets, in multisets it holds that $\{a,a\}\ne \{a\}$.

Definition 2

(Sub-multiset) Suppose $A=\langle {\mathscr {A}},f\rangle$ and $B=\langle {\mathscr {B}},g\rangle$ are multisets. A is a sub-multiset of B, $A\subseteq B$, if for all $a\in {\mathscr {A}}$ it holds that $f(a)\le g(a)$ (and < for proper sub-multiset $A\subset B$).

Definition 3

(Multi-powerset) Suppose $A=\langle {\mathscr {A}},f\rangle$ is a multiset. The multi-powerset of A, ${\mathcal {P}}(\langle {\mathscr {A}},f\rangle )$, is defined by taking all sub-multiplicities.

Consider as an example multiset $A=\{a,b,b\}$. Its power set is given by ${\{\emptyset ,\{a\},\{b\},\{a,b\},\{a,b\},\{b,b\},\{a,b,b\}\}}$.

Definition 4

(Support) Suppose $A=\langle {\mathscr {A}},f\rangle$ is a multiset. The subset ${\mathbf {A}}$ of ${\mathscr {A}}$ is called the support of A if $a\in {\mathbf {A}}$ iff $f(a)>0$.

Definition 5

(Positive multiset) Suppose $A=\langle {\mathscr {A}},f\rangle$ is a multiset. The positive multiset$A^{+}$ of A is defined as $\langle {\mathbf {A}},f^{+}\rangle$ such that ${\mathbf {A}}$ is the support of A and $f^{+}(a){:=}f(a)$.

For construction of the canonical model in the completeness proof of ${\mathsf {MEOL}}^s$ (see Appendix 2), define the union of two multisets:

Definition 6

(Union multisets) Suppose ${\mathbf {A}}=\langle A,f\rangle$ and ${\mathbf {B}}=\langle A,g\rangle$ are multisets. Their union, ${\mathbf {A}}\cup {\mathbf {B}}$, is ${\mathbf {C}}=\langle A,h\rangle$ where for all $a\in A$ it holds that $h(a)=\mathsf {max}(f(a),g(a))$.

2. Proofs

Theorem 1

(Soundness of ${\mathsf {MEOL}}^s$) ${\mathsf {MEOL}}^s$ is sound: for all$\varphi \in {\mathscr {F}}^s$ and all$\varPhi \subseteq {\mathscr {F}}^s$ it holds that

$$\begin{aligned} \hbox {if }\varPhi \vdash \varphi \hbox { then }\varPhi \models \varphi \end{aligned}$$

Proof

We prove soundness by induction on the length k of derivation. For the base case, $\varphi$ must be an axiom of ${\mathsf {MEOL}}^s$. As most axioms of ${\mathsf {MEOL}}^s$ (Definition 9) match with the semantic conditions (Definition 4), it follows directly for those axioms that $\models \varphi$. Additionally, Multiset follows semantically by Property 2 and Neighborhood holds semantically by truth definition of $N_ij$ (Definition 7). For the induction step, suppose that for $\varphi$ derived at step k, it holds that if $\varPhi \vdash \varphi$ then $\varPhi \models \varphi$. Consider $\gamma$ that is derived by applying one of the rules at the $k+1$-step. For each rule it follows by the induction hypothesis and the truth definitions of $\rightarrow$, $\square _i$ and $K_i$ (Definition 7) that respectively $\models \gamma$, $\models \square _i\varphi$ and $\models K_i\varphi$.□

Theorem 2

(Completeness of ${\mathsf {MEOL}}^s$) ${\mathsf {MEOL}}^s$ is strongly complete. That is, for all$\varphi \in {\mathscr {F}}^s$ and all$\varPhi \subseteq {\mathscr {F}}^s$ it holds that

$$\begin{aligned} \hbox {if } \varPhi \models \varphi \hbox { then } \varPhi \vdash \varphi \end{aligned}$$

Proof

We prove by way of a canonical model construction. Given sets of atomic propostions $\varPhi$, agents ${\mathcal {A}}$, and a limit L, define the canonical model $\varOmega {:=}({\mathcal {W}}^{\varOmega },{\mathcal {V}},O_i,\le _i,\sim _i)$ by setting

$$\begin{aligned} W^{\varOmega }&{:=}\{\varGamma \subseteq {\mathscr {F}}^s\mid \varGamma \text { is maximal consistent}\}\\ {\mathcal {V}}(\varGamma )&{:=}\{p\in \varPhi \mid p\in \varGamma \}\\ O_i(\varGamma )&{:=}\bigcup _{O_i(o_{\psi },n)\in \varGamma }(o_{\psi },n)\\ \varGamma \le _i\varDelta&\text { iff }\{\varphi \mid \square _i\varphi \in \varGamma \}\subseteq \varDelta \\ \varGamma \sim _i\varDelta&\text { iff }\{\varphi \mid K_i\varphi \in \varGamma \}\subseteq \varDelta \end{aligned}$$

