# nLab the logic S5(m)

The epistemic logics and

# The epistemic logics $S5$ and $S5_{(m)}$

## Idea

(Note that $S5 = S5_{(1)}$. See S5 modal logic)

Although better than $K$ (resp. $K(m)$), and $T$ (resp. $T(m)$), $S4$ (resp. $S4(m)$) is still not considered adequate for knowledge representation. Because of this further axioms have been put forward, too many to be mentioned here. We will limit ourselves, (at this stage in the development of these entries, at least) to introducing one more axiom, that is a bit more contentious, (but is nice from the nPOV.)

## The axioms $B_i$

The axiom denoted $B_i$, (and again refer to Kracht for some indication of possible reasons) is

• $p \to K_i M_i p$.

The interpretation is the ‘if $p$ is true then agent $i$ knows that it is possible that $p$ is true.’

Another axiom that is relevant here is:

## The axioms $(5)_i$

• $\neg K_i p \to K_i \neg K_i p$

Axiom (5) interprets as saying ‘If agent $i$ does not know that $p$ holds, then (s)he knows that (s)he does not know’. This is termed ‘negative introspection’.

## The logics $S5$ and $S5_{(m)}$

• $S5$ - this logic is obtained from $S4$ by adding the axiom $B$.

• $S5_{(m)}$ starting from $S4_{(m)}$ the logic S4(m), add, for each $i = 1,\ldots, m$ the axiom $B_i$.

In either case the same result can be obtained by adding in $5$ or $(5)_i$ in place of the corresponding $B$.

###### Critique

This is highly doubtful as a property of knowledge when applied to human beings. If an agent is ignorant of the truth of an assertion, it is very often unlikely that (s)he knows this ignorance.

## Equivalence frames

With $T_{(m)}$, the models corresponded to frames where each relation $R_i$ was reflexive. With $S4_{(m)}$, the frames needed to be transitive as well. Here we consider the class $\mathcal{S}5(m)$ of models with frames, where each $R_i$ is an equivalence relation. These are sometimes called equivalence frames.

###### Theorem (Soundness of $S5_{(m)}$)

$S5_{(m)}\vdash \phi \Rightarrow \mathcal{S}5(m)\models \phi.$

###### Proof

(We show this for $m = 1$.) We have already shown (here in the logic S4(m)) that the older axioms $T$ and $(4)$ hold so it remains to show if we have a frame, $\mathfrak{M}= ((W,R),V)$, where $R$ is an equivalence relation on $W$ then $\mathfrak{M}\models B$.

We suppose the we have a state $w$ so that $\mathfrak{W},w \models p$. Now we need to find out if $\mathfrak{W},w \models K M p$, so we note that

1. $\mathfrak{W},w \models K M p$ if and only if $\forall u$ with $R w u$, \$$\mathfrak{W},u \models M p$, but

2. that holds if $\forall u$ with $R w u$, there is some $v$ with $R u v$ and $\mathfrak{W},v \models p$.

However whatever $u$ we have with $R w u$, we have $R u w$ as $R$ is symmetric, and we know, by assumption, that $\mathfrak{W},w \models p$, so we have what we need.

## References

More on $S5$, $S5_{(m)}$ and their applications in Artificial Intelligence can be found in

• J.- J. Ch. Meyer and W. Van der Hoek, Epistemic logic for AI and Computer Science, Cambridge Tracts in Theoretical Computer Science, vol. 41, 1995.

General books on modal logics which treat these logics thoroughly in the general context include e

• [[Patrick Blackburn]], M. de Rijke and [[Yde Venema]], Modal Logic, Cambridge Tracts in Theoretical Computer Science, vol. 53, 2001.

• [[Marcus Kracht]], Tools and Techniques in Modal Logic, Studies in Logic and the Foundation of Mathematics, 142, Elsevier, 1999.

[[!redirects the logic S5(m)]] [[!redirects S5(m)]]

Last revised on July 21, 2022 at 20:55:25. See the history of this page for a list of all contributions to it.