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In my modal logic book it's written that, for each frame $F(S,R)$ the accessibility relation $R$ is reflexive IF AND ONLY IF the scheme T:$\square A \implies A$ is valid in $F$.

Even if I can easily prove that reflexivity $\implies$ T, I can't prove that T $\implies$ reflexivity.

As a counterexample I show a model $M(S,R,V)$ where $S=\{\alpha,\beta\}$, $R=\{(\alpha,\beta),(\beta,\alpha)\}$ and $V(A)=\{\alpha,\beta\}$. In this model T is true for both $\alpha$ and $\beta$, but R is not reflexive. What am I doing wrong? what is the right way to demonstrate T $\implies$ reflexivity?

Giacomo Tagliabue
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1 Answers1

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In your model $T$ is true, but in your frame it is not valid. A scheme is valid if it true in the frame for every possible valuation.

The standard proof is via the contrapositive. Assume that $F(S,R)$ is not reflexive, i.e. there is some node $s\in S$ such that $\lnot sRs$. Take the valuation $V(p)=S\setminus\{s\}$. Then $(S,R,V),s\Vdash\square p$ but of course $(S,R,V),s\nVdash p$. Therefore $(S,R,V)\nvDash\square p\to p$ and thus $(S,R)\nvDash\square p\to p$.

Apostolos
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