We will first prove that $\varOmega$ is a model for ${\mathsf {MEOL}}^s$, by showing that it satisfies all semantic properties. We use standard methods to show that $\le _i$ is a preorder and that $\sim _i$ is an equivalence relation.

To prove indefeasibility we show $\varGamma \le _i\varDelta \rightarrow \varGamma \sim _i\varDelta$ for all $i\in {\mathcal {A}}$. Suppose not. Then there exists $i,\varGamma ,\varDelta$ such that $\varGamma \le _i\varDelta$ while $\varGamma \not \sim _i\varDelta$. This implies that there is a $\varphi$ such that $K_i\varphi \in \varGamma$ while $\varphi \not \in \varDelta$. By the axiom Indefeasibility, it follows that $\square _i\varphi \in \varGamma$. Since $\varGamma \le _i\varDelta$ this implies $\varphi \in \varDelta$. By contradiction we conclude that indefeasibility holds.

To prove local connectedness we show $\varGamma \sim _i\varDelta \rightarrow (\varGamma \le _i\varDelta \vee \varDelta \le _i\varGamma )$ for all $i\in {\mathcal {A}}$. Suppose not. Then there exists $i,\varGamma ,\varDelta$ such that $\varGamma \sim _i\varDelta$ while $\varGamma \not \le _i\varDelta$ and $\varDelta \not \le _i\varGamma$. This implies that there are $\gamma ,\varphi$ such that $\square _i\gamma \in \varGamma$$(*)$ while $\gamma \not \in \varDelta$, and $\square _i \varphi \in \varDelta$ while $\varphi \not \in \varGamma$. As $\gamma \not \in \varDelta$: $\lnot \gamma \in \varDelta$ and hence $\lnot K_i\varphi \in \varGamma$. As $\varphi \not \in \varGamma$: $\lnot \varphi \in \varGamma$ and hence $\lnot K_i\gamma \in \varGamma$ (because $\sim _i$ is reflexive). Hence, $\lnot (K_i\varphi \vee K_i\gamma )\in \varGamma$. By axiom Local Connectedness, it follows that either i) $\lnot K_i(\varphi \vee \square _i\gamma )\in \varGamma$ or ii) $\lnot K_i(\gamma \vee \square _i\varphi )\in \varGamma$. In case i), then $K_i(\lnot \varphi \wedge \lnot \square _i\gamma )\in \varGamma$ which implies $\lnot \square _i\gamma \in \varGamma$. This contradicts to $(*)$. In case ii), $K_i(\lnot \gamma \wedge \lnot \square _i\varphi )\in \varGamma$ which implies $\lnot \gamma \in \varGamma$. So $\lnot \square _i\gamma \in \varGamma$ (because $\le _i$ is reflexive). This contradicts to $(*)$. Thus, local connectedness holds in the model.

Now we will prove observation introspection: $\varGamma \sim _i\varDelta \rightarrow O_i(\varGamma )=O_i(\varDelta )$ for all $i\in {\mathcal {A}}$. Suppose not, then there exists $i,\varGamma ,\varDelta$ such that $\varGamma \sim _i\varDelta$ although there is a pair $(o_{\psi },n)$ such that either $(o_{\psi },n)\subseteq O_i(\varGamma )$ but $(o_{\psi },n)\not \subseteq O_i(\varDelta )$, or $(o_{\psi },n)\not \subseteq O_i(\varGamma )$ and ${(o_{\psi },n)\subseteq O_i(\varDelta )}$. Suppose the former holds. Then $O_i(o_{\psi },n)\in \varGamma$. By axiom Observation Introspection: $K_iO_i(o_{\psi },n)\in \varGamma$. Since $\varGamma \sim _i\varDelta$ it follows that $O_i(o_{\psi },n)\in \varDelta$ and thus $(o_{\psi },n)\subseteq O_i(\varDelta )$. By contradiction and similar argument for the other case, we conclude that observation introspection holds.

