nLab
the logic T(m)

Idea

The flavor of modal logic called $T$ is propositional logic equipped with a single modality usually written ”$\square$” subject to the rules that for all propositions $p,q:\mathrm{Prop}$ we have

1. $\square K:\square (p\to q)\to (\square p\to \square q)$ (K modal logic)

2. $\square T:\square p\to p$ (T modal logic)

Often one considers adding one more axiom:

• $\square 4:\square p\to \square \square p$. (S4 modal logic).

In some applications especially in Artificial Intelligence, $\square p$ 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 multiagent system where there may be $m$ different ‘agents’. In that setting, ${\square }_{i}p$ 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 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 $\varphi$ and also that agent knows $\varphi \to \psi$, 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, … each has its own history).

The axioms $T_i$

This is found in two equivalent forms

• $p\to M_{i}p$

and

• $K_{i}p\to p$.

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

The logics $T$ and $T_{(m)}$

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

Semantics

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

Proposition

$𝔉\vDash T$ if and only if $𝔉\vDash \forall w\in W,Rww$.

So the frames that support models for the logic $T$ are exactly the reflexive frames.

Proof

Suppose $𝔉$ is a reflexive frame and take an arbitrary valuation $V$ on $𝔉$ and a state $w$ in $𝔉$ so that $(𝔉,V),w\vDash p$. We use the first form of $T$ above, and have to prove that $Mp$ holds at $w$, i.e., that $p$ holds at some state ‘accessible’ from $w$, but as $R$ is reflexive, $w$ is accessible from itself, … .

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 $Mp$ does not. (Recall the semantics of $M$: $𝔐,w\vDash M\varphi$ if and only if, for some $v\in W$ such that $Rwv$, $𝔐,v\vDash \varphi$.)

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) $\{x\in W\mid Rwx\}\cap V(p)=\varnothing$. 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, $\neg (w\vDash M\varphi )$, so suppose that $v$ is any $R$-successor of $w$, i.e., $Rwv$, then $w\ne v$, so $\neg (v\vDash p)$, hence $\neg (w\vDash Mp)$ as required.$\blacksquare$