# Show that the canonical modal for the modal logic s4.3 has no branching to the right

$$S4.3 = S4 + \Box(\Box p \to q) \lor \Box(\Box q \to p)$$

We may use that the canonical modal of S4.3 is reflexive and transitive.

A reflexive frame has no branching to the right if $$\forall x \forall y \forall z ((Rxy \land Rxz) \to (Ryx \lor Rzy))$$

In the canonical modal a relation is defined as: RAB iff $$\Box \phi \in A \to \phi \in B, \phi \in B \to \Diamond \phi \in A.$$

For reflexive we have the formulas: $p \to \Diamond p$ and $\Box p \to p$. For transitive: $\Diamond \Diamond p \to \Diamond p$ and $\Box p \to \Box \Box p$.

You would first have to assume that for $M = (W, R, V)$,the canonical modal of S4.3, that for $\Gamma, \Delta, E \in W: R\Gamma\Delta \land R\Gamma E$ from where you would want to derive $R\Delta E \lor RE\Delta$.

I thought that a good start would be assuming $\Gamma \vdash_{S4.3} \Box(\Box \phi \to \psi)$. But this leads me nowhere so far. Could someone help?

## 1 Answer

Sorry for the late reply. The property $\forall x \forall y \forall z ((Rxy \land Rxz) \to (Ryx \lor Rzy))$ says that whenever there are transition from $x$ to $y$ and $z$, there is either transition from $y$ to $z$, or vice versa. So, let us assume, that in a canonical (which is reflexive and transitive) model $M$ $(w,v) \in R$ and $(w,u) \in R$, but $(u,v) \not \in R$ and $(v,u) \not \in R$. Since $(u,v) \not \in R$, then, according to the definition of $R$ for canonical models you provided, there is some $\varphi$ such that $M,v \models \square \varphi$ but $M,u \not \models \varphi$. By the same reasoning, for some $\psi$: $M,u \models \square \psi$ but $M,v \not \models \psi$. In other words, since states in the model are maximal consistent sets, $M,v \models \neg (\square \varphi \rightarrow \psi)$ and $M,u \models \neg (\square \psi \rightarrow \varphi)$. Since $(w,v) \in R$ and $(w,u) \in R$, then, by definition of $R$, $M,w \models \diamond \neg (\square \varphi \rightarrow \psi)$ and $M,w \models \diamond \neg (\square \psi \rightarrow \varphi)$. By closure, we have $M,w \models \diamond \neg (\square \varphi \rightarrow \psi) \wedge \diamond \neg (\square \psi \rightarrow \varphi)$, which is $M,w \models \neg [\square (\square \varphi \rightarrow \psi) \vee \square (\square \psi \rightarrow \varphi)]$ by DeMorgan. Contradiction, thus $(v,u) \in R$ or $(u,v) \in R$.