To finish the proof, we need to show that the Truth Lemma holds: for each $\varGamma \in W^{\varOmega }$ and each $\varphi \in {\mathscr {F}}$, we have $\varphi \in \varGamma$ iff $\varGamma \models _{\varOmega }\varphi$. We prove this by induction on the construction of $\varphi$. For the base cases, i) $\varphi =\top$ or ii) $\varphi =p$. Since $\varGamma$ is a maximal consistent set, $\top \in \varGamma$ is always true, so by truth definition, it follows that $\varGamma \models _{\varOmega }\top$. Likewise, $p\in \varGamma$ iff $p\in {\mathcal {V}}(\varGamma )$ iff $\varGamma \models _{\varOmega }p$. Then for the induction steps, assume the induction hypothesis: for all $\gamma ,\chi$ of less complexity than $\varphi$ the Truth Lemma holds. For case i) $\varphi =\lnot \gamma$ it holds that $\lnot \gamma \in \varGamma$ iff $\gamma \not \in \varGamma$ iff $\varGamma \nvDash _{\varOmega }\gamma$ iff $\varGamma \models _{\varOmega }\lnot \gamma$. For case ii) $\varphi =\gamma \wedge \chi$, $\gamma \wedge \chi \in \varGamma$ iff $\gamma \in \varGamma$ and $\chi \in \varGamma$ iff $\varGamma \models _{\varOmega }\gamma$ and $\varGamma \models _{\varOmega }\chi$ iff $\varGamma \models \gamma \wedge \chi$. For case iii) $\varphi =O_i(o_{\psi },n)$ it holds that $O_i(o_{\psi },n)\in \varGamma$ iff $(o_{\psi },n)\subseteq O_i(\varGamma )$ iff $\varGamma \models _{\varOmega } O_i(o_{\psi },n)$. We continue with case iv) $\varphi =N_ij$: $N_ij\in \varGamma$ iff $\varGamma \vdash N_ij$ iff $N_ij$ (by Neighborhood axiom) iff $\varGamma \models _{\varOmega } N_ij$. For case v) $\varphi =\square _i\gamma$, $\square _i\gamma \in \varGamma$ iff for all $\varDelta \ge _i\varGamma$ it holds that $\gamma \in \varDelta$ (by definition of canonical model) iff $\varDelta \models _{\varOmega }\gamma$ for all $\varDelta \ge _i\varGamma$ iff $\varGamma \models _{\varOmega }\square _i\gamma$ (by truth definition). Finally for case vii) $\varphi =K_i\gamma$, $K_i\gamma \in \varGamma$ iff $\gamma \in \varGamma$ for all $\varDelta \sim _i\varGamma$ iff $\varDelta \models _{\varOmega }\gamma$ for all $\varDelta \sim _i\varGamma$ iff $\varGamma \models _{\varOmega } K_i\gamma$.□

Theorem 3

(Soundness and completeness of ${\mathsf {MEOL}}$) For all$\varphi \in {\mathscr {F}}$ it holds that:

$$\begin{aligned} \models \varphi \hbox { iff }\vdash \varphi \end{aligned}$$

Proof

The proof system of ${\mathsf {MEOL}}$ contains reduction axioms such that for each $\varphi \in {\mathscr {F}}$, there is a reduced $\varphi ^{\dagger }\in {\mathscr {F}}^s$ such that

$$\begin{aligned} \vdash \varphi \leftrightarrow \varphi ^{\dagger } \end{aligned}$$

We will prove that the reduction axioms hold also semantically in the models of ${\mathsf {MEOL}}$:

$$\begin{aligned} \models \varphi \leftrightarrow \varphi ^{\dagger } \end{aligned}$$

