I have been reading about epistemic logic and there is something that I can´t understand very well. Some texts says that an aproach to the epistemic logic is to view it as a modal logic and interpret the necessity operator (◻) as the epistemic operator (K). My question is, If I interchange the operators in a modal logic system to epistemic operators, i.e., ◻ for K and ◊ for B (belief), will that be a system of epistemic logic itself?

Thanks in advance for your help


1 Answer 1


Epistemic logic (as well as doxastic logic) is multimodal, in the sense that there is a modality of each type for every agent. Let us then extend the box and diamond notation with an index.

If $\Box_i p$ stands for "$i$ knows $p$," (epistemic interpretation) then $\Diamond_i p$ stands for "$p$ is compatible with the sum of $i\,$'s knowledge," or "$i$ considers $p$ possible," rather than "$i$ believes $p$."

If, on the other hand, $\Box_i p$ stands for "$i$ believes $p$," (doxastic interpretation) then $\Diamond_i p$ stands for "$p$ is compatible with the sum of $i\,$'s beliefs."

Giving $\Box$ an epistemic interpretation, while at the same time giving $\Diamond$ a doxastic interpretation, has the drawback that it is no longer the case that $\Box_i p$ if and only if $\neg \Diamond_i \neg p$.

One the one hand, it is useful to relate various modal logics by identifying the "boxes and diamonds" of each logic. For each "box" one can define a "diamond" via $$\Diamond p \text{ if and only if } \neg \Box \neg p,$$ though the "diamond" modality may not be as heavily used in some logics. See, for instance the summary table at the beginning of this article, where no "diamond" is reported for doxastic logic. (Of course, it can be defined, as in Fitting and Mendelsohn, First-Order Modal Logic, where $\Box_i$ is $\mathcal{B}_i$ and $\Diamond_i$ is $\mathcal{C}_i$.)

On the other hand, when dealing with more than one type of modality, it may be inconvenient not to call them by their proper names. First, you may have more than two modalities, as the "until" of temporal logic or the "it is common knowledge that" or "every agent in this group knows that" of epistemic logic. (See Fagin et al., Reasoning about Knowledge.)

Second, when allowing fundamentally different modalities to coexist, as when combining "knows" and "believes", one "box" is not necessarily definable in terms of the other. In fact, the standard approach to defining the semantics of a logic of knowledge and belief is to have a Kripke structure with two accessibility relations. (See, for instance, Meyer, Modal Epistemic and Doxastic Logic in Handbook of Philosophical Logic.)

The two accessibility relations are not fully independent. For instance, it is reasonable to assume $\mathcal{K}_i p \rightarrow \mathcal{B}_i p$. On the other hand, in most epistemic logics $\mathcal{K_i} \mathcal{K}_j p$ entails $\mathcal{K}_i p$, but in no reasonable doxastic logic does $\mathcal{B_i} \mathcal{B}_j p$ entail $\mathcal{B}_i p$.

  • $\begingroup$ Does the validity of $\Box_i P \iff \neg \Diamond_i \neg p$ depends on the interpretation of the doxastic operator? if is not the case, a valid epistemic logic consider the K operator and its dual and the B operator and its dual? $\endgroup$ May 10, 2017 at 23:37
  • $\begingroup$ The validity of that equivalence is typically established by defining $\Diamond$ in terms of $\Box$. You could define a modal logic in which you have both epistemic and doxastic modalities. You'd throw in some axioms that connect the two, as in "If $i$ knows $p$, then $i$ believes $p$. $\endgroup$ May 11, 2017 at 0:07
  • $\begingroup$ If I understood correctly, then we could conclude that it is possible to define $\Box_i P \iff \neg \Diamond_i \neg p$ and $ \Diamond_i p \iff \neg \Box_i \neg p$ to construct some kind of modal logic, but we must bear in mind that we have to modify some aspects in order to ensure the "functionality" of that system of modal logic. Is that correct? $\endgroup$ May 11, 2017 at 3:04
  • $\begingroup$ In a vague sense, yes. I've expanded my answer to make things a bit more precise and to add a few references if you want to delve deeper into the subject. $\endgroup$ May 11, 2017 at 16:24
  • $\begingroup$ Oh, now I can understand much better the topic, thanks. $\endgroup$ May 11, 2017 at 19:47

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