Show that the canonical modal for the modal logic K4.3 has no branching to the right

Question

K4.3 is the normal logic axiomatized by 4 ($\Diamond \Diamond p \to \Diamond p$) and $$ \Diamond p\land\Diamond q \to\Diamond(p\land\Diamond q)\lor\Diamond(p\land q)\lor\Diamond(q\land\Diamond p)\ $$ or by 4 and $$ \Box(p\land\Box p \to q) \lor \Box(q\land\Box q \to p)\ $$

(These two forms of .3-axioms are given seperately and it remains to show they are equal. So it might be better to use only one of them in a proof.)

I have shown that the canonical modal for the modal logic S4.3 (given by $S4 + \Box(\Box p \to q) \lor \Box(\Box q \to p)$) has no branching to the right (the proof is similar to the one in Show that the canonical modal for the modal logic s4.3 has no branching to the right).

However I find it difficult to deal with the precondition $u \neq v$ , where w, u and v are points in the canonical model for K4.3 such that $$ Rwu, Rwv, \lnot Ruv, \lnot Rvu, u \neq v. $$ Since $\lnot Ruv$, there is $\varphi$ such that $ M,u \models \Box \varphi$ but $M,v \not \models \varphi$. Since $\lnot Rvu$, there is $\psi$ such that $ M,v \models \Box \psi$ but $M,u \not \models \psi$. And since $u \neq v$, there is $\chi$ such that $\chi\in u$ but $\lnot\chi \in v$.

It seems that I should consider three formulas $\varphi$, $\psi$ and $\chi$ rather than two. I've tried to use them to construct other two fprmulas, but found it difficult.

This question is in Blackburn, Patrick, Maarten de Rijke, and Yde Venema, 2001, Modal Logic, Cambridge: Cambridge University Press, p.211, Exercise 4.3.3.

Thank you so much! ^_^

Answer 1

The answer of @blueberry_pie is correct. And I do the proof using $\Box (p\wedge \Box p\to q)\vee \Box (q\wedge \Box q\to p)\notin w$ as (.3').

Define $\mathbf{K4.3'}=\mathbf{K4}+(.3')$. Let $(W^{\mathbf{K4.3'}},R^{\mathbf{K4.3'}},V^{\mathbf{K4.3'}})$ be the canonical model for $\mathbf{K4.3'}$. Assume $R^{\mathbf{K4.3'}}ww',R^{\mathbf{K4.3'}}ww''$, and $w'\neq w''$, not $R^{\mathbf{K4.3'}}w'w''$, not $R^{\mathbf{K4.3'}}w''w'$. We want to get a contradiction. As not $R^{\mathbf{K4.3'}}w'w''$, there exists $\psi $ such that $\Box \psi \in w'$ and $\psi \notin w''$(then $\neg \psi \in w''$). Similaely, there exists $\phi $ such that $\Box \phi \in w''$ and $\phi \notin w'$(then $\neg \phi \in w'$). As $w'\neq w''$, there exists $\chi $ such that $\chi \in w'$ and $\chi \notin w''$. Then $\neg \chi \in w''$. Note $\Box \gamma \vdash_{\mathbf{K}}\Box (\gamma \vee \pi )$. We have $((\chi \vee \psi )\wedge \Box (\chi \vee \psi )\to (\neg \chi \vee \phi ))\notin w'$, and $((\neg \chi \vee \phi )\wedge \Box (\neg \chi \vee \phi )\to (\chi \vee \psi ))\notin w''$. Then $\Diamond \neg ((\chi \vee \psi )\wedge \Box (\chi \vee \psi )\to (\neg \chi \vee \phi ))\wedge \Diamond \neg ((\neg \chi \vee \phi )\wedge \Box (\neg \chi \vee \phi )\to (\chi \vee \psi ))\in w$. Let $\alpha =(\chi \vee \psi ), \beta =(\neg \chi \vee \phi )$. Then $\Box (\alpha \wedge \Box \alpha \to \beta )\vee \Box (\beta \wedge \Box \beta \to \alpha )\notin w$. Contradictory!