We will demonstrate the proof only for some of the more complex formulas, as the techniques are standard. We show Observation Dynamics: $\models [\alpha ,\sigma ]O_i(o_{\psi },n)\leftrightarrow {({\mathsf {pre}}_{\sigma }\rightarrow ({\mathsf {post}}_{\sigma }\rightarrow O_i(o_{\psi },n)))}$. First from left to right: assume for an arbitrary $({\mathcal {M}},w)$ that ${({\mathcal {M}},w)\models [\alpha ,\sigma ]O_i(o_{\psi },n)}$. We know that either $({\mathcal {M}},w)\nvDash {\mathsf {pre}}_{\sigma }$ or $({\mathcal {M}},w)\models {\mathsf {pre}}_{\sigma }$. Suppose the former holds, then it directly follows that $({\mathcal {M}},w)\models {\mathsf {pre}}_{\sigma }\rightarrow {({\mathsf {post}}_{\sigma }\rightarrow O_i(o_{\psi },n))}$. Now suppose the latter holds. Assume for contradiction that $({\mathcal {M}},w)\nvDash {\mathsf {pre}}_{\sigma }\rightarrow {({\mathsf {post}}_{\sigma }\rightarrow O_i(o_{\psi },n))}$. Then $({\mathcal {M}},w)\models {\mathsf {pre}}_{\sigma } \wedge {\mathsf {post}}_{\sigma } \wedge \lnot O_i(o_{\psi },n))$ ($\star$). We know by truth definition of ${\mathsf {MEOL}}$ that since $({\mathcal {M}},w)\models {\mathsf {pre}}_{\sigma }$ and $({\mathcal {M}},w)\models [\alpha ,\sigma ]O_i(o_{\psi },n)$, it holds that ${\mathcal {M}}\otimes \alpha ,(w,\sigma )\models O_i(o_{\psi },n)$. Hence $(o_{\psi },n)\in O_i^{\alpha }(w,\sigma )$. This implies by definition of $O_i^{\alpha }$ that either (i) $({\mathcal {M}},w)\models {\mathsf {post}}_{\sigma }\rightarrow O_i(o_{\psi },n)$, which contradicts to ($\star$), or (ii) not $({\mathcal {M}},w)\nvDash {\mathsf {post}}_{\sigma }\rightarrow O_i(o_{\psi },n)$, and $({\mathcal {M}},w)\models O_i(o_{\psi },n)$. This again contradicts to ($\star$). Hence, $({\mathcal {M}},w)\models {\mathsf {pre}}_{\sigma }\rightarrow ({\mathsf {post}}_{\sigma }\rightarrow O_i(o_{\psi },n))$. Now from right to left, by contraposition: assume that $({\mathcal {M}},w)\nvDash [\alpha ,\sigma ]O_i(o_{\psi },n)$. Then $({\mathcal {M}},w)\models {\mathsf {pre}}_{\sigma }$ and not ${\mathcal {M}}\otimes \alpha ,(w,\sigma )\models O_i(o_{\psi },n)$. This implies that $(w,\sigma )\not \in [\![O_i(o_{\psi },n)]\!]^{\alpha }$. Now assume for contradiction that $({\mathcal {M}},w)\models {\mathsf {post}}(\sigma )\rightarrow O_i(o_{\psi },n)$. By $({\mathcal {M}},w)\models {\mathsf {pre}}_{\sigma }$ and by definition of $O_i^{\alpha }$, $(o_{\psi },n)\in O_i^{\alpha }(w,\sigma )$. Contradiction. Hence, $({\mathcal {M}},w)\models \lnot ({\mathsf {post}}_{\sigma }\rightarrow O_i(o_{\psi },n)$. And therefore, $({\mathcal {M}},w)\nvDash {\mathsf {pre}}_{\sigma }\rightarrow ({\mathsf {post}}_{\sigma }\rightarrow O_i(o_{\psi },n))$.

We continue proving the Knowledge Dynamics I: $\models [\alpha ,\sigma ]\square _i\varphi \leftrightarrow ({\mathsf {pre}}_{\sigma }\rightarrow \square _i[\alpha ,\sigma ]\varphi )$ for $\alpha \in \{o_{\psi }+_i,O_i(o_{\psi },n)!_i,\lnot O_i(o_{\psi },1)!_i\}$ for every $\psi \in \Psi$. From left to right: suppose ${({\mathcal {M}},w)\models [\alpha ,\sigma ]\square _i\varphi }$. Then ${\mathcal {M}}\otimes \alpha ,(w,\sigma )\models \square _i\varphi$. This means that for all $(v,\sigma )\ge _i (w,\sigma )$: ${\mathcal {M}}\otimes \alpha ,(w,\sigma )\models \varphi$. By definition of ${\mathcal {M}}\otimes \alpha$, for all $v\ge _i w$: $({\mathcal {M}},v)\models [\alpha ,\sigma ]\varphi$. Hence $({\mathcal {M}},w)\models \square _i[\alpha ,\sigma ]\varphi$, which implies $({\mathcal {M}},w)\models {\mathsf {pre}}_{\sigma }\rightarrow \square _i[\alpha ,\sigma ]\varphi$. From right to left, suppose $({\mathcal {M}},w)\models {\mathsf {pre}}_{\sigma }\rightarrow \square _i[\alpha ,\sigma ]\varphi$. Now if $({\mathcal {M}},w)\models {\mathsf {pre}}_{\sigma }$, then $({\mathcal {M}},w)\models \square _i[\alpha ,\sigma ]\varphi$. By same steps as above, but inversed, it follows that ${\mathcal {M}}\otimes \alpha ,(w,\sigma )\models \square _i\varphi$ which implies that $({\mathcal {M}},w)\models [\alpha ,\sigma ]\square _i\varphi$. If $({\mathcal {M}},w)\nvDash {\mathsf {pre}}_{\sigma }$, then we follow the usual steps.

