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\begin{document}

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\section*{the logic T(m)}

\hypertarget{the_epistemic_logics__and_}{}\section*{{The epistemic logics $T$ and $T_{(m)}$}}\label{the_epistemic_logics__and_}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{the_axioms_}{The axioms $T_i$}\dotfill \pageref*{the_axioms_} \linebreak
\noindent\hyperlink{the_logics__and_}{The logics $T$ and $T_{(m)}$}\dotfill \pageref*{the_logics__and_} \linebreak
\noindent\hyperlink{semantics}{Semantics}\dotfill \pageref*{semantics} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

The flavor of \emph{[[modal logic]]} called $T$ is [[propositional logic]] equipped with a single [[modality]] usually written `'$\Box$`` subject to the rules that for all [[propositions]] $p, q \colon Prop$ we have

\begin{enumerate}%
\item $\Box K \colon \Box(p \to q) \to (\Box p \to \Box q)$ ([[K modal logic]])


\item $\Box T \colon \Box p \to p$ (T modal logic)



\end{enumerate}
Often one considers adding one more axiom:

\begin{itemize}%
\item $\Box 4 \colon \Box p \to \Box \Box p$. ([[S4 modal logic]]).

\end{itemize}
In some applications especially in Artificial Intelligence, $\Box p$ is interpreted as representing a statement that some (fixed) `agent' knows the proposition $p$. It is then a short step to handling the idea of knowledge in a [[knowledge in a multiagent system|multiagent system]] where there may be $m$ different `agents'. In that setting, $\Box_i p$ interprets as `agent $i$ knows that $p$'. This leads to modelling of the passage of the interchange of information between neighbours, e.g. we might have three agents and the proposition that `agent 2 knows that agent 3 does not know $p$'.

The [[the logic K(m)|basic epistemic logics]], $K$ and $K_{(m)}$, do not reflect much of our intuition of `knowledge'. The $K$-axiom merely says that, if an agent knows $\phi$ and also that agent knows $\phi\to \psi$, then the agent knows $\psi$. There are a series of additional axioms proposed as being appropriate for knowledge, (although, it seems, each has their supporters and detractors!) These are called $T$, $(4)$, and $B$ (and please don't ask why, \ldots{} each has its own history).

\hypertarget{the_axioms_}{}\subsection*{{The axioms $T_i$}}\label{the_axioms_}

This is found in two equivalent forms

\begin{itemize}%
\item $p\to M_i p$

\end{itemize}
and

\begin{itemize}%
\item $K_i p\to p$.

\end{itemize}
The first interprets as \emph{if $p$ is true, then agent $i$ considers it possible} and the second as \emph{atomic statements known by agent $i$ are true} .

\hypertarget{the_logics__and_}{}\subsection*{{The logics $T$ and $T_{(m)}$}}\label{the_logics__and_}

These logics are generated by $K$ (resp. $K_{(m)}$) and the axiom $T$, (resp. axioms $T_i$ for each $i = 1,\ldots m$.

\hypertarget{semantics}{}\subsection*{{Semantics}}\label{semantics}

First looking in the monomodal case, suppose that we have a frame $\mathfrak{F} = (W,R)$ then

\begin{uprop}
$\mathfrak{F}\models T$ if and only if $\mathfrak{F} \models  \forall w\in W,\; R w w$.

\end{uprop}
So the frames that support models for the logic $T$ are exactly the reflexive frames.

\begin{uproof}
Suppose $\mathfrak{F}$ is a reflexive frame and take an arbitrary valuation $V$ on $\mathfrak{F}$ and a state $w$ in $\mathfrak{F}$ so that $(\mathfrak{F},V),w\models p$. We use the first form of $T$ above, and have to prove that $M p$ holds at $w$, i.e., that $p$ holds at some state `accessible' from $w$, but as $R$ is reflexive, $w$ is accessible from itself, \ldots{} .

For the converse, we will suppose $R$ is not reflexive, so there is some state, $w \in W$ which is not $R$-related to itself. We will falsify the formula $T$ if we can find a valuation $V$ and a state $v$ such that $p$ holds at $v$ but $M p$ does not. (Recall the semantics of $M$: $\mathfrak{M},w \models M \phi$ if and only if, for some $v \in W$ such that $R w v$, $\mathfrak{M},v \models \phi$.)

We need a state with this property and we only know about one namely $w$, so that is the obvious to try! We need a valuation such that 1) $w\in V(p)$ and 2) $\{x\in W\mid R w x\}\cap V(p)= \empty$. If we set $V(p) = \{w\}$ this works since $w$ is not related to itself. (Other values of $V$ are irrelevant.) If $w$ has no $R$-successors, then we are finished since clearly in that case, $\neg(w \models M \phi)$, so suppose that $v$ is any $R$-successor of $w$, i.e., $R w v$, then $w\neq v$, so $\neg(v\models p)$, hence $\neg(w\models M p)$ as required.$\blacksquare$

\end{uproof}
[[!redirects the logic T(m)]] [[!redirects The logic T(m)]] [[!redirects T(m)]]

[[!redirects T modal logic]]



\end{document}