Answer 2

The proof works similarly as in the case of K4.3: Since $M, u \models \square \phi \wedge \neg\psi$, it follows that $M, w \models \Diamond(\square\phi \wedge \neg \psi)$. Analogously, we have $M,w \models \Diamond(\square\psi \wedge \neg \phi)$. So $w$ satisfies the consequent of the instance of the .3-axiom, which results from the substitution that maps $p$ to $(\square \phi \wedge \neg \psi)$ and $q$ to $\square \psi \wedge \neg \phi$.

If

• $M, w \models \Diamond((\square \phi \wedge \neg \psi) \wedge \Diamond(\square \psi \wedge \neg \phi))$,

there are points $x, y$ with $x\models \square \phi, x Ry, y \models \neg \phi$. Contradiction.

If

• $M, w \models \Diamond ((\square \phi \wedge \neg \psi) \wedge (\square \psi \wedge \neg \phi))$,

there is a point $x$ with $wRx, x \models \neg(\phi \wedge \square \phi \rightarrow \psi ) \wedge \neg( \psi \wedge \square\psi \rightarrow \psi)$. But this contradicts the fact that $w$ satisfies

• $\square(p \wedge \square p \rightarrow q) \lor \square(q \wedge \square q \rightarrow p)$.

If

• $M, w \models \Diamond(\Diamond(\square \phi \wedge \neg \psi) \wedge (\square \psi \wedge \neg \phi)$,

there are points $x,y$ with $x \models \square \psi, xRy, y \models \neg \psi$. Contradiction.

Answer 3

It seems that I've done withe the .3-axiom of the form $$ \Diamond p\land\Diamond q \to\Diamond(p\land\Diamond q)\lor\Diamond(p\land q)\lor\Diamond(q\land\Diamond p)\ $$. But I don't know how to use the other one yet.

Proof:

Assume not. Then there are w, u and v in the canonical model such that $$Rwu, Rwv, \lnot Ruv, \lnot Rvu, u \neq v.$$

Since $\lnot Ruv$, there is $\varphi$ such that $M,u \models \Box \varphi$ but $M,v \models \Box \varphi$. Since $\lnot Rvu$, there is $\psi$ such that $M,v \models \Box \psi$ but $M,v \models \Box \psi$. And since $u \neq v$ , there is $\chi$ such that $\chi \in u$ but $\lnot \chi \in v$.

Then $M,u \models \Box \varphi \land \lnot \psi \land \chi$, and $M,v \models \Box \psi \land \lnot \varphi \land \lnot \chi$.

And since $Rwu$ and $Rwv$, $M,w \models \Diamond (\Box \varphi \land \lnot \psi \land \chi) \land \Diamond (\Box \psi \land \lnot \varphi \land \lnot \chi)$

So $w$ satisfies the consequent of the instance of the .3-axiom, which results from the substitution that maps $p$ to $(\Box \varphi \land \lnot \psi \land \chi)$ and $q$ to $(\Box \psi \land \lnot \varphi \land \lnot \chi)$.

If

• $M,w \models \Diamond (\Diamond (\Box \varphi \land \lnot \psi \land \chi) \land (\Box \psi \land \lnot \varphi \land \lnot \chi)),$

then there are $x$, $y$ such that $M,x \models \Box \psi$ and $Rxy$ and $M,y \models \lnot \psi$. Contradiction.

If

• $M,w \models \Diamond ((\Box \varphi \land \lnot \psi \land \chi) \land \Diamond (\Box \psi \land \lnot \varphi \land \lnot \chi)),$

we can see a contradiction analogously.

And if

• $M,w \models \Diamond ((\Box \varphi \land \lnot \psi \land \chi) \land (\Box \psi \land \lnot \varphi \land \lnot \chi)),$

there is $z$ such that $M,z \models \chi$ and $M,z \models \lnot \chi$. Contradiction.

So our assumption is false.