Next, we prove Knowledge Dynamics II: $\models [\alpha ,\sigma ]\square _i\varphi \leftrightarrow ({\mathsf {pre}}_{\sigma }\rightarrow (\square _i(\gamma \rightarrow [\alpha ,\sigma ]\varphi )\wedge \lnot \gamma \rightarrow (\square _i[\alpha ,\sigma ]\varphi \wedge K_i(\gamma \rightarrow [\alpha ,\sigma ]\varphi ))))$. First, from left to right. Suppose $({\mathcal {M}},w)\models [\alpha ,\sigma ]\square _i\varphi$ ($*$). As before, when $({\mathcal {M}},w)\nvDash {\mathsf {pre}}_{\sigma }$ then we’re done. Suppose $({\mathcal {M}},w)\models {\mathsf {pre}}_{\sigma }$ and for contradiction, suppose $({\mathcal {M}},w)\nvDash (\square _i(\gamma \rightarrow [\alpha ,\sigma ]\varphi )\wedge \lnot \gamma \rightarrow (\square _i[\alpha ,\sigma ]\varphi \wedge K_i(\gamma \rightarrow [\alpha ,\sigma ]\varphi )))$. Suppose $({\mathcal {M}},w)\nvDash \square _i(\gamma \rightarrow [\alpha ,\sigma ]\varphi )$. Then there is a v such that $w\le _i v$ and $({\mathcal {M}},v)\models \gamma$ while $({\mathcal {M}},v)\nvDash [\alpha ,\sigma ]\varphi$, thus ${\mathcal {M}}\otimes \alpha ,(v,\sigma )\nvDash \varphi$. However, by our assumption ($*$), we know that ${\mathcal {M}}\otimes \alpha ,(w,\sigma )\models \square _i\varphi$. Note that $(w,\sigma )\le _i(v,\sigma )$ (by definition of $\le _i^{\alpha }$), so ${{\mathcal {M}}\otimes \alpha ,(v,\sigma )\models \varphi }$. Contradiction. Now suppose $({\mathcal {M}},w)\nvDash \lnot \gamma \rightarrow (\square _i[\alpha ,\sigma ]\varphi \wedge {K_i(\gamma \rightarrow [\alpha ,\sigma ]\varphi ))}$. Then ${({\mathcal {M}},w)\models \lnot \gamma }$ but $({\mathcal {M}},w)\nvDash \square _i[\alpha ,\sigma ]\varphi$ or $({\mathcal {M}},w)\nvDash K_i(\gamma \rightarrow [\alpha ,\sigma ]\varphi ))$. In the first case, there must be some $v\ge _i w$ such that $({\mathcal {M}},v)\nvDash [\alpha ,\sigma ]\varphi$ and thus ${\mathcal {M}}\otimes \alpha ,(v,\sigma )\nvDash \varphi$. Note that again $(w,\sigma )\le _i(v,\sigma )$, which implies together with ($*$) that ${\mathcal {M}}\otimes \alpha ,(v,\sigma )\models \varphi$. Contradiction. In the second case, there must be a $v\sim _i w$ such that $({\mathcal {M}},v)\models \gamma$ and $({\mathcal {M}},v)\nvDash [\alpha ,\sigma ]\varphi$. Hence ${\mathcal {M}}\otimes \alpha ,(v,\sigma )\nvDash \varphi$. As $({\mathcal {M}},w)\models \lnot \gamma$ and $({\mathcal {M}},v)\models \gamma$ and $w\sim _i v$, it follows that $(w,\sigma )\le _i(v,\sigma )$. In the same way as before, this leads to a contradiction. Hence $({\mathcal {M}},w)\models (\square _i(\gamma \rightarrow [\alpha ,\sigma ]\varphi )\wedge \lnot \gamma \rightarrow (\square _i[\alpha ,\sigma ]\varphi \wedge K_i(\gamma \rightarrow [\alpha ,\sigma ]\varphi )))$. For right to left, suppose the right side of the equivalence relation holds at $({\mathcal {M}},w)$. For purpose of contradiction, suppose $({\mathcal {M}},w)\nvDash [\alpha ,\sigma ]\square _i\varphi$. This implies $({\mathcal {M}},w)\models {\mathsf {pre}}_{\sigma }$ and ${\mathcal {M}}\otimes \alpha ,(w,\sigma )\nvDash \square _i\varphi$. So there is a $(v,\sigma )$ such that $(w,\sigma )\le _i(v,\sigma )$ while ${\mathcal {M}}\otimes \alpha ,(v,\sigma )\nvDash \varphi$. By definition of $(w,\sigma )\le _i(v,\sigma )$, it follows that either i) $\gamma \not \in {\mathcal {V}}(w)$ and $\gamma \in {\mathcal {V}}(v)$ and $w\sim _i v$, or ii) $\gamma \not \in {\mathcal {V}}(w)$ and $w\sim _i v$, or iii) $\gamma \in {\mathcal {V}}(v)$ and $w\sim _i v$. In case of i), then $({\mathcal {M}},w)\models \lnot \gamma$, so $({\mathcal {M}},w)\models K_i(\gamma \rightarrow [\alpha ,\sigma ]\varphi )$. Since $w\sim _i v$ and $({\mathcal {M}},v)\models \gamma$, this implies that $({\mathcal {M}},w)\models [\alpha ,\sigma ]\varphi$ and thus that ${\mathcal {M}}\otimes \alpha ,(v,\sigma )\models \varphi$. Contradiction. In case of ii), then $w\le _i v$ and thus $({\mathcal {M}},v)\models [\alpha ,\sigma ]\varphi$. Again this implies that ${\mathcal {M}}\otimes \alpha ,(v,\sigma )\models \varphi$. Contradiction. In case of iii), it also follows that $({\mathcal {M}},v)\models [\alpha ,\sigma ]\varphi$ and hence that ${\mathcal {M}}\otimes \alpha ,(v,\sigma )\models \varphi$. So we get a contradiction once again. This proves that $({\mathcal {M}},w)\models [\alpha ,\sigma ]\square _i\varphi$ and thereby finishes the proof for Knowledge Dynamics II.

Finally, we prove Knowledge Dynamics III: $\models [\alpha ,\sigma ]K_i\varphi \leftrightarrow ({\mathsf {pre}}_{\sigma }\rightarrow K_i[\alpha ,\sigma ]\varphi )$. Left to right: suppose $({\mathcal {M}},w)\models [\alpha ,\sigma ]K_i\varphi$, so ${\mathcal {M}}\otimes \alpha ,(w,\sigma )\models K_i\varphi$. By definition of ${\mathcal {M}}[ \alpha ,\sigma ]$, this implies that for all $(v,\sigma )$ such that $(w,\sigma )\sim _i (v,\sigma )$ it holds that ${\mathcal {M}}\otimes \alpha ,(v,\sigma )\models \varphi$. Then, for all v such that $w\sim _i v$ it holds that $({\mathcal {M}},w)\models [\alpha ,\sigma ]\varphi$ and thus ${({\mathcal {M}},w)\models K_i[\alpha ,\sigma ]\varphi }$. From right to left: suppose $({\mathcal {M}},w)\models {\mathsf {pre}}_{\sigma }\rightarrow K_i[\alpha ,\sigma ]\varphi$. If $({\mathcal {M}},w)\models {\mathsf {pre}}_{\sigma }$, then ${({\mathcal {M}},w)\models K_i[\alpha ,\sigma ]\varphi }$. Thus, for all v such that $w\sim _i v$ it holds that $({\mathcal {M}},w)\models [\alpha ,\sigma ]\varphi$. So for all $(v,\sigma )$ s.t. $(w,\sigma )\sim _i (v,\sigma )$: ${\mathcal {M}}\otimes \alpha ,(v,\sigma )\models \varphi$, which implies ${\mathcal {M}}\otimes \alpha ,(w,\sigma )\models K_i\varphi$. Therefore, $({\mathcal {M}},w)\models [\alpha ,\sigma ]K_i\varphi$. If $({\mathcal {M}},w)\nvDash {\mathsf {pre}}_{\sigma }$, then the proof continues as before.

By Theorems 1 and 2, this implies soundness and completeness of ${\mathsf {MEOL}